{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0001","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b340dadcd7a2bc923a4ee826dc5a2ad62715b4b71effc8645c503af0592d748","family_id":"qf-mod47-taylor-qualification","seed":202611163001,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[2,1],[3,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[2, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([2, 3], 0, 1/2, 1, [2, 3, 0], [2, 3/2], 7/2, 7/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([2, 3], 0, 1/2, 1, [2, 3, 0], [2, 3/2], 7/2, 7/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0002","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"069e163ca04a3d7261efdff0a626508e981bbca3efe0717574f3ed16df7eb6a2","family_id":"qf-mod47-taylor-qualification","seed":202611163002,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[3,1],[5,1],[7,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[3, 5, 7] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([3, 5, 7], 0, 1/2, 2, [3, 5, 14, 0], [3, 5/2, 7/4], 29/4, 29/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([3, 5, 7], 0, 1/2, 2, [3, 5, 14, 0], [3, 5/2, 7/4], 29/4, 29/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0003","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2bdb439aff45a320418b7e1b882a117bee38cef2e017452d589fde8624261ff1","family_id":"qf-mod47-taylor-qualification","seed":202611163003,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[4,1],[7,1],[10,1],[13,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[4, 7, 10, 13] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([4, 7, 10, 13], 0, 1/2, 3, [4, 7, 20, 78, 0], [4, 7/2, 5/2, 13/8], 93/8, 93/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([4, 7, 10, 13], 0, 1/2, 3, [4, 7, 20, 78, 0], [4, 7/2, 5/2, 13/8], 93/8, 93/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0004","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4d9ddb60f81ae45e800b7aa69c1c3e502375df5f59884ba2f847151c7ca74e6f","family_id":"qf-mod47-taylor-qualification","seed":202611163004,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[5,1],[9,1],[13,1],[17,1],[4,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[5, 9, 13, 17, 4] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([5, 9, 13, 17, 4], 0, 1/2, 4, [5, 9, 26, 102, 96, 0], [5, 9/2, 13/4, 17/8, 1/4], 121/8, 121/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([5, 9, 13, 17, 4], 0, 1/2, 4, [5, 9, 26, 102, 96, 0], [5, 9/2, 13/4, 17/8, 1/4], 121/8, 121/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0005","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9eb2dc14b2130a9af7884bf9d31570dba5a102c881f853b09ff151ee1a26992","family_id":"qf-mod47-taylor-qualification","seed":202611163005,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[6,1],[11,1],[16,1],[4,1],[9,1],[14,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[6, 11, 16, 4, 9, 14] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([6, 11, 16, 4, 9, 14], 0, 1/2, 5, [6, 11, 32, 24, 216, 1680, 0], [6, 11/2, 4, 1/2, 9/16, 7/16], 17, 17, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([6, 11, 16, 4, 9, 14], 0, 1/2, 5, [6, 11, 32, 24, 216, 1680, 0], [6, 11/2, 4, 1/2, 9/16, 7/16], 17, 17, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0006","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"024683fa8eed41d3f496ea18cc55786c5fd0cebce9054a040465e3d204d25cc0","family_id":"qf-mod47-taylor-qualification","seed":202611163006,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[7,1],[13,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[7, 13] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([7, 13], 0, 1/2, 1, [7, 13, 0], [7, 13/2], 27/2, 27/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([7, 13], 0, 1/2, 1, [7, 13, 0], [7, 13/2], 27/2, 27/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0007","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d587ad16146e8690ea16e4112a4e93b15289e59f2e0e355e9fff748cb56694b1","family_id":"qf-mod47-taylor-qualification","seed":202611163007,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[8,1],[15,1],[5,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[8, 15, 5] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([8, 15, 5], 0, 1/2, 2, [8, 15, 10, 0], [8, 15/2, 5/4], 67/4, 67/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([8, 15, 5], 0, 1/2, 2, [8, 15, 10, 0], [8, 15/2, 5/4], 67/4, 67/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0008","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14d67be0914260bca3683dedfe8c6d6032a41d01b101009fa7aff7b541c93d01","family_id":"qf-mod47-taylor-qualification","seed":202611163008,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[9,1],[17,1],[8,1],[16,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[9, 17, 8, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([9, 17, 8, 16], 0, 1/2, 3, [9, 17, 16, 96, 0], [9, 17/2, 2, 2], 43/2, 43/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([9, 17, 8, 16], 0, 1/2, 3, [9, 17, 16, 96, 0], [9, 17/2, 2, 2], 43/2, 43/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0009","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"355c69a204c7663a02d54661bc8b30fe1e3199fd4af1b0032361b791294a1edd","family_id":"qf-mod47-taylor-qualification","seed":202611163009,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[10,1],[2,1],[11,1],[3,1],[12,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[10, 2, 11, 3, 12] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([10, 2, 11, 3, 12], 0, 1/2, 4, [10, 2, 22, 18, 288, 0], [10, 1, 11/4, 3/8, 3/4], 119/8, 119/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([10, 2, 11, 3, 12], 0, 1/2, 4, [10, 2, 22, 18, 288, 0], [10, 1, 11/4, 3/8, 3/4], 119/8, 119/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0010","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0fde7f5704394bf06cdc88454fd0d298cf449b56246b1547cb6cef4107e2d6d3","family_id":"qf-mod47-taylor-qualification","seed":202611163010,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[11,1],[4,1],[14,1],[7,1],[17,1],[10,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[11, 4, 14, 7, 17, 10] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([11, 4, 14, 7, 17, 10], 0, 1/2, 5, [11, 4, 28, 42, 408, 1200, 0], [11, 2, 7/2, 7/8, 17/16, 5/16], 75/4, 75/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([11, 4, 14, 7, 17, 10], 0, 1/2, 5, [11, 4, 28, 42, 408, 1200, 0], [11, 2, 7/2, 7/8, 17/16, 5/16], 75/4, 75/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0011","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"abaa2dfdd8be44f22a299153721172ebd20269d8ae88976519917b9b9164bf25","family_id":"qf-mod47-taylor-qualification","seed":202611163011,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[12,1],[6,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[12, 6] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([12, 6], 0, 1/2, 1, [12, 6, 0], [12, 3], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([12, 6], 0, 1/2, 1, [12, 6, 0], [12, 3], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0012","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"564bb104bcf17a6ce37a5ede39d94cdeb2cb6923390a6fa3593433496b1b0816","family_id":"qf-mod47-taylor-qualification","seed":202611163012,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[13,1],[8,1],[3,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[13, 8, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([13, 8, 3], 0, 1/2, 2, [13, 8, 6, 0], [13, 4, 3/4], 71/4, 71/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([13, 8, 3], 0, 1/2, 2, [13, 8, 6, 0], [13, 4, 3/4], 71/4, 71/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0013","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"30db47c229bc12eb8dfe4c35b37eaf9e547a03c118ee3d3d4d9394023a95010b","family_id":"qf-mod47-taylor-qualification","seed":202611163013,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[14,1],[10,1],[6,1],[2,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[14, 10, 6, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([14, 10, 6, 2], 0, 1/2, 3, [14, 10, 12, 12, 0], [14, 5, 3/2, 1/4], 83/4, 83/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([14, 10, 6, 2], 0, 1/2, 3, [14, 10, 12, 12, 0], [14, 5, 3/2, 1/4], 83/4, 83/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0014","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5dce26686d8fc8265d337bdbd479c0d8e47bb58b96ab5e228b170663ffb52a9f","family_id":"qf-mod47-taylor-qualification","seed":202611163014,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[15,1],[12,1],[9,1],[6,1],[3,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[15, 12, 9, 6, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([15, 12, 9, 6, 3], 0, 1/2, 4, [15, 12, 18, 36, 72, 0], [15, 6, 9/4, 3/4, 3/16], 387/16, 387/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([15, 12, 9, 6, 3], 0, 1/2, 4, [15, 12, 18, 36, 72, 0], [15, 6, 9/4, 3/4, 3/16], 387/16, 387/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0015","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"452d8b758b6b68b0a362c10fc2fc4bba96618290f9cb19df57f9581603e0b6b5","family_id":"qf-mod47-taylor-qualification","seed":202611163015,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[16,1],[14,1],[12,1],[10,1],[8,1],[6,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[16, 14, 12, 10, 8, 6] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([16, 14, 12, 10, 8, 6], 0, 1/2, 5, [16, 14, 24, 60, 192, 720, 0], [16, 7, 3, 5/4, 1/2, 3/16], 447/16, 447/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([16, 14, 12, 10, 8, 6], 0, 1/2, 5, [16, 14, 24, 60, 192, 720, 0], [16, 7, 3, 5/4, 1/2, 3/16], 447/16, 447/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0016","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db4f4da85af08c6304a94d97cda43cfac028c359a4566cc14acdc6b2398c28b3","family_id":"qf-mod47-taylor-qualification","seed":202611163016,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[17,1],[16,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[17, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([17, 16], 0, 1/2, 1, [17, 16, 0], [17, 8], 25, 25, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([17, 16], 0, 1/2, 1, [17, 16, 0], [17, 8], 25, 25, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0017","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fbb50b985ae9cf57e50542f18a0c972dc35c65f434a44c7a2338ca948047ee6d","family_id":"qf-mod47-taylor-qualification","seed":202611163017,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[1,1],[1,1],[1,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[1, 1, 1] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([1, 1, 1], 0, 1/2, 2, [1, 1, 2, 0], [1, 1/2, 1/4], 7/4, 7/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([1, 1, 1], 0, 1/2, 2, [1, 1, 2, 0], [1, 1/2, 1/4], 7/4, 7/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0018","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"90253dabc53554c0e7da86ac375fbd3795f2233d4be632e6d5014516b96dd33d","family_id":"qf-mod47-taylor-qualification","seed":202611163018,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[2,1],[3,1],[4,1],[5,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[2, 3, 4, 5] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([2, 3, 4, 5], 0, 1/2, 3, [2, 3, 8, 30, 0], [2, 3/2, 1, 5/8], 41/8, 41/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([2, 3, 4, 5], 0, 1/2, 3, [2, 3, 8, 30, 0], [2, 3/2, 1, 5/8], 41/8, 41/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0019","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"51b34909e795953fe7eb9f79dad84fbbe5a890c0d863cef1acd0517b3520a6a2","family_id":"qf-mod47-taylor-qualification","seed":202611163019,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[3,1],[5,1],[7,1],[9,1],[11,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[3, 5, 7, 9, 11] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([3, 5, 7, 9, 11], 0, 1/2, 4, [3, 5, 14, 54, 264, 0], [3, 5/2, 7/4, 9/8, 11/16], 145/16, 145/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([3, 5, 7, 9, 11], 0, 1/2, 4, [3, 5, 14, 54, 264, 0], [3, 5/2, 7/4, 9/8, 11/16], 145/16, 145/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0020","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"38e43782af78a35feb5b0f0177c72e3e3f59e23196d64585949af45709baf391","family_id":"qf-mod47-taylor-qualification","seed":202611163020,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[4,1],[7,1],[10,1],[13,1],[16,1],[2,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[4, 7, 10, 13, 16, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([4, 7, 10, 13, 16, 2], 0, 1/2, 5, [4, 7, 20, 78, 384, 240, 0], [4, 7/2, 5/2, 13/8, 1, 1/16], 203/16, 203/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([4, 7, 10, 13, 16, 2], 0, 1/2, 5, [4, 7, 20, 78, 384, 240, 0], [4, 7/2, 5/2, 13/8, 1, 1/16], 203/16, 203/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0021","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3689521f26188b9bfb79a026df13ff799821d9b74ad5219a92cd44d481f2455e","family_id":"qf-mod47-taylor-qualification","seed":202611163021,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[5,1],[9,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[5, 9] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([5, 9], 0, 1/2, 1, [5, 9, 0], [5, 9/2], 19/2, 19/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([5, 9], 0, 1/2, 1, [5, 9, 0], [5, 9/2], 19/2, 19/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0022","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d860b00aef574125c64d6e65aa4691a4fee8f5e2a84812070cac91fca4da4870","family_id":"qf-mod47-taylor-qualification","seed":202611163022,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[6,1],[11,1],[16,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[6, 11, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([6, 11, 16], 0, 1/2, 2, [6, 11, 32, 0], [6, 11/2, 4], 31/2, 31/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([6, 11, 16], 0, 1/2, 2, [6, 11, 32, 0], [6, 11/2, 4], 31/2, 31/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0023","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d52139fee46501fe8749f2c9f74c2ed1c64c1dbb44a73e01c046a90e1000902","family_id":"qf-mod47-taylor-qualification","seed":202611163023,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[7,1],[13,1],[2,1],[8,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[7, 13, 2, 8] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([7, 13, 2, 8], 0, 1/2, 3, [7, 13, 4, 48, 0], [7, 13/2, 1/2, 1], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([7, 13, 2, 8], 0, 1/2, 3, [7, 13, 4, 48, 0], [7, 13/2, 1/2, 1], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0024","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ed03aff51fdc8c3a0c94da15aed786a3ace0bea26d715e7bf7b0bd726591fa32","family_id":"qf-mod47-taylor-qualification","seed":202611163024,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[8,1],[15,1],[5,1],[12,1],[2,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[8, 15, 5, 12, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([8, 15, 5, 12, 2], 0, 1/2, 4, [8, 15, 10, 72, 48, 0], [8, 15/2, 5/4, 3/2, 1/8], 147/8, 147/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([8, 15, 5, 12, 2], 0, 1/2, 4, [8, 15, 10, 72, 48, 0], [8, 15/2, 5/4, 3/2, 1/8], 147/8, 147/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0025","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"213d4e3ebd8cf7d6d56da84f1ee82ed4e569da3cc2f51807576fbe95293fbc13","family_id":"qf-mod47-taylor-qualification","seed":202611163025,"semantic_template":"qf-mod47-taylor-qualification/{\"center\":[0,1],\"coefficients\":[[9,1],[17,1],[8,1],[16,1],[7,1],[15,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses","stem":"Coefficients=[9, 17, 8, 16, 7, 15] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).","answer":{"kind":"expression","value":"([9, 17, 8, 16, 7, 15], 0, 1/2, 5, [9, 17, 16, 96, 168, 1800, 0], [9, 17/2, 2, 2, 7/16, 15/32], 717/32, 717/32, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)","explanation":"Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([9, 17, 8, 16, 7, 15], 0, 1/2, 5, [9, 17, 16, 96, 168, 1800, 0], [9, 17/2, 2, 2, 7/16, 15/32], 717/32, 717/32, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)."},"skills":["An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0026","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9477cd1161606ed2ed01027539055a10b97c67c19878a7c3aa996172dab83d85","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163026,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[2,1],[3,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[2, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(1, 1, [2, 3/2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (1, 1, [2, 3/2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0027","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6f4ba0821c5aaace9421a81a4fb32e73ad1fd5dc64a6c0b9627618bcb12c7e81","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163027,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[3,1],[5,1],[7,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[3, 5, 7] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(2, 2, [3, 5/2, 7/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (2, 2, [3, 5/2, 7/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0028","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d645cd23b254b21bf0cd7363ea1cbc3e30478a282ddb7ce0e41d73aca071b06d","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163028,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[4,1],[7,1],[10,1],[13,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[4, 7, 10, 13] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(3, 3, [4, 7/2, 5/2, 13/8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (3, 3, [4, 7/2, 5/2, 13/8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0029","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a2bf44dd11b326fd990ec5a7fec8ee5777def03de5c5d4d81d9667f6d29f6611","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163029,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[5,1],[9,1],[13,1],[17,1],[4,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[5, 9, 13, 17, 4] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(4, 4, [5, 9/2, 13/4, 17/8, 1/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (4, 4, [5, 9/2, 13/4, 17/8, 1/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0030","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b818842943383b09b9c236131977d9a7e2745e4369db00d0983f47131291505c","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163030,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[6,1],[11,1],[16,1],[4,1],[9,1],[14,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[6, 11, 16, 4, 9, 14] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(5, 5, [6, 11/2, 4, 1/2, 9/16, 7/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (5, 5, [6, 11/2, 4, 1/2, 9/16, 7/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0031","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6bf037bd95f84578851a21689ccbe5547b1bcf4f2fcfd207f2f484fe67cc0c20","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163031,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[7,1],[13,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[7, 13] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(1, 1, [7, 13/2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (1, 1, [7, 13/2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0032","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"589335bda00c18cf647a5803494ffe3d23efd7ae9d48a77b17a84bd847c98b9e","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163032,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[8,1],[15,1],[5,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[8, 15, 5] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(2, 2, [8, 15/2, 5/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (2, 2, [8, 15/2, 5/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0033","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7fef736216d47bb8a88ad48396d27653ca7891ea6331401de2b3f9fb756c39fb","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163033,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[9,1],[17,1],[8,1],[16,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[9, 17, 8, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(3, 3, [9, 17/2, 2, 2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (3, 3, [9, 17/2, 2, 2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0034","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"89f39ee14fb7cafb2b382a1d227047803051e3c031423560d0267dc070b2311b","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163034,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[10,1],[2,1],[11,1],[3,1],[12,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[10, 2, 11, 3, 12] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(4, 4, [10, 1, 11/4, 3/8, 3/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (4, 4, [10, 1, 11/4, 3/8, 3/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0035","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06a152079e3cb5b575f3d20e53ee30783ab2b5d4b5dd8c9dcfe3ed1bc08cd322","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163035,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[11,1],[4,1],[14,1],[7,1],[17,1],[10,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[11, 4, 14, 7, 17, 10] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(5, 5, [11, 2, 7/2, 7/8, 17/16, 5/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (5, 5, [11, 2, 7/2, 7/8, 17/16, 5/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0036","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eeb2e8ac701e62546281b06bdbcdb49128e605a159d542b420ee8077c50a5202","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163036,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[12,1],[6,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[12, 6] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(1, 1, [12, 3], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (1, 1, [12, 3], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0037","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b602bb404ead6b6dec5d747bb1d19a33ef667207417610b6bc40feb8bb1914f","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163037,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[13,1],[8,1],[3,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[13, 8, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(2, 2, [13, 4, 3/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (2, 2, [13, 4, 3/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0038","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4b2b88c15a508ffb6c78bd542cffcac70cca2c17c243b5e9da2050bcd9910113","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163038,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[14,1],[10,1],[6,1],[2,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[14, 10, 6, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(3, 3, [14, 5, 3/2, 1/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (3, 3, [14, 5, 3/2, 1/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0039","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"049a2c8f10ab23975c186e842ffa735f9a356302f5beda77bb941d3f594f00c3","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163039,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[15,1],[12,1],[9,1],[6,1],[3,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[15, 12, 9, 6, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(4, 4, [15, 6, 9/4, 3/4, 3/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (4, 4, [15, 6, 9/4, 3/4, 3/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0040","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"86aa7a483022bab48e0f473a4fa37b333bacac6fe8d5211d12a2eae8740c24c0","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163040,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[16,1],[14,1],[12,1],[10,1],[8,1],[6,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[16, 14, 12, 10, 8, 6] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(5, 5, [16, 7, 3, 5/4, 1/2, 3/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (5, 5, [16, 7, 3, 5/4, 1/2, 3/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0041","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d139e1538f824fc399fff45ec4c6267587e3d7da90949786cc4e5e001562d93e","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163041,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[17,1],[16,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[17, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(1, 1, [17, 8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (1, 1, [17, 8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0042","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e14f3ed0093781bb765f4addbca36d52a843e075a2a581f611c3a7c3b7949ba6","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163042,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[1,1],[1,1],[1,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[1, 1, 1] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(2, 2, [1, 1/2, 1/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (2, 2, [1, 1/2, 1/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0043","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eda95b1d8d5c1c1fb4bfad2226d1dc9aeaa46d6bc7d0464658309d3f5315bc16","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163043,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[2,1],[3,1],[4,1],[5,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[2, 3, 4, 5] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(3, 3, [2, 3/2, 1, 5/8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (3, 3, [2, 3/2, 1, 5/8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0044","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a880d2dda5ba96f8cab7f80e0506a6d939a4dbe8f9f2fea6f298e413ad38fbec","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163044,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[3,1],[5,1],[7,1],[9,1],[11,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[3, 5, 7, 9, 11] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(4, 4, [3, 5/2, 7/4, 9/8, 11/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (4, 4, [3, 5/2, 7/4, 9/8, 11/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0045","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa86e73f563d7df1a540679aeb9bf3434533c89bdd34512c4be7e39c74ebf744","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163045,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[4,1],[7,1],[10,1],[13,1],[16,1],[2,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[4, 7, 10, 13, 16, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(5, 5, [4, 7/2, 5/2, 13/8, 1, 1/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (5, 5, [4, 7/2, 5/2, 13/8, 1, 1/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0046","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"16a109bf6ea0caa23483dd0010b66ee847df68bf02ef504eb675a6c026203972","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163046,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[5,1],[9,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[5, 9] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(1, 1, [5, 9/2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (1, 1, [5, 9/2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0047","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3dec0f412902d182e51b0863d8f46e4be2356de7e0bb71f7967f05fb122e550f","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163047,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[6,1],[11,1],[16,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[6, 11, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(2, 2, [6, 11/2, 4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (2, 2, [6, 11/2, 4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0048","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e32f2b44cdd1aad7c0cfd622b2562adbb67b3a98ecf7234a0b5d060d3f52e9fb","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163048,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[7,1],[13,1],[2,1],[8,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[7, 13, 2, 8] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(3, 3, [7, 13/2, 1/2, 1], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (3, 3, [7, 13/2, 1/2, 1], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0049","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bec4fd6b776bb7ffb56513a7f42e4822d66e0d9c20567bb3fd145aec732d1638","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163049,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[8,1],[15,1],[5,1],[12,1],[2,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[8, 15, 5, 12, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(4, 4, [8, 15/2, 5/4, 3/2, 1/8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (4, 4, [8, 15/2, 5/4, 3/2, 1/8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0050","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"57543a082c85bc82f8c251619e721307b4c7713dbd94e3249e278d93c0f81793","family_id":"qf-mod47-taylor-zero-remainder","seed":202611163050,"semantic_template":"qf-mod47-taylor-zero-remainder/{\"center\":[0,1],\"coefficients\":[[9,1],[17,1],[8,1],[16,1],[7,1],[15,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"A Polynomial of Degree at Most the Taylor Order Has Zero Remainder","stem":"Coefficients=[9, 17, 8, 16, 7, 15] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).","answer":{"kind":"expression","value":"(5, 5, [9, 17/2, 2, 2, 7/16, 15/32], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)","explanation":"Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (5, 5, [9, 17/2, 2, 2, 7/16, 15/32], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)."},"skills":["A Polynomial of Degree at Most the Taylor Order Has Zero Remainder"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0051","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a1dfe331699748d5d3a127324bf660f9b59079fb5eb8397fe33b75851c8d58f9","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163051,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[2,1],[3,1],[2,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[2, 3, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(2, 1, 4, 1/4, Yes, 1/2, 1/2, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (2, 1, 4, 1/4, Yes, 1/2, 1/2, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0052","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff449cf695e7c5905e11f5e6add4e9e89a54a36ccd6158bb2d00fd2a5562a44d","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163052,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[3,1],[5,1],[7,1],[3,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[3, 5, 7, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(3, 2, 18, 1/4, Yes, 3/8, 3/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (3, 2, 18, 1/4, Yes, 3/8, 3/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0053","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f1ceb7e4a31ba89195de7038e050265076d0c7b1c3f86c9a71a494722b2b9406","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163053,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[4,1],[7,1],[10,1],[13,1],[4,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[4, 7, 10, 13, 4] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(4, 3, 96, 1/4, Yes, 1/4, 1/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (4, 3, 96, 1/4, Yes, 1/4, 1/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0054","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"55623d16b5afd58a14186e2f5ff4e3cd0bc17fc3d8857160f63e4f152f62deaf","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163054,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[5,1],[9,1],[13,1],[17,1],[4,1],[5,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[5, 9, 13, 17, 4, 5] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(5, 4, 600, 1/4, Yes, 5/32, 5/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (5, 4, 600, 1/4, Yes, 5/32, 5/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0055","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db7dba84475b0d83f82b0f4f90cb9a91cfefa07cb43ad4872c8c3a4d357985fa","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163055,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[6,1],[11,1],[16,1],[4,1],[9,1],[14,1],[6,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[6, 11, 16, 4, 9, 14, 6] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(6, 5, 4320, 1/4, Yes, 3/32, 3/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (6, 5, 4320, 1/4, Yes, 3/32, 3/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0056","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ee0ac554e83c63c3b2f9ab80c0fd4168830c159231bd3c6b9656aa203467e5c0","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163056,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[7,1],[13,1],[7,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[7, 13, 7] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(2, 1, 14, 1/4, Yes, 7/4, 7/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (2, 1, 14, 1/4, Yes, 7/4, 7/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0057","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4ce4679cf09ca350106828a4ac8d18ac1af99b8a2acf6a71d23e13d2f473471","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163057,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[8,1],[15,1],[5,1],[8,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[8, 15, 5, 8] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(3, 2, 48, 1/4, Yes, 1, 1, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (3, 2, 48, 1/4, Yes, 1, 1, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0058","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6edd46bc88b3b470d85d6e36c2a3b8c0d35eda0efa540bf6bd6289082bdf2f7e","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163058,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[9,1],[17,1],[8,1],[16,1],[9,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[9, 17, 8, 16, 9] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(4, 3, 216, 1/4, Yes, 9/16, 9/16, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (4, 3, 216, 1/4, Yes, 9/16, 9/16, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0059","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4599459ffa3837936035afd50535b2e105e9a5845f34e16ec934b98ccb1115fe","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163059,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[10,1],[2,1],[11,1],[3,1],[12,1],[10,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[10, 2, 11, 3, 12, 10] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(5, 4, 1200, 1/4, Yes, 5/16, 5/16, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (5, 4, 1200, 1/4, Yes, 5/16, 5/16, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0060","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d362fdef80a67e55c7eeb1d49ea57fb088f9a9e7b680bafdea62c47026e1d8f8","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163060,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[11,1],[4,1],[14,1],[7,1],[17,1],[10,1],[11,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[11, 4, 14, 7, 17, 10, 11] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(6, 5, 7920, 1/4, Yes, 11/64, 11/64, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (6, 5, 7920, 1/4, Yes, 11/64, 11/64, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0061","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65237ece99e838fdd4ce872817189fbe9e344013dbe688a958100296ac5f9196","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163061,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[12,1],[6,1],[12,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[12, 6, 12] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(2, 1, 24, 1/4, Yes, 3, 3, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (2, 1, 24, 1/4, Yes, 3, 3, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0062","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b36ad57c8e0f522d73fbac4fa60d0fab064aa5f26435fa3f0cbba4e6fcd9b3a4","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163062,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[13,1],[8,1],[3,1],[13,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[13, 8, 3, 13] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(3, 2, 78, 1/4, Yes, 13/8, 13/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (3, 2, 78, 1/4, Yes, 13/8, 13/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0063","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b320bc3101d9855be593c0d334702ec048b62cc397f1711759fe3e7015ed001","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163063,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[14,1],[10,1],[6,1],[2,1],[14,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[14, 10, 6, 2, 14] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(4, 3, 336, 1/4, Yes, 7/8, 7/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (4, 3, 336, 1/4, Yes, 7/8, 7/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0064","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3dddf7e31fd3cefbfb7524baf4c7f77502ba6d0df11b44ba70e8b6f7bba6f5aa","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163064,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[15,1],[12,1],[9,1],[6,1],[3,1],[15,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[15, 12, 9, 6, 3, 15] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(5, 4, 1800, 1/4, Yes, 15/32, 15/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (5, 4, 1800, 1/4, Yes, 15/32, 15/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0065","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e6d071f24badbd9268af62d16a0861b20fe5091f8048dc4af8e13e4b6d71d6b","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163065,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[16,1],[14,1],[12,1],[10,1],[8,1],[6,1],[16,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[16, 14, 12, 10, 8, 6, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(6, 5, 11520, 1/4, Yes, 1/4, 1/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (6, 5, 11520, 1/4, Yes, 1/4, 1/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0066","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a72641931892d479a999e91e4c4d37017229fd8d5084c4c8ad819e6b4afa8926","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163066,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[17,1],[16,1],[17,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[17, 16, 17] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(2, 1, 34, 1/4, Yes, 17/4, 17/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (2, 1, 34, 1/4, Yes, 17/4, 17/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0067","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ff59341804512866477dd45467e4f130b7976e7cf3e01e59f833f19a5832ef5","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163067,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[1,1],[1,1],[1,1],[18,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[1, 1, 1, 18] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(3, 2, 108, 1/4, Yes, 9/4, 9/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (3, 2, 108, 1/4, Yes, 9/4, 9/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0068","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01245a002ccc8b875afb62b20f502a63fca6f89b75a7f2e90ddd4bd210d2eaed","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163068,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[2,1],[3,1],[4,1],[5,1],[19,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[2, 3, 4, 5, 19] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(4, 3, 456, 1/4, Yes, 19/16, 19/16, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (4, 3, 456, 1/4, Yes, 19/16, 19/16, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0069","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c52b9c775061b0c359b5de058a24d7df0d105cf3c48924a776c44efd455611b","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163069,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[3,1],[5,1],[7,1],[9,1],[11,1],[20,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[3, 5, 7, 9, 11, 20] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(5, 4, 2400, 1/4, Yes, 5/8, 5/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (5, 4, 2400, 1/4, Yes, 5/8, 5/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0070","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca9af3ff1ef83e5772bfebab480968019fa4ba2139292f59a9f52d318f15a3e0","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163070,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[4,1],[7,1],[10,1],[13,1],[16,1],[2,1],[21,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[4, 7, 10, 13, 16, 2, 21] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(6, 5, 15120, 1/4, Yes, 21/64, 21/64, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (6, 5, 15120, 1/4, Yes, 21/64, 21/64, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0071","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"964f72559f78f282888e56130c5d16d74f51c12b5e8b706615cdddee03150f6a","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163071,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[5,1],[9,1],[22,1]],\"order\":1,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[5, 9, 22] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(2, 1, 44, 1/4, Yes, 11/2, 11/2, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (2, 1, 44, 1/4, Yes, 11/2, 11/2, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0072","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d0027930ce09d1ac8293e1ac6b2b356cb9aea83214bc56a2ebda75f84bbd5001","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163072,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[6,1],[11,1],[16,1],[23,1]],\"order\":2,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[6, 11, 16, 23] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(3, 2, 138, 1/4, Yes, 23/8, 23/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (3, 2, 138, 1/4, Yes, 23/8, 23/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0073","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1029bf2967c62609661b2fed0471c4a1452c6ed843f98c90ba12dae6c3f99100","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163073,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[7,1],[13,1],[2,1],[8,1],[24,1]],\"order\":3,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[7, 13, 2, 8, 24] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(4, 3, 576, 1/4, Yes, 3/2, 3/2, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (4, 3, 576, 1/4, Yes, 3/2, 3/2, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0074","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5699c6897a57106ecb797ff8585e1b2229ca129779b8b9d69b885a6d3279bf89","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163074,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[8,1],[15,1],[5,1],[12,1],[2,1],[25,1]],\"order\":4,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[8, 15, 5, 12, 2, 25] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(5, 4, 3000, 1/4, Yes, 25/32, 25/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (5, 4, 3000, 1/4, Yes, 25/32, 25/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0075","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd11006c62817d0efa45c608ed1def76c3b14080342134b9d78613540e182b4f","family_id":"qf-mod47-taylor-lagrange-witness","seed":202611163075,"semantic_template":"qf-mod47-taylor-lagrange-witness/{\"center\":[0,1],\"coefficients\":[[9,1],[17,1],[8,1],[16,1],[7,1],[15,1],[26,1]],\"order\":5,\"x\":[1,2]}"},"fields":{"title":"A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness","stem":"Coefficients=[9, 17, 8, 16, 7, 15, 26] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).","answer":{"kind":"expression","value":"(6, 5, 18720, 1/4, Yes, 13/32, 13/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)","explanation":"Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (6, 5, 18720, 1/4, Yes, 13/32, 13/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)."},"skills":["A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0076","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd417885b33a407eb8cb5647663dcc6e2fdb027cf30149d31c97f2de8c3cc67c","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163076,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[4,1],\"center\":[0,1],\"coefficients\":[[2,1],[3,1],[2,1]],\"mode\":\"valid_bound\",\"order\":1,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[2, 3, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=4≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(valid_bound, 1/2, 4, 4, 1/2, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 1/2, 4, 4, 1/2, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0077","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2ee740413e1cb1c79dc9f7b72c23b0e3bbecc25396e4fb64f30aa11c30da78b3","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163077,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[17,1],\"center\":[0,1],\"coefficients\":[[3,1],[5,1],[7,1],[3,1]],\"mode\":\"undersized_M\",\"order\":2,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[3, 5, 7, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=17≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(undersized_M, 3/8, 18, 17, 17/48, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 3/8, 18, 17, 17/48, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0078","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"31f87954d08d58f68daf6cdb308d1fa916727f8d8f8902b0361c0e67da754fe8","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163078,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[96,1],\"center\":[0,1],\"coefficients\":[[4,1],[7,1],[10,1],[13,1],[4,1]],\"mode\":\"valid_bound\",\"order\":3,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[4, 7, 10, 13, 4] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=96≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(valid_bound, 1/4, 96, 96, 1/4, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 1/4, 96, 96, 1/4, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0079","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f813004d85fc612cdfadbc5cf6aaea87a44bf87d11b9b716045e0af49d54e02","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163079,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[599,1],\"center\":[0,1],\"coefficients\":[[5,1],[9,1],[13,1],[17,1],[4,1],[5,1]],\"mode\":\"undersized_M\",\"order\":4,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[5, 9, 13, 17, 4, 5] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=599≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(undersized_M, 5/32, 600, 599, 599/3840, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 5/32, 600, 599, 599/3840, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0080","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87446d493871bed2df24f204e18ace0a7a6048314007e7e63a07d98aba00ca81","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163080,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[4320,1],\"center\":[0,1],\"coefficients\":[[6,1],[11,1],[16,1],[4,1],[9,1],[14,1],[6,1]],\"mode\":\"valid_bound\",\"order\":5,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[6, 11, 16, 4, 9, 14, 6] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=4320≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(valid_bound, 3/32, 4320, 4320, 3/32, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 3/32, 4320, 4320, 3/32, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0081","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e9cab425cd3833ee53f96a3c95aaec7d64e3f65107b4303ac6622978b339cc7","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163081,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[13,1],\"center\":[0,1],\"coefficients\":[[7,1],[13,1],[7,1]],\"mode\":\"undersized_M\",\"order\":1,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[7, 13, 7] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=13≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(undersized_M, 7/4, 14, 13, 13/8, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 7/4, 14, 13, 13/8, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0082","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d6fb598d24c4e40655c050b4a7f1a13221ebf9595b02f45afff330faf6b27a5","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163082,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[48,1],\"center\":[0,1],\"coefficients\":[[8,1],[15,1],[5,1],[8,1]],\"mode\":\"valid_bound\",\"order\":2,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[8, 15, 5, 8] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=48≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(valid_bound, 1, 48, 48, 1, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 1, 48, 48, 1, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0083","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c5dd538fa791b6171db13b8f0262076f1044eefbd2bc35dc4347474e97c69393","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163083,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[215,1],\"center\":[0,1],\"coefficients\":[[9,1],[17,1],[8,1],[16,1],[9,1]],\"mode\":\"undersized_M\",\"order\":3,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[9, 17, 8, 16, 9] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=215≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(undersized_M, 9/16, 216, 215, 215/384, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 9/16, 216, 215, 215/384, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0084","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e2b9af353c804e3136c39970acc50993da3f6c466db06f4091c6f4997da88cd4","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163084,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[1200,1],\"center\":[0,1],\"coefficients\":[[10,1],[2,1],[11,1],[3,1],[12,1],[10,1]],\"mode\":\"valid_bound\",\"order\":4,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[10, 2, 11, 3, 12, 10] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=1200≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(valid_bound, 5/16, 1200, 1200, 5/16, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 5/16, 1200, 1200, 5/16, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0085","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f8f899b949a17135218df13e78b48b96a293e92be208e6cbbc50bb9b993d46b","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163085,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[7919,1],\"center\":[0,1],\"coefficients\":[[11,1],[4,1],[14,1],[7,1],[17,1],[10,1],[11,1]],\"mode\":\"undersized_M\",\"order\":5,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[11, 4, 14, 7, 17, 10, 11] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=7919≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(undersized_M, 11/64, 7920, 7919, 7919/46080, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 11/64, 7920, 7919, 7919/46080, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0086","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11a85dcd415358f07d2498522f8a38a84daab83089821e14fad50c8dfa9f8802","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163086,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[24,1],\"center\":[0,1],\"coefficients\":[[12,1],[6,1],[12,1]],\"mode\":\"valid_bound\",\"order\":1,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[12, 6, 12] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=24≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(valid_bound, 3, 24, 24, 3, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 3, 24, 24, 3, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0087","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"61b34d4065db3c373a3677daa2319e8821f9926efeb4503ff48fd4182e8bae1c","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163087,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[77,1],\"center\":[0,1],\"coefficients\":[[13,1],[8,1],[3,1],[13,1]],\"mode\":\"undersized_M\",\"order\":2,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[13, 8, 3, 13] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=77≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(undersized_M, 13/8, 78, 77, 77/48, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 13/8, 78, 77, 77/48, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0088","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff2540f4f36fa652b371088b68853bd261ca93e97254ef6984122043128c368d","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163088,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[336,1],\"center\":[0,1],\"coefficients\":[[14,1],[10,1],[6,1],[2,1],[14,1]],\"mode\":\"valid_bound\",\"order\":3,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[14, 10, 6, 2, 14] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=336≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(valid_bound, 7/8, 336, 336, 7/8, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 7/8, 336, 336, 7/8, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0089","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5b15f0c49f038dc053cb717ec234adb059b9b1d3f7e5a594f6bcbe60e93d7c7f","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163089,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[1799,1],\"center\":[0,1],\"coefficients\":[[15,1],[12,1],[9,1],[6,1],[3,1],[15,1]],\"mode\":\"undersized_M\",\"order\":4,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[15, 12, 9, 6, 3, 15] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=1799≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(undersized_M, 15/32, 1800, 1799, 1799/3840, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 15/32, 1800, 1799, 1799/3840, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0090","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f2f56233062a0307278b38afbb9864c87b3d7752a43065e5c1a5ac760786f381","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163090,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[11520,1],\"center\":[0,1],\"coefficients\":[[16,1],[14,1],[12,1],[10,1],[8,1],[6,1],[16,1]],\"mode\":\"valid_bound\",\"order\":5,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[16, 14, 12, 10, 8, 6, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=11520≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(valid_bound, 1/4, 11520, 11520, 1/4, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 1/4, 11520, 11520, 1/4, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0091","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"25f748418ef3a394723195e25bb5ae8dd79a0cbcf00a8f8dae3ce9c5add9bf02","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163091,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[33,1],\"center\":[0,1],\"coefficients\":[[17,1],[16,1],[17,1]],\"mode\":\"undersized_M\",\"order\":1,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[17, 16, 17] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=33≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(undersized_M, 17/4, 34, 33, 33/8, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 17/4, 34, 33, 33/8, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0092","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f2ab35cd5b779548227321dac5e3ccdfd96bb49b348144278302eae928981afe","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163092,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[108,1],\"center\":[0,1],\"coefficients\":[[1,1],[1,1],[1,1],[18,1]],\"mode\":\"valid_bound\",\"order\":2,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[1, 1, 1, 18] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=108≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(valid_bound, 9/4, 108, 108, 9/4, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 9/4, 108, 108, 9/4, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0093","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e623d6fe7e97766d6082f4f7c96e7ada11adcdb8fee867c81ca4d3fbbb727ed4","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163093,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[455,1],\"center\":[0,1],\"coefficients\":[[2,1],[3,1],[4,1],[5,1],[19,1]],\"mode\":\"undersized_M\",\"order\":3,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[2, 3, 4, 5, 19] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=455≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(undersized_M, 19/16, 456, 455, 455/384, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 19/16, 456, 455, 455/384, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0094","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"837d82fb04d75f10a4022e3c260915dd35460772cafb90575478edde67611e30","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163094,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[2400,1],\"center\":[0,1],\"coefficients\":[[3,1],[5,1],[7,1],[9,1],[11,1],[20,1]],\"mode\":\"valid_bound\",\"order\":4,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[3, 5, 7, 9, 11, 20] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=2400≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(valid_bound, 5/8, 2400, 2400, 5/8, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 5/8, 2400, 2400, 5/8, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0095","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5ce778c6a4fb9a6dfaac3dfb77f225b508f88d362cbe8b05756ffdf78c90e5bc","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163095,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[15119,1],\"center\":[0,1],\"coefficients\":[[4,1],[7,1],[10,1],[13,1],[16,1],[2,1],[21,1]],\"mode\":\"undersized_M\",\"order\":5,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[4, 7, 10, 13, 16, 2, 21] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=15119≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(undersized_M, 21/64, 15120, 15119, 15119/46080, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 21/64, 15120, 15119, 15119/46080, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0096","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d286c03e8dd884466ffeebbb6d5e72792f4af1c7b6d91b0e1fcd6d48f1df3395","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163096,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[44,1],\"center\":[0,1],\"coefficients\":[[5,1],[9,1],[22,1]],\"mode\":\"valid_bound\",\"order\":1,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[5, 9, 22] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=44≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(valid_bound, 11/2, 44, 44, 11/2, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 11/2, 44, 44, 11/2, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0097","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a663682a58f3ccbb09fe4069b8eb262e7087b3fce295607b1329f93f71669e1b","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163097,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[137,1],\"center\":[0,1],\"coefficients\":[[6,1],[11,1],[16,1],[23,1]],\"mode\":\"undersized_M\",\"order\":2,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[6, 11, 16, 23] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=137≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(undersized_M, 23/8, 138, 137, 137/48, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 23/8, 138, 137, 137/48, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0098","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"72d1b3a13c347bf428a36486fab0198126286429194cbced41a29850b538c698","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163098,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[576,1],\"center\":[0,1],\"coefficients\":[[7,1],[13,1],[2,1],[8,1],[24,1]],\"mode\":\"valid_bound\",\"order\":3,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[7, 13, 2, 8, 24] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=576≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(valid_bound, 3/2, 576, 576, 3/2, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 3/2, 576, 576, 3/2, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0099","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6eee9f23d631c45b967c960fe877ff75a5ec312e4202c7ade7edaf521496b8d4","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163099,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[2999,1],\"center\":[0,1],\"coefficients\":[[8,1],[15,1],[5,1],[12,1],[2,1],[25,1]],\"mode\":\"undersized_M\",\"order\":4,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[8, 15, 5, 12, 2, 25] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=2999≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(undersized_M, 25/32, 3000, 2999, 2999/3840, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 25/32, 3000, 2999, 2999/3840, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261116-r03-taylor-remainder-theorem-0100","batch_id":"batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ebee5c06e87f95c7f8c771b29866c4cb068d77072ba7cec023f94bfb00dd6c46","family_id":"qf-mod47-taylor-bound-boundary","seed":202611163100,"semantic_template":"qf-mod47-taylor-bound-boundary/{\"M\":[18720,1],\"center\":[0,1],\"coefficients\":[[9,1],[17,1],[8,1],[16,1],[7,1],[15,1],[26,1]],\"mode\":\"valid_bound\",\"order\":5,\"x\":[1,2]}"},"fields":{"title":"An Actual Error Inequality Is Different from a Qualified Derivative Bound","stem":"Coefficients=[9, 17, 8, 16, 7, 15, 26] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=18720≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).","answer":{"kind":"expression","value":"(valid_bound, 13/32, 18720, 18720, 13/32, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)","explanation":"The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 13/32, 18720, 18720, 13/32, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)."},"skills":["An Actual Error Inequality Is Different from a Qualified Derivative Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}