{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r01-0001","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83154f3d5f316c303291491ef7a94385e5b71611bb43f594df5689ea26e39d14","family_id":"qf-mod50-newton-step","seed":202611191001,"semantic_template":"newton-step|{\"coefficients\":[[-1,1],[0,1],[1,1]],\"x\":[2,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-1, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=2. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-1, 0, 1], 2, 3, 4, updated, 5/4, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-1, 0, 1], 2, 3, 4, updated, 5/4, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0002","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d44dc2de95619dd54b820afe724b99e4703079097d7dc8d1475fc34f5a8a202c","family_id":"qf-mod50-newton-step","seed":202611191002,"semantic_template":"newton-step|{\"coefficients\":[[-2,1],[0,1],[1,1]],\"x\":[3,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-2, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=3. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-2, 0, 1], 3, 7, 6, updated, 11/6, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-2, 0, 1], 3, 7, 6, updated, 11/6, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0003","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eece726a3698109c08e0201ef0437f2e705173114da8015a241457d2fd92101f","family_id":"qf-mod50-newton-step","seed":202611191003,"semantic_template":"newton-step|{\"coefficients\":[[-3,1],[0,1],[1,1]],\"x\":[4,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-3, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=4. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-3, 0, 1], 4, 13, 8, updated, 19/8, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-3, 0, 1], 4, 13, 8, updated, 19/8, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0004","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"be72f6ffd9fdfb07f5d7c8477cd7212e4614ccaf780ff27f17c57ba8c6996e9d","family_id":"qf-mod50-newton-step","seed":202611191004,"semantic_template":"newton-step|{\"coefficients\":[[-4,1],[0,1],[1,1]],\"x\":[5,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-4, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=5. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-4, 0, 1], 5, 21, 10, updated, 29/10, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-4, 0, 1], 5, 21, 10, updated, 29/10, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0005","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f158703d6f5d69e810f0ae76cf9542891d8fd7d885124ee9fb1e8e63d4009fb","family_id":"qf-mod50-newton-step","seed":202611191005,"semantic_template":"newton-step|{\"coefficients\":[[-5,1],[0,1],[1,1]],\"x\":[1,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-5, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=1. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-5, 0, 1], 1, -4, 2, updated, 3, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-5, 0, 1], 1, -4, 2, updated, 3, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0006","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f5ef99051ee6d93ce6e14c1107358b0fbd900a063706a0dabd79cb3dde0ac125","family_id":"qf-mod50-newton-step","seed":202611191006,"semantic_template":"newton-step|{\"coefficients\":[[-6,1],[0,1],[1,1]],\"x\":[2,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-6, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=2. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-6, 0, 1], 2, -2, 4, updated, 5/2, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-6, 0, 1], 2, -2, 4, updated, 5/2, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0007","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"09f5d0388e95d2b2e80915b856dd5bae158f4b886b3c2955c131aa72cb897ebc","family_id":"qf-mod50-newton-step","seed":202611191007,"semantic_template":"newton-step|{\"coefficients\":[[-7,1],[0,1],[1,1]],\"x\":[3,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-7, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=3. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-7, 0, 1], 3, 2, 6, updated, 8/3, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-7, 0, 1], 3, 2, 6, updated, 8/3, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0008","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d50b0638dcd96854c87595b27a6e906269eeb8ea1374d956a4b0a84d17f1fc78","family_id":"qf-mod50-newton-step","seed":202611191008,"semantic_template":"newton-step|{\"coefficients\":[[-8,1],[0,1],[1,1]],\"x\":[4,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-8, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=4. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-8, 0, 1], 4, 8, 8, updated, 3, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-8, 0, 1], 4, 8, 8, updated, 3, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0009","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e7c189eda02bc4c80ea6a076cfdfb86cc927571d01fb25f9610e57dc0f14154","family_id":"qf-mod50-newton-step","seed":202611191009,"semantic_template":"newton-step|{\"coefficients\":[[-9,1],[0,1],[1,1]],\"x\":[5,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-9, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=5. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-9, 0, 1], 5, 16, 10, updated, 17/5, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-9, 0, 1], 5, 16, 10, updated, 17/5, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0010","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e2989adcf402f03bb764f94f6de917322a6b4698499a7f85390ec67bab8525b5","family_id":"qf-mod50-newton-step","seed":202611191010,"semantic_template":"newton-step|{\"coefficients\":[[-10,1],[0,1],[1,1]],\"x\":[1,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-10, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=1. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-10, 0, 1], 1, -9, 2, updated, 11/2, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-10, 0, 1], 1, -9, 2, updated, 11/2, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0011","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"052a3013121f7425aed6be9745448e04a42ca1e67c38f8f1eff8710c65e23c3c","family_id":"qf-mod50-newton-step","seed":202611191011,"semantic_template":"newton-step|{\"coefficients\":[[-11,1],[0,1],[1,1]],\"x\":[2,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-11, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=2. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-11, 0, 1], 2, -7, 4, updated, 15/4, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-11, 0, 1], 2, -7, 4, updated, 15/4, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0012","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13f27880b53f7e871d2874f5232572d747219351586207ec9ba4d6b73134acb1","family_id":"qf-mod50-newton-step","seed":202611191012,"semantic_template":"newton-step|{\"coefficients\":[[-12,1],[0,1],[1,1]],\"x\":[3,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-12, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=3. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-12, 0, 1], 3, -3, 6, updated, 7/2, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-12, 0, 1], 3, -3, 6, updated, 7/2, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0013","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9beb87693f246b0c428c81fecbafc9e92b92f82cf5954d1db437646d49b9004b","family_id":"qf-mod50-newton-step","seed":202611191013,"semantic_template":"newton-step|{\"coefficients\":[[-13,1],[0,1],[1,1]],\"x\":[4,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-13, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=4. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-13, 0, 1], 4, 3, 8, updated, 29/8, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-13, 0, 1], 4, 3, 8, updated, 29/8, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0014","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1940bbc66fca7f5c18cab78cc0d201f2e316334f5120c0662d08f4a0022c270f","family_id":"qf-mod50-newton-step","seed":202611191014,"semantic_template":"newton-step|{\"coefficients\":[[-14,1],[0,1],[1,1]],\"x\":[5,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-14, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=5. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-14, 0, 1], 5, 11, 10, updated, 39/10, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-14, 0, 1], 5, 11, 10, updated, 39/10, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0015","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ce3b453d6acbf4189d675bac9659c1094aa679e29ce2f6776d59b2444226f49","family_id":"qf-mod50-newton-step","seed":202611191015,"semantic_template":"newton-step|{\"coefficients\":[[-15,1],[0,1],[1,1]],\"x\":[1,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-15, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=1. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-15, 0, 1], 1, -14, 2, updated, 8, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-15, 0, 1], 1, -14, 2, updated, 8, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0016","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc775fb4db9b078e8499a74f031b9ba8f04aa7947a933174b38f558b96b81375","family_id":"qf-mod50-newton-step","seed":202611191016,"semantic_template":"newton-step|{\"coefficients\":[[-16,1],[0,1],[1,1]],\"x\":[2,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-16, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=2. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-16, 0, 1], 2, -12, 4, updated, 5, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-16, 0, 1], 2, -12, 4, updated, 5, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0017","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2bd2517267f71ed6defff1626573acffe54c3ae0cb52ef6ac07e4896bb39721d","family_id":"qf-mod50-newton-step","seed":202611191017,"semantic_template":"newton-step|{\"coefficients\":[[-17,1],[0,1],[1,1]],\"x\":[3,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-17, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=3. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-17, 0, 1], 3, -8, 6, updated, 13/3, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-17, 0, 1], 3, -8, 6, updated, 13/3, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0018","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91fb1142664120e7bb8ebca91d25b68748439dd0098b800c970656ec136650da","family_id":"qf-mod50-newton-step","seed":202611191018,"semantic_template":"newton-step|{\"coefficients\":[[-18,1],[0,1],[1,1]],\"x\":[4,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-18, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=4. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-18, 0, 1], 4, -2, 8, updated, 17/4, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-18, 0, 1], 4, -2, 8, updated, 17/4, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0019","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49e2c3992cf011a1908dbcc291a95aa75270bc20f85afd2fac3baf73b9c79f36","family_id":"qf-mod50-newton-step","seed":202611191019,"semantic_template":"newton-step|{\"coefficients\":[[-19,1],[0,1],[1,1]],\"x\":[5,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-19, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=5. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-19, 0, 1], 5, 6, 10, updated, 22/5, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-19, 0, 1], 5, 6, 10, updated, 22/5, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0020","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7e4d83fa8ef25ca95289190ac5ce8966cf06c5033a6dc0d0c7137da58d300dc0","family_id":"qf-mod50-newton-step","seed":202611191020,"semantic_template":"newton-step|{\"coefficients\":[[-20,1],[0,1],[1,1]],\"x\":[1,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-20, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=1. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-20, 0, 1], 1, -19, 2, updated, 21/2, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-20, 0, 1], 1, -19, 2, updated, 21/2, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0021","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a1aeb788084e30306837149e0a90c2a95342ed87ef97113d086b80eaf8042b7e","family_id":"qf-mod50-newton-step","seed":202611191021,"semantic_template":"newton-step|{\"coefficients\":[[-21,1],[0,1],[1,1]],\"x\":[2,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-21, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=2. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-21, 0, 1], 2, -17, 4, updated, 25/4, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-21, 0, 1], 2, -17, 4, updated, 25/4, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0022","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4d345c71dfb392e6999cb16c4261d2fa5a1d28da98a86c049f76642629ef32a","family_id":"qf-mod50-newton-step","seed":202611191022,"semantic_template":"newton-step|{\"coefficients\":[[-22,1],[0,1],[1,1]],\"x\":[3,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-22, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=3. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-22, 0, 1], 3, -13, 6, updated, 31/6, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-22, 0, 1], 3, -13, 6, updated, 31/6, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0023","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"afca52fcf97ddf1ce4c665ff6ceb12a5bd353efac962c25a3c161fe92e1544aa","family_id":"qf-mod50-newton-step","seed":202611191023,"semantic_template":"newton-step|{\"coefficients\":[[-23,1],[0,1],[1,1]],\"x\":[4,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-23, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=4. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-23, 0, 1], 4, -7, 8, updated, 39/8, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-23, 0, 1], 4, -7, 8, updated, 39/8, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0024","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"278720d2ab4f07f749321cfce00cd401feb4c5270f7edb2e9f3d37000898ea7f","family_id":"qf-mod50-newton-step","seed":202611191024,"semantic_template":"newton-step|{\"coefficients\":[[-24,1],[0,1],[1,1]],\"x\":[5,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-24, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=5. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-24, 0, 1], 5, 1, 10, updated, 49/10, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-24, 0, 1], 5, 1, 10, updated, 49/10, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0025","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"574a1a392ebd100e849e996830a733368e61f0fe413598feabcfcaf10db84631","family_id":"qf-mod50-newton-step","seed":202611191025,"semantic_template":"newton-step|{\"coefficients\":[[-25,1],[0,1],[1,1]],\"x\":[1,1]}"},"fields":{"title":"An Exact Root Stops before Any Newton Division","stem":"Coefficients in low-to-high degree order are=[-25, 0, 1]; all iterations use exact rational arithmetic. The current iterate is x=1. Give (coefficients_low_to_high, current_iterate, function_value, derivative_value, status, next_iterate_or_null, root_first_convention).","answer":{"kind":"expression","value":"([-25, 0, 1], 1, -24, 2, updated, 13, root_test_precedes_division)","explanation":"Evaluate f and f_prime exactly. Test f(x)=0 first; a root stops with next=null even when its derivative is0. Otherwise require a nonzero derivative and use the tangent-line zero x−f/f_prime. This is a qualified step, not a convergence or global-root certificate. The exact result, retaining the requested order, is ([-25, 0, 1], 1, -24, 2, updated, 13, root_test_precedes_division)."},"skills":["An Exact Root Stops before Any Newton Division"],"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-pulse-20261119-r01-0026","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f27a2f6b5193195866c853738883e50a8fd3a1def07918abf30afa990646aa28","family_id":"qf-mod50-newton-finite-orbit","seed":202611191026,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-1,1],[0,1],[1,1]],\"steps\":1,\"x0\":[2,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-1, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=2 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-1, 0, 1], 1, [[0, 2, 3, 4, updated, 5/4]], step_budget_exhausted, 5/4, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-1, 0, 1], 1, [[0, 2, 3, 4, updated, 5/4]], step_budget_exhausted, 5/4, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0027","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"73ffb58631dace30cd9465c25213081a2cbdd9f0f9eda4a2ffd3afbe6b43146a","family_id":"qf-mod50-newton-finite-orbit","seed":202611191027,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-2,1],[0,1],[1,1]],\"steps\":1,\"x0\":[3,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-2, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=3 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-2, 0, 1], 1, [[0, 3, 7, 6, updated, 11/6]], step_budget_exhausted, 11/6, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-2, 0, 1], 1, [[0, 3, 7, 6, updated, 11/6]], step_budget_exhausted, 11/6, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0028","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1274118e542e06caed7d39d0914008d6c3fa809afa83f44d4f4522f645fd768b","family_id":"qf-mod50-newton-finite-orbit","seed":202611191028,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-3,1],[0,1],[1,1]],\"steps\":1,\"x0\":[4,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-3, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=4 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-3, 0, 1], 1, [[0, 4, 13, 8, updated, 19/8]], step_budget_exhausted, 19/8, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-3, 0, 1], 1, [[0, 4, 13, 8, updated, 19/8]], step_budget_exhausted, 19/8, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0029","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6827475c53dd450ec22685526cb53a19de22e193f01c93d2e8d9a9036553d721","family_id":"qf-mod50-newton-finite-orbit","seed":202611191029,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-4,1],[0,1],[1,1]],\"steps\":1,\"x0\":[5,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-4, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=5 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-4, 0, 1], 1, [[0, 5, 21, 10, updated, 29/10]], step_budget_exhausted, 29/10, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-4, 0, 1], 1, [[0, 5, 21, 10, updated, 29/10]], step_budget_exhausted, 29/10, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0030","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0271a6519a32008c40f9632dc122ebf0215e515e70e306f398429aada9cfdb75","family_id":"qf-mod50-newton-finite-orbit","seed":202611191030,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-5,1],[0,1],[1,1]],\"steps\":1,\"x0\":[1,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-5, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=1 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-5, 0, 1], 1, [[0, 1, -4, 2, updated, 3]], step_budget_exhausted, 3, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-5, 0, 1], 1, [[0, 1, -4, 2, updated, 3]], step_budget_exhausted, 3, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0031","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"478bf6bd1bf0b81b0f7ce92f576c9925e54d67c3b798fb5b6daba7ac66063c95","family_id":"qf-mod50-newton-finite-orbit","seed":202611191031,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-6,1],[0,1],[1,1]],\"steps\":1,\"x0\":[2,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-6, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=2 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-6, 0, 1], 1, [[0, 2, -2, 4, updated, 5/2]], step_budget_exhausted, 5/2, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-6, 0, 1], 1, [[0, 2, -2, 4, updated, 5/2]], step_budget_exhausted, 5/2, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0032","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3667d80828895fb239d039952f6fb7391b4c14c3e6e62c76829bef609bda2e25","family_id":"qf-mod50-newton-finite-orbit","seed":202611191032,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-7,1],[0,1],[1,1]],\"steps\":1,\"x0\":[3,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-7, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=3 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-7, 0, 1], 1, [[0, 3, 2, 6, updated, 8/3]], step_budget_exhausted, 8/3, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-7, 0, 1], 1, [[0, 3, 2, 6, updated, 8/3]], step_budget_exhausted, 8/3, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0033","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6f390ecc6e96cf8e317deaeebd5a1e136b4dcc8c23906234d1e183fcc4fe90e","family_id":"qf-mod50-newton-finite-orbit","seed":202611191033,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-8,1],[0,1],[1,1]],\"steps\":1,\"x0\":[4,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-8, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=4 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-8, 0, 1], 1, [[0, 4, 8, 8, updated, 3]], step_budget_exhausted, 3, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-8, 0, 1], 1, [[0, 4, 8, 8, updated, 3]], step_budget_exhausted, 3, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0034","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf0b444292218db7a393c1b9e1116550b19405cc629d8c3f0b6a96cd5ef2c44d","family_id":"qf-mod50-newton-finite-orbit","seed":202611191034,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-9,1],[0,1],[1,1]],\"steps\":1,\"x0\":[5,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-9, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=5 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-9, 0, 1], 1, [[0, 5, 16, 10, updated, 17/5]], step_budget_exhausted, 17/5, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-9, 0, 1], 1, [[0, 5, 16, 10, updated, 17/5]], step_budget_exhausted, 17/5, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0035","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b881ac9fc57bbf654938028ff1a0134ef8b7982d552dcae7082cc1fd332c302","family_id":"qf-mod50-newton-finite-orbit","seed":202611191035,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-10,1],[0,1],[1,1]],\"steps\":1,\"x0\":[1,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-10, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=1 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-10, 0, 1], 1, [[0, 1, -9, 2, updated, 11/2]], step_budget_exhausted, 11/2, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-10, 0, 1], 1, [[0, 1, -9, 2, updated, 11/2]], step_budget_exhausted, 11/2, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0036","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a885e6f1868975516de35945ada85aacdad7f08ba456c00f3855760b57233d52","family_id":"qf-mod50-newton-finite-orbit","seed":202611191036,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-11,1],[0,1],[1,1]],\"steps\":1,\"x0\":[2,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-11, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=2 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-11, 0, 1], 1, [[0, 2, -7, 4, updated, 15/4]], step_budget_exhausted, 15/4, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-11, 0, 1], 1, [[0, 2, -7, 4, updated, 15/4]], step_budget_exhausted, 15/4, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0037","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"090761591637bb368de11f46d6bda809875eed815366b32cde09b9ca017afb8f","family_id":"qf-mod50-newton-finite-orbit","seed":202611191037,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-12,1],[0,1],[1,1]],\"steps\":1,\"x0\":[3,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-12, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=3 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-12, 0, 1], 1, [[0, 3, -3, 6, updated, 7/2]], step_budget_exhausted, 7/2, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-12, 0, 1], 1, [[0, 3, -3, 6, updated, 7/2]], step_budget_exhausted, 7/2, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0038","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"37899c6ec2ea6a6ccc58dae28e8c59220eebc3e70caff67637dea5cd2c615162","family_id":"qf-mod50-newton-finite-orbit","seed":202611191038,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-13,1],[0,1],[1,1]],\"steps\":1,\"x0\":[4,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-13, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=4 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-13, 0, 1], 1, [[0, 4, 3, 8, updated, 29/8]], step_budget_exhausted, 29/8, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-13, 0, 1], 1, [[0, 4, 3, 8, updated, 29/8]], step_budget_exhausted, 29/8, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0039","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68614fcc41b26b9b11287885ead9dc18cb7ddef4497046fa442435c23540349f","family_id":"qf-mod50-newton-finite-orbit","seed":202611191039,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-14,1],[0,1],[1,1]],\"steps\":1,\"x0\":[5,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-14, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=5 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-14, 0, 1], 1, [[0, 5, 11, 10, updated, 39/10]], step_budget_exhausted, 39/10, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-14, 0, 1], 1, [[0, 5, 11, 10, updated, 39/10]], step_budget_exhausted, 39/10, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0040","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4d48abd4ee28a4c8fa73f58df68341fbfcaef3dee9f0732aeb392c53042a1c60","family_id":"qf-mod50-newton-finite-orbit","seed":202611191040,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-15,1],[0,1],[1,1]],\"steps\":1,\"x0\":[1,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-15, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=1 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-15, 0, 1], 1, [[0, 1, -14, 2, updated, 8]], step_budget_exhausted, 8, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-15, 0, 1], 1, [[0, 1, -14, 2, updated, 8]], step_budget_exhausted, 8, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0041","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d33f8c020e8de29c9eb03fbb17b76af1424491152bdb8a93b5e30168bda1602","family_id":"qf-mod50-newton-finite-orbit","seed":202611191041,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-16,1],[0,1],[1,1]],\"steps\":1,\"x0\":[2,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-16, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=2 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-16, 0, 1], 1, [[0, 2, -12, 4, updated, 5]], step_budget_exhausted, 5, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-16, 0, 1], 1, [[0, 2, -12, 4, updated, 5]], step_budget_exhausted, 5, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0042","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"550bc2e44b71301f5572ccc36de7c6e726f70687b80d4f1ab2c28e17df0f0861","family_id":"qf-mod50-newton-finite-orbit","seed":202611191042,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-17,1],[0,1],[1,1]],\"steps\":1,\"x0\":[3,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-17, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=3 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-17, 0, 1], 1, [[0, 3, -8, 6, updated, 13/3]], step_budget_exhausted, 13/3, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-17, 0, 1], 1, [[0, 3, -8, 6, updated, 13/3]], step_budget_exhausted, 13/3, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0043","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0153119a824e5440bc65ff3da60ab92f60e8ce4693aafcf1f6b9d88f95cf93de","family_id":"qf-mod50-newton-finite-orbit","seed":202611191043,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-18,1],[0,1],[1,1]],\"steps\":1,\"x0\":[4,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-18, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=4 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-18, 0, 1], 1, [[0, 4, -2, 8, updated, 17/4]], step_budget_exhausted, 17/4, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-18, 0, 1], 1, [[0, 4, -2, 8, updated, 17/4]], step_budget_exhausted, 17/4, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0044","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"afaa2dc6b9cab9cdcc8d0f1c5b0efd68302dccef4589d8e4aab7b16563016243","family_id":"qf-mod50-newton-finite-orbit","seed":202611191044,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-19,1],[0,1],[1,1]],\"steps\":1,\"x0\":[5,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-19, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=5 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-19, 0, 1], 1, [[0, 5, 6, 10, updated, 22/5]], step_budget_exhausted, 22/5, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-19, 0, 1], 1, [[0, 5, 6, 10, updated, 22/5]], step_budget_exhausted, 22/5, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0045","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e1c7a780680d905032a5f956a86713ed9f07efaae8183ed76e67fb934ef5384e","family_id":"qf-mod50-newton-finite-orbit","seed":202611191045,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-20,1],[0,1],[1,1]],\"steps\":1,\"x0\":[1,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-20, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=1 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-20, 0, 1], 1, [[0, 1, -19, 2, updated, 21/2]], step_budget_exhausted, 21/2, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-20, 0, 1], 1, [[0, 1, -19, 2, updated, 21/2]], step_budget_exhausted, 21/2, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0046","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7187beabe28aaec05b082f26de6cc9787ee61e060d750670346cd79790a34dcb","family_id":"qf-mod50-newton-finite-orbit","seed":202611191046,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-21,1],[0,1],[1,1]],\"steps\":1,\"x0\":[2,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-21, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=2 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-21, 0, 1], 1, [[0, 2, -17, 4, updated, 25/4]], step_budget_exhausted, 25/4, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-21, 0, 1], 1, [[0, 2, -17, 4, updated, 25/4]], step_budget_exhausted, 25/4, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0047","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a37f3b83ff30a4fabca8c27c4035aba6994e916748a37f9c704459196d787ab","family_id":"qf-mod50-newton-finite-orbit","seed":202611191047,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-22,1],[0,1],[1,1]],\"steps\":1,\"x0\":[3,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-22, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=3 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-22, 0, 1], 1, [[0, 3, -13, 6, updated, 31/6]], step_budget_exhausted, 31/6, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-22, 0, 1], 1, [[0, 3, -13, 6, updated, 31/6]], step_budget_exhausted, 31/6, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0048","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"394bbf9b50d3427d5d932e00914749e68d4580bab379427a7991d18afe48f65e","family_id":"qf-mod50-newton-finite-orbit","seed":202611191048,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-23,1],[0,1],[1,1]],\"steps\":1,\"x0\":[4,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-23, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=4 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-23, 0, 1], 1, [[0, 4, -7, 8, updated, 39/8]], step_budget_exhausted, 39/8, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-23, 0, 1], 1, [[0, 4, -7, 8, updated, 39/8]], step_budget_exhausted, 39/8, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0049","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"35f0b44d849d26da0917adb7e88e5ce76dccfc895204d8c9e1c29c8fe34f5b28","family_id":"qf-mod50-newton-finite-orbit","seed":202611191049,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-24,1],[0,1],[1,1]],\"steps\":1,\"x0\":[5,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-24, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=5 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-24, 0, 1], 1, [[0, 5, 1, 10, updated, 49/10]], step_budget_exhausted, 49/10, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-24, 0, 1], 1, [[0, 5, 1, 10, updated, 49/10]], step_budget_exhausted, 49/10, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0050","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"facd91b28a88042f91d12f7d081dfc8642db207570f4aa8695e7da14d17b0105","family_id":"qf-mod50-newton-finite-orbit","seed":202611191050,"semantic_template":"newton-finite-orbit|{\"coefficients\":[[-25,1],[0,1],[1,1]],\"steps\":1,\"x0\":[1,1]}"},"fields":{"title":"A Finite Newton Orbit Records Steps without Certifying Convergence","stem":"Coefficients in low-to-high degree order are=[-25, 0, 1]; all iterations use exact rational arithmetic. Use initial iterate=1 and attempted-step budget=1. Give (coefficients_low_to_high, step_budget, step_records, status, final_iterate, convergence_or_rate_proof_claim, finite_scope).","answer":{"kind":"expression","value":"([-25, 0, 1], 1, [[0, 1, -24, 2, updated, 13]], step_budget_exhausted, 13, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)","explanation":"Recalculate f, f_prime and every qualified update. Each record gives index, current x, f, f_prime, status and next or null. Exact roots stop early; a nonroot zero derivative is rejected. A finite residual pattern proves neither a general limit nor quadratic convergence, a global basin or a forward-error bound. The exact result, retaining the requested order, is ([-25, 0, 1], 1, [[0, 1, -24, 2, updated, 13]], step_budget_exhausted, 13, No, finite_exact_orbit_is_not_a_convergence_or_quadratic_rate_certificate)."},"skills":["A Finite Newton Orbit Records Steps without Certifying Convergence"],"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-pulse-20261119-r01-0051","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b6ec858be52a367ff233e2efdde74cab813e6f0f9c74cf4305255cdd9ad1a988","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191051,"semantic_template":"newton-square-root-orbit|{\"radicand\":[2,1],\"steps\":1,\"x0\":[2,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=2, positive initial iterate=2, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(2, 2, 1, [[0, 2, 2, updated, 3/2]], step_budget_exhausted, 3/2, 1/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (2, 2, 1, [[0, 2, 2, updated, 3/2]], step_budget_exhausted, 3/2, 1/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0052","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cff21cc5b4ca898946a7f4c1a16289d2e33c045e7b78e11e42c0b874c3c3de7c","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191052,"semantic_template":"newton-square-root-orbit|{\"radicand\":[3,1],\"steps\":1,\"x0\":[3,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=3, positive initial iterate=3, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(3, 3, 1, [[0, 3, 6, updated, 2]], step_budget_exhausted, 2, 1, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (3, 3, 1, [[0, 3, 6, updated, 2]], step_budget_exhausted, 2, 1, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0053","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d2b3b6032f332ffc2676f888c6c26c664962fb473aa29a0a1451369903bcb1e3","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191053,"semantic_template":"newton-square-root-orbit|{\"radicand\":[4,1],\"steps\":1,\"x0\":[4,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=4, positive initial iterate=4, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(4, 4, 1, [[0, 4, 12, updated, 5/2]], step_budget_exhausted, 5/2, 9/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (4, 4, 1, [[0, 4, 12, updated, 5/2]], step_budget_exhausted, 5/2, 9/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0054","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65b34ce3c27ff28302af2db645d993a2009c2afb47a210ab1df8fcd1baabcb6e","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191054,"semantic_template":"newton-square-root-orbit|{\"radicand\":[5,1],\"steps\":1,\"x0\":[5,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=5, positive initial iterate=5, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(5, 5, 1, [[0, 5, 20, updated, 3]], step_budget_exhausted, 3, 4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (5, 5, 1, [[0, 5, 20, updated, 3]], step_budget_exhausted, 3, 4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0055","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"64a29a7c4146b79cf9112f7c3e447b0849407608f8b5eccfa639279b8fa1d622","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191055,"semantic_template":"newton-square-root-orbit|{\"radicand\":[6,1],\"steps\":1,\"x0\":[1,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=6, positive initial iterate=1, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(6, 1, 1, [[0, 1, -5, updated, 7/2]], step_budget_exhausted, 7/2, 25/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (6, 1, 1, [[0, 1, -5, updated, 7/2]], step_budget_exhausted, 7/2, 25/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0056","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8681d4fb23c49641f30cd041a20df1742093befc57f45730c8ee8b9a5accb475","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191056,"semantic_template":"newton-square-root-orbit|{\"radicand\":[7,1],\"steps\":1,\"x0\":[2,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=7, positive initial iterate=2, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(7, 2, 1, [[0, 2, -3, updated, 11/4]], step_budget_exhausted, 11/4, 9/16, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (7, 2, 1, [[0, 2, -3, updated, 11/4]], step_budget_exhausted, 11/4, 9/16, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0057","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68cc4d48c44ccf6c19689c03daf93464fcb1bfc72b4426b7bd60fee251874934","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191057,"semantic_template":"newton-square-root-orbit|{\"radicand\":[8,1],\"steps\":1,\"x0\":[3,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=8, positive initial iterate=3, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(8, 3, 1, [[0, 3, 1, updated, 17/6]], step_budget_exhausted, 17/6, 1/36, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (8, 3, 1, [[0, 3, 1, updated, 17/6]], step_budget_exhausted, 17/6, 1/36, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0058","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5ff1949815995ce1e17afb273219df29c0606af467485cd598355bb28d83b0e2","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191058,"semantic_template":"newton-square-root-orbit|{\"radicand\":[9,1],\"steps\":1,\"x0\":[4,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=9, positive initial iterate=4, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(9, 4, 1, [[0, 4, 7, updated, 25/8]], step_budget_exhausted, 25/8, 49/64, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (9, 4, 1, [[0, 4, 7, updated, 25/8]], step_budget_exhausted, 25/8, 49/64, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0059","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a2153eec62b40596b45340b7e7892440b572691fc68af9adce69c4eeb0412e0b","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191059,"semantic_template":"newton-square-root-orbit|{\"radicand\":[10,1],\"steps\":1,\"x0\":[5,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=10, positive initial iterate=5, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(10, 5, 1, [[0, 5, 15, updated, 7/2]], step_budget_exhausted, 7/2, 9/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (10, 5, 1, [[0, 5, 15, updated, 7/2]], step_budget_exhausted, 7/2, 9/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0060","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"92279163fcd2b1b0a69b6cfdb8274261148019824127a8ef314489525a21aed8","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191060,"semantic_template":"newton-square-root-orbit|{\"radicand\":[11,1],\"steps\":1,\"x0\":[1,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=11, positive initial iterate=1, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(11, 1, 1, [[0, 1, -10, updated, 6]], step_budget_exhausted, 6, 25, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (11, 1, 1, [[0, 1, -10, updated, 6]], step_budget_exhausted, 6, 25, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0061","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe766fbccebac0bc6cb4674aacec2eaa4970d278bffb689a1900b2012c6ec872","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191061,"semantic_template":"newton-square-root-orbit|{\"radicand\":[12,1],\"steps\":1,\"x0\":[2,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=12, positive initial iterate=2, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(12, 2, 1, [[0, 2, -8, updated, 4]], step_budget_exhausted, 4, 4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (12, 2, 1, [[0, 2, -8, updated, 4]], step_budget_exhausted, 4, 4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0062","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a90563d1752a0222b7243573140bf7543065e38aa5ae80eb358aae66bf9fc04","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191062,"semantic_template":"newton-square-root-orbit|{\"radicand\":[13,1],\"steps\":1,\"x0\":[3,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=13, positive initial iterate=3, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(13, 3, 1, [[0, 3, -4, updated, 11/3]], step_budget_exhausted, 11/3, 4/9, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (13, 3, 1, [[0, 3, -4, updated, 11/3]], step_budget_exhausted, 11/3, 4/9, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0063","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c481fa7877c3844271f57cf7ed91ea21ed19928dd2e211aa93ae512c55acd4b0","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191063,"semantic_template":"newton-square-root-orbit|{\"radicand\":[14,1],\"steps\":1,\"x0\":[4,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=14, positive initial iterate=4, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(14, 4, 1, [[0, 4, 2, updated, 15/4]], step_budget_exhausted, 15/4, 1/16, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (14, 4, 1, [[0, 4, 2, updated, 15/4]], step_budget_exhausted, 15/4, 1/16, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0064","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3646a08808fd87fb22e4062575a2e74411906b2b959e9282b63d31989edda62a","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191064,"semantic_template":"newton-square-root-orbit|{\"radicand\":[15,1],\"steps\":1,\"x0\":[5,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=15, positive initial iterate=5, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(15, 5, 1, [[0, 5, 10, updated, 4]], step_budget_exhausted, 4, 1, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (15, 5, 1, [[0, 5, 10, updated, 4]], step_budget_exhausted, 4, 1, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0065","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7befc31ed33929d259099326c5c74a7458306b61f42c0bc567a3de063949b724","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191065,"semantic_template":"newton-square-root-orbit|{\"radicand\":[16,1],\"steps\":1,\"x0\":[1,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=16, positive initial iterate=1, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(16, 1, 1, [[0, 1, -15, updated, 17/2]], step_budget_exhausted, 17/2, 225/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (16, 1, 1, [[0, 1, -15, updated, 17/2]], step_budget_exhausted, 17/2, 225/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0066","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"566355854962179f0375557972be118870e357520a7d29efd93c5bf6460dfde2","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191066,"semantic_template":"newton-square-root-orbit|{\"radicand\":[17,1],\"steps\":1,\"x0\":[2,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=17, positive initial iterate=2, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(17, 2, 1, [[0, 2, -13, updated, 21/4]], step_budget_exhausted, 21/4, 169/16, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (17, 2, 1, [[0, 2, -13, updated, 21/4]], step_budget_exhausted, 21/4, 169/16, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0067","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a6a5025b9d2c53572d5885a6a43732e35e80cd5311a5f1fa0346672a61661ee","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191067,"semantic_template":"newton-square-root-orbit|{\"radicand\":[18,1],\"steps\":1,\"x0\":[3,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=18, positive initial iterate=3, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(18, 3, 1, [[0, 3, -9, updated, 9/2]], step_budget_exhausted, 9/2, 9/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (18, 3, 1, [[0, 3, -9, updated, 9/2]], step_budget_exhausted, 9/2, 9/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0068","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f25a19306a2e21cbe59af8ebfcc15c403db47ec05a6fe92fc429891ca132ee4e","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191068,"semantic_template":"newton-square-root-orbit|{\"radicand\":[19,1],\"steps\":1,\"x0\":[4,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=19, positive initial iterate=4, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(19, 4, 1, [[0, 4, -3, updated, 35/8]], step_budget_exhausted, 35/8, 9/64, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (19, 4, 1, [[0, 4, -3, updated, 35/8]], step_budget_exhausted, 35/8, 9/64, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0069","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5dc376e062c37d1304f15ec3b229743e0865cf6dc57e796579961ffeecf64327","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191069,"semantic_template":"newton-square-root-orbit|{\"radicand\":[20,1],\"steps\":1,\"x0\":[5,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=20, positive initial iterate=5, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(20, 5, 1, [[0, 5, 5, updated, 9/2]], step_budget_exhausted, 9/2, 1/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (20, 5, 1, [[0, 5, 5, updated, 9/2]], step_budget_exhausted, 9/2, 1/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0070","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"47417a358cd72f335622030af18173a6d21a7689f2a5dd52a5af943319c65434","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191070,"semantic_template":"newton-square-root-orbit|{\"radicand\":[21,1],\"steps\":1,\"x0\":[1,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=21, positive initial iterate=1, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(21, 1, 1, [[0, 1, -20, updated, 11]], step_budget_exhausted, 11, 100, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (21, 1, 1, [[0, 1, -20, updated, 11]], step_budget_exhausted, 11, 100, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0071","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"752cfc53d485d1b17b0c9624ee4bafe00468e991d2d98b8767d947227d1c857f","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191071,"semantic_template":"newton-square-root-orbit|{\"radicand\":[22,1],\"steps\":1,\"x0\":[2,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=22, positive initial iterate=2, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(22, 2, 1, [[0, 2, -18, updated, 13/2]], step_budget_exhausted, 13/2, 81/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (22, 2, 1, [[0, 2, -18, updated, 13/2]], step_budget_exhausted, 13/2, 81/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0072","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa3f841b6b7896f18c093b33e7ae4e90293c23301141a30323b5afce692464fa","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191072,"semantic_template":"newton-square-root-orbit|{\"radicand\":[23,1],\"steps\":1,\"x0\":[3,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=23, positive initial iterate=3, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(23, 3, 1, [[0, 3, -14, updated, 16/3]], step_budget_exhausted, 16/3, 49/9, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (23, 3, 1, [[0, 3, -14, updated, 16/3]], step_budget_exhausted, 16/3, 49/9, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0073","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e5ff497561bcf14ea89400423d6c404bfa6ff2d54b70386499065df6ea9db7b5","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191073,"semantic_template":"newton-square-root-orbit|{\"radicand\":[24,1],\"steps\":1,\"x0\":[4,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=24, positive initial iterate=4, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(24, 4, 1, [[0, 4, -8, updated, 5]], step_budget_exhausted, 5, 1, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (24, 4, 1, [[0, 4, -8, updated, 5]], step_budget_exhausted, 5, 1, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0074","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf1fc1f3f86ffb496e39ae11da54f3b651f07d6c42bd3aeba532f71caab7f346","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191074,"semantic_template":"newton-square-root-orbit|{\"radicand\":[25,1],\"steps\":1,\"x0\":[5,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=25, positive initial iterate=5, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(25, 5, 1, [[0, 5, 0, root_found, null]], root_found, 5, 0, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (25, 5, 1, [[0, 5, 0, root_found, null]], root_found, 5, 0, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0075","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2ef78c4c17297fa3946c8f5e146b5778ddb2fd7debf4ef8863acc6a07d0ac5f5","family_id":"qf-mod50-newton-square-root-orbit","seed":202611191075,"semantic_template":"newton-square-root-orbit|{\"radicand\":[26,1],\"steps\":1,\"x0\":[1,1]}"},"fields":{"title":"Positive Square-Root Iterates Use Exact Rational Newton Steps","stem":"Use positive radicand A=26, positive initial iterate=1, and attempted-step budget=1; f(x)=x^2−A. Give (radicand, initial_iterate, step_budget, step_records, status, final_iterate, final_absolute_residual, general_Newton_proof_claim, finite_scope).","answer":{"kind":"expression","value":"(26, 1, 1, [[0, 1, -25, updated, 27/2]], step_budget_exhausted, 27/2, 625/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)","explanation":"At an exact root stop before dividing; otherwise update to (x+A/x)/2. Each record gives index, current x, signed residual, status and next or null. The final absolute residual is |x^2−A|, not an unprovided forward error to sqrt(A) or a general Newton convergence proof. The exact result, retaining the requested order, is (26, 1, 1, [[0, 1, -25, updated, 27/2]], step_budget_exhausted, 27/2, 625/4, No, finite_square_root_orbit_does_not_prove_a_general_Newton_theorem)."},"skills":["Positive Square-Root Iterates Use Exact Rational Newton Steps"],"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-pulse-20261119-r01-0076","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"acac475e4dfa0a888dd0b538040b99900bd9bebb965950c8e09eb37bc96e313c","family_id":"qf-mod50-newton-boundary","seed":202611191076,"semantic_template":"newton-boundary|{\"mode\":\"zero_derivative\",\"model\":{\"coefficients\":[[1,1],[0,1],[1,1]],\"x\":[0,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=zero_derivative, complete model={\"coefficients\":[1,0,1],\"x\":0}. Give (mode, current_iterate, function_value, derivative_value, status, update_qualification_claim, boundary).","answer":{"kind":"expression","value":"(zero_derivative, 0, 1, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (zero_derivative, 0, 1, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0077","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fbe537c518a18039e216c8f9584e5ee838fc6539b0222d9ee48c8188c53fcfae","family_id":"qf-mod50-newton-boundary","seed":202611191077,"semantic_template":"newton-boundary|{\"mode\":\"finite_trace_only\",\"model\":{\"coefficients\":[[-2,1],[0,1],[1,1]],\"steps\":1,\"x0\":[3,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=finite_trace_only, complete model={\"coefficients\":[-2,0,1],\"x0\":3,\"steps\":1}. Give (mode, step_budget, step_records, status, convergence_or_rate_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_trace_only, 1, [[0, 3, 7, 6, updated, 11/6]], step_budget_exhausted, No, finite_residual_behavior_does_not_prove_convergence_or_rate)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (finite_trace_only, 1, [[0, 3, 7, 6, updated, 11/6]], step_budget_exhausted, No, finite_residual_behavior_does_not_prove_convergence_or_rate)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0078","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"367a96868e0d2a10f2eb10c656683be867c3b3d19698ac87a5fbd4195d3ff518","family_id":"qf-mod50-newton-boundary","seed":202611191078,"semantic_template":"newton-boundary|{\"mode\":\"root_stop\",\"model\":{\"coefficients\":[[-3,1],[1,1]],\"x\":[3,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=root_stop, complete model={\"coefficients\":[-3,1],\"x\":3}. Give (mode, current_iterate, function_value, derivative_value, root_first_status, root_stop_qualified).","answer":{"kind":"expression","value":"(root_stop, 3, 0, 1, root_found_before_derivative_division, Yes)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (root_stop, 3, 0, 1, root_found_before_derivative_division, Yes)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0079","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aeb999db82f1f505833f9ca73494275df2f6662a34b17c4db1d962c675133cf4","family_id":"qf-mod50-newton-boundary","seed":202611191079,"semantic_template":"newton-boundary|{\"mode\":\"residual_forward_confusion\",\"model\":{\"approximation\":[6,1],\"intercept\":[-4,1],\"slope\":[5,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=residual_forward_confusion, complete model={\"slope\":5,\"intercept\":-4,\"approximation\":6}. Give (mode, slope, exact_root, absolute_residual, forward_error, residual_equals_forward_error, scale_relation).","answer":{"kind":"expression","value":"(residual_forward_confusion, 5, 4/5, 26, 26/5, No, residual_equals_abs_slope_times_forward_error_not_forward_error_itself)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (residual_forward_confusion, 5, 4/5, 26, 26/5, No, residual_equals_abs_slope_times_forward_error_not_forward_error_itself)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0080","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6bedb060527ac302bf2f03c2a349d4ccad5da985e3babb93849d039158afc8be","family_id":"qf-mod50-newton-boundary","seed":202611191080,"semantic_template":"newton-boundary|{\"mode\":\"zero_derivative\",\"model\":{\"coefficients\":[[5,1],[0,1],[1,1]],\"x\":[0,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=zero_derivative, complete model={\"coefficients\":[5,0,1],\"x\":0}. Give (mode, current_iterate, function_value, derivative_value, status, update_qualification_claim, boundary).","answer":{"kind":"expression","value":"(zero_derivative, 0, 5, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (zero_derivative, 0, 5, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0081","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9035fbf7284810e8227be5cb053d9d2482e6a16a7bed9a1d02ae8a1d4f3c8f60","family_id":"qf-mod50-newton-boundary","seed":202611191081,"semantic_template":"newton-boundary|{\"mode\":\"finite_trace_only\",\"model\":{\"coefficients\":[[-6,1],[0,1],[1,1]],\"steps\":1,\"x0\":[2,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=finite_trace_only, complete model={\"coefficients\":[-6,0,1],\"x0\":2,\"steps\":1}. Give (mode, step_budget, step_records, status, convergence_or_rate_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_trace_only, 1, [[0, 2, -2, 4, updated, 5/2]], step_budget_exhausted, No, finite_residual_behavior_does_not_prove_convergence_or_rate)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (finite_trace_only, 1, [[0, 2, -2, 4, updated, 5/2]], step_budget_exhausted, No, finite_residual_behavior_does_not_prove_convergence_or_rate)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0082","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0edc931ec5d928a3a22dbfacc7e427787fa907a8f95b0007c009a07a4c03afbe","family_id":"qf-mod50-newton-boundary","seed":202611191082,"semantic_template":"newton-boundary|{\"mode\":\"root_stop\",\"model\":{\"coefficients\":[[-7,1],[1,1]],\"x\":[7,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=root_stop, complete model={\"coefficients\":[-7,1],\"x\":7}. Give (mode, current_iterate, function_value, derivative_value, root_first_status, root_stop_qualified).","answer":{"kind":"expression","value":"(root_stop, 7, 0, 1, root_found_before_derivative_division, Yes)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (root_stop, 7, 0, 1, root_found_before_derivative_division, Yes)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0083","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8f97deabadc76416fe1de050d5115dcc1a3c18438c7895d866dea58d3eb763e3","family_id":"qf-mod50-newton-boundary","seed":202611191083,"semantic_template":"newton-boundary|{\"mode\":\"residual_forward_confusion\",\"model\":{\"approximation\":[10,1],\"intercept\":[-8,1],\"slope\":[9,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=residual_forward_confusion, complete model={\"slope\":9,\"intercept\":-8,\"approximation\":10}. Give (mode, slope, exact_root, absolute_residual, forward_error, residual_equals_forward_error, scale_relation).","answer":{"kind":"expression","value":"(residual_forward_confusion, 9, 8/9, 82, 82/9, No, residual_equals_abs_slope_times_forward_error_not_forward_error_itself)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (residual_forward_confusion, 9, 8/9, 82, 82/9, No, residual_equals_abs_slope_times_forward_error_not_forward_error_itself)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0084","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"abc6f30d61f7fb977f26220cae3bac803b0dbed6a8f2998bfa25e20cd73de796","family_id":"qf-mod50-newton-boundary","seed":202611191084,"semantic_template":"newton-boundary|{\"mode\":\"zero_derivative\",\"model\":{\"coefficients\":[[9,1],[0,1],[1,1]],\"x\":[0,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=zero_derivative, complete model={\"coefficients\":[9,0,1],\"x\":0}. Give (mode, current_iterate, function_value, derivative_value, status, update_qualification_claim, boundary).","answer":{"kind":"expression","value":"(zero_derivative, 0, 9, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (zero_derivative, 0, 9, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0085","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6d14caa71ab3465898ad2aec9a884a4dd53a8ef4dbcbb88f48af6c7a9225138","family_id":"qf-mod50-newton-boundary","seed":202611191085,"semantic_template":"newton-boundary|{\"mode\":\"finite_trace_only\",\"model\":{\"coefficients\":[[-10,1],[0,1],[1,1]],\"steps\":1,\"x0\":[1,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=finite_trace_only, complete model={\"coefficients\":[-10,0,1],\"x0\":1,\"steps\":1}. Give (mode, step_budget, step_records, status, convergence_or_rate_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_trace_only, 1, [[0, 1, -9, 2, updated, 11/2]], step_budget_exhausted, No, finite_residual_behavior_does_not_prove_convergence_or_rate)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (finite_trace_only, 1, [[0, 1, -9, 2, updated, 11/2]], step_budget_exhausted, No, finite_residual_behavior_does_not_prove_convergence_or_rate)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0086","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ba278818ab49e8c215058d0ef6d11c7c7eeebdeae5680fbc5041e037ba7d0e3","family_id":"qf-mod50-newton-boundary","seed":202611191086,"semantic_template":"newton-boundary|{\"mode\":\"root_stop\",\"model\":{\"coefficients\":[[-11,1],[1,1]],\"x\":[11,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=root_stop, complete model={\"coefficients\":[-11,1],\"x\":11}. Give (mode, current_iterate, function_value, derivative_value, root_first_status, root_stop_qualified).","answer":{"kind":"expression","value":"(root_stop, 11, 0, 1, root_found_before_derivative_division, Yes)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (root_stop, 11, 0, 1, root_found_before_derivative_division, Yes)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0087","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eb30c4b8e888347dc27d4bd3a9142c948867342cf9d9cbc8c3d296974d162dbf","family_id":"qf-mod50-newton-boundary","seed":202611191087,"semantic_template":"newton-boundary|{\"mode\":\"residual_forward_confusion\",\"model\":{\"approximation\":[14,1],\"intercept\":[-12,1],\"slope\":[13,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=residual_forward_confusion, complete model={\"slope\":13,\"intercept\":-12,\"approximation\":14}. Give (mode, slope, exact_root, absolute_residual, forward_error, residual_equals_forward_error, scale_relation).","answer":{"kind":"expression","value":"(residual_forward_confusion, 13, 12/13, 170, 170/13, No, residual_equals_abs_slope_times_forward_error_not_forward_error_itself)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (residual_forward_confusion, 13, 12/13, 170, 170/13, No, residual_equals_abs_slope_times_forward_error_not_forward_error_itself)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0088","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27bc1618d833ceeeaad51abad4e97686e8d5e0f38102220173326677d6134e56","family_id":"qf-mod50-newton-boundary","seed":202611191088,"semantic_template":"newton-boundary|{\"mode\":\"zero_derivative\",\"model\":{\"coefficients\":[[13,1],[0,1],[1,1]],\"x\":[0,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=zero_derivative, complete model={\"coefficients\":[13,0,1],\"x\":0}. Give (mode, current_iterate, function_value, derivative_value, status, update_qualification_claim, boundary).","answer":{"kind":"expression","value":"(zero_derivative, 0, 13, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (zero_derivative, 0, 13, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0089","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"69e0c80c67787be41e4a49bfafc0b4384b0dbebb60edf0efb69bdc5739650b02","family_id":"qf-mod50-newton-boundary","seed":202611191089,"semantic_template":"newton-boundary|{\"mode\":\"finite_trace_only\",\"model\":{\"coefficients\":[[-14,1],[0,1],[1,1]],\"steps\":1,\"x0\":[5,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=finite_trace_only, complete model={\"coefficients\":[-14,0,1],\"x0\":5,\"steps\":1}. Give (mode, step_budget, step_records, status, convergence_or_rate_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_trace_only, 1, [[0, 5, 11, 10, updated, 39/10]], step_budget_exhausted, No, finite_residual_behavior_does_not_prove_convergence_or_rate)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (finite_trace_only, 1, [[0, 5, 11, 10, updated, 39/10]], step_budget_exhausted, No, finite_residual_behavior_does_not_prove_convergence_or_rate)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0090","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4687253381e2fbfad9ccb5de1ec639c2597d775d80394d4c39a25151451a7166","family_id":"qf-mod50-newton-boundary","seed":202611191090,"semantic_template":"newton-boundary|{\"mode\":\"root_stop\",\"model\":{\"coefficients\":[[-15,1],[1,1]],\"x\":[15,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=root_stop, complete model={\"coefficients\":[-15,1],\"x\":15}. Give (mode, current_iterate, function_value, derivative_value, root_first_status, root_stop_qualified).","answer":{"kind":"expression","value":"(root_stop, 15, 0, 1, root_found_before_derivative_division, Yes)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (root_stop, 15, 0, 1, root_found_before_derivative_division, Yes)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0091","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2838dfab9d1890b326afeab1de0480dd6034acec91bf8f284ca914aaec820d20","family_id":"qf-mod50-newton-boundary","seed":202611191091,"semantic_template":"newton-boundary|{\"mode\":\"residual_forward_confusion\",\"model\":{\"approximation\":[18,1],\"intercept\":[-16,1],\"slope\":[17,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=residual_forward_confusion, complete model={\"slope\":17,\"intercept\":-16,\"approximation\":18}. Give (mode, slope, exact_root, absolute_residual, forward_error, residual_equals_forward_error, scale_relation).","answer":{"kind":"expression","value":"(residual_forward_confusion, 17, 16/17, 290, 290/17, No, residual_equals_abs_slope_times_forward_error_not_forward_error_itself)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (residual_forward_confusion, 17, 16/17, 290, 290/17, No, residual_equals_abs_slope_times_forward_error_not_forward_error_itself)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0092","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c2d92e7a5efffbbc911c32bbd8984eff3f5a97bcedd6fa104bc1a606839a1902","family_id":"qf-mod50-newton-boundary","seed":202611191092,"semantic_template":"newton-boundary|{\"mode\":\"zero_derivative\",\"model\":{\"coefficients\":[[17,1],[0,1],[1,1]],\"x\":[0,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=zero_derivative, complete model={\"coefficients\":[17,0,1],\"x\":0}. Give (mode, current_iterate, function_value, derivative_value, status, update_qualification_claim, boundary).","answer":{"kind":"expression","value":"(zero_derivative, 0, 17, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (zero_derivative, 0, 17, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0093","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d08fdc5c8e64be42459af1d12d9f10a7f25101a3f84627d36a35a0332c00cce0","family_id":"qf-mod50-newton-boundary","seed":202611191093,"semantic_template":"newton-boundary|{\"mode\":\"finite_trace_only\",\"model\":{\"coefficients\":[[-18,1],[0,1],[1,1]],\"steps\":1,\"x0\":[4,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=finite_trace_only, complete model={\"coefficients\":[-18,0,1],\"x0\":4,\"steps\":1}. Give (mode, step_budget, step_records, status, convergence_or_rate_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_trace_only, 1, [[0, 4, -2, 8, updated, 17/4]], step_budget_exhausted, No, finite_residual_behavior_does_not_prove_convergence_or_rate)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (finite_trace_only, 1, [[0, 4, -2, 8, updated, 17/4]], step_budget_exhausted, No, finite_residual_behavior_does_not_prove_convergence_or_rate)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0094","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e0f9eed440e01da3b6ee3b709b80ec69514aa697f65cfe6c246b6b669ec05b29","family_id":"qf-mod50-newton-boundary","seed":202611191094,"semantic_template":"newton-boundary|{\"mode\":\"root_stop\",\"model\":{\"coefficients\":[[-19,1],[1,1]],\"x\":[19,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=root_stop, complete model={\"coefficients\":[-19,1],\"x\":19}. Give (mode, current_iterate, function_value, derivative_value, root_first_status, root_stop_qualified).","answer":{"kind":"expression","value":"(root_stop, 19, 0, 1, root_found_before_derivative_division, Yes)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (root_stop, 19, 0, 1, root_found_before_derivative_division, Yes)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0095","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c6d847c25353b3eabf704ec3fc88f0f3a5d0738d1ec27d8fe8f23de59be53305","family_id":"qf-mod50-newton-boundary","seed":202611191095,"semantic_template":"newton-boundary|{\"mode\":\"residual_forward_confusion\",\"model\":{\"approximation\":[22,1],\"intercept\":[-20,1],\"slope\":[21,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=residual_forward_confusion, complete model={\"slope\":21,\"intercept\":-20,\"approximation\":22}. Give (mode, slope, exact_root, absolute_residual, forward_error, residual_equals_forward_error, scale_relation).","answer":{"kind":"expression","value":"(residual_forward_confusion, 21, 20/21, 442, 442/21, No, residual_equals_abs_slope_times_forward_error_not_forward_error_itself)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (residual_forward_confusion, 21, 20/21, 442, 442/21, No, residual_equals_abs_slope_times_forward_error_not_forward_error_itself)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0096","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9083c8674acf9d9f43c5564d8bdcda2a374f2b8d665e648d320a9ba2ee87cccf","family_id":"qf-mod50-newton-boundary","seed":202611191096,"semantic_template":"newton-boundary|{\"mode\":\"zero_derivative\",\"model\":{\"coefficients\":[[21,1],[0,1],[1,1]],\"x\":[0,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=zero_derivative, complete model={\"coefficients\":[21,0,1],\"x\":0}. Give (mode, current_iterate, function_value, derivative_value, status, update_qualification_claim, boundary).","answer":{"kind":"expression","value":"(zero_derivative, 0, 21, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (zero_derivative, 0, 21, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0097","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"58dfea185cf0077a6bffe366f4489e753a456df32300f29d0ba913bdc74c5a0c","family_id":"qf-mod50-newton-boundary","seed":202611191097,"semantic_template":"newton-boundary|{\"mode\":\"finite_trace_only\",\"model\":{\"coefficients\":[[-22,1],[0,1],[1,1]],\"steps\":1,\"x0\":[3,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=finite_trace_only, complete model={\"coefficients\":[-22,0,1],\"x0\":3,\"steps\":1}. Give (mode, step_budget, step_records, status, convergence_or_rate_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_trace_only, 1, [[0, 3, -13, 6, updated, 31/6]], step_budget_exhausted, No, finite_residual_behavior_does_not_prove_convergence_or_rate)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (finite_trace_only, 1, [[0, 3, -13, 6, updated, 31/6]], step_budget_exhausted, No, finite_residual_behavior_does_not_prove_convergence_or_rate)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0098","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"155ec524ae7f1d93b444f92b4c8a5ae0b0eb969365eb0e994eb7f9d8115f78da","family_id":"qf-mod50-newton-boundary","seed":202611191098,"semantic_template":"newton-boundary|{\"mode\":\"root_stop\",\"model\":{\"coefficients\":[[-23,1],[1,1]],\"x\":[23,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=root_stop, complete model={\"coefficients\":[-23,1],\"x\":23}. Give (mode, current_iterate, function_value, derivative_value, root_first_status, root_stop_qualified).","answer":{"kind":"expression","value":"(root_stop, 23, 0, 1, root_found_before_derivative_division, Yes)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (root_stop, 23, 0, 1, root_found_before_derivative_division, Yes)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0099","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f6c0a0cff615b9c4c4d608d72cdf8ec29f7cb3acb10b7b39c78ea3b641aa16f7","family_id":"qf-mod50-newton-boundary","seed":202611191099,"semantic_template":"newton-boundary|{\"mode\":\"residual_forward_confusion\",\"model\":{\"approximation\":[26,1],\"intercept\":[-24,1],\"slope\":[25,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=residual_forward_confusion, complete model={\"slope\":25,\"intercept\":-24,\"approximation\":26}. Give (mode, slope, exact_root, absolute_residual, forward_error, residual_equals_forward_error, scale_relation).","answer":{"kind":"expression","value":"(residual_forward_confusion, 25, 24/25, 626, 626/25, No, residual_equals_abs_slope_times_forward_error_not_forward_error_itself)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (residual_forward_confusion, 25, 24/25, 626, 626/25, No, residual_equals_abs_slope_times_forward_error_not_forward_error_itself)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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-pulse-20261119-r01-0100","batch_id":"batch-20261015-advance-20261119-r01-newton-method-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c315104977b41ddadbaac5a9fc7cf49d5d9653d2c0ca3c212d96c4973604124","family_id":"qf-mod50-newton-boundary","seed":202611191100,"semantic_template":"newton-boundary|{\"mode\":\"zero_derivative\",\"model\":{\"coefficients\":[[25,1],[0,1],[1,1]],\"x\":[0,1]}}"},"fields":{"title":"Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct","stem":"Use mode=zero_derivative, complete model={\"coefficients\":[25,0,1],\"x\":0}. Give (mode, current_iterate, function_value, derivative_value, status, update_qualification_claim, boundary).","answer":{"kind":"expression","value":"(zero_derivative, 0, 25, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)","explanation":"A nonroot with zero derivative has no Newton update and is not silently repaired. An exact root stops even if the derivative is0. A finite trace is not a convergence/rate proof. For a nonzero affine slope, residual=abs(slope)*forward_error; numerical equality in one row does not identify the two quantities universally. The exact result, retaining the requested order, is (zero_derivative, 0, 25, 0, undefined_zero_derivative, No, division_by_zero_is_rejected_not_repaired)."},"skills":["Root-First Stopping, Zero Derivatives and Residual Scaling Stay Distinct"],"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."}}