{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0001","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5747e4f0ae8091fb322f7bd8731adc856a799001aac8d106d7ef92dd70807a96","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032001,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[1,-2],\"limit\":[0,1],\"n\":1,\"point\":0,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[0, 1], coefficients=[1, -2], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=1, point=0. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(1, 0, 1/3, 0, 1/3)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (1, 0, 1/3, 0, 1/3)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0002","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cbad1a460310c4cd2dc7eb648c3a2f69c223cbbd68e0d44544884461c63f6e66","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032002,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[2,-3],\"limit\":[1,2],\"n\":2,\"point\":1,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[1, 2], coefficients=[2, -3], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=2, point=1. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(2, 1, 5/4, 2, -3/4)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (2, 1, 5/4, 2, -3/4)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0003","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0314891c051f172e9a44ee1b68d48c2380a22de86537c3035b8bf3f4bc86a59f","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032003,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[3,-4],\"limit\":[2,3],\"n\":3,\"point\":0,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[2, 3], coefficients=[3, -4], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=3, point=0. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(3, 0, 13/5, 2, 3/5)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (3, 0, 13/5, 2, 3/5)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0004","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bedd74faa5153b33ba2548143cc57af7e1b2b24c7525c71f2e6db9b770396812","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032004,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[4,-5],\"limit\":[3,4],\"n\":4,\"point\":1,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[3, 4], coefficients=[4, -5], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=4, point=1. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(4, 1, 19/6, 4, -5/6)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (4, 1, 19/6, 4, -5/6)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0005","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ab0841a5b34cf395e9e4daad4d22694d519c29881bdc62aef0aa50d7ad440001","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032005,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[5,-6],\"limit\":[4,5],\"n\":5,\"point\":0,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[4, 5], coefficients=[5, -6], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=5, point=0. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(5, 0, 33/7, 4, 5/7)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (5, 0, 33/7, 4, 5/7)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0006","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a33e955d3dcf022ec1d4479f6a1fb7fe4e7b66b2db29227d4accda530537ee0c","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032006,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[6,-7],\"limit\":[5,6],\"n\":6,\"point\":1,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[5, 6], coefficients=[6, -7], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=6, point=1. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(6, 1, 41/8, 6, -7/8)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (6, 1, 41/8, 6, -7/8)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0007","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"121df2b081f906e3a6a66b7a7aadc45eb213ed067a695bfa6782a3736bcdecbf","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032007,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[7,-8],\"limit\":[6,7],\"n\":7,\"point\":0,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[6, 7], coefficients=[7, -8], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=7, point=0. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(7, 0, 61/9, 6, 7/9)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (7, 0, 61/9, 6, 7/9)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0008","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9c39ee6337b14e448b53afb3fe56f68f7a243d3e21d2b87a134cf5b011961c6e","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032008,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[8,-9],\"limit\":[7,8],\"n\":8,\"point\":1,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[7, 8], coefficients=[8, -9], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=8, point=1. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(8, 1, 71/10, 8, -9/10)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (8, 1, 71/10, 8, -9/10)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0009","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d41358ab8b816f14b4f322665d6b3e1c80be918011292cd2e5c7b511ef31b67c","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032009,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[9,-10],\"limit\":[8,9],\"n\":9,\"point\":0,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[8, 9], coefficients=[9, -10], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=9, point=0. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(9, 0, 97/11, 8, 9/11)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (9, 0, 97/11, 8, 9/11)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0010","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9df40b6cb829b8784a0984b10143f771a1401f03dcce2d8a53636738cc6c2f1b","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032010,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[10,-11],\"limit\":[9,10],\"n\":10,\"point\":1,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[9, 10], coefficients=[10, -11], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=10, point=1. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(10, 1, 109/12, 10, -11/12)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (10, 1, 109/12, 10, -11/12)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0011","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"29c243df353cc973faff4835255d1e9a75d8ce52303d33a0678e81b714766796","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032011,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[11,-12],\"limit\":[10,11],\"n\":11,\"point\":0,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[10, 11], coefficients=[11, -12], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=11, point=0. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(11, 0, 141/13, 10, 11/13)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (11, 0, 141/13, 10, 11/13)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0012","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa1760348ede643cbdfcf565a3a6903897efa905d9b656bbdc7884b7736eff8e","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032012,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[12,-13],\"limit\":[11,12],\"n\":12,\"point\":1,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[11, 12], coefficients=[12, -13], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=12, point=1. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(12, 1, 155/14, 12, -13/14)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (12, 1, 155/14, 12, -13/14)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0013","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"487eb114ecf32352cea468aecdeb768e975d08ea03e224de878e2b7af7aa9a39","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032013,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[13,-14],\"limit\":[12,13],\"n\":13,\"point\":0,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[12, 13], coefficients=[13, -14], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=13, point=0. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(13, 0, 193/15, 12, 13/15)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (13, 0, 193/15, 12, 13/15)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0014","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a42208c36bb496e6896bb7b5dd5d4de2d75d81a2b83cbc80a53aac60a39a6ff","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032014,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[14,-15],\"limit\":[13,14],\"n\":14,\"point\":1,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[13, 14], coefficients=[14, -15], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=14, point=1. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(14, 1, 209/16, 14, -15/16)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (14, 1, 209/16, 14, -15/16)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0015","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d3b785e427936ef879de87ae0d2d28919c4842066d3f0dbc2c4d099263cddfd","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032015,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[15,-16],\"limit\":[14,15],\"n\":15,\"point\":0,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[14, 15], coefficients=[15, -16], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=15, point=0. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(15, 0, 253/17, 14, 15/17)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (15, 0, 253/17, 14, 15/17)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0016","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3079868e605df72a382fa117c893ebe98daa8da700bde9d58412399741af6e0a","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032016,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[16,-17],\"limit\":[15,16],\"n\":16,\"point\":1,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[15, 16], coefficients=[16, -17], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=16, point=1. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(16, 1, 271/18, 16, -17/18)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (16, 1, 271/18, 16, -17/18)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0017","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b857bc5a7c28ea576fc44d3c83a7977a2042793095b28d485ff9f650b81d2f93","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032017,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[17,-18],\"limit\":[16,17],\"n\":17,\"point\":0,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[16, 17], coefficients=[17, -18], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=17, point=0. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(17, 0, 321/19, 16, 17/19)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (17, 0, 321/19, 16, 17/19)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0018","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7359c881619cb99e7a3448ff96a958e5b456f99f25b99075c995ddf067320cd5","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032018,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[18,-19],\"limit\":[17,18],\"n\":18,\"point\":1,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[17, 18], coefficients=[18, -19], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=18, point=1. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(18, 1, 341/20, 18, -19/20)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (18, 1, 341/20, 18, -19/20)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0019","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5ed36e2226d852f48bc6df12d65fa835ab48a82b90916eb4f7cc23c6d1c3acc8","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032019,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[19,-20],\"limit\":[18,19],\"n\":19,\"point\":0,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[18, 19], coefficients=[19, -20], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=19, point=0. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(19, 0, 397/21, 18, 19/21)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (19, 0, 397/21, 18, 19/21)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0020","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"beba96a36d4c095bd5ee1e0fcf04d039b8f6e039a248ccf7ab15503564ea513e","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032020,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[20,-21],\"limit\":[19,20],\"n\":20,\"point\":1,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[19, 20], coefficients=[20, -21], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=20, point=1. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(20, 1, 419/22, 20, -21/22)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (20, 1, 419/22, 20, -21/22)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0021","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7d999b7f1e303321513ae5fd1742479254972182d238c008d40e502c0110e21b","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032021,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[21,-22],\"limit\":[20,21],\"n\":21,\"point\":0,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[20, 21], coefficients=[21, -22], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=21, point=0. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(21, 0, 481/23, 20, 21/23)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (21, 0, 481/23, 20, 21/23)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0022","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ac75d2f910c9fc65a034454af1fc06bce565421036323ce4238969368de2e200","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032022,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[22,-23],\"limit\":[21,22],\"n\":22,\"point\":1,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[21, 22], coefficients=[22, -23], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=22, point=1. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(22, 1, 505/24, 22, -23/24)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (22, 1, 505/24, 22, -23/24)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0023","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de4eeba78214abb8c7d47163efbf077fcb45bbd3f4bd6581f4f263246a8d80a4","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032023,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[23,-24],\"limit\":[22,23],\"n\":23,\"point\":0,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[22, 23], coefficients=[23, -24], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=23, point=0. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(23, 0, 573/25, 22, 23/25)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (23, 0, 573/25, 22, 23/25)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0024","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c13fe6634d844c5047cce9e6ecab33047ddd9ab2a61b10918b8260462344dbbc","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032024,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[24,-25],\"limit\":[23,24],\"n\":24,\"point\":1,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[23, 24], coefficients=[24, -25], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=24, point=1. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(24, 1, 599/26, 24, -25/26)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (24, 1, 599/26, 24, -25/26)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0025","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d48ffae663546f28265a15d56f8f63fd8956104740f3e452a1539544f74c6199","family_id":"qf-mod34-pointwise-finite-domain-evaluate","seed":202611032025,"semantic_template":"qf-mod34-pointwise-finite-domain-evaluate/{\"coefficients\":[25,-26],\"limit\":[24,25],\"n\":25,\"point\":0,\"shift\":2}"},"fields":{"title":"Keep the Signed Function Error Separate from Its Absolute Bound","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[24, 25], coefficients=[25, -26], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. n=25, point=0. Give (n,point,f_n_x,limit_x,error), where error=f_n(x)−limit[x] is signed.","answer":{"kind":"expression","value":"(25, 0, 673/27, 24, 25/27)","explanation":"Evaluate the exact rational tail at the selected point. A negative coefficient gives a negative signed error; do not replace that field with its absolute value. One value is not the full pointwise-limit proof. The exact result, retaining the requested order, is (25, 0, 673/27, 24, 25/27)."},"skills":["Keep the Signed Function Error Separate from Its Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}} {"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r02-pointwise-convergence-of-functions-0026","batch_id":"batch-20261014-advance-20261103-r02-pointwise-convergence-of-functions-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f5558015770729964f055b42b17ef35528eff3a773003a6ad5067325a8037c5","family_id":"qf-mod34-pointwise-epsilon-witness","seed":202611032026,"semantic_template":"qf-mod34-pointwise-epsilon-witness/{\"coefficients\":[1,-2],\"epsilon\":[1,10],\"limit\":[0,1],\"point\":0,\"shift\":2}"},"fields":{"title":"A Strict Pointwise Tail Witness Can Depend on the Point","stem":"Use the same complete finite domain Fin(2)={0,…,1}, with aligned limit values=[0, 1], coefficients=[1, -2], positive shift=2, and f_n(x)=limit[x]+coefficients[x]/(n+shift) for every n≥0. Values and tail bounds are exact rationals. Pointwise witnesses may depend on both x and epsilon. A finite maximum of all pointwise witnesses gives a common witness only on this finite domain; it is not a general infinite-domain implication. A finite observed prefix alone is not a convergence certificate. Positive epsilon=1/10, point=0. Give (point,epsilon,N,error_at_N,strict_tail?,forall_n_ge_N), with error_at_N the absolute error bound.","answer":{"kind":"expression","value":"(0, 1/10, 9, 1/11, Yes, forall_n_ge_N)","explanation":"Choose the least nonnegative N with |coefficient[point]|/(N+shift)