# Exact Polynomial Remainders, Lagrange Witnesses and Qualified Bounds

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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations · 2026-10-15 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.

Field order uses a source-bound presentation supplement; canonical JSONL and mathematical answers are unchanged.

[Field-order and source-binding data](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[2, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([2, 3], 0, 1/2, 1, [2, 3, 0], [2, 3/2], 7/2, 7/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([2, 3], 0, 1/2, 1, [2, 3, 0], [2, 3/2], 7/2, 7/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0001/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 2. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[3, 5, 7] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([3, 5, 7], 0, 1/2, 2, [3, 5, 14, 0], [3, 5/2, 7/4], 29/4, 29/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([3, 5, 7], 0, 1/2, 2, [3, 5, 14, 0], [3, 5/2, 7/4], 29/4, 29/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0002/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 3. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[4, 7, 10, 13] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([4, 7, 10, 13], 0, 1/2, 3, [4, 7, 20, 78, 0], [4, 7/2, 5/2, 13/8], 93/8, 93/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([4, 7, 10, 13], 0, 1/2, 3, [4, 7, 20, 78, 0], [4, 7/2, 5/2, 13/8], 93/8, 93/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0003/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 4. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[5, 9, 13, 17, 4] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([5, 9, 13, 17, 4], 0, 1/2, 4, [5, 9, 26, 102, 96, 0], [5, 9/2, 13/4, 17/8, 1/4], 121/8, 121/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([5, 9, 13, 17, 4], 0, 1/2, 4, [5, 9, 26, 102, 96, 0], [5, 9/2, 13/4, 17/8, 1/4], 121/8, 121/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0004/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 5. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[6, 11, 16, 4, 9, 14] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([6, 11, 16, 4, 9, 14], 0, 1/2, 5, [6, 11, 32, 24, 216, 1680, 0], [6, 11/2, 4, 1/2, 9/16, 7/16], 17, 17, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([6, 11, 16, 4, 9, 14], 0, 1/2, 5, [6, 11, 32, 24, 216, 1680, 0], [6, 11/2, 4, 1/2, 9/16, 7/16], 17, 17, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0005/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 6. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[7, 13] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([7, 13], 0, 1/2, 1, [7, 13, 0], [7, 13/2], 27/2, 27/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([7, 13], 0, 1/2, 1, [7, 13, 0], [7, 13/2], 27/2, 27/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0006/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 7. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[8, 15, 5] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([8, 15, 5], 0, 1/2, 2, [8, 15, 10, 0], [8, 15/2, 5/4], 67/4, 67/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([8, 15, 5], 0, 1/2, 2, [8, 15, 10, 0], [8, 15/2, 5/4], 67/4, 67/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0007/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 8. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[9, 17, 8, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([9, 17, 8, 16], 0, 1/2, 3, [9, 17, 16, 96, 0], [9, 17/2, 2, 2], 43/2, 43/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([9, 17, 8, 16], 0, 1/2, 3, [9, 17, 16, 96, 0], [9, 17/2, 2, 2], 43/2, 43/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0008/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 9. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[10, 2, 11, 3, 12] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([10, 2, 11, 3, 12], 0, 1/2, 4, [10, 2, 22, 18, 288, 0], [10, 1, 11/4, 3/8, 3/4], 119/8, 119/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([10, 2, 11, 3, 12], 0, 1/2, 4, [10, 2, 22, 18, 288, 0], [10, 1, 11/4, 3/8, 3/4], 119/8, 119/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0009/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 10. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[11, 4, 14, 7, 17, 10] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([11, 4, 14, 7, 17, 10], 0, 1/2, 5, [11, 4, 28, 42, 408, 1200, 0], [11, 2, 7/2, 7/8, 17/16, 5/16], 75/4, 75/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([11, 4, 14, 7, 17, 10], 0, 1/2, 5, [11, 4, 28, 42, 408, 1200, 0], [11, 2, 7/2, 7/8, 17/16, 5/16], 75/4, 75/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0010/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 11. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[12, 6] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([12, 6], 0, 1/2, 1, [12, 6, 0], [12, 3], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([12, 6], 0, 1/2, 1, [12, 6, 0], [12, 3], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0011/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 12. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[13, 8, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([13, 8, 3], 0, 1/2, 2, [13, 8, 6, 0], [13, 4, 3/4], 71/4, 71/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([13, 8, 3], 0, 1/2, 2, [13, 8, 6, 0], [13, 4, 3/4], 71/4, 71/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0012/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 13. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[14, 10, 6, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([14, 10, 6, 2], 0, 1/2, 3, [14, 10, 12, 12, 0], [14, 5, 3/2, 1/4], 83/4, 83/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([14, 10, 6, 2], 0, 1/2, 3, [14, 10, 12, 12, 0], [14, 5, 3/2, 1/4], 83/4, 83/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0013/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 14. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[15, 12, 9, 6, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([15, 12, 9, 6, 3], 0, 1/2, 4, [15, 12, 18, 36, 72, 0], [15, 6, 9/4, 3/4, 3/16], 387/16, 387/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([15, 12, 9, 6, 3], 0, 1/2, 4, [15, 12, 18, 36, 72, 0], [15, 6, 9/4, 3/4, 3/16], 387/16, 387/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0014/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 15. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[16, 14, 12, 10, 8, 6] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([16, 14, 12, 10, 8, 6], 0, 1/2, 5, [16, 14, 24, 60, 192, 720, 0], [16, 7, 3, 5/4, 1/2, 3/16], 447/16, 447/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([16, 14, 12, 10, 8, 6], 0, 1/2, 5, [16, 14, 24, 60, 192, 720, 0], [16, 7, 3, 5/4, 1/2, 3/16], 447/16, 447/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0015/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 16. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[17, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([17, 16], 0, 1/2, 1, [17, 16, 0], [17, 8], 25, 25, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([17, 16], 0, 1/2, 1, [17, 16, 0], [17, 8], 25, 25, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0016/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 17. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[1, 1, 1] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([1, 1, 1], 0, 1/2, 2, [1, 1, 2, 0], [1, 1/2, 1/4], 7/4, 7/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([1, 1, 1], 0, 1/2, 2, [1, 1, 2, 0], [1, 1/2, 1/4], 7/4, 7/4, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0017/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 18. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[2, 3, 4, 5] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([2, 3, 4, 5], 0, 1/2, 3, [2, 3, 8, 30, 0], [2, 3/2, 1, 5/8], 41/8, 41/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([2, 3, 4, 5], 0, 1/2, 3, [2, 3, 8, 30, 0], [2, 3/2, 1, 5/8], 41/8, 41/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0018/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 19. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[3, 5, 7, 9, 11] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([3, 5, 7, 9, 11], 0, 1/2, 4, [3, 5, 14, 54, 264, 0], [3, 5/2, 7/4, 9/8, 11/16], 145/16, 145/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([3, 5, 7, 9, 11], 0, 1/2, 4, [3, 5, 14, 54, 264, 0], [3, 5/2, 7/4, 9/8, 11/16], 145/16, 145/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0019/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 20. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[4, 7, 10, 13, 16, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([4, 7, 10, 13, 16, 2], 0, 1/2, 5, [4, 7, 20, 78, 384, 240, 0], [4, 7/2, 5/2, 13/8, 1, 1/16], 203/16, 203/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([4, 7, 10, 13, 16, 2], 0, 1/2, 5, [4, 7, 20, 78, 384, 240, 0], [4, 7/2, 5/2, 13/8, 1, 1/16], 203/16, 203/16, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0020/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 21. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[5, 9] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([5, 9], 0, 1/2, 1, [5, 9, 0], [5, 9/2], 19/2, 19/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([5, 9], 0, 1/2, 1, [5, 9, 0], [5, 9/2], 19/2, 19/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0021/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 22. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[6, 11, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([6, 11, 16], 0, 1/2, 2, [6, 11, 32, 0], [6, 11/2, 4], 31/2, 31/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([6, 11, 16], 0, 1/2, 2, [6, 11, 32, 0], [6, 11/2, 4], 31/2, 31/2, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0022/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 23. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[7, 13, 2, 8] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([7, 13, 2, 8], 0, 1/2, 3, [7, 13, 4, 48, 0], [7, 13/2, 1/2, 1], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([7, 13, 2, 8], 0, 1/2, 3, [7, 13, 4, 48, 0], [7, 13/2, 1/2, 1], 15, 15, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0023/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 24. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[8, 15, 5, 12, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([8, 15, 5, 12, 2], 0, 1/2, 4, [8, 15, 10, 72, 48, 0], [8, 15/2, 5/4, 3/2, 1/8], 147/8, 147/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([8, 15, 5, 12, 2], 0, 1/2, 4, [8, 15, 10, 72, 48, 0], [8, 15/2, 5/4, 3/2, 1/8], 147/8, 147/8, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0024/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 25. An Exact Center Jet and Taylor Error Do Not Prove Interval-Wide Hypotheses

Coefficients=[9, 17, 8, 16, 7, 15] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (coefficients, center, x, order, center_derivatives, Taylor_terms, Taylor_value, function_value, remainder, boundary).

View answer and explanation([9, 17, 8, 16, 7, 15], 0, 1/2, 5, [9, 17, 16, 96, 168, 1800, 0], [9, 17/2, 2, 2, 7/16, 15/32], 717/32, 717/32, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet)

Retain center derivatives through order+1, every Taylor term, their sum, the exact function value and their difference. A finite jet at the center is not the general interval-wide differentiability or derivative-bound obligation. The exact result, retaining the requested order, is ([9, 17, 8, 16, 7, 15], 0, 1/2, 5, [9, 17, 16, 96, 168, 1800, 0], [9, 17/2, 2, 2, 7/16, 15/32], 717/32, 717/32, 0, interval_wide_n_plus_one_differentiability_is_separate_from_center_jet).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0025/)

Taylor remainder theorem · qf-mod47-taylor-qualification · candidate





## 26. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[2, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(1, 1, [2, 3/2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (1, 1, [2, 3/2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0026/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 27. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[3, 5, 7] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(2, 2, [3, 5/2, 7/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (2, 2, [3, 5/2, 7/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0027/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 28. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[4, 7, 10, 13] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(3, 3, [4, 7/2, 5/2, 13/8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (3, 3, [4, 7/2, 5/2, 13/8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0028/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 29. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[5, 9, 13, 17, 4] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(4, 4, [5, 9/2, 13/4, 17/8, 1/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (4, 4, [5, 9/2, 13/4, 17/8, 1/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0029/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 30. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[6, 11, 16, 4, 9, 14] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(5, 5, [6, 11/2, 4, 1/2, 9/16, 7/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (5, 5, [6, 11/2, 4, 1/2, 9/16, 7/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0030/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 31. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[7, 13] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(1, 1, [7, 13/2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (1, 1, [7, 13/2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0031/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 32. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[8, 15, 5] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(2, 2, [8, 15/2, 5/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (2, 2, [8, 15/2, 5/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0032/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 33. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[9, 17, 8, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(3, 3, [9, 17/2, 2, 2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (3, 3, [9, 17/2, 2, 2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0033/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 34. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[10, 2, 11, 3, 12] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(4, 4, [10, 1, 11/4, 3/8, 3/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (4, 4, [10, 1, 11/4, 3/8, 3/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0034/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 35. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[11, 4, 14, 7, 17, 10] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(5, 5, [11, 2, 7/2, 7/8, 17/16, 5/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (5, 5, [11, 2, 7/2, 7/8, 17/16, 5/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0035/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 36. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[12, 6] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(1, 1, [12, 3], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (1, 1, [12, 3], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0036/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 37. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[13, 8, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(2, 2, [13, 4, 3/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (2, 2, [13, 4, 3/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0037/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 38. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[14, 10, 6, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(3, 3, [14, 5, 3/2, 1/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (3, 3, [14, 5, 3/2, 1/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0038/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 39. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[15, 12, 9, 6, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(4, 4, [15, 6, 9/4, 3/4, 3/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (4, 4, [15, 6, 9/4, 3/4, 3/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0039/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 40. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[16, 14, 12, 10, 8, 6] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(5, 5, [16, 7, 3, 5/4, 1/2, 3/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (5, 5, [16, 7, 3, 5/4, 1/2, 3/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0040/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 41. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[17, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(1, 1, [17, 8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (1, 1, [17, 8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0041/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 42. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[1, 1, 1] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(2, 2, [1, 1/2, 1/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (2, 2, [1, 1/2, 1/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0042/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 43. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[2, 3, 4, 5] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(3, 3, [2, 3/2, 1, 5/8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (3, 3, [2, 3/2, 1, 5/8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0043/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 44. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[3, 5, 7, 9, 11] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(4, 4, [3, 5/2, 7/4, 9/8, 11/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (4, 4, [3, 5/2, 7/4, 9/8, 11/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0044/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 45. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[4, 7, 10, 13, 16, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(5, 5, [4, 7/2, 5/2, 13/8, 1, 1/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (5, 5, [4, 7/2, 5/2, 13/8, 1, 1/16], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0045/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 46. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[5, 9] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(1, 1, [5, 9/2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (1, 1, [5, 9/2], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0046/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 47. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[6, 11, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(2, 2, [6, 11/2, 4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (2, 2, [6, 11/2, 4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0047/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 48. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[7, 13, 2, 8] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(3, 3, [7, 13/2, 1/2, 1], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (3, 3, [7, 13/2, 1/2, 1], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0048/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 49. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[8, 15, 5, 12, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(4, 4, [8, 15/2, 5/4, 3/2, 1/8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (4, 4, [8, 15/2, 5/4, 3/2, 1/8], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0049/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 50. A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Coefficients=[9, 17, 8, 16, 7, 15] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, Taylor_terms, remainder, zero_remainder, certificate).

View answer and explanation(5, 5, [9, 17/2, 2, 2, 7/16, 15/32], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

Require degree≤order after removing trailing zeros. Retain all terms through the requested order, including zero terms, and the exact zero remainder. This finite polynomial identity does not establish analyticity of an arbitrary function. The exact result, retaining the requested order, is (5, 5, [9, 17/2, 2, 2, 7/16, 15/32], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0050/)

Taylor remainder theorem · qf-mod47-taylor-zero-remainder · candidate





## 51. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[2, 3, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(2, 1, 4, 1/4, Yes, 1/2, 1/2, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (2, 1, 4, 1/4, Yes, 1/2, 1/2, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0051/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 52. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[3, 5, 7, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(3, 2, 18, 1/4, Yes, 3/8, 3/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (3, 2, 18, 1/4, Yes, 3/8, 3/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0052/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 53. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[4, 7, 10, 13, 4] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(4, 3, 96, 1/4, Yes, 1/4, 1/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (4, 3, 96, 1/4, Yes, 1/4, 1/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0053/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 54. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[5, 9, 13, 17, 4, 5] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(5, 4, 600, 1/4, Yes, 5/32, 5/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (5, 4, 600, 1/4, Yes, 5/32, 5/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0054/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 55. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[6, 11, 16, 4, 9, 14, 6] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(6, 5, 4320, 1/4, Yes, 3/32, 3/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (6, 5, 4320, 1/4, Yes, 3/32, 3/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0055/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 56. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[7, 13, 7] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(2, 1, 14, 1/4, Yes, 7/4, 7/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (2, 1, 14, 1/4, Yes, 7/4, 7/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0056/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 57. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[8, 15, 5, 8] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(3, 2, 48, 1/4, Yes, 1, 1, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (3, 2, 48, 1/4, Yes, 1, 1, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0057/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 58. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[9, 17, 8, 16, 9] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(4, 3, 216, 1/4, Yes, 9/16, 9/16, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (4, 3, 216, 1/4, Yes, 9/16, 9/16, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0058/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 59. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[10, 2, 11, 3, 12, 10] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(5, 4, 1200, 1/4, Yes, 5/16, 5/16, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (5, 4, 1200, 1/4, Yes, 5/16, 5/16, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0059/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 60. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[11, 4, 14, 7, 17, 10, 11] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(6, 5, 7920, 1/4, Yes, 11/64, 11/64, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (6, 5, 7920, 1/4, Yes, 11/64, 11/64, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0060/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 61. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[12, 6, 12] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(2, 1, 24, 1/4, Yes, 3, 3, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (2, 1, 24, 1/4, Yes, 3, 3, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0061/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 62. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[13, 8, 3, 13] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(3, 2, 78, 1/4, Yes, 13/8, 13/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (3, 2, 78, 1/4, Yes, 13/8, 13/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0062/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 63. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[14, 10, 6, 2, 14] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(4, 3, 336, 1/4, Yes, 7/8, 7/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (4, 3, 336, 1/4, Yes, 7/8, 7/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0063/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 64. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[15, 12, 9, 6, 3, 15] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(5, 4, 1800, 1/4, Yes, 15/32, 15/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (5, 4, 1800, 1/4, Yes, 15/32, 15/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0064/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 65. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[16, 14, 12, 10, 8, 6, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(6, 5, 11520, 1/4, Yes, 1/4, 1/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (6, 5, 11520, 1/4, Yes, 1/4, 1/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0065/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 66. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[17, 16, 17] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(2, 1, 34, 1/4, Yes, 17/4, 17/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (2, 1, 34, 1/4, Yes, 17/4, 17/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0066/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 67. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[1, 1, 1, 18] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(3, 2, 108, 1/4, Yes, 9/4, 9/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (3, 2, 108, 1/4, Yes, 9/4, 9/4, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0067/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 68. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[2, 3, 4, 5, 19] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(4, 3, 456, 1/4, Yes, 19/16, 19/16, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (4, 3, 456, 1/4, Yes, 19/16, 19/16, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0068/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 69. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[3, 5, 7, 9, 11, 20] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(5, 4, 2400, 1/4, Yes, 5/8, 5/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (5, 4, 2400, 1/4, Yes, 5/8, 5/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0069/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 70. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[4, 7, 10, 13, 16, 2, 21] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(6, 5, 15120, 1/4, Yes, 21/64, 21/64, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (6, 5, 15120, 1/4, Yes, 21/64, 21/64, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0070/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 71. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[5, 9, 22] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(2, 1, 44, 1/4, Yes, 11/2, 11/2, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (2, 1, 44, 1/4, Yes, 11/2, 11/2, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0071/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 72. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[6, 11, 16, 23] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(3, 2, 138, 1/4, Yes, 23/8, 23/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (3, 2, 138, 1/4, Yes, 23/8, 23/8, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0072/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 73. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[7, 13, 2, 8, 24] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(4, 3, 576, 1/4, Yes, 3/2, 3/2, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (4, 3, 576, 1/4, Yes, 3/2, 3/2, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0073/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 74. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[8, 15, 5, 12, 2, 25] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(5, 4, 3000, 1/4, Yes, 25/32, 25/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (5, 4, 3000, 1/4, Yes, 25/32, 25/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0074/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 75. A Constant Next Derivative Gives a Strictly Intermediate Lagrange Witness

Coefficients=[9, 17, 8, 16, 7, 15, 26] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Give (degree, order, constant_next_derivative, intermediate_point, strictly_between, Lagrange_remainder, actual_remainder, remainders_equal, certificate).

View answer and explanation(6, 5, 18720, 1/4, Yes, 13/32, 13/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness)

Require degree=order+1 and x distinct from the center. The next derivative is constant, so the midpoint lies strictly between center and x in either orientation and gives the exact Lagrange remainder. Do not extend this construction to an arbitrary nonconstant next derivative. The exact result, retaining the requested order, is (6, 5, 18720, 1/4, Yes, 13/32, 13/32, Yes, constant_n_plus_one_derivative_gives_exact_strictly_intermediate_Lagrange_witness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0075/)

Taylor remainder theorem · qf-mod47-taylor-lagrange-witness · candidate





## 76. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[2, 3, 2] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=4≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(valid_bound, 1/2, 4, 4, 1/2, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 1/2, 4, 4, 1/2, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0076/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 77. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[3, 5, 7, 3] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=17≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(undersized_M, 3/8, 18, 17, 17/48, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 3/8, 18, 17, 17/48, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0077/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 78. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[4, 7, 10, 13, 4] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=96≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(valid_bound, 1/4, 96, 96, 1/4, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 1/4, 96, 96, 1/4, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0078/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 79. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[5, 9, 13, 17, 4, 5] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=599≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(undersized_M, 5/32, 600, 599, 599/3840, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 5/32, 600, 599, 599/3840, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0079/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 80. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[6, 11, 16, 4, 9, 14, 6] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=4320≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(valid_bound, 3/32, 4320, 4320, 3/32, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 3/32, 4320, 4320, 3/32, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0080/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 81. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[7, 13, 7] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=13≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(undersized_M, 7/4, 14, 13, 13/8, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 7/4, 14, 13, 13/8, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0081/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 82. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[8, 15, 5, 8] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=48≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(valid_bound, 1, 48, 48, 1, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 1, 48, 48, 1, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0082/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 83. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[9, 17, 8, 16, 9] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=215≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(undersized_M, 9/16, 216, 215, 215/384, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 9/16, 216, 215, 215/384, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0083/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 84. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[10, 2, 11, 3, 12, 10] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=1200≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(valid_bound, 5/16, 1200, 1200, 5/16, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 5/16, 1200, 1200, 5/16, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0084/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 85. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[11, 4, 14, 7, 17, 10, 11] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=7919≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(undersized_M, 11/64, 7920, 7919, 7919/46080, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 11/64, 7920, 7919, 7919/46080, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0085/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 86. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[12, 6, 12] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=24≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(valid_bound, 3, 24, 24, 3, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 3, 24, 24, 3, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0086/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 87. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[13, 8, 3, 13] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=77≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(undersized_M, 13/8, 78, 77, 77/48, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 13/8, 78, 77, 77/48, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0087/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 88. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[14, 10, 6, 2, 14] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=336≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(valid_bound, 7/8, 336, 336, 7/8, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 7/8, 336, 336, 7/8, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0088/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 89. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[15, 12, 9, 6, 3, 15] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=1799≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(undersized_M, 15/32, 1800, 1799, 1799/3840, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 15/32, 1800, 1799, 1799/3840, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0089/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 90. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[16, 14, 12, 10, 8, 6, 16] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=11520≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(valid_bound, 1/4, 11520, 11520, 1/4, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 1/4, 11520, 11520, 1/4, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0090/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 91. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[17, 16, 17] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=33≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(undersized_M, 17/4, 34, 33, 33/8, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 17/4, 34, 33, 33/8, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0091/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 92. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[1, 1, 1, 18] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=108≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(valid_bound, 9/4, 108, 108, 9/4, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 9/4, 108, 108, 9/4, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0092/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 93. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[2, 3, 4, 5, 19] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=455≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(undersized_M, 19/16, 456, 455, 455/384, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 19/16, 456, 455, 455/384, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0093/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 94. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[3, 5, 7, 9, 11, 20] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=2400≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(valid_bound, 5/8, 2400, 2400, 5/8, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 5/8, 2400, 2400, 5/8, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0094/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 95. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[4, 7, 10, 13, 16, 2, 21] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=15119≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(undersized_M, 21/64, 15120, 15119, 15119/46080, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 21/64, 15120, 15119, 15119/46080, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0095/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 96. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[5, 9, 22] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=1; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=44≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(valid_bound, 11/2, 44, 44, 11/2, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 11/2, 44, 44, 11/2, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0096/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 97. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[6, 11, 16, 23] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=2; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=137≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(undersized_M, 23/8, 138, 137, 137/48, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 23/8, 138, 137, 137/48, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0097/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 98. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[7, 13, 2, 8, 24] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=3; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=576≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(valid_bound, 3/2, 576, 576, 3/2, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 3/2, 576, 576, 3/2, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0098/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 99. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[8, 15, 5, 12, 2, 25] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=4; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=undersized_M, supplied M=2999≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(undersized_M, 25/32, 3000, 2999, 2999/3840, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (undersized_M, 25/32, 3000, 2999, 2999/3840, No, No, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0099/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate





## 100. An Actual Error Inequality Is Different from a Qualified Derivative Bound

Coefficients=[9, 17, 8, 16, 7, 15, 26] are in ascending power order, with exact rationals and trailing zero coefficients removed for degree. Center=0, point=1/2, order=5; keep terms through that order, and define remainder=f(x)−Taylor_value. Mode=valid_bound, supplied M=18720≥0. Give (mode, remainder, derivative_supremum, M, proposed_error_bound, actual_error_inequality, qualified_bound_certificate, general_theorem_or_analyticity_claim, boundary).

actual_error_inequality differs from qualified_bound_certificate: at x=center even an undersized M can satisfy0≤0 without qualifying the derivative bound.

View answer and explanation(valid_bound, 13/32, 18720, 18720, 13/32, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity)

The degree=order+1 polynomial has a constant next derivative, whose absolute value is the exact interval supremum D. valid_bound requires M≥D; undersized_M requires M<D. Test the actual absolute error against M*|x−center|^(order+1)/(order+1)! separately. At x=center an undersized M can still satisfy0≤0 without qualifying the derivative bound. Neither a finite jet nor a few samples prove the general remainder theorem or analyticity. The exact result, retaining the requested order, is (valid_bound, 13/32, 18720, 18720, 13/32, Yes, Yes, No, exact_constant_n_plus_one_derivative_controls_the_bound_finite_jet_does_not_prove_general_theorem_or_analyticity).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261116-r03-taylor-remainder-theorem-0100/)

Taylor remainder theorem · qf-mod47-taylor-bound-boundary · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261116-r03-taylor-remainder-theorem-foundations/validation-report.json)