Question 26487: A Polynomial of Degree at Most the Taylor Order Has Zero Remainder

Original question and givens

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).

• G001 — 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).

Definition and scope

Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.

• mko-taylor-remainder-theorem

Step-by-step derivation

source roles = [taylor-zero-remainder]

derived = [[degree, order, Taylor_terms, remainder, zero_remainder, certificate]]

Taylor ORIGINAL x minus declared center := sub(1/2, 0)

Taylor ORIGINAL x minus declared center = 1/2

(0)^0

Taylor ORIGINAL derivative0 center power0 term0 = 1

Taylor ORIGINAL derivative0 falling factor times power term0 := mul(1, 1)

Taylor ORIGINAL derivative0 falling factor times power term0 = 1

Taylor ORIGINAL derivative0 ORIGINAL coefficient term0 := mul(13, 1)

Taylor ORIGINAL derivative0 ORIGINAL coefficient term0 = 13

Taylor ORIGINAL derivative0 ascending derivative sum term0 := sum(0, 13)

Taylor ORIGINAL derivative0 ascending derivative sum term0 = 13

(0)^1

Taylor ORIGINAL derivative0 center power1 term1 = 0

Taylor ORIGINAL derivative0 falling factor times power term1 := mul(1, 0)

Taylor ORIGINAL derivative0 falling factor times power term1 = 0

Taylor ORIGINAL derivative0 ORIGINAL coefficient term1 := mul(8, 0)

Taylor ORIGINAL derivative0 ORIGINAL coefficient term1 = 0

Taylor ORIGINAL derivative0 ascending derivative sum term1 := sum(13, 0)

Taylor ORIGINAL derivative0 ascending derivative sum term1 = 13

(0)^2

Taylor ORIGINAL derivative0 center power2 term2 = 0

Taylor ORIGINAL derivative0 falling factor times power term2 := mul(1, 0)

Taylor ORIGINAL derivative0 falling factor times power term2 = 0

Taylor ORIGINAL derivative0 ORIGINAL coefficient term2 := mul(3, 0)

Taylor ORIGINAL derivative0 ORIGINAL coefficient term2 = 0

Taylor ORIGINAL derivative0 ascending derivative sum term2 := sum(13, 0)

Taylor ORIGINAL derivative0 ascending derivative sum term2 = 13

source roles = [0, 0, [13, 8, 3]]

derived = [[[0, 13, 1, 1, 13], [1, 8, 1, 0, 0], [2, 3, 1, 0, 0]], 13]

(1/2)^0

Taylor ORIGINAL displacement power0 = 1

Taylor ORIGINAL derivative times displacement0 := mul(13, 1)

Taylor ORIGINAL derivative times displacement0 = 13

Taylor ORIGINAL term0 divide factorial := div(13, 1)

Taylor ORIGINAL term0 divide factorial = 13

(0)^0

Taylor ORIGINAL derivative1 center power0 term1 = 1

Taylor ORIGINAL derivative1 falling factor times power term1 := mul(1, 1)

Taylor ORIGINAL derivative1 falling factor times power term1 = 1

Taylor ORIGINAL derivative1 ORIGINAL coefficient term1 := mul(8, 1)

Taylor ORIGINAL derivative1 ORIGINAL coefficient term1 = 8

Taylor ORIGINAL derivative1 ascending derivative sum term1 := sum(0, 8)

Taylor ORIGINAL derivative1 ascending derivative sum term1 = 8

(0)^1

Taylor ORIGINAL derivative1 center power1 term2 = 0

Taylor ORIGINAL derivative1 falling factor times power term2 := mul(2, 0)

Taylor ORIGINAL derivative1 falling factor times power term2 = 0

Taylor ORIGINAL derivative1 ORIGINAL coefficient term2 := mul(3, 0)

Taylor ORIGINAL derivative1 ORIGINAL coefficient term2 = 0

Taylor ORIGINAL derivative1 ascending derivative sum term2 := sum(8, 0)

Taylor ORIGINAL derivative1 ascending derivative sum term2 = 8

source roles = [1, 0, [13, 8, 3]]

derived = [[[1, 8, 1, 1, 8], [2, 3, 2, 0, 0]], 8]

(1/2)^1

Taylor ORIGINAL displacement power1 = 1/2

Taylor ORIGINAL derivative times displacement1 := mul(8, 1/2)

Taylor ORIGINAL derivative times displacement1 = 4

Taylor ORIGINAL term1 divide factorial := div(4, 1)

Taylor ORIGINAL term1 divide factorial = 4

(0)^0

Taylor ORIGINAL derivative2 center power0 term2 = 1

Taylor ORIGINAL derivative2 falling factor times power term2 := mul(2, 1)

Taylor ORIGINAL derivative2 falling factor times power term2 = 2

Taylor ORIGINAL derivative2 ORIGINAL coefficient term2 := mul(3, 2)

Taylor ORIGINAL derivative2 ORIGINAL coefficient term2 = 6

Taylor ORIGINAL derivative2 ascending derivative sum term2 := sum(0, 6)

Taylor ORIGINAL derivative2 ascending derivative sum term2 = 6

source roles = [2, 0, [13, 8, 3]]

derived = [[[2, 3, 2, 1, 6]], 6]

(1/2)^2

Taylor ORIGINAL displacement power2 = 1/4

Taylor ORIGINAL derivative times displacement2 := mul(6, 1/4)

Taylor ORIGINAL derivative times displacement2 = 3/2

Taylor ORIGINAL term2 divide factorial := div(3/2, 2)

Taylor ORIGINAL term2 divide factorial = 3/4

Taylor ORIGINAL partial sum0 := sum(0, 13)

Taylor ORIGINAL partial sum0 = 13

Taylor ORIGINAL partial sum1 := sum(13, 4)

Taylor ORIGINAL partial sum1 = 17

Taylor ORIGINAL partial sum2 := sum(17, 3/4)

Taylor ORIGINAL partial sum2 = 71/4

(1/2)^0

Taylor ORIGINAL full function at x ORIGINAL argument power0 = 1

Taylor ORIGINAL full function at x coefficient term0 := mul(13, 1)

Taylor ORIGINAL full function at x coefficient term0 = 13

Taylor ORIGINAL full function at x ascending partial sum0 := sum(0, 13)

Taylor ORIGINAL full function at x ascending partial sum0 = 13

(1/2)^1

Taylor ORIGINAL full function at x ORIGINAL argument power1 = 1/2

Taylor ORIGINAL full function at x coefficient term1 := mul(8, 1/2)

Taylor ORIGINAL full function at x coefficient term1 = 4

Taylor ORIGINAL full function at x ascending partial sum1 := sum(13, 4)

Taylor ORIGINAL full function at x ascending partial sum1 = 17

(1/2)^2

Taylor ORIGINAL full function at x ORIGINAL argument power2 = 1/4

Taylor ORIGINAL full function at x coefficient term2 := mul(3, 1/4)

Taylor ORIGINAL full function at x coefficient term2 = 3/4

Taylor ORIGINAL full function at x ascending partial sum2 := sum(17, 3/4)

Taylor ORIGINAL full function at x ascending partial sum2 = 71/4

Taylor ORIGINAL f(x) minus Taylor value := sub(71/4, 71/4)

Taylor ORIGINAL f(x) minus Taylor value = 0

source roles = [[13, 8, 3], 0, 1/2, 2, [13, 8, 6], [13, 4, 3/4]]

derived = [71/4, 71/4, 0, 2]

Reverse checks

• S052 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S051 · premise_and_result_audit — Recompute Taylor ORIGINAL f(x) minus Taylor value with exact rational arithmetic, not rounded decimals.
• S050 · premise_and_result_audit — Recompute Taylor ORIGINAL full function at x ascending partial sum2 with exact rational arithmetic, not rounded decimals.
• S049 · premise_and_result_audit — Recompute Taylor ORIGINAL full function at x coefficient term2 with exact rational arithmetic, not rounded decimals.
• S048 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S047 · premise_and_result_audit — Recompute Taylor ORIGINAL full function at x ascending partial sum1 with exact rational arithmetic, not rounded decimals.
• S046 · premise_and_result_audit — Recompute Taylor ORIGINAL full function at x coefficient term1 with exact rational arithmetic, not rounded decimals.
• S045 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S044 · premise_and_result_audit — Recompute Taylor ORIGINAL full function at x ascending partial sum0 with exact rational arithmetic, not rounded decimals.
• S043 · premise_and_result_audit — Recompute Taylor ORIGINAL full function at x coefficient term0 with exact rational arithmetic, not rounded decimals.
• S042 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S041 · premise_and_result_audit — Recompute Taylor ORIGINAL partial sum2 with exact rational arithmetic, not rounded decimals.
• S040 · premise_and_result_audit — Recompute Taylor ORIGINAL partial sum1 with exact rational arithmetic, not rounded decimals.
• S039 · premise_and_result_audit — Recompute Taylor ORIGINAL partial sum0 with exact rational arithmetic, not rounded decimals.
• S038 · premise_and_result_audit — Recompute Taylor ORIGINAL term2 divide factorial with exact rational arithmetic, not rounded decimals.
• S037 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative times displacement2 with exact rational arithmetic, not rounded decimals.
• S036 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S035 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S034 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative2 ascending derivative sum term2 with exact rational arithmetic, not rounded decimals.
• S033 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative2 ORIGINAL coefficient term2 with exact rational arithmetic, not rounded decimals.
• S032 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative2 falling factor times power term2 with exact rational arithmetic, not rounded decimals.
• S031 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S030 · premise_and_result_audit — Recompute Taylor ORIGINAL term1 divide factorial with exact rational arithmetic, not rounded decimals.
• S029 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative times displacement1 with exact rational arithmetic, not rounded decimals.
• S028 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S027 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S026 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative1 ascending derivative sum term2 with exact rational arithmetic, not rounded decimals.
• S025 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative1 ORIGINAL coefficient term2 with exact rational arithmetic, not rounded decimals.
• S024 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative1 falling factor times power term2 with exact rational arithmetic, not rounded decimals.
• S023 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S022 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative1 ascending derivative sum term1 with exact rational arithmetic, not rounded decimals.
• S021 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative1 ORIGINAL coefficient term1 with exact rational arithmetic, not rounded decimals.
• S020 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative1 falling factor times power term1 with exact rational arithmetic, not rounded decimals.
• S019 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S018 · premise_and_result_audit — Recompute Taylor ORIGINAL term0 divide factorial with exact rational arithmetic, not rounded decimals.
• S017 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative times displacement0 with exact rational arithmetic, not rounded decimals.
• S016 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S015 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S014 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative0 ascending derivative sum term2 with exact rational arithmetic, not rounded decimals.
• S013 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative0 ORIGINAL coefficient term2 with exact rational arithmetic, not rounded decimals.
• S012 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative0 falling factor times power term2 with exact rational arithmetic, not rounded decimals.
• S011 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S010 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative0 ascending derivative sum term1 with exact rational arithmetic, not rounded decimals.
• S009 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative0 ORIGINAL coefficient term1 with exact rational arithmetic, not rounded decimals.
• S008 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative0 falling factor times power term1 with exact rational arithmetic, not rounded decimals.
• S007 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S006 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative0 ascending derivative sum term0 with exact rational arithmetic, not rounded decimals.
• S005 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative0 ORIGINAL coefficient term0 with exact rational arithmetic, not rounded decimals.
• S004 · premise_and_result_audit — Recompute Taylor ORIGINAL derivative0 falling factor times power term0 with exact rational arithmetic, not rounded decimals.
• S003 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S002 · premise_and_result_audit — Recompute Taylor ORIGINAL x minus declared center with exact rational arithmetic, not rounded decimals.
• S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.

Result

(2, 2, [13, 4, 3/4], 0, Yes, degree_at_most_n_polynomial_has_zero_remainder)

The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.

This question review and source binding

Review date: 2026-10-26

Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.