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R0001-01376 · q-20261007-advance-20261008-r06-antiderivative-0076
On all R, f has ascending coefficients [1, 2], starting with its constant term. Let F0 be its zero-constant primitive. Find F0(0)+(0), retaining the actual additive constant. Give a reduced fraction with positive denominator, retaining /1 for integers.
G001 — On all R, f has ascending coefficients [1, 2], starting with its constant term. Let F0 be its zero-constant primitive. Find F0(0)+(0), retaining the actual additive constant. Give a reduced fraction with positive denominator, retaining /1 for integers.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.
Before
domain=R; coefficient lists are in ascending powers
After
constant term first; rational pairs=[numerator,denominator]
Conditions: none
Fix coefficient order;do not read ascending coefficients as descending or swap numerator and denominator.
Rule: polynomial_representation · Uses: G001
Before
f coefficients ascending=[1, 2]
After
seek F₀ with constant coefficient0 and F₀′=f
Conditions: none
Antiderivatives are not generally unique; this task additionally fixes the constant coefficient at0.
Rule: canonical_primitive_domain · Uses: G001, S001
Before
primitive coefficient at degree1 := div(1, 1)
After
primitive coefficient at degree1 = 1
Conditions: none
Since (x^(k+1))′=(k+1)x^k,divide the coefficient by the positive nonzero integer k+1.
Rule: exact_rational_arithmetic · Uses: G001, S002
Before
differentiate back coefficient0 := mul(1, 1)
After
differentiate back coefficient0 = 1
Conditions: none
Differentiation multiplies back by k+1 and must recover the original coefficient.
Rule: exact_rational_arithmetic · Uses: G001, S003
Before
primitive coefficient at degree2 := div(2, 2)
After
primitive coefficient at degree2 = 1
Conditions: none
Since (x^(k+1))′=(k+1)x^k,divide the coefficient by the positive nonzero integer k+1.
Rule: exact_rational_arithmetic · Uses: G001, S004
Before
differentiate back coefficient1 := mul(1, 2)
After
differentiate back coefficient1 = 2
Conditions: none
Differentiation multiplies back by k+1 and must recover the original coefficient.
Rule: exact_rational_arithmetic · Uses: G001, S005
Before
[0, 1, 1]
After
all derivative coefficients agree with f
Conditions: none
Coefficient agreement proves the polynomial identity on all reals,not merely equality at a few points.
Rule: global_polynomial_identity · Uses: G001, S006
Before
constant coefficient+C := sum(0, 0)
After
constant coefficient+C = 0
Conditions: none
Change only the constant coefficient;the derivative remains the same f.
Rule: exact_rational_arithmetic · Uses: G001, S007
Before
F(x) multiply at degree2 := mul(0, 0)
After
F(x) multiply at degree2 = 0
Conditions: none
In Horner order, multiply by the same evaluation point first.
Rule: exact_rational_arithmetic · Uses: G001, S008
Before
F(x) add coefficient2 := sum(0, 1)
After
F(x) add coefficient2 = 1
Conditions: none
Then add the exact rational coefficient for this degree.
Rule: exact_rational_arithmetic · Uses: G001, S009
Before
F(x) multiply at degree1 := mul(1, 0)
After
F(x) multiply at degree1 = 0
Conditions: none
In Horner order, multiply by the same evaluation point first.
Rule: exact_rational_arithmetic · Uses: G001, S010
Before
F(x) add coefficient1 := sum(0, 1)
After
F(x) add coefficient1 = 1
Conditions: none
Then add the exact rational coefficient for this degree.
Rule: exact_rational_arithmetic · Uses: G001, S011
Before
F(x) multiply at degree0 := mul(1, 0)
After
F(x) multiply at degree0 = 0
Conditions: none
In Horner order, multiply by the same evaluation point first.
Rule: exact_rational_arithmetic · Uses: G001, S012
Before
F(x) add coefficient0 := sum(0, 0)
After
F(x) add coefficient0 = 0
Conditions: none
Then add the exact rational coefficient for this degree.
Rule: exact_rational_arithmetic · Uses: G001, S013
Before
x=0; additive constant=0
After
F(x)=0/1
Conditions: none
Use a reduced positive-denominator fraction,retaining /1 even for integers.
Rule: primitive_value_format · Uses: G001, S014
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 F(x) add coefficient0 with exact rational arithmetic, not rounded decimals.S013 · premise_and_result_audit — Recompute F(x) multiply at degree0 with exact rational arithmetic, not rounded decimals.S012 · premise_and_result_audit — Recompute F(x) add coefficient1 with exact rational arithmetic, not rounded decimals.S011 · premise_and_result_audit — Recompute F(x) multiply at degree1 with exact rational arithmetic, not rounded decimals.S010 · premise_and_result_audit — Recompute F(x) add coefficient2 with exact rational arithmetic, not rounded decimals.S009 · premise_and_result_audit — Recompute F(x) multiply at degree2 with exact rational arithmetic, not rounded decimals.S008 · premise_and_result_audit — Recompute constant coefficient+C 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 differentiate back coefficient1 with exact rational arithmetic, not rounded decimals.S005 · premise_and_result_audit — Recompute primitive coefficient at degree2 with exact rational arithmetic, not rounded decimals.S004 · premise_and_result_audit — Recompute differentiate back coefficient0 with exact rational arithmetic, not rounded decimals.S003 · premise_and_result_audit — Recompute primitive coefficient at degree1 with exact rational arithmetic, not rounded decimals.S002 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.0/1
The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.
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.
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