S01 · implies
Before
f'=3x^2,\quad f''=6x,\quad f'''=6
After
(f(1),f'(1),f''(1))=(1,3,6)
Conditions: none
Rule: polynomial · Uses: G1
Leaf basis: polynomial
Every required order exists on the entire interval, and the third derivative is constant.
An exact Lagrange remainder for a cubic
On this page
derivation-mko-taylor-remainder-theorem-01 · ocme-new142-reader-adoption-20261021
Goal and scope
For f(x)=x^3, a=1, n=2, x=2, audit the exact remainder and a qualified derivative bound.
A complete degree-n+1 polynomial, not a proof of a general interval theorem from a finite center jet.
Givens and symbols
algebrapolynomialtaylor_def
G1— f(x)=x^3, a=1, n=2, evaluation point 2, connecting interval [1,2].
R_2— remainder f-TD,M— exact derivative supremum and candidate bound
Rules and sources
mko_definition· cited — Declare center a, point x, order n and the complete polynomial coefficients, then compute the Taylor polynomial and exact remainder directly. Degree≤n gives zero remainder. For degree n+1 the next derivative is constant, and a strictly intermediate midpoint supplies an exact Lagrange witness when x differs from a. Boundary families accept a qualified M at least the exact derivative supremum D or an undersized witness0≤M<D, rejecting negative M and unknown modes. A finite center jet does not prove the general remainder theorem, equality with a Taylor series or analyticity. · mko-taylor-remainder-theoremalgebra· prerequisite — Use real-field identities; division requires a nonzero denominator and multiplication by a positive number preserves inequalities.polynomial· prerequisite — Real polynomials are continuous and differentiate termwise: the derivative of x^k is k*x^(k-1), and constants differentiate to zero.taylor_def· prerequisite — T_n at center a sums f^(k)(a)*(x-a)^k/k! from zero through n; the remainder is f-T_n; a derivative supremum is its magnitude's least upper bound on the full connecting interval.
Main derivation
S02 · definition
Before
T_2(x)=\sum_{k=0}^2\frac{f^{(k)}(1)}{k!}(x-1)^kAfter
T_2(x)=1+3(x-1)+3(x-1)^2
Conditions: none
Rule: taylor_def · Uses: G1, S01, taylor_def
Leaf basis: taylor_def
Divide by each factorial and retain the center shift.
S03 · equivalent
Before
f(2)=8,\quad T_2(2)=7
After
R_2(2)=f(2)-T_2(2)=1
Conditions: none
Rule: algebra · Uses: S02
Leaf basis: algebra
Evaluate the function and polynomial independently before forming the signed remainder.
S04 · implies
Before
\xi=(1+2)/2=3/2\in(1,2)
After
\frac{f'''(\xi)}{3!}(2-1)^3=\frac6{6}=1=R_2(2)Conditions: none
Rule: algebra · Uses: G1, S01, S03
Leaf basis: algebra
The constant third derivative makes every interior point valid; explicitly use the strict midpoint as a Lagrange witness.
S05 · definition
Before
D=\sup_{t\in[1,2]}|f'''(t)|After
D=6,\quad M=6\ge D
Conditions: none
Rule: taylor_def · Uses: S01, taylor_def
Leaf basis: taylor_def
The derivative is constantly six on the whole interval, so the supremum is exact.
S06 · implies
Before
|R_2(2)|\le\frac M{3!}|2-1|^3After
1\le1\quad(M=6)
Conditions: none
Rule: algebra · Uses: S03, S04, S05
Leaf basis: algebra
The qualified derivative bound gives a remainder estimate attained with equality.
S07 · implies
Before
M=3<D=6
After
\frac M{3!}|2-1|^3=1/2<|R_2(2)|=1Conditions: none
Rule: algebra · Uses: S03, S05
Leaf basis: algebra
The nonnegative undersized M is not a derivative bound and actually violates the error inequality at this point.
S08 · implies
Before
x=a=1
After
R_2(1)=0;\quad M=0\text{ gives }0\le0\text{ but is not a bound on }|f'''|Conditions: none
Rule: algebra · Uses: S02, S05
Leaf basis: algebra
At the center a zero factor may accidentally satisfy the inequality with invalid M; this does not qualify the derivative bound. A degree≤n polynomial separately has identically zero remainder.
Substeps and lemmas
No substeps in this guide.
Reverse audits
S08· premise_audit · S02, S05 — Separate bound qualification from an accidentally true pointwise inequality.S07· premise_audit · S03, S05 — A negative M would also be invalid, but is not being mixed into this nonnegative undersized witness.S06· premise_audit · S03, S04, S05 — The premise M≥D is separate from the final numerical comparison.S05· premise_audit · S01, taylor_def — This is not a maximum from the center or a few samples.S04· premise_audit · G1, S01, S03 — A strict midpoint exists here because the evaluation point differs from the center.S03· inverse_equivalence · S02 — The remainder here is f-T, not its negative.S02· premise_audit · G1, S01, taylor_def — The quadratic coefficient is 6/2!=3.S01· premise_audit · G1 — The full polynomial supplies both the center jet and interval-wide third derivative.
Error controls and negative controls
Interval obligation
M must control the required derivative on the entire connecting interval.
Sign and factorial
Keep R=f-T and divide by (n+1)!.
- A finite second-order jet automatically supplies a third-derivative bound on any connecting interval. →
reject— Every f_K(x)=1+3*(x-1)+3*(x-1)^2+K*(x-1)^3 has the same order-two jet (1,3,6) at one, but third derivative 6K is arbitrarily large. - M=3 bounds the third derivative. →
reject— The third derivative has constant magnitude 6. - The error inequality at x=a proves M is qualified. →
reject— A zero displacement can hide an invalid bound, as S08 shows.
Result and limitations
The remainder is 1, an interior witness is 3/2, M=6 is valid, and M=3 fails here.
Verification and source binding
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