A Taylor polynomial retaining the center and factorials

Goal and scope

Construct T_2 at center a=1 for the specified cubic polynomial.

Construct a finite Taylor polynomial only; a finite jet is no general remainder or analyticity certificate.

Givens and symbols

• algebra
• polynomial
• definition_here

• G1 — f(x)=1+2(x-1)+3(x-1)^2-(x-1)^3, a=1, n=2.

• h=x-1 — center shift
• T_n — finite Taylor polynomial

Rules and sources

• mko_definition · cited — Fix center a and nonnegative order n and require every f^(k)(a) for0≤k≤n to exist. T_n(x)=sum over k=0..n of f^(k)(a)*(x−a)^k/k!; at n=0 this is constant f(a). The polynomial itself gives no remainder bound or guarantee that an infinite Taylor series converges or equals f. A polynomial f of degree at most n reconstructs exactly; this does not make an arbitrary smooth function analytic. · mko-taylor-polynomial
• algebra · 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.
• definition_here · prerequisite — The Taylor polynomial of order n at a sums f^(k)(a)*(x-a)^k/k! from k=0 through n and requires the complete center jet; at n=0 it is f(a).

Main derivation

h=x-1

f(x)=1+2h+3h^2-h^3

f'(x)=2+6h-3h^2

f(1)=1,\quad f'(1)=2

f''(x)=6-6h

f''(1)=6

T_2(x)=\sum_{k=0}^{2}\frac{f^{(k)}(1)}{k!}(x-1)^k

T_2(x)=1+2h+\frac6{2!}h^2

2!=2

T_2(x)=1+2(x-1)+3(x-1)^2

f(x)-T_2(x)

f(x)-T_2(x)=-(x-1)^3

f^{(3)}(1)=-6

T_3(x)=T_2(x)-h^3=f(x)

n=0\text{ as a separate order boundary}

T_0(x)=f(1)=1

Substeps and lemmas

No substeps in this guide.

Reverse audits

• S08 · premise_audit · S02, S05 — Order zero does not mean there are no terms.
• S07 · premise_audit · G1 — Exact reconstruction for degree≤n is not extrapolated to arbitrary smooth functions.
• S06 · inverse_equivalence · G1, S05 — This identity does not claim a general Taylor remainder theorem.
• S05 · inverse_equivalence · S04 — Multiplying coefficient 3 by 2! recovers the second derivative 6.
• S04 · premise_audit · S02, S03 — The quadratic coefficient is not the unscaled derivative 6.
• S03 · premise_audit · S02 — No lower order is omitted in favor of only the highest one.
• S02 · premise_audit · S01 — Every derivative is evaluated at the same center 1.
• S01 · premise_audit · G1 — h is not x; the shift -1 cannot be omitted.

Error controls and negative controls

Factorials

Divide each k-th derivative by k!, including 0!=1.

No implicit remainder promise

The finite polynomial alone promises no interval error bound.

• The quadratic coefficient is 6. → reject — Division by 2! gives 3.
• The same jet can be moved directly to center zero. → reject — The jet is bound to center 1; a new center requires reevaluation.
• T_2 necessarily equals the full f. → reject — The difference here is a nonzero cubic term.

Result and limitations

T_2=1+2(x-1)+3(x-1)^2, with difference -(x-1)^3; T_3 equals this f.

Verification and source binding

1 declared checks for this guide; only the listed expressions are covered.

• source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
• adopted source SHA-256: 933a8e53367e588e2e0bd950532d7fde4ed554f553b2e996d3de73948108fed0
• original source SHA-256: 933a8e53367e588e2e0bd950532d7fde4ed554f553b2e996d3de73948108fed0
• Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

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