LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-01

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Taylor polynomials match the prescribed derivatives at the centre

Statement

For $0 \leq r \leq n$ , $(T_{n, a} f)^{(r)} (x) = \sum_{j = r}^{n} \frac{f^{(j)} (a)}{ι ((j - r)!)} (x - a)^{j - r} .$ Consequently $(T_{n, a} f)^{(r)} (a) = f^{(r)} (a)$ , and $R_{n, a} f$ and its derivatives through order $n$ vanish at $a$ .

Facts & Assumptions

Given: The Taylor polynomial of Taylor polynomials and their remainders.

[L1]

Natural powers differentiate as in For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term; applying the chain rule to $x \mapsto x - a$ , whose derivative is $1$ , gives the same shifted-power formula; and finite sums differentiate termwise by The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ and Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ .

[L2]

Falling factorials cancel factorials according to The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ; finite sums reindex by Laws of finite sums and finite products.

Proof

technique · induction

1.1

At $r = 0$ the formula is the definition.

basegiven

1.2

Assuming the formula at $r < n$ , differentiate termwise. The term indexed $j$ acquires $ι (j - r)$ , which cancels $ι ((j - r)!)$ to $ι ((j - r - 1)!)$ ; the $j = r$ constant term disappears. This is the formula at $r + 1$ .

ihL1L2algebra

2.1

At $x = a$ , only the $j = r$ term survives and equals $f^{(r)} (a)$ . Subtraction from $f^{(r)} (a)$ gives the remainder assertion.

step 1.2L1algebra

3.1

The formula and both consequences hold through order $n$ .

step 1.1step 2.1discharge-induction∎

Depends on

Used by

Dependency tree · two levels

44 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 lecture notes (standard reference, not scraped)
J. Lebl, Basic Analysis I, Taylor's theorem and related calculus (standard reference, not scraped)