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The Taylor polynomial of (1x)1(1-x)^{-1} at 00 has the exact geometric remainder xn+1/(1x)x^{n+1}/(1-x)

Example

For f(x)=1/(1x)f(x)=1/(1-x) and nNn\in\mathbb N, Tn,0f(x)=j=0nxj,Rn,0f(x)=xn+11xT_{n,0}f(x)=\sum_{j=0}^{n}x^j,\qquad R_{n,0}f(x)=\frac{x^{n+1}}{1-x} whenever x1x\ne1.

Facts & Assumptions

Given: The geometric function.

[L1]

Finite geometric sums follow from Laws of finite sums and finite products. Derivative algebra, the chain rule, and the natural-power derivative give the successive derivatives of (1x)1(1-x)^{-1}; factorial arithmetic is preserved by the canonical embedding (Sums, scalar multiples, products and quotients: (f+g)(c)=f(c)+g(c)(f+g)'(c) = f'(c) + g'(c), (αf)(c)=αf(c)(\alpha f)'(c) = \alpha f'(c), (fg)(c)=f(c)g(c)+f(c)g(c)(fg)'(c) = f'(c)g(c) + f(c)g'(c), and (f/g)(c)=(f(c)g(c)f(c)g(c))/g(c)2(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2} when g(c)0g(c) \ne 0, The chain rule, in one line from Carathéodory: if gg is differentiable at cc and ff is differentiable at g(c)g(c), then fgf \circ g is differentiable at cc with (fg)(c)=f(g(c))g(c)(f \circ g)'(c) = f'(g(c))\,g'(c), For a natural n1n \ge 1 the function xxnx \mapsto x^{n} is differentiable everywhere with derivative ι(n)xn1\iota(n)\,x^{\,n-1}; for n=0n = 0 it is the constant 11, with derivative 00; for a natural n1n \ge 1 the function xxnx \mapsto x^{-n} is differentiable at every x0x \ne 0 with derivative ι(n)xn1-\iota(n)\,x^{-n-1}; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, The factorial n!n! and the falling factorial nkn^{\underline{k}}, defined by recursion in N\mathbb{N}, The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field, Canonical naturals are positive and strictly increasing), and induction is The principle of mathematical induction.

Verification

technique · direct
1.1

Induction gives f(j)(x)=ι(j!)(1x)j1f^{(j)}(x)=\iota(j!)(1-x)^{-j-1}, hence f(j)(0)/ι(j!)=1f^{(j)}(0)/\iota(j!)=1.

L1
2.1

Multiplying j=0nxj\sum_{j=0}^{n}x^j by 1x1-x telescopes to 1xn+11-x^{n+1}. Subtracting from 1/(1x)1/(1-x) gives the stated remainder.

step 1.1L1algebra
3.1

This exact expression agrees with the qualitative estimate supplied by [L2] on every closed interval avoiding 11.

step 2.1L2

Depends on

Used by

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