Integral remainder in Taylor theorem

If

f

has

k + 1

continuous derivatives, its Taylor remainder about

a

is

R_{k} (x) = \frac{1}{k !} \int_{a}^{x} (x - t)^{k} f^{(k + 1)} (t) d t .

Table of contents
- Logarithmic series from an integral Taylor remainder Integral remainder in Taylor theorem

Logarithmic series from an integral Taylor remainder ( $- lo g (1 - t) = \sum_{k = 1}^{\infty} t^{k} / k, 0 \leq t < 1$ )

The integral remainder in Taylor theorem for

lo g (1 - t)

gives the exact finite identity

- lo g (1 - t) = \sum_{k = 1}^{N} \frac{t ^{k}}{k} + \int_{0}^{t} \frac{( t - x ) ^{N}}{( 1 - x ) ^{N + 1}} d x .

For

0 \leq x \leq t < 1

,

(t - x) / (1 - x) \leq t

, so the nonnegative remainder is at most

t^{N + 1} / (1 - t)

and tends to zero. At

t = 1/2

it is at most

2^{- N}

, proving both convergence to

lo g 2

and an explicit error bound without assuming the logarithmic power series beforehand.

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