Section 8.8, Q. 4

Find the Taylor series expansion of the function \ln (1+x) about the point x=1.

(This question was asked at Friday’ tutorial but, with one eye on the answer given, I was unable to do it. Having looked at the problem again I’m sure that the question should have been:)

Find the Taylor series expansion of the function \ln x about the point x=1.

(I have indicated this issue to Prof. Stynes)

Solution

The Taylor series of any infinitely differentiable function about a point x=a is given by the power series:

f(x)\approx \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n

Computing the first few derivatives of f(x)=\ln x:

\left.f^{(0)}(x)=\ln x\right|_{x=1}=0

\left.f'(x)=x^{-1}\right|_{x=1}=1

\left.f''(x)=(-1)x^{-2}\right|_{x=1}=-1

\left.f'''(x)=(-1)(-2)x^{-3}\right|_{x=1}=2

\left.f^{(iv)}(x)=(-1)(-2)(-3)x^{-4}\right|_{x=1}=-6

\vdots

\left.f^{(n)}(x)=(-1)^{n+1}(n-1)!x^{-n}\right|_{x=1}=(-1)^{n+1}(n-1)!

This is valid for n\geq 1. At n=0, f^{(0)}(1)=f(1)=0. Hence we have;

f(x)\approx \sum_{n=1}^\infty \frac{(-1)^{n+1}(n-1)!}{n!}(x-1)^n

f(x)\approx \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^n \Box

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