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;

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

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