Power series of $\ln x$ near $2$

Question

I wanted to find the Taylor series of $f(x)=\ln x$ around $2$, is below correct?

Observe that:

(1).$f(2)=\ln 2$ and $f(x)=\ln 2$

(2).$f'(2)=\frac 12$ and $f'(x)=\frac 1x$

(3). $f''(2) =-1\times 2^{-2}$ and $f''(x)=-x^{-2}$

(4). $f'''(2)=-1\times-2\times2^{-3}$ and $f'''(x)=2x^{-3}$

So by induction, in general we get :

$f^{(n)}(2)=(-1)^{n-1}2^{-n}(n-1)! $

And the series is $\ln 2+ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2^nn}(x-2)^n$

And is there any other ways to get the series?

Answer 1

Hint

Suppose you let $x=2t+2$; so, you are considering $$f=\log(2+2t)\implies f'=\frac12\times \frac 1{1+t}$$ Consider now the Taylor expansion of $$\frac 1{1+t}=\sum_{n=0}^\infty (-1)^n t^n$$ and integrate to get.$$f=\frac12\sum_{n=0}^\infty \frac{(-1)^n}{n+1} t^{n+1}+C$$ The integration constant is $C=\log(2)$; replace now $t$ by $\frac{x-2}2$.

Answer 2

There are other ways to get the series as well, for example consider \begin{align*} \ln x & = \int_1^x \frac{1}{t} \, dt\\ & = \int_1^x \frac{1}{2+t-2} \, dt\\ & =\frac{1}{2} \int_1^x \frac{1}{1+\frac{t-2}{2}} \, dt\\ \end{align*} Now use geometric series to expand the integrand and then integrate term wise to get the series.