Find the Taylor series of $f(x)=\arctan (x) $ around $c=1$

Question

Find the Taylor series of $f(x)=\arctan (x) $ around $c=1$. For which $x$ does it converge to $f(x)$?

This is what I have been able to do so far

$$f(x)=\arctan(x)=\int \frac{1}{1+x^2}\, dx=\int \frac{1}{1+x^2+1-1}\,dx$$

Any hints please???

Answer 1

Hint: The $n$-th degree Taylor polynomial about $c$ is:

$$T_n(x) = f(c) + \frac{f^{\prime}(c)}{1!} (x-c) + \frac{f^{\prime\prime}(c)}{2!} (x-c)^2 + \cdots + \frac{f^{(n)}(x)}{n!} (x-c)^n \approx f(x)$$

for $x$ near $c$.

The taylor series is formed by taking $T_\infty$.

Answer 2

I bet the OP was asking to find the values of the derivatives in $x=1$.

This is clearly the same as finding the derivatives in $x=1$ for $f(x)=\frac{1}{x^2+1}$. Since: $$\frac{1}{x^2+1}=\frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right),\tag{1}$$ we have that: $$\frac{d^n}{dx^n}\frac{1}{x^2+1}=\frac{n!(-1)^n}{2i}\left(\frac{1}{(x-i)^{n+1}}-\frac{1}{(x+i)^{n+1}}\right)\tag{2}$$ so: $$ f^{(n)}(1) = n!(-1)^n\cdot \Im\left(\frac{1}{(1-i)^{n+1}}\right)\tag{3}$$ and since $(1-i)=\sqrt{2}e^{-i\pi/4}$ we have: $$ f^{(n)}(1) = n! (-1)^n \sin\left(\frac{\pi(n+1)}{4}\right)2^{-\frac{n+1}{2}}\tag{4}$$ and: $$ \frac{1}{x^2+1}=\sum_{k=0}^{+\infty}\frac{f^{(k)}(1)}{k!}(x-1)^k = \sum_{k=0}^{+\infty} (-1)^k \sin\left(\frac{\pi(k+1)}{4}\right)2^{-\frac{k+1}{2}}(x-1)^k.\tag{5} $$ The radius of convergence is given, as usual, by the distance from the closest singularity, the simple pole in $x=i$, giving $\rho=\sqrt{2}$. Integrating termwise we get: $$ \arctan x = \frac{\pi}{4}+\sum_{k=0}^{+\infty} \frac{(-1)^k}{k+1} \sin\left(\frac{\pi(k+1)}{4}\right)2^{-\frac{k+1}{2}}(x-1)^{k+1}.\tag{6} $$