friends.
As stated on the title, my question is: find the power series representation for $ f(x) = \arctan (e^x) $ and its interval of convergence.
This question got me a bit confused due to the presence of the $ e^x $.
The first path I tried was taking the derivative of the function but I realized that it wouldn't take me anywhere. Then, I decided to do the following:
Let $ g(x) = \arctan (x) $. Then:
$ g'(x) = \frac{1}{1+x^2} $ and so $ g'(x) = \sum_{n=0}^{\infty} (-1)^n \cdot x^{2n} $ for $ -1 < x < 1 $.
Integrating both sides we get:
$ \arctan(x) = \sum_{n=0}^{\infty} (-1)^n \cdot \frac{x^{2n+1}}{2n+1} + C $ and because $ \arctan(0) = 0 $, $ C $ also equals 0.
With that in mind, replacing $ x $ with $ e^x $ we have:
$ \arctan(e^x) = \sum_{n=0}^{\infty} (-1)^n \cdot \frac{ (e^x)^{2n+1}}{2n+1} $
and now for the interval of convergence, using the Ratio Test:
$ \lim_{n \to \infty} \left| \frac{(e^x)^{2n+3} }{2n+3} \cdot \frac{2n+1}{(e^x)^{2n+1}} \right| = |e^{2x}| $
and so for the series to converge we must have:
$ -1 < e^{2x} < 1 $
Now, I have two questions:
Can I switch $ x $ with $ e^x $ without any problems?
And if so and assuming my solution and my ratio test are correct, is there any to simplify my interval of convergence?
Really appreciate any help.
Thanks in advance.
Pedro.
Note that $f(0)=\frac{\pi}{4}$ and that $$f'(x)=\frac{e^x}{1+e^{2x}}=\frac{1}{2\cosh x}$$ The nearest singularities to zero are $\pm i\pi/2$, so the radius of convergence of the power series expansion of $f$ around $0$ is $\pi/2$. Now, since $f'$ is even it has a power series expansion of the form $f'(x)=\sum_{n=0}^\infty a_nx^{2n}$. Using the fact that $\cosh(x)f'(x)=\frac{1}{2}$ we see that the coefficients $(a_n)$ can be inductively calculated by the formula: $$a_0=\frac{1}{2},\qquad a_n=-\sum_{k=0}^{n-1}\frac{a_k}{(2n-2k)!}$$ In particular, $$ a_0=\frac{1}{2},a_1=-\frac{1}{4},a_2=\frac{5}{48},a_3=-\frac{61}{1440} $$ and then $$f(x)=\frac{\pi}{4}+\sum_{n=0}^\infty\frac{a_n}{2n+1}x^{2n+1}.$$ That is $$f(x)=\frac{\pi }{4}+\frac{x}{2}-\frac{x^3}{12}+\frac{x^5}{48}-\frac{61 x^7}{10080}+\cdots$$