# Power series of $f(x)=\ln(1+\sin(x))$ around $x_0=0$

At class we were given the following question:

Let $$f(x)=\ln(1+\sin(x))$$. Expand $$f(x)$$ into a power series around $$x_0=0$$, a.k.a. find $$\sum_{0}^{\infty} a_n x^n$$ that converge to $$f(x)$$ at $$(-\delta, \delta)$$.

We saw at class how to use the multiplication rule to expand functions such as $$f(x)g(x)$$ (e.g. $$\sin(x)\cos(x)$$ etc.), but idk how to deal with nested functions.

I thought of using the definition $$\ln(1+x) = \sum_{n=0}^{\infty} \frac{(-1)^nx^{n+1}}{(n+1)}$$ and then substitute $$x=\sin(x)$$ when $$\sin(x)=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$$

But I'm not sure that that's the appropriate approach. I'd be thankful for your help!

• I doubt there's a closed form for this, so I suppose you're only supposed to compute the first few terms. If I'm right about that, then your method looks fine to me. Commented Jan 30, 2019 at 16:54

Define $$f(x)=\ln(1+\sin(x))$$. Then $$\begin{split} f^\prime(x)&=\frac {\cos x}{1+\sin x}\\ &=\cot\left(\frac x 2 +\frac \pi 4\right) \,\,\, \left (\text{since }\cot\frac u 2=\frac {\sin u}{1-\cos u}\right)\\ \end{split}$$ Using the power series for $$\cot$$: $$\begin{split} f^\prime(x) &= \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} B_{2 n} \, \left ( \frac x 2 +\frac \pi 4\right )^{2 n - 1} } {\left({2 n}\right)!}\\ &= \frac 1 {\frac x 2 +\frac \pi 4} + \sum_{n \mathop = 1}^\infty \frac {\left({- 1}\right)^n 2^{2 n} B_{2 n}} {\left({2 n}\right)!} \sum_{k=0}^{2n-1}{ {2n-1} \choose k}\left(\frac{\pi}{4}\right)^{2n-1-k}\frac{x^k}{2^k}\\ &= \frac 2 {x +\frac \pi 2} + \sum_{k \mathop = 0}^\infty\left( \sum_{n=\lceil \frac {k+1} 2 \rceil}^{+\infty}{ {2n-1} \choose k}\left(\frac{\pi}{4}\right)^{2n-1-k}\frac{1}{2^k}\frac {\left({- 1}\right)^n 2^{2 n} B_{2 n}} {\left({2 n}\right)!}\right)x^k\\ \end{split}$$ So, using the fact that $$f(0)=0$$, $$\begin{split} f(x) =& 2\ln\left(1 + \frac 2 \pi x\right)+\sum_{k \mathop = 0}^\infty\left( \sum_{n=\lceil \frac {k+1} 2 \rceil}^{+\infty}{ {2n-1} \choose k}\left(\frac{\pi}{4}\right)^{2n-1-k}\frac{1}{2^k}\frac {\left({- 1}\right)^n 2^{2 n} B_{2 n}} {\left({2 n}\right)!}\right)\frac{x^{k+1}}{k+1} \end{split}$$ and you can expand the log as a power series as well, giving you one power series for the whole thing. Yep, it's pretty ugly, but at least there seems to be a closed form for the coefficients (as long as you can compute Bernoulli numbers).