Series expansion of $\sqrt{\log(1+x)}$ at $x=0$ Mathematica gives the following series expansion of $\sqrt{\log(1+x)}$ at $x=0$.
$$
x^{1/2}-\frac{1}{4}x^{3/2}+\frac{13}{96}x^{5/2}-\cdots
$$
You can find it from Wolfram alpha too.
How can I obtain the expansion?
Obviously Taylor expansion is impossible because $\sqrt{\log(1+x)}$ is not analytic at $x=0$.
Taylor expansion of the $\log(1+x)$ at $x=1$ is possible. But I don't know how to take sequre root on the expanded series.
I think I have not learned about square root of a series from calculs or analysis course.
From what material can I study about such things?
 A: $\log(1+x)$ is analytic at $x=0$, and its Taylor series is $x-x^2/2+x^3/3-...$
Take out the common factor $x(1-x/2+x^2/3-...)$
When you take the square-root, the first factor gives $x^{1/2}$ of course, and the second factor gives an ordinary Taylor series.  Do you need help finding its square root?
A: We have
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
and recall that
$$(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)x^2}{2}+\frac{\alpha(\alpha-1)(\alpha-2)x^3}{3}+\cdots$$
so for $\alpha=\frac{1}{2}$ we have
$$\sqrt{\log(1+x)}=\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4)\right)^{1/2}=\sqrt{x}\left(1+(\underbrace{-\frac{x}{2}+\frac{x^2}{3}+O(x^3))}_{=u}\right)^{1/2}\\=\sqrt{x}(1+\frac{1}{2}u-\frac{1}{8}u^2+O(u^3))=\sqrt{x}(1-\frac{x}{4}+\frac{x^2}{6}-\frac{1}{8}\frac{x^2}{4}+O(x^3))\\=\sqrt{x}(1-\frac{x}{4}+\frac{13x^2}{96}+O(x^3))$$
A: The function
$$g(x):={\log(1+x)\over x}=1-{1\over2} x+{1\over3}x^2-\ldots$$
has a removable singularity at $0$ and takes the value $1$ there. It follows that in some neighborhood $U$ of $0$ the function $g$ has an analytic square root $f$ with $f(0)=1$, and that we may write
$$\sqrt{\log(1+x)}=\sqrt{\mathstrut x}\  f(x)\qquad (x\in U)\ ,\tag{1}$$
where now the  ambiguity in the given expression resides in the factor $\sqrt{\mathstrut x}$. The function
$$f(x)=\sum_{k\geq0} a_k x^k,\quad a_0=1,$$
 satisfies
$$x\>\bigl(f(x)\big)^2=\log(1+x)\qquad(x\in U)\ ,$$
or
$$\sum_{r\geq0}\left(\sum_{k+l=r} a_k a_l\right) x^{r+1}=\sum_{r\geq0}{(-1)^r\over r+1}x^{r+1}\qquad(x\in U)\ .$$
This implies
$$2a_0a_r +\sum_{k=1}^{r-1} a_k a_{r-k} ={(-1)^r\over r+1}\qquad(r\geq1)\ ,$$
from which we obtain the following recursion formula for the coefficients $a_r\ $:
$$a_r={1\over2}\left({(-1)^r\over r+1}-\sum_{k=1}^{r-1} a_k a_{r-k}\right)\qquad(r\geq1)\ .$$
Plugging the first few $a_r$ into $(1)$ we therefore have
$$\sqrt{\log(1+x)}=\sqrt{\mathstrut x}\ \left(1-{1\over4}x+{13\over96} x^2-{35\over 384}x^3+{6271\over 92\,160} x^4-\ldots\right)\quad.$$
