# 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?

• At $x=0$. What does that mean?. – d80d2729a352b1366139fc119d3345 Jul 13 '13 at 9:12
• @boywholived, I can't understand your question. What is your point? The expansion is an asymptotic expansion for $\sqrt{\log(1+x)}$ when $x \to +0$. – asofas Jul 13 '13 at 9:31
• Here is a similar problem. – Mhenni Benghorbal Jul 13 '13 at 9:34
• I hope you got the idea from the link I gave you which you consider the function $\sqrt{x}\sqrt{\ln(1+x)}$. – Mhenni Benghorbal Jul 13 '13 at 9:51
• @Mhenni Benghorbal, yes I got how it works. But what is the motivation of the idea multiplying $\sqrt{x}$ at first? I'd like to know if there are some general rule or technic for this kind of problem. – asofas Jul 13 '13 at 9:54

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 $$\sqrt{\log(1+x)}=\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4)\right)^{1/2}=\sqrt{x}\left(1+(\underbrace{-\frac{x}{2}+\frac{x^2}{3}+O(x^3))}_{=u}\right)^{1/2}\\=\sqrt{x}(1+\frac{1}{2}u-\frac{1}{8}u^2+O(u^3))=\sqrt{x}(1-\frac{x}{4}+\frac{x^2}{6}-\frac{1}{8}\frac{x^2}{4}+O(x^3))\\=\sqrt{x}(1-\frac{x}{4}+\frac{13x^2}{96}+O(x^3))$$

$\log(1+x)$ 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?

• Yes, I need the help! I can't understand what "the second factor gives an ordinary Taylor series" means. – asofas Jul 13 '13 at 9:34
• Try $\sqrt{1-x/2+x^2/3-...}=1+ax+bx^2+cx^3+...$ Square both sides $1-x/2+x^2/3-...=(1+ax+bx^2+cx^3+...)(1+ax+bx^2+cx^3+...)$ Expand the right-hand side, and equate coefficients. You will find $a$ first, from the coefficients of $x$, then use that and the coefficients of $x^2$ to find $b$. – Empy2 Jul 13 '13 at 9:54
• Then factoring $x$ out and making a constant term seem like a crucial step to use this method of taking the square root. Is it right? – asofas Jul 13 '13 at 9:59
• Yes, I think so. – Empy2 Jul 13 '13 at 10:03

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 $$\sqrt{\log(1+x)}=\sqrt{\mathstrut x}\ \left(1-{1\over4}x+{13\over96} x^2-{35\over 384}x^3+{6271\over 92\,160} x^4-\ldots\right)\quad.$$

• What is your motivation of factoring out $\sqrt{x}$ at first? And how could you sure $f(x)$ would be analytic before completing the rest of the process by finding $a_i$? – asofas Jul 13 '13 at 12:00
• @asofas: See my edit. – Christian Blatter Jul 13 '13 at 12:18
• I got it. Thank you for the kindly explanation! – asofas Jul 13 '13 at 12:34