Series expansion of $\sqrt{\log(1+x)}$ at $x=0$

Question

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?

Answer 1

$\log(1+x)$ 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?

Answer 2

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

Answer 3

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

Answer 4

Let $f(x)=\sqrt{\ln(1+x)}\,$ for $x>0$. Then, by virtue of Faa di Bruno formula and some properties of the partial Bell polynomials, \begin{align*} f^{(n)}(x)&=\sum_{k=0}^n\biggl\langle\frac12\biggr\rangle_k[\ln(1+x)]^{1/2-k} B_{n,k}\biggl(\frac{0!}{1+x},-\frac{1!}{(1+x)^2},\dotsc,(-1)^{n-k}\frac{(n-k)!}{(1+x)^{n-k+1}}\biggr)\\ &=\sum_{k=0}^n\biggl\langle\frac12\biggr\rangle_k[\ln(1+x)]^{1/2-k}\frac{(-1)^{n+k}}{(1+x)^n} B_{n,k}(0!,1!,\dotsc,(n-k)!)\\ &=\sum_{k=0}^n\biggl\langle\frac12\biggr\rangle_k[\ln(1+x)]^{1/2-k}\frac{(-1)^{n+k}}{(1+x)^n} (-1)^{n-k}s(n,k)\\ &=\frac{\sqrt{\ln(1+x)}\,}{(1+x)^n}\sum_{k=0}^n(-1)^{k-1}\frac{(2k-3)!!}{2^k}s(n,k)\frac{1}{[\ln(1+x)]^{k}}, \end{align*} where $s(n,k)$ denotes the Stirling numbers of the first kind. Utilizing this general formula for the $n$th derivative of $f(x)=\sqrt{\ln(1+x)}\,$, one can derive Taylor series expansion of the function $f(x)=\sqrt{\ln(1+x)}\,$ around any point $x_0>0$.

References

1. L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions, Revised and Enlarged Edition, D. Reidel Publishing Co., 1974; available online at https://doi.org/10.1007/978-94-010-2196-8.
2. F. Qi, D.-W. Niu, D. Lim, and Y.-H. Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, J. Math. Anal. Appl. 491 (2020), no. 2, Article 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.

Answer 5

This series expansion can be evaluated by first writing $$ \log^{1/2}(1+x)=\left(\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}x^k\right)^{1/2}=\sqrt x\left(\sum_{k=0}^\infty\frac{(-1)^k}{k+1}x^k\right)^{1/2}. $$ Applying the well-known expansion for powers of power series we obtain $$ \log^{1/2}(1+x)=\sqrt x\sum_{m=0}^\infty c_m x^m, $$ where $c_0=1$ and $$ c_m=\frac{1}{m}\sum_{k=1}^m(k/2-m+k)\frac{(-1)^k}{k+1}c_{m-k}. $$ This series converges on $x\in[0,1)$ and is implemented in Mathematica as

c[0] := 1;
c[m_] := 1/m Sum[(k/2 - m + k) (-1)^k/(k + 1) c[m - k], {k, 1, m}];
g[x_, M_] := Sqrt[x] Sum[c[m] x^m, {m, 0, M}];

Here is a plot of $\log^{1/2}(1+x)$ compared to the truncated series expansion for $M=1$ and $M=6$:

Finally, here is a list of the first several coefficients $c_m$:

$$ \left( \begin{array}{cc} m & c_m \\ 0 & 1 \\ 1 & -\frac{1}{4} \\ 2 & \frac{13}{96} \\ 3 & -\frac{35}{384} \\ 4 & \frac{6271}{92160} \\ 5 & -\frac{2211}{40960} \\ 6 & \frac{2760011}{61931520} \\ 7 & -\frac{1875133}{49545216} \\ 8 & \frac{557576779}{16986931200} \\ 9 & -\frac{13761972821}{475634073600} \\ 10 & \frac{3244313727791}{125567395430400} \\ \end{array} \right) $$

Answer 6

It is well known that \begin{equation} \frac{\ln(1+x)}{x}=\sum_{k=0}^\infty(-1)^k\frac{x^k}{k+1}, \quad |x|<1. \end{equation} This means that \begin{equation} \lim_{x\to0}\biggl[\frac{\ln(1+x)}{x}\biggr]^{(k)} =(-1)^k\frac{k!}{k+1}, \quad k\ge0. \end{equation} By virtue of the Faa di Bruno formula, we obtain \begin{align*} \Biggl[\sqrt{\frac{\ln(1+x)}{x}}\,\Biggr]^{(n)} &=\sum_{k=0}^n\biggl\langle\frac12\biggr\rangle_k \biggl[\frac{\ln(1+x)}{x}\biggr]^{1/2-k} B_{n,k}\Biggl(\biggl[\frac{\ln(1+x)}{x}\biggr]', \biggl[\frac{\ln(1+x)}{x}\biggr]'', \dotsc, \biggl[\frac{\ln(1+x)}{x}\biggr]^{(n-k+1)}\Biggr)\\ &\to\sum_{k=0}^n\biggl\langle\frac12\biggr\rangle_k B_{n,k}\biggl(-\frac{1!}{2}, \frac{2!}{3}, \dotsc, (-1)^{n-k+1}\frac{(n-k+1)!}{n-k+2}\biggr), \quad x\to0\\ &=(-1)^n\sum_{k=0}^n(-1)^{k-1}\frac{(2k-3)!!}{2^k} B_{n,k}\biggl(\frac{1!}{2}, \frac{2!}{3}, \dotsc, \frac{(n-k+1)!}{n-k+2}\biggr)\\ &=(-1)^n\sum_{k=0}^n(-1)^{k-1}\frac{(2k-3)!!}{2^k} \frac{(-1)^{n-k}}{k!}\sum_{m=0}^k(-1)^m\binom{k}{m}\frac{s(n+m,m)}{\binom{n+m}{m}}\\ &=-\sum_{k=0}^n\frac{(2k-3)!!}{(2k)!!} \sum_{m=0}^k(-1)^m\binom{k}{m}\frac{s(n+m,m)}{\binom{n+m}{m}}, \end{align*} where we used the formula \begin{equation}\label{Bell-Stir1st=eq} B_{n,k}\biggl(\frac{1!}2,\frac{2!}3,\dotsc,\frac{(n-k+1)!}{n-k+2}\biggr) =\frac{(-1)^{n-k}}{k!}\sum_{m=0}^k(-1)^m\binom{k}{m}\frac{s(n+m,m)}{\binom{n+m}{m}}. \end{equation} Consequently, we arrive at \begin{equation} \sqrt{\frac{\ln(1+x)}{x}}\,=-\sum_{n=0}^\infty\Biggl[\sum_{k=0}^n\frac{(2k-3)!!}{(2k)!!} \sum_{m=0}^k(-1)^m\binom{k}{m}\frac{s(n+m,m)}{\binom{n+m}{m}}\Biggr]\frac{x^n}{n!} \end{equation} and \begin{equation} \sqrt{|\ln(1+x)|}\,=-\sqrt{|x|}\,\sum_{n=0}^\infty\Biggl[\sum_{k=0}^n\frac{(2k-3)!!}{(2k)!!} \sum_{m=0}^k(-1)^m\binom{k}{m}\frac{s(n+m,m)}{\binom{n+m}{m}}\Biggr]\frac{x^n}{n!}. \end{equation}

References

1. Feng Qi and Aying Wan, A closed-form expression of a remarkable sequence of polynomials originating from a family of entire functions connecting the Bessel and Lambert functions, Sao Paulo Journal of Mathematical Sciences 16 (2022), no. 2, 1238--1248; available online at https://doi.org/10.1007/s40863-021-00235-2.
2. F. Qi, Diagonal recurrence relations for the Stirling numbers of the first kind, Contrib. Discrete Math. 11 (2016), no. 1, 22--30; available online at https://doi.org/10.11575/cdm.v11i1.62389.
3. F. Qi, Explicit formulas for computing Bernoulli numbers of the second kind and Stirling numbers of the first kind, Filomat 28 (2014), no. 2, 319--327; available online at https://doi.org/10.2298/FIL1402319O.
4. F. Qi and B.-N. Guo, A diagonal recurrence relation for the Stirling numbers of the first kind, Appl. Anal. Discrete Math. 12 (2018), no. 1, 153--165; available online at https://doi.org/10.2298/AADM170405004Q.
5. Feng Qi, Da-Wei Niu, Dongkyu Lim, and Yong-Hong Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, Journal of Mathematical Analysis and Applications 491 (2020), no. 2, Paper No. 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.