How to find MacLaurin series for $\sqrt{\frac{\sin(x)}{x}}$?

Question

I’m asked to find the terms up to $x^2$ for the MacLaurin series of $\sqrt{\frac{\sin(x)}{x}}$.

I get $\frac{d}{dx}\left(\sqrt{\frac{\sin(x)}{x}}\right) = \frac{\frac{\cos(x)}{x}-\frac{\sin(x)}{x^2}}{2\sqrt{\frac{\sin(x)}{x}}}$, which is undefined for $x = 0$.

But the same happens at the second derivative; undefined for $x = 0$.

Do I just take the limit of the derivative? How do I get around this?

Hints appreciated, no solution please.

Answer 1

$$\frac{\sin x}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}-\cdots.$$

So if $\sqrt{\frac{\sin x}{x}}=1+ax+bx^2+\cdots$ then:

$$\frac{\sin x}{x}=\left(1+ax+bx^2+\cdots\right)^2$$

So $2a=0$ and $2b+a^2=-\frac{1}{6}$. So what are $a,b$?

Answer 2

The series for $\sin(x) $ is $x - x^3 / 3! + x^5 / 5! - \ldots$. So that for the quotient is

$$ f(x) = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \ldots. $$ Now, can you find a quadratic polynomial whose square is those first three terms? If you can, you're done.

Answer 3

You just have to compose series (expanded up to order $2$):

• $\dfrac{\sin x}x=\dfrac{x-\dfrac{x^3}6+o\bigl(x^3\bigr)}x=1-\dfrac{x^2}6+o\bigl(x^2\bigr)$
• $\sqrt{\mathstrut1-u}=1-\dfrac u2-\dfrac{u^2}8+o\bigl(u^2\bigr),$

whence (the composition has to be truncated at order $2$) $$\sqrt{\frac{\sin x}x}=1-\frac{x^2}{12}+o\bigl(x^2\bigr).$$

Answer 4

Since in a neighbourhood of the origin we have $$ \frac{\sin x}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}-\ldots \tag{1}$$ and $\sqrt{\frac{\sin x}{x}}$ is an even analytic function in a neighbourhood of the origin, $$\sqrt{\frac{\sin x}{x}}= 1 + a x^2 + b x^4+\ldots \tag{2}$$ where by squaring the RHS of $(2)$ we have to recover the RHS of $(1)$. That implies: $$ \sqrt{\frac{\sin x}{x}}=1-\frac{x^2}{12}+o(x^3).\tag{3} $$

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We also have a straightforward generalization of this approach. Assuming $$ \sqrt{\frac{\sin x}{x}} = \sum_{n\geq 0}\frac{(-1)^n a_n}{n!} x^{2n}\tag{4}$$ we have: $$ \sum_{k=0}^{m}\frac{a_k a_{m-k}}{k!(m-k)!} = \frac{1}{(2m+1)!}\tag{5}$$ hence the coefficients $a_1,a_2,a_3,\ldots$ can be computed through a simple recursion, given $a_0=1$.

Answer 5

Let \begin{equation}\label{S(n,k,x)-satisfy-eq} R(n,k,r)=\frac1{k!}\sum_{j=0}^k(-1)^{k-j}\binom{k}{j}(r+j)^n \end{equation} for $r\in\mathbb{R}$ and $n\ge k\ge0$. The quantities $R(n,k,r)$ are called weighted Stirling numbers of the second kind in Carlitz's paper [1] below.

A general answer to this question is the following theorem which is stated in Theorem 1 of the paper [2] below.

Theorem. For $r\in\mathbb{R}$, assume that the value of the power function $\bigl(\frac{\sin x}{x}\bigr)^{r}$ at the point $x=0$ is $1$.

1. For $r\ge0$, the series expansion \begin{equation}\label{recip-sin-ser-closed-eq}\tag{SEH} \biggl(\frac{\sin x}{x}\biggr)^r=1+\sum_{m=1}^{\infty}(-1)^m\Biggl[\sum_{k=1}^{2m}\frac{(-r)_k}{k!} \sum_{j=1}^k(-1)^j\binom{k}{j} \frac{R\bigl(2m+j,j,-\frac{j}2\bigr)}{\binom{2m+j}{j}}\Biggr]\frac{(2x)^{2m}}{(2m)!} \end{equation} is convergent in $x\in\mathbb{R}$, where the rising factorial $(r)_k$ is defined by \begin{equation*}%\label{rising-Factorial} (r)_k=\prod_{\ell=0}^{k-1}(r+\ell) = \begin{cases} r(r+1)\dotsm(r+k-1), & k\ge1;\\ 1, & k=0. \end{cases} \end{equation*}
2. For $r<0$, the series expansion \eqref{recip-sin-ser-closed-eq} is convergent in $x\in(-\pi,\pi)$.

References

1. L. Carlitz, Weighted Stirling numbers of the first and second kind, I, Fibonacci Quart. 18 (1980), no. 2, 147--162.
2. Feng Qi, Series expansions for any real powers of (hyperbolic) sine functions in terms of weighted Stirling numbers of the second kind, arXiv (2022), available online at https://arxiv.org/abs/2204.05612 or https://doi.org/10.48550/arXiv.2204.05612.