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

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?

• Do you know the MacLaurin series of $\sin(x)$ and of $\sqrt{1+x}$? – 永劫回帰 Nov 9 '16 at 18:08
• it's undefined at $x=0$, so are its derivatives. I doubt you can find its MacLaurin series. Maybe do a Taylor series around $x=\varepsilon$ and take limit $\varepsilon$ approaching zero. – Aydin Gerek Nov 9 '16 at 18:09
• Are you sure this singularity is not removable? As $\frac{sin(x)}{x}$ has a removable singularity at $x=0$ via $1$ seen by L'Hospial for example. – Fritz Nov 9 '16 at 18:09
• You indeed just take the limit $x \to 0$. – TastyRomeo Nov 9 '16 at 18:10

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

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 $a,b$?

• Gonna call this the best answer, certainly the one that was most intuitive to me. Thanks! – hsherl Nov 9 '16 at 19:19

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.

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 $2$) $$\sqrt{\frac{\sin x}x}=1-\frac{x^2}{12}+o\bigl(x^2\bigr).$$

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: $$\sqrt{\frac{\sin x}{x}}=1-\frac{x^2}{12}+o(x^3).\tag{3}$$

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 $a_0=1$.