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.
Best Answer
$$\frac{\sin x}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}-\cdots.$$
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 $a,b$?