what is the taylor expansion of $\ln(\sin x/x)$

Question

what is the Taylor expansion of the function $$f(x) = \ln \frac{\sin x}x$$ around the point $x=0$? Ignore powers of $x$ which are greater than $6$.

Here is my method: $$\ln(1+x)=x-\frac{x^2}2 + \frac{x^3}3 -\frac{x^4}4,$$ so we should get the function $$g(x)=\frac{\sin x}x$$ in the form of $1 + x$ to get the answer … or is there another way to consider the Taylor expansion of $\ln (u)$ by considering $f$ to be the answer of $$\frac{\sin x}x\cdots$$ Which way is the best?? What is the final answer???

Answer 1

$$\begin{align}\log(\frac{\sin x}{x}) = & \log\prod_{n=1}^{\infty}\left( 1 - \frac{x^2}{n^2\pi^2}\right) = \sum_{n=1}^{\infty}\log\left(1 - \frac{x^2}{n^2\pi^2}\right)\\ \stackrel{\color{blue}{^{[1]}}}{=} & -\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{x}{n\pi}\right)^{2k} \stackrel{\color{blue}{^{[2]}}}{=} -\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{k}\left(\frac{x}{n\pi}\right)^{2k}\\ \stackrel{\color{blue}{^{[3]}}}{=} & -\sum_{k=1}^{\infty}\frac{\zeta(2k)}{k\pi^{2k}} x^{2k} = -\frac{1}{6}x^2 -\frac{1}{180}x^4 - \frac{1}{2835} x^6 - \frac{1}{37800} x^8 - \cdots \end{align}$$

Notes

$\color{blue}{[1]}$ the power series converges when $|x| < \pi$.
$\color{blue}{[2]}$ legal to change order of summation because all terms have same sign.
$\color{blue}{[3]}$ $\zeta(z) = \sum_{n=1}^{\infty} n^{-z}$ is the Riemann zeta function. When $z$ is an even integer, it can be expressed in terms of Bernoulli number $B_{2k}$: $$\zeta(2k) = \frac{(-1)^{k+1} B_{2k} (2\pi)^{2k}}{2(2k)!}$$ Look at the wiki pages for more relations.

Answer 2

As you suggest, it will be useful to know the expansion $$\ln(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\cdots\tag{1}$$ (at least for $|t|\lt 1$).

We also need the power series expansion for $\sin x$, which is $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots.$$ So the function which is $1$ at $x=0$ and $\frac{\sin x}{x}$ elsewhere has power series expansion $$1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots.\tag{2}$$ Now substitute (2) for $1+t$ in (1). Thus $t=-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots$.

Make sure not to compute too far. So throw away all terms in powers of $x$ greater than $6$.

Because the expression (2) has no $x$ term, you will only need to compute to the $\frac{t^3}{3}$ term in (1). So the calculation is less time-consuming than it seems.

Getting a general term for the power series expansion looks, however, very difficult. Luckily you were not asked to find many terms. You will end up with something of the shape $ax^2+bx^4+cx^6$.

Answer 3

Your instincts are correct. Write

$$\frac{\sin{x}}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots$$

We wish to maintain the series to $O(x^6)$. Then

$$\begin{align}\ln{\left ( \frac{\sin{x}}{x}\right)} &= \left (\frac{\sin{x}}{x} - 1\right ) - \frac12 \left (\frac{\sin{x}}{x} - 1\right )^2 + \frac13 \left (\frac{\sin{x}}{x} - 1\right )^3-\cdots \\ &= \left ( - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots \right ) - \frac12 \left ( - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots \right )^2+\\ &\frac13 \left ( - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots \right )^3-\cdots \\ &= - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots - \frac{x^4}{72} \left (1 - \frac{x^2}{10}+\cdots \right ) - \frac{x^6}{648}+\cdots \\ &= - \frac{x^2}{6} - \frac{x^4}{180}-\frac{x^6}{2835}+\cdots \end{align}$$

Note that

$$-\frac{1}{180} = \frac{1}{120} - \frac{1}{72}$$ $$-\frac{1}{2835} = -\frac{1}{5040}+\frac{1}{720}-\frac{1}{648}$$