Finding the Taylor series for $\ln(\sin(x))$

sequences-and-seriestaylor expansion

I'm trying to calculate the Taylor series for $\ln(\sin(x))$, though I'm not sure how to correctly expand the series.

when $a = \frac{\pi}{4}$

Here's what I have so far:

$$\ln(\sin(x))= \ln\left(x-\frac{x^3}{3!}+ \cdots+ (-1)^{n-1}\frac{x^{2n-2}}{(2n-2)!}\right)$$

What are some tips for expanding logarithmic series, for I have many exercises very similar to this and need an approach, though I'm stuck on converting them into series, your help is much appreciated!

As far as I'm aware, I've noticed two approaches thus far.

  1. $\frac{dy}{dx}=\frac{1}{\sin(x)}\cdot \cos(x) = \cot(x) \implies \int\cot(x)$

Given the taylor series is:

If we let $(x-a) = h$

$f(a+h) = f(a) + f'(a)h+\frac{f''(a)}{2!}h^2 …$

Then the first few series should be:

$\ln(\sin(x)) = \ln(\sin(\frac{\pi}{4}))+\cot(\frac{\pi}{4})(x-\frac{\pi}{4})…$?

Best Answer

There can be two approaches here, in both, you shall have to make direct use of Taylor's formula:

\begin{align*}f\left( x \right) & = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}{{\left( {x - a} \right)}^n}} \\ & = f\left( a \right) + f'\left( a \right)\left( {x - a} \right) + \frac{{f''\left( a \right)}}{{2!}}{\left( {x - a} \right)^2} + \frac{{f'''\left( a \right)}}{{3!}}{\left( {x - a} \right)^3} + \cdots \end{align*}

Method-1:

Take $ f(x) = ln(y) $ with $ y= \sin (x)$ and $ a = \sin(\pi / 4)$ and expand directly

Method-2:

Make use of the fact that:

$$\frac{d}{dx} \ln (\sin x ) =\frac{1}{\sin x} \cdot \cos x = \cot (x)$$

So, you can directly evaluate with $f(x) = \cot x $ and $ a = \pi / 4 $ and integrate from $0$ to $x$

I trust that you will be able to perform the calculation now.

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