How Can I calculate the Taylor series of $\ln(\cos(x))$?

Question

I found it in an exercise about limits and i don't know how to solve it. If it Is possible I would like to understand the steps to calculate It. Many thanks in advance.

Answer 1

A quick hack is often to partially express some function in terms of a Taylor approximation about $x_0$, since higher order terms of $x$ go to zero if we are considering limits for $(x-x_0) \rightarrow 0$. To really answer your question we need to know what the original question was, that is, about which point do you want the expansion? Let us assume around $0$. Then we have the Maclaurin series: $$\cos(x)=1 - \frac{1}{2}x^2 + \mathcal{O} (x^4)$$ You can add more terms if you need to. Now we write: $$ \ln(1+(- \frac{1}{2}x^2) ) = \dots$$ Do you know the standard Maclaurin series for this function?

Hint: it is of the form $\ln(1+u)$

Answer 2

You are given $f(x)=\log(\cos x)$. Then, by the chain rule, $f'(x)=-{\sin x\over\cos x}=-\tan x$. Since you know that $\tan' x=1+\tan^2 x$ you now can "mechanically" compute all desired derivatives $f^{(j)}(x)$, $\>j\geq0$. For the Taylor series of the given $f$ computed at $x=0$ you have to evaluate these derivatives at $x=0$. The resulting numbers do not follow a simple law that you might recognize.

Answer 3

If $$\ell(1+u)=u-\frac{u^2}{2}+\frac{u^3}{3}-\cdots$$ and $$c(x)=1-\frac{x^2}{2}+\frac{x^4}{4}-\cdots$$ try $$\ell\bigl( c(x)-1 \bigr)$$

Answer 4

Since $\ln\cos x=-\int_0^x\tan t dt$, this problem reduces to knowing how to express the coefficients in the Taylor series of $\tan x$ in terms of the up/down numbers.

Answer 5

Since $$\cos(x)=\prod_{n\geq 0}\left(1-\frac{4x^2}{\pi^2(2n+1)^2}\right)$$ we have $$ -\log\cos x = \sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m\pi^{2m}(2n+1)^{2m}}=\sum_{m\geq 1}\frac{4^m x^{2m}}{m\pi^{2m}}\left[\zeta(2m)-\frac{\zeta(2m)}{4^m}\right] $$ and $$ \log\cos x = \sum_{m\geq 1}(1-4^m)\frac{\zeta(2m)}{m\pi^{2m}} x^{2m}. $$