Taylor-polynomial of $f(x)=\log(\cos(x))$

Question

$f: (-\frac{\pi}{2}, \frac{\pi}{2}) \rightarrow \mathbb{R}, f(x) = \log(\cos x)$

Count the Taylor-polynomial $T_{2}(f, 0)(x)$ of the second degree of $f$ in $x_{0} = 0$

Alright because it was saying second degree, I knew we will need one derivation of $f$ at least (is that correct?). (Is it allowed to use more derivations if the task says second degree by the way?)

$f^{0}(x) = \frac{\ln(\cos x)}{\ln(10)}$

$f^{1}(x) = -\frac{\sin x}{\cos x\cdot \ln(10)}$

Put that into the Taylor formula: $(T_{n,x_{0}}f)(x)=\sum_{k=0}^{\infty}\frac{1}{k!}f^{k}(x_{0})(x-x_{0})^{k}$

$\Rightarrow$

$$\log(\cos x) = \frac{\frac{\ln(\cos(0))}{\ln(10)}}{0!}*0(x-0)^{0} + \frac{-\frac{\sin(1)}{\cos(1)\cdot \ln(10)}}{1!}*0(x-0)^{1} + ..$$

$$\log(\cos x)=0$$

No wonder if we do it at $x_{0} = 0$ that it will all result in $0$...

But I must have done something wrong, don't think it's as easy as I did above, or is it? :o

Because now I cannot really create a series with this solution.

Answer 1

For the derivatives one get:

$$f'(x_0)=-\tan(x)$$ $$f''(x_0)=-\frac{1}{\cos^2(x)}$$ (Use the chainrule with $u=\log(x), v=\cos(x)$)

Then you get: $$T_{f,0,2}(x)=\sum\limits_{j=1}^2 \frac{f^{(j)}(0)}{n!}\cdot x^n$$ $$T_{f,0,2}(x)=0-\tan(0)(x-0)-\frac{1}{2}x^2$$ $$T_{f,0,2}(x)=-\frac{1}{2}x^2$$

Answer 2

We may compute the whole Taylor series from the Weierstrass product: $$ \cos(x)=\prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right) \tag{1}$$ and the Taylor series of $\log(1-x)$, leading to: $$ \log\cos(x) = -\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m\pi^{2m}(2n+1)^{2m}}=-\sum_{m\geq 1}\frac{(4^m-1)\,\zeta(2m)}{m\,\pi^{2m}}\,x^{2m}.\tag{2}$$ Since $\zeta(2)=\frac{\pi^2}{6}$ and $\zeta(4)=\frac{\pi^4}{90}$, in a neighbourhood of the origin we have: $$ \log\cos(x) = -\frac{x^2}{2}-\frac{x^4}{12}+O(x^6).\tag{3}$$