[Math] Finding the Taylor series of $\cos(x)$ centered around $\pi$

Question

When finding the Taylor Series of $\cos(x)$ it always seems to be centered around $0$. Would centering it around another number – $\pi$, for example – produce a different Taylor Series that is also equal to $\cos(x)$?

Answer 1

I think that the choice of $\pi$ is nor the most interesting. Let us consider the exapnsion around $x=a$.

This would give you $$\cos(x)=\sum_{n=0}^\infty \frac{b_n}{n!} (x-a)^n \qquad \text{with} \qquad b_n=\cos \left(a+n\frac{\pi }{2}\right)$$

So, if $n$ is even $(n=2m)$, $b_m=(-1)^{m+1} \cos(a)$ and if $n$ is odd $(n=2m+1)$, $b_m=(-1)^{m} \sin(a)$.

So, you can see that $a=\pi$ or $a=\frac \pi 2$ are very particular cases.