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

A Taylor series expanded around $x=0$ is actually called a Maclaurin series. Notice if you center around $x=-\pi$, then $\cos(x+\pi)=-\cos(x)$ so you can just use the Maclaurin series here as well. However, the general formula for the Taylor series allows you to center it around any point, and in fact if you want to approximate it at a value far from $0$, you should use such a form. However, because of the periodicity of $\cos(x)$ you can shift your center in most cases to near zero. If you wanted to expand near $x=\pi/2$ then you would realize that $\cos(x+\pi/2)=\sin(x)$, and so on and so forth. It should be noted that such an approximation is not preferred when it is expanded about a not very nice trigonometric argument (ie, a constant where we cannot easily calculate what it is expanded about) because, in general, the Taylor series about $x=c$ is

$$\cos(c)-(x-c)\sin(c)+\ldots$$

which you can already see requires we be able to evaluate cosine and sine. See this answer for the general formula for a Taylor series.

Answer 2

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.

Answer 3

Based on the formula we have: $$f\left(x\right)=\sum_{n=0}^{∞}f^{\left(n\right)}\left(x_{0}\right)\frac{\left(x-x_{0}\right)^{n}}{n!}$$$$=\cos\left(\pi\right)\frac{\left(x-\pi\right)^{0}}{0!}-\sin\left(\pi\right)\frac{\left(x-\pi\right)^{1}}{1!}-\cos\left(\pi\right)\frac{\left(x-\pi\right)^{2}}{2!}+\sin\left(\pi\right)\frac{\left(x-\pi\right)^{3}}{3!}+\cos\left(\pi\right)\frac{\left(x-\pi\right)^{4}}{4!}+...=-\frac{\left(x-\pi\right)^{0}}{0!}+\frac{\left(x-\pi\right)^{2}}{2!}-\frac{\left(x-\pi\right)^{4}}{4!}+...$$$$=\sum_{n=0}^{∞}\left(-1\right)^{\left(n+1\right)}\frac{\left(x-\pi\right)^{2n}}{\left(2n\right)!}$$

but generally for any real $x_0$ we have: $$\cos\left(ax\right)=\sum_{n=0}^{∞}f^{\left(n\right)}\left(x_{0}\right)\frac{\left(x-x_{0}\right)^{n}}{n!}=\sum_{n=0}^{∞}a^{n}\cos\left(ax_{0}+\frac{n\pi}{2}\right)\frac{\left(x-x_{0}\right)^{n}}{\left(n\right)!}$$

where $$a^{n}\cos\left(ax_{0}+\frac{n\pi}{2}\right)=\frac{d^{n}}{dx^{n}}\cos\left(x\right)$$