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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 $\cos(x)$?

Burt
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3 Answers3

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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.

Claude Leibovici
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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.

Kraigolas
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  • So it would be a different, equivalent series (which is not as practical) or it wouldn't? – Burt Jan 10 '20 at 03:45
  • @burt if you are using a finite series to approximate, it would be different. If it was the same series then there would be no point in expanding about a different point because they'd all give the same answer. And in fact, it would not be equivalent. However, if you used the infinite series, then you could use them equivalently. For example, if your series was centered around $x=c$, then substituting $x+c$ in every place in the series would give you an identical series to the infinite series of $\cos(x)$ centered at zero. – Kraigolas Jan 10 '20 at 04:02
  • In short, if you're using an infinite series, the center doesn't matter because the series is exactly $\cos(x)$ for all $x$, so you're better off using the easiest center (and the least confusing because you don't have to worry about shifting your variable $x$ to account for where the series is centered). However, if you're approximating by only grabbing a few terms, you want the center to be as close to the point you are approximating as possible. – Kraigolas Jan 10 '20 at 04:04
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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 $$a^{n}\cos\left(ax_{0}+\frac{n\pi}{2}\right)=\frac{d^{n}}{dx^{n}}\cos\left(x\right)$$