I'm a little bit confused as to how to go about this question. I've looked extensively out on the net about it but the examples always tend to be at $a = 0$ which is a pretty easy example.

How do I write the power series for $\cos(x)$ evaluated at $a = \frac{\pi}{3}$?

What is the technique I should use? I can't seem to find an example that helps me. The preceding question is to find the first few terms of the Taylor series, so I assume from there I am supposed to be able to find the power series (in summation notation)?


2 Answers 2


The Taylor expansion of $f$, centered in $a$, of order $n$, is: $$P_n (f, a, x)= \sum_{k = 0}^n \frac{f^{(k)} (a)}{k!} (x - a)^k $$

Where $f^{(k)}$ denotes the $k$th-derivative, and we have the convention $f^{(0)} = f $. For example: $$P_{3} \left(\cos, \frac{\pi}{3}, x \right) = \cos \frac{\pi}{3} - \sin \frac{\pi}{3} \left(x - \frac{\pi}{3} \right) - \frac{\cos \frac{\pi}{3}}{2} \left( x - \frac{\pi}{3} \right)^2 + \frac{\sin \frac{\pi}{3}}{6} \left( x - \frac{\pi}{3} \right)^3 $$


A Taylor series is an infinite series of $f(x)$ centered at $a$. It is defined as: $$\sum_{n = 0}^\infty \frac{f^{(n)} (a)}{n!} (x - a)^n $$

Where $f^{(n)}(a)$ is the $n_{th}$ derivative of the function $f(a)$. There is also a special type of Taylor series, known as a Maclaurin series. This is simply a Taylor series that is centered at 0. This is defined in the same way: $$\sum_{n = 0}^\infty \frac{f^{(n)} (0)}{n!} (x - 0)^n = \sum_{n-0}^\infty \frac{f^{(n)}(a)x^n}{n!}$$ To find the Taylor series of a function using the definition of aTaylor series, the first thing you have to do is find a series that arithmetically (not using derivatives) gets the same values for any given iteration as $f^{(n)}(x)$. The simplest method of doing this is to to just take the first few derivatives, plug in a, and find a pattern that reproduces the same values. if $f(x) = cos(x); a = \frac{pi}{3}$, then:

\begin{align*} &f(x)=cos(x) \quad && \implies \quad f\left(\frac{\pi}{3}\right) = cos\left(\frac{\pi}{3}\right) = &\frac{1}{2} \\ &f'(x) = -sin(x) &&\implies \quad f'\left(\frac{\pi}{3}\right) = sin\left(\frac{\pi}{3}\right)=&-\frac{\sqrt{3}}{2} \\ &f''(x) = -cos(x) \quad && \implies \quad f''\left(\frac{\pi}{3}\right) =-cos\left(\frac{\pi}{3}\right) = &-\frac{1}{2} \\ &f'''(x)=sin(x) \quad && \implies \quad f'''\left(\frac{\pi}{3}\right) = sin\left(\frac{\pi}{3}\right) = &\frac{\sqrt{3}}{2} \\ &f^{(4)}(x)=cos(x) \quad && \implies \quad f^{(4)}\left(\frac{\pi}{3}\right) = cos\left(\frac{\pi}{3}\right) = &\frac{1}{2} \end{align*} Now let's break this into parts. It seems to switch from positive to negative to negative to positive, etc. This tells me it is going to be some form of and alternating series, which means that part of it will look something like $(-1)^{n}$. Next I notice that the numerator switches between $1$ and $\sqrt{3}$, but the denominator stays the same. Well, one way to go from anything to $1$ is to give it an exponent of $0$, so what exponent involving $n$ can we give $\sqrt{3}$ such that the net exponent will switch between $0$ and $1$? Well, one thing I can use is $(-1)^n$ which switches things between positive and negative. If I want to switch between a number and $0$, I can simply say $\text{half of that number} \pm \text{half the number}$. This means that if I want to switch back and forth between $(\sqrt{3})^0$ and $(\sqrt{3})^1$, I would express it as


Now I said before that the Alternating series part of it would look something like $(-1)^n$ but that would switch every iteration, and we need something that switches every two iterations. This actually becomes incredibly complicated, and the only way I found it, was through this question from Math.SE. Instead of $(-1)^n$, it's going to look like this: $$\frac{1-\mathrm i}2\cdot\mathrm i^n+\frac{1+\mathrm i}2\cdot(-\mathrm i)^n$$ Now since the denominator of $f^{(n)}(a)$ stays at $2$, we can leave it that way in the closed form of the series. We have now found that the closed form of $f^{(n)}(a)$ for $cos(x)$ at $a=\frac{\pi}{3}$ is equal to:

$$\frac{\left(\frac{1-\mathrm i}2\cdot\mathrm i^n+\frac{1+\mathrm i}2\cdot(-\mathrm i)^n\right) \left(\sqrt{3}\right)^{\left(\frac{1}{2}+\frac{(-1)^n}{2}\right)}}{2}$$

Now we plug this into our original definition of Taylor series:

$$\sum_{n = 0}^\infty \frac{f^{(n)} (a)}{n!} (x - a)^n$$

$$= \require{enclose} \enclose{box}{\sum_{n = 0}^\infty \frac{\left(\frac{1-\mathrm i}2\cdot\mathrm i^n+\frac{1+\mathrm i}2\cdot(-\mathrm i)^n\right) \left(\sqrt{3}\right)^{\left(\frac{1}{2}+\frac{(-1)^n}{2}\right)}\left(x-\frac{\pi}{3}\right)^n}{2(n!)}}$$

The other two ways to express it are:

$$\require{enclose} \enclose{box}{\sum_{n = 0}^\infty \frac{\sqrt2\cdot\cos\left(n\frac\pi2-\frac\pi4\right) \left(\sqrt{3}\right)^{\left(\frac{1}{2}+\frac{(-1)^n}{2}\right)}\left(x-\frac{\pi}{3}\right)^n}{2(n!)}}$$


$$ \require{enclose} \enclose{box}{\sum_{n = 0}^\infty \frac{(-1)^{\lceil n/2\rceil} \left(\sqrt{3}\right)^{\left(\frac{1}{2}+\frac{(-1)^n}{2}\right)}\left(x-\frac{\pi}{3}\right)^n}{2(n!)}} $$

I'm gonna go out on a limb and say there is absolutely no way your Calc II professor gave that to you for homework. You will also probably never ever see it on any test ever (at least in Calc II).

  • $\begingroup$ It's on a take-home quiz, and I was completely baffled to be honest. Maybe I misunderstood the question but it said first to find the first five non-zero terms, then represent the power series in summation notation. Just couldn't tell from the direction what 'power series' meant. $\endgroup$ May 12, 2014 at 4:19
  • $\begingroup$ The first five non-zero terms part of that would be Ivo's answer, and is something I would expect in a Calc II class. The Power series in summation notation would bring you to my answer. I'm surprised that the second part was on the quiz, but I guess that's why the teacher let you take it home. What i'm used to seeing is "find the Maclaurin series of $\cos(x)$" which is just $\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}$ and can be found easily using the definition of taylor series. When it's centered at $a=\frac{\pi}{3}$, not so simple... $\endgroup$
    – Ephraim
    May 12, 2014 at 11:16

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