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Is it possible to expand $\cos^n(x)$ as a function of $\cos(kx)$? i.e. if $\cos^n (x)$ can be expanded as the following series

$$ \cos^n (x) = \sum_{k=0}^{n} a_k \cos{kx} $$

then what are the constants $a_k$? If not, is there any way to recover $a_0$ for any $n$?

Chris
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  • Use http://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-eit-cos-t-i-sin-t – lab bhattacharjee Mar 30 '17 at 15:37
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    Just consider that $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$, apply the binomial theorem to $(e^{ix}+e^{-ix})^n$ and collect back together terms like $e^{ikx}$ and $e^{-ikx}$. – Jack D'Aurizio Mar 30 '17 at 15:37
  • @JackD'Aurizio brilliant, thanks! – Chris Mar 30 '17 at 15:43
  • This seems closely related to the problem of Chebyshev polynomials (https://en.wikipedia.org/wiki/Chebyshev_polynomials). – Connor Harris Mar 30 '17 at 15:48
  • @ConnorHarris I think the chebyshev polynomials finds $\cos(nx)$ in terms of $\cos(x)$ whilst I'm trying to find the "inverse" of the $T_n$ function – Chris Mar 30 '17 at 15:52

3 Answers3

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For odd $n$,

$$\cos^n x=\left(\frac{e^{ix}+e^{-ix}}2\right)^n=\frac1{2^n}\sum_{k=0}^n\binom nke^{ikx}e^{-i(n-k)x}=\frac1{2^n}\sum_{k=0}^n\binom nke^{i(2k-n)x}\\ =\frac2{2^n}\sum_{k=(n+1)/2}^n\binom nk\cos(2k-n)x.$$

For even $n$, the development is very similar, but for the middle term.

  • excellent - thanks, that gives me an equation to find the $a_0$ term so I can figure out the integral of $\cos^n (x)$ between 0 and $2\pi$ for all n. – Chris Mar 30 '17 at 16:18
  • @Chris: you should have asked that only. –  Mar 30 '17 at 17:19
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Since $f(x) = \cos^n x$ is even and smooth on $[-\pi,\pi]$ is has a Fourier cosine series $f(x) = {c_0 \over 2} + \sum_k c_k \cos (kx)$ where $c_k = {2 \over \pi} \int_0^\pi \cos^n x \cos(kx) dx$.

Since $\cos^n x = {1 \over 2^n} (e^{ix}+e^{-ix})^n$, we see that for $k >n$ we have $c_k = 0$.

We see that $a_0 = {c_0 \over 2} = {1 \over \pi} \int_0^\pi \cos^n x dx$.

It should be clear (by integrating over $[0, {\pi \over 2}]$ and $[ {\pi \over 2},1]$ separately) that for odd $n$ that $a_0 = 0$.

When $n$ is even, combine the binomial theorem with Euler's formula to get $a_0 = {1 \over 2^n \pi} \int_0^\pi \binom{n}{k}e^{i(n-2k)x} dx = {1 \over 2^n} \binom{n}{n \over 2}$.

copper.hat
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You can use the Fourier series expansion. Since the function $cos^n(x)$ is even it is written in even terms of the Fourier series, i.e. $cos(kx)$ in the domain of $[-\pi,\pi]$. The Fourier series uses orthogonal functions $cos(kx)$ and $sins(kx)$ meaning that using the integral as an inner product you will have: $$\int\limits_{-\pi}^{\pi}cos(qx)cos(px)dx = \pi\delta_{qp},\quad q,p\geq1$$ $$\int\limits_{-\pi}^{\pi}sin(qx)sin(px)dx = \pi\delta_{qp},\quad q,p\geq1$$ $$\int\limits_{-\pi}^{\pi}sin(qx)cos(px)dx = 0$$ In order to find the $a_k$ coefficient in your sum it is needed to calculate the integral $$a_k = 1/\pi\int\limits_{-\pi}^{\pi}cos^n(x)cos(kx)dx$$

More about Fourier series is found in here.

MohammadSh
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