# Powers of Cosine as a Linear Combination of a Sequence

• Sep 11th 2010, 11:01 AM
mathematicalbagpiper
Powers of Cosine as a Linear Combination of a Sequence
So I was reading through my Applied Math textbook and came across a proof which they omitted some (I guess what was supposed to be obvious) details but I can't see where they're getting it from.

We're given a sequence defined as follows:

$\displaystyle e_0(x) = \frac{1}{\sqrt{\pi}}$

$\displaystyle e_k(x) = \sqrt{\frac{2}{\pi}}\cos(kx), k\geq1$

They showed that this forms an orthonormal basis, which I understand how they did it, but I'm not getting how they're going from using the following identity:

$\displaystyle (\cos x)^k = (\frac{e^{ix}+e^{-ix}}{2})^k$

to claiming that this demonstrates any power of $\displaystyle (\cos x)^k$ is a linear combination of the elements from $\displaystyle \{e_k(x)\}$.

How does this identity get used to show that? I'm just really curious.
• Sep 11th 2010, 05:51 PM
Ackbeet
Use the binomial theorem as follows:

$\displaystyle \displaystyle{\left(\frac{e^{ix}+e^{-ix}}{2}\right)^{\!\!k}=\frac{1}{2^{k}}\sum_{j=0}^{ k}\binom{k}{j}(e^{ix})^{k-j}(e^{-ix})^{j}=\frac{1}{2^{k}}\sum_{j=0}^{k}\binom{k}{j} e^{ix(k-j)}\,e^{-ijx}=\frac{1}{2^{k}}\sum_{j=0}^{k}\binom{k}{j}e^{i x(k-2j)}.}$

Now we can use Euler to say that this equals

$\displaystyle \displaystyle{\frac{1}{2^{k}}\sum_{j=0}^{k}\binom{ k}{j}\left[\cos(x(k-2j))+i\sin(x(k-2j))\right],}$ which we can break up as

$\displaystyle \displaystyle{\frac{1}{2^{k}}\left[\sum_{j=0}^{k}\binom{k}{j}\cos(x(k-2j))+i\sum_{j=0}^{k}\binom{k}{j}\sin(x(k-2j))\right].}$

Since the original LHS equals this last expression, we can say that the imaginary part of the RHS must vanish, since the LHS is real. That is, we can say that

$\displaystyle \displaystyle{(\cos(x))^{k}=\frac{1}{2^{k}}\sum_{j =0}^{k}\binom{k}{j}\cos(x(k-2j)).}$

You're almost there now. All you need do to finish is to note that the cosine function is even. That means some of the terms in the sum can be rewritten in terms of positive multiples of $\displaystyle x$ inside the argument. Moreover, the constant term, if there is one (which will happen precisely when $\displaystyle k$ is even), you can take out of the sum, if you wish, and include in the multiple of $\displaystyle e_{0}$.

Does this help?
• Sep 11th 2010, 05:54 PM
mathematicalbagpiper
Helps a bunch. Thanks!
• Sep 11th 2010, 05:56 PM
Ackbeet
You're welcome. Hey, love the screen name. I'm in the Manchester Regional Police and Fire Pipe Band (CT). I play snare.