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Which is the correct way for expressing the $n$th power of a cosine as a series of cosines without any exponent?

By using the Euler's formula

$\cos^n{(\theta)}=\left( \frac{e^{j\theta}+e^{-j\theta}}{2} \right)^n= \frac{1}{2^n}\left( e^{j\theta}+e^{-j\theta} \right)^n=\frac{1}{2^n}\left( z+z^{-1} \right)^n.$

with $z=e^{j\theta}$.

Since the term $\left( z+z^{-1} \right)^n$ is the $n$th power of a binomial, I could express it using the binomial identity, thus

$\cos^n{(\theta)} = \frac{1}{2^n} \displaystyle\sum_{k=0}^n \binom{n}{k} z^k(z^{-1})^{n-k} = \frac{1}{2^n} \displaystyle\sum_{k=0}^n \binom{n}{k} z^{2k-n}.$

If we expand the expression above, we obtain

$\cos^n{(\theta)}=\frac{1}{2^n}\Bigg ( z^{-n} + \binom{n}{1}z^{-(n-2)} + \binom{n}{2}z^{-(n-4)} + \dots + \binom{n}{2}z^{n-4} + \binom{n}{1}z^{(n-2)} + z^n \Bigg )$

which can be rewritten as

$\cos^n{(\theta)}=\frac{1}{2^n} \Bigg ( (z^{-n} + z^n) + \binom{n}{1} \left(z^{-(n-2)} + z^{(n-2)}\right) + \binom{n}{2}\left( z^{-(n-4)} + z^{(n-4)}\right) + \dots \Bigg ) $

Finally, since $z=e^{j\theta}$

$\cos^n{(\theta)}=\frac{1}{2^n} \Bigg ( (e^{-jn\theta} + e^{jn\theta}) + \binom{n}{1} \left(e^{j(n-2)\theta} + e^{-j(n-2)\theta}\right) + \binom{n}{2}\left( e^{j(n-4)\theta} + e^{-j(n-4)\theta}\right) + \dots \Bigg )$

By applying the Euler's formula once again, we obtain

$\cos^n{(\theta)}=\frac{2}{2^{n}} \sum_{k=0}^n \binom{n}{k} \cos{((n-2k)\theta)}.$

Unfortunately, if I plug n=2, I obtain

$\cos^2{(\theta)}= \cos{(2\theta)} + 1.$

instead of the well-known result

$\cos^2{(\theta)}= \frac{1}{2}(\cos{(2\theta)} + 1).$

a) Why my result is scaled by a factor of 2 ? b) Is the correct general formula $\cos^n{(\theta)}=\frac{1}{2^{n}} \sum_{k=0}^n \binom{n}{k} \cos{((n-2k)\theta)}.$ c)If so, why ?

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    $\begingroup$ You paired the terms up $k=0$ with $k=n$ etc so when you rewrite it in summation notation the sum shouldn't be over all $0 \leq k \leq n$ anymore. Also for $n$ even there is a term in the middle with no pair. $\endgroup$
    – podiki
    Nov 8, 2021 at 22:33
  • $\begingroup$ The right way to do this is a (finite) Fourier series. $\endgroup$
    – Will Jagy
    Nov 8, 2021 at 22:37
  • $\begingroup$ And this can be used to compute the integral $$ \int \cos^n(x) dx $$ $\endgroup$ Nov 8, 2021 at 22:43
  • $\begingroup$ @podiki Yeah, I noticed the even $n$ constant, which is $\binom {n}{n/2}cos(0)$. I was trying to be as general as possible. Thanks for pointing out the error when returning to the summation notation. I will try to fix it and update the question. $\endgroup$ Nov 8, 2021 at 22:49
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    $\begingroup$ Also, you never get out of summation notation. $\endgroup$ Nov 9, 2021 at 1:48

2 Answers 2

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Here is a way to fix the factor of $2$ while still summing from $0$ to $n$ (setting aside the question of whether that is the best way to do the sum). You have $$ \cos^n(\theta) = \frac{1}{2^n}\left( z^{-n} + \binom n1 z^{-(n-2)} + \binom n2 z^{-(n-4)} + \cdots + \binom n2 z^{n-4} + \binom n1 z^{n-2} + z^n \right). $$ Reversing the order of the sum, $$ \cos^n(\theta) = \frac{1}{2^n}\left( z^n + \binom n1 z^{n-2} + \binom n2 z^{n-4} + \cdots + \binom n2 z^{-(n-4)} + \binom n1 z^{-(n-2)} + z^{-n} \right). $$ Adding termwise, \begin{align} 2 \cos^n(\theta) &= \frac{1}{2^n}\bigg( \left(z^{-n} + z^n\right) + \binom n1 \left(z^{-(n-2)} + z^{n-2}\right) + \binom n2 \left(z^{-(n-4)} + z^{n-4}\right) + \cdots \\ &\qquad\qquad + \binom n2 \left(z^{n-4} + z^{-(n-4)}\right) + \binom n1 \left(z^{n-2} + z^{-(n-2)}\right) + \left(z^n + z^{-n}\right) \bigg) \\ &=\frac{2}{2^n}\bigg(\cos(n\theta) + \binom n1 \cos((n-2)\theta) + \binom n2 \cos((n-4)\theta) + \cdots \\ &\qquad\qquad + \binom n2 \cos(-(n-4)\theta) + \binom n1 \cos(-(n-2)\theta) + \cos(-n\theta) \bigg) . \end{align}

Canceling one factor of $2$ on each side, this simplifies to $$ \cos^n(\theta) = \frac{1}{2^n} \sum_{k=0}^n \binom nk \cos{((n-2k)\theta)}. $$ For $n = 2,$ this gives \begin{align} \cos^2(\theta) &= \frac 14\left(\binom 20 \cos(2\theta) + \binom 21 \cos(0) + \binom 22 \cos(-2\theta) \right) \\ &= \frac 12\left(\cos(2\theta) + 1\right) \end{align} as expected.

Your mistake was you did not write the end of the sum after the three dots, so you did not notice that you had moved terms from the right end of the sum to the left end without replacing them, so about half the terms in your summation were not matched by terms in the $+ \cdots +$ notation. Adding two copies of the sum in reverse order, you don't have to move any terms so you won't make this mistake.

Note that it's conventional to combine the $\cos(m\theta)$ and $\cos(-m\theta)$ terms in the sum and have about half as many terms (exactly half as many for odd powers) at the cost of (usually) having separate summations for even and odd powers of the cosine.

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  • $\begingroup$ Thanks a lot, this was the answer I was looking for! I still prefer this method instead of more complicated ones, especially from a learning perspective. $\endgroup$ Nov 9, 2021 at 8:37
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We start from $$ s_n = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} z^{2k-n} $$ Case 1: Let $n = 2m$ $$ s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{2m} \binom{2m}{k} z^{2(k-m)} $$

$$ s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{m} \binom{2m}{k} z^{2(k-m)} + \frac{1}{2^{2m}} \sum_{k=m+1}^{2m} \binom{2m}{k} z^{2(k-m)} $$

$$ s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{m} \binom{2m}{k} z^{2(k-m)} + \frac{1}{2^{2m}} \sum_{l=0}^{m-1} \binom{2m}{2m - l} z^{2(m - l)} $$

$$ s_{n} = \frac{n!}{2^{n}(n/2)!^2} + \frac{1}{2^{n}} \sum_{k=0}^{n/2-1} \binom{n}{k} (z^{2k-n} + z^{n - 2k}) $$

$$ cos^{n}(\theta) = \frac{n!}{2^{n}(n/2)!^2} + \frac{2}{2^{n}} \sum_{k=0}^{n/2-1} \binom{n}{k} cos((2k-n)\theta) $$

$$ cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n/2-1} \binom{n}{k} cos((2k-n)\theta) + \frac{1}{2^{n}}\binom{n}{n/2} \cos((2(n/2)-n)\theta) + \frac{1}{2^{n}} \sum_{l=n/2+1}^{n} \binom{n}{l} \cos((2l-n)\theta) $$

$$ cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} cos((2k-n)\theta) $$

Case 2: Let $n = 2m - 1$ $$ s_{2m - 1} = \frac{1}{2^{2m - 1}} \sum_{k=0}^{2m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1} $$

$$ s_{2m - 1} = \frac{1}{2^{2m - 1}} \sum_{k=0}^{m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1} + \frac{1}{2^{2m - 1}} \sum_{k=m}^{2m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1} $$

$$ s_{2m - 1} = \frac{1}{2^{2m - 1}} \sum_{k=0}^{m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1} + \frac{1}{2^{2m - 1}} \sum_{k = 0}^{m} \binom{2m - 1}{k} z^{2(m-k) - 1} $$

$$ s_{n} = \frac{1}{2^{n}} \sum_{k=0}^{(n - 1)/2} \binom{n}{k} (z^{2k-n} + z^{n-2k}) $$

$$ cos^{n}(\theta) = \frac{2}{2^{n}} \sum_{k=0}^{(n - 1)/2} \binom{n}{k} cos((2k-n)\theta) $$

$$ cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{(n - 1)/2} \binom{n}{k} cos((2k-n)\theta) + \frac{1}{2^{n}} \sum_{l=(n + 1)/2}^{n} \binom{n}{l} cos((2l-n)\theta) $$

$$ cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} cos((2k-n)\theta) $$

Therefore, mixing the both cases we get

$$ cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} cos((2k-n)\theta) $$

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