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Is there a closed form formula or trigonometric relation for $\cos^{n}(ax+b)$?

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    $\begingroup$ This is already a closed formula $\endgroup$ Feb 16, 2021 at 23:34
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    $\begingroup$ There are other ways to write this, but I'm afraid that they are actually more cumbersome than this. $\endgroup$ Feb 16, 2021 at 23:36
  • $\begingroup$ My purpose for this question to prove that for even and odd values of $n$, the period of this function is $\frac{\pi}{a}$ and $\frac{2\pi}{a}$ respectively so I will have to find a way to get a relation for $\cos^{n}(ax+b)$. Perhaps using exponentials or taylor series? @DonThousand $\endgroup$
    – UserX
    Feb 16, 2021 at 23:38

2 Answers 2

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Let's forget $a$ and $b$, as it's not the problem here.

$$\cos^n x=\left(\frac{e^{ix}+e^{-ix}}{2}\right)^n=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}e^{kix}e^{-(n-k)ix}\\=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}e^{(2k-n)ix}\\=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}(\cos(2k-n)x+i\sin(2k-n)x)\\ =\frac1{2^n}\sum_{k=0}^n\binom{n}{k}\cos(2k-n)x$$

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  • $\begingroup$ Alright that looks really nice thank you. For further inspection, how would you tackle the issue of finding the period when there's binomial? $\endgroup$
    – UserX
    Feb 16, 2021 at 23:43
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    $\begingroup$ @SofiaYaseveya The binomial plays little role here, what is important is the period of the terms. $\endgroup$ Feb 16, 2021 at 23:44
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    $\begingroup$ I think recasting the original closed form as a sum makes it harder to reason about the periodicity. $\endgroup$
    – Rob Arthan
    Feb 16, 2021 at 23:45
  • $\begingroup$ @RobArthan If think so. I didn't see the comments until I wrote the answer though. $\endgroup$ Feb 16, 2021 at 23:47
  • $\begingroup$ I see, perhaps it would've required a once in a lifetime computation. However, I am still satisfied with this beautiful answer. $\endgroup$
    – UserX
    Feb 16, 2021 at 23:47
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As mentioned in your comment, if you are only looking to find the period of $cos^n(ax+b)$, it wouldn't be necessary to write a closed form formula or expression for it. When $n$ is even, We see that $(\cos(\pi - (ax+b)))^n = (\cos(\pi + (ax+b)))^n$ and this turns the period of the function into $\pi/a$, whereas when $n$ is odd, we don't have this luxury anymore. Instead we have $\cos^n(ax+b)$ going its full $2\pi/a$ cycle.

Edit: As @Rob Arthan pointed out in his comment, given $f(x)$ with period $T$ and some function $g(x)$; $h(x) = g(f(x))$ has period $T/k$ for some integer $k$. It is also possible to observe that $T$ is a period of $h$, but often depending upon the function $g$ we can do better.

In our case $f = \cos(ax+b)$ and $g(x) = x^n$. When $n$ is odd, $g(x)$ is injective, and so every input to $g$ has a unique associated output, which means there is no room in $[0,2\pi/a]$ (the original period of $f$) for repetition of the same output values. This is why, when $n$ is odd, the period of $h$ is the same as the period of $f$. However, when $g$ is not injective, we have multiple inputs being mapped to the same output. In the case of an even $n$, we have exactly two inputs in a given period being mapped to the same output, which essentially means we are repeating the graph of $f$ twice in $[0,2\pi/a]$ upon composing, which means the period is halved simply giving us a period of $\pi/a$.

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    $\begingroup$ Perhaps you should explain that you are using the fact that the function $x \mapsto x^n : [-1, 1] \to [-1, 1]$ is 1-1 when $n$ is odd and 2-1 when $n$ is even (so that the periods cannot be proper divisors of $\pi/a$ or $2\pi/a$). $\endgroup$
    – Rob Arthan
    Feb 16, 2021 at 23:49
  • $\begingroup$ I didn't quite understand the reasoning of $n$ being even implying $(\cos(\pi+(ax+b)))^{n}$. I am though, interested in your answer if you could please clarify more. $\endgroup$
    – UserX
    Feb 16, 2021 at 23:51
  • $\begingroup$ If $f$ is periodic with period $t$, and $g$ is any function (in this case $y \mapsto y^n$), then $x \mapsto g(f(x))$ is either constant or periodic with a periodic with period $t/k$ for some positive integer $k$. You should find this easy to prove from the definition of a periodic function. $\endgroup$
    – Rob Arthan
    Feb 16, 2021 at 23:56
  • $\begingroup$ I can't seem to understand your intuition :( @RobArthan $\endgroup$
    – UserX
    Feb 16, 2021 at 23:59
  • $\begingroup$ If $f(x) = f(x + t)$, then $g(f(x)) = g(f(x + t))$. $\endgroup$
    – Rob Arthan
    Feb 17, 2021 at 0:00

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