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If it is easier, you can do it the other way around, by writing $\cos(\pi t)$ in terms of $\cos(\frac{\pi t}{2^n})$. I just wanted to know if there was a nice closed form solution to a problem like this, and, if so, how many terms scale up with how large $n$ scales up.

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    $\begingroup$ I would expect many square roots. $\endgroup$
    – Szeto
    Jun 23, 2018 at 3:16
  • $\begingroup$ Would going the other way around create a "cleaner" solution? $\endgroup$
    – wjmccann
    Jun 23, 2018 at 3:25
  • $\begingroup$ You mean express $\cos(\frac{\pi t}{2^n})$ in terms of $\cos(\pi t)$? $\endgroup$
    – Szeto
    Jun 23, 2018 at 3:39
  • $\begingroup$ Yes, or writing $\cos(\pi t)$ in terms of $\cos(\frac{\pi t}{2^n})$ $\endgroup$
    – wjmccann
    Jun 23, 2018 at 3:42

1 Answer 1

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Writing $\cos(\pi t)$ in terms of $\cos(\frac{\pi t}{2^n})$ ... the same thing as writing $\cos(2^n\theta)$ in terms of $\cos(\theta)$ ... is a Chebyshev polynomial

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  • $\begingroup$ And inverting that would mean taking the square root of the average with 1, $n$ times. $\endgroup$
    – ccorn
    Jun 23, 2018 at 11:20

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