Prove-that-for-any-real-number-x-and-odd-positive-integer-n-cos-n-x-1-2-n-1-k-0-n-1-2-C-k-n-cos-n-2k-x-

Question Number 170800 by thfchristopher last updated on 31/May/22

Prove that, for any real number x and odd positive integer n,

\cos^{n} x = \frac{1}{2^{n + 1}} \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} \cos (n - 2 k) x

Answered by aleks041103 last updated on 04/Jun/22

c o s x = \frac{e^{i x} + e^{- i x}}{2}

{c o s}^{n} x = \frac{1}{2^{n}} {(e^{i x} + e^{- i x})}^{n} =

= \frac{1}{2^{n}} (\underset{k = 0}{\sum^{n}} C_{k}^{n} e^{i k x} e^{- i (n - k) x}) =

= \frac{1}{2^{n}} (\underset{k = 0}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)})

\underset{k = 0}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)} =

= \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} e^{i x (2 k - n)} + \underset{k = (n + 1) / 2}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)}

\underset{k = (n + 1) / 2}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)}

l e t n - 2 j = 2 k - n

\Rightarrow 2 j = 2 n - 2 k \Rightarrow j = n - k, k = n - j

j_{1} = n - \frac{n + 1}{2} = \frac{n - 1}{2}

j_{2} = n - n = 0

\Rightarrow \underset{k = (n + 1) / 2}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)} = \underset{j = 0}{\sum^{(n - 1) / 2}} C_{n - j}^{n} e^{- i x (2 j - n)}

b u t C_{n - j}^{n} = C_{j}^{n}

c h a n g i n g j t o k

\Rightarrow \underset{k = (n + 1) / 2}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)} = \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} e^{- i x (2 k - n)}

\Rightarrow \underset{k = 0}{\sum^{n}} C_{k}^{n} e^{i x (2 k - n)} =

= \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} e^{i x (2 k - n)} + \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} e^{- i x (2 k - n)} =

= \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} (e^{i x (2 k - n)} + e^{- i x (2 k - n)}) =

= 2 \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} c o s ((2 k - n) x)

\Rightarrow {c o s}^{n} x = \frac{1}{2^{n - 1}} \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} c o s ((2 k - n) x)

c o s i s e v e n f u n c t i o n

{c o s}^{n} x = \frac{1}{2^{n - 1}} \underset{k = 0}{\sum^{(n - 1) / 2}} C_{k}^{n} c o s ((n - 2 k) x)