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Question Number 144000 by bluberry508 last updated on 20/Jun/21

prove that

\forall_{m} \in N, a_{k}, b_{k} \in R

\cos^{2 m} x = \underset{k = 1}{\sum^{m}} a_{k} \cos 2 k x

\cos^{2 m - 1} x = \underset{k = 1}{\sum^{m}} b_{k} \cos (2 k - 1) x

and find expr of a_{k}, b_{k} in terms of k .

Answered by mathmax by abdo last updated on 20/Jun/21

\cos^{2 m} x = {(\frac{e^{ix} + e^{- ix}}{2})}^{2 m} = \frac{1}{2^{2 m}} \sum_{k = 0}^{2 m} C_{2 m}^{k} {(e^{ix})}^{k} {(e^{- ix})}^{2 m - k}

= \frac{1}{2^{2 m}} \sum_{k = 0}^{2 m} e^{ikx} . e^{- i (2 m - k) x}

= \frac{1}{2^{2 m}} \sum_{k = 0}^{2 m} C_{2 m}^{k} e^{i (2 k - 2 m) x} = \frac{1}{2^{2 m}} \sum_{k = 0}^{2 m} C_{2 m}^{k} e^{2 i (k - m) x}

= \frac{1}{2^{2 m}} \sum_{k = 0}^{2 m} C_{2 m}^{k} \cos 2 (k - m) x + \frac{i}{2^{2 m}} \sum_{k 0}^{2 m} C_{2 m}^{k} \sin 2 (k - m) x

but \cos^{2 m} x is real \Rightarrow \cos^{2 m} x = \frac{1}{2^{2 m}} \sum_{k = 0}^{2 m} C_{2 m}^{k} \cos 2 (k - m) x

=_{k - m = j} \frac{1}{2^{2 m}} \sum_{j = - m}^{m} C_{2 m}^{m + j} \cos 2 jx

= \frac{1}{2^{2 m}} \sum_{j = - m}^{- 1} C_{2 m}^{m + j} \cos (2 jx) + \frac{C_{2 m}^{m}}{2^{2 m}} + \sum_{j = 1}^{m} C_{2 m}^{m + j} \cos (2 jx)

=_{j = - k} \frac{1}{2^{2 m}} \sum_{k = 1}^{m} C_{2 m}^{m - k} \cos (2 kx) + \frac{C_{2 m}^{m}}{2^{2 m}} + \frac{1}{2^{2 m}} \sum_{k = 1}^{m} C_{2 m}^{m - k} \cos (2 kx)

= 1 + \frac{2}{2^{2 m}} \sum_{k = 1}^{m} C_{2 m}^{m - k} \cos (2 kx) = \frac{C_{2 m}^{m}}{2^{2 m}} + \frac{1}{2^{2 m - 1}} \sum_{k = 1}^{m} C_{2 m}^{m - k} \cos (2 kx)

\Rightarrow a_{k} = \sum_{k = 1}^{m} \frac{C_{2 m}^{m - k}}{2^{2 m - 1}} for k ⩾ 1 and a_{0} = \frac{C_{2 m}^{m}}{2^{2 m}}

\cos^{2 m} x = a_{0} + \sum_{k = 1}^{m} a_{k} \cos (2 kx)

Commented by bluberry508 last updated on 20/Jun/21

wow \dots I {haven}^{'} t ever expected this view

thank you sir

Commented by Dwaipayan Shikari last updated on 20/Jun/21

S o r r y i h a v e m a r k e d y o u r c o m m e n t a s i n a p p r o p i a t e

b y m i s t a k e .

Commented by Rasheed.Sindhi last updated on 20/Jun/21

A n d I^{'} v e c l e a r e d t h e r e d m a r k b y

l i k i n g t h e p o s t!

Commented by Dwaipayan Shikari last updated on 20/Jun/21

T h a n k s s i r

Commented by mathmax by abdo last updated on 20/Jun/21

you are welcome