prove that ∀_m ∈N , a_k ,b_k ∈R cos^(2m) x =Σ_(k=1) ^m a_k cos 2kx cos^(2m−1) x=Σ_(k=1) ^m b_k cos (2k−1)x and find expr of a_k ,b


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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

	$wow . . . I {haven}^{'} t ever expected this view$ $thank you sir$
	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

	$you are welcome$
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