Π_(k=1) ^∞ cos((x/2^k )) = ?


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Question Number 119479 by 675480065 last updated on 24/Oct/20

$\underset{k = 1}{\prod^{\infty}} \cos (\frac{x}{2^{k}}) = ?$
	Answered by Olaf last updated on 25/Oct/20

	$\sin 2 θ = 2 \sin θ \cos θ$ $\cos θ = \frac{1}{2} . \frac{\sin 2 θ}{\sin θ}, θ \neq k π$ $\Rightarrow \cos (\frac{x}{2^{k}}) = \frac{1}{2} . \frac{\sin (\frac{x}{2^{k - 1}})}{\sin (\frac{x}{2^{k}})}$ $And \underset{k = 1}{\prod^{n}} \cos (\frac{x}{2^{k}}) = \frac{1}{2^{n}} \underset{k = 1}{\prod^{n}} \frac{\sin (\frac{x}{2^{k - 1}})}{\sin (\frac{x}{2^{k}})}$ $\underset{k = 1}{\prod^{n}} \cos (\frac{x}{2^{k}}) = \frac{1}{2^{n}} . \frac{\sin x}{\sin (\frac{x}{2^{n}})}$ $(telescopic product)$ $\underset{k = 1}{\prod^{\infty}} \cos (\frac{x}{2^{k}}) = \sin x . \lim_{n \to \infty} \frac{1}{2^{n} \sin (\frac{x}{2^{n}})}$ $2^{n} \sin (\frac{x}{2^{n}}) \underset{\infty}{\sim} 2^{n} \times \frac{x}{2^{n}} = x$ $\underset{k = 1}{\prod^{\infty}} \cos (\frac{x}{2^{k}}) = \frac{\sin x}{x}$
	Commented by 675480065 last updated on 25/Oct/20

	$Thanks sir .$ $can i apply complex numbers to it ?$
	Answered by Bird last updated on 25/Oct/20

	$l e t A_{n} = \prod_{k = 1}^{n} c o s (\frac{x}{2^{k}}) s n d$ $B_{n} = \prod_{k = 1}^{n} s i n (\frac{x}{2^{k}}) w e h s v e$ $A_{n} . B_{n} = \prod_{k = 1}^{n} c o s (\frac{x}{2^{k}}) s i n (\frac{x}{2^{k}})$ $= \frac{1}{2^{n}} \prod_{k = 1}^{n} s i n (\frac{x}{2^{k - 1}})$ $= \frac{1}{2^{n}} \prod_{k = 0}^{n - 1} s i n (\frac{x}{2^{k}})$ $= \frac{1}{2^{n}} \frac{s i n x}{s i n (\frac{x}{2^{n}})} \times \prod_{k = 1}^{n} s i n (\frac{x}{2^{k}})$ $= \frac{s i n x}{2^{n} s i n (\frac{x}{2^{n}})} \times B_{n} w e h s v e B_{n} \neq 0 \Rightarrow$ $A_{n} = \frac{s i n x}{2^{n} s i n (\frac{x}{2^{n}})} b u t$ $s i n (\frac{x}{2^{n}}) \sim \frac{x}{2^{n}} \Rightarrow 2^{n} s i n (\frac{x}{2^{n}}) \sim x (n \to \infty)$ $\Rightarrow {l i m}_{n \to \infty} A_{n} = \frac{s i n x}{x} \Rightarrow$ $\prod_{k = 1}^{\infty} c o s (\frac{x}{2^{k}}) = \frac{s i n x}{x}$
	Commented by 675480065 last updated on 25/Oct/20

	$Thanks Sir .$ $Plz i want to learn this topic .$ $how can i get notes on it sir .$ $pls help me if u have$ $Thanks$
	Answered by Bird last updated on 25/Oct/20
	$A_n =Π_(k=1) ^n cos((x/2^k )) ⇒ A_n =Π_(k=1) ^n ((e^(i(x/2^k )) +e^(−((ix)/2^k )) )/2) =(1/2^n )Π_(k=1) ^n e^((ix)/2^k ) Π_(k=1) ^n (1+e^(−((2ix)/2^k )) ) =(1/2^n )e^(ixΣ_(k=1) ^n (1/2^k )) Π_(k=1) ^n (1+cos(((2x)/2^k ))−isin(((2x)/2^k ))) =(1/2^n ) e^(ixΣ_(k=0) ^(n−1) (1/(2^(k+1) ))) ×Π_(k=1) ^n (1+cos((x/2^(k−1) ))−isin((x/2^(k−1) ))) =(1/2^n ) e^(((ix)/2)×(1/(1−(1/2)))) ×Π_(k=0) ^(n−1) (1+cos((x/2^k ))−isin((x/2^k ))) =(1/2^n ) e^(ix) ×Π_(k=0) ^(n−1) {2cos^2 ((x/2^(k+1) ))−2isin((x/2^(k+1) ))cos((x/2^(k+1) ))} ....be continued..i think this method give the answer ...$
	$A_{n} = \prod_{k = 1}^{n} c o s (\frac{x}{2^{k}}) \Rightarrow$ $A_{n} = \prod_{k = 1}^{n} \frac{e^{i \frac{x}{2^{k}}} + e^{- \frac{i x}{2^{k}}}}{2}$ $= \frac{1}{2^{n}} \prod_{k = 1}^{n} e^{\frac{i x}{2^{k}}} \prod_{k = 1}^{n} (1 + e^{- \frac{2 i x}{2^{k}}})$ $= \frac{1}{2^{n}} e^{i x \sum_{k = 1}^{n} \frac{1}{2^{k}}} \prod_{k = 1}^{n} (1 + c o s (\frac{2 x}{2^{k}}) - i s i n (\frac{2 x}{2^{k}}))$ $= \frac{1}{2^{n}} e^{i x \sum_{k = 0}^{n - 1} \frac{1}{2^{k + 1}}} \times \prod_{k = 1}^{n} (1 + c o s (\frac{x}{2^{k - 1}}) - i s i n (\frac{x}{2^{k - 1}}))$ $= \frac{1}{2^{n}} e^{\frac{i x}{2} \times \frac{1}{1 - \frac{1}{2}}} \times \prod_{k = 0}^{n - 1} (1 + c o s (\frac{x}{2^{k}}) - i s i n (\frac{x}{2^{k}}))$ $= \frac{1}{2^{n}} e^{i x} \times \prod_{k = 0}^{n - 1} {2 {c o s}^{2} (\frac{x}{2^{k + 1}}) - 2 i s i n (\frac{x}{2^{k + 1}}) c o s (\frac{x}{2^{k + 1}})}$ $. . . . b e c o n t i n u e d . . i t h i n k t h i s$ $m e t h o d g i v e t h e a n s w e r . . .$
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