k-1-cos-x-2-k-

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} = \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}})}

\dots . 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 \dots

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