Evaluate-lim-n-k-0-n-1-cos-2-k-pi-2-n-1-

Question Number 209356 by mnjuly1970 last updated on 08/Jul/24

E v a l u a t e :

\lim_{n \to \infty} \underset{k = 0}{\prod^{n - 1}} c o s (\frac{2^{k} . π}{2^{n} - 1}) = ?

Answered by mr W last updated on 08/Jul/24

P_{n} = \underset{k = 0}{\prod^{n - 1}} \cos \frac{2^{k} π}{2^{n} - 1} = \cos \frac{π}{2^{n} - 1} \cos \frac{2 π}{2^{n} - 1} \dots \cos \frac{2^{n - 1} π}{2^{n} - 1}

2 \sin \frac{π}{2^{n} - 1} P_{n} = 2 \sin \frac{π}{2^{n} - 1} \cos \frac{π}{2^{n} - 1} \cos \frac{2 π}{2^{n} - 1} \dots \cos \frac{2^{n - 1} π}{2^{n} - 1}

2 \sin \frac{π}{2^{n} - 1} P_{n} = \sin \frac{2 π}{2^{n} - 1} \cos \frac{2 π}{2^{n} - 1} \dots \cos \frac{2^{n - 1} π}{2^{n} - 1}

2^{2} \sin \frac{π}{2^{n} - 1} P_{n} = 2 \sin \frac{2 π}{2^{n} - 1} \cos \frac{2 π}{2^{n} - 1} \dots \cos \frac{2^{n - 1} π}{2^{n} - 1}

2^{2} \sin \frac{π}{2^{n} - 1} P_{n} = \sin \frac{2^{2} π}{2^{n} - 1} \dots \cos \frac{2^{n - 1} π}{2^{n} - 1}

\dots \dots

2^{n - 1} \sin \frac{π}{2^{n} - 1} P_{n} = \sin \frac{2^{n - 1} π}{2^{n} - 1} \cos \frac{2^{n - 1} π}{2^{n} - 1}

2^{n} \sin \frac{π}{2^{n} - 1} P_{n} = 2 \sin \frac{2^{n - 1} π}{2^{n} - 1} \cos \frac{2^{n - 1} π}{2^{n} - 1}

2^{n} \sin \frac{π}{2^{n} - 1} P_{n} = \sin \frac{2^{n} π}{2^{n} - 1}

\Rightarrow P_{n} = \frac{\sin \frac{2^{n} π}{2^{n} - 1}}{2^{n} \sin \frac{π}{2^{n} - 1}} = \frac{\sin (π + \frac{π}{2^{n} - 1})}{2^{n} \sin \frac{π}{2^{n} - 1}}

= - \frac{\sin \frac{π}{2^{n} - 1}}{2^{n} \sin \frac{π}{2^{n} - 1}} = - \frac{1}{2^{n}}

\lim_{n \to \infty} P_{n} = 0

Commented by mnjuly1970 last updated on 08/Jul/24

t h a n k s a l o t s i r W

s o e x c e l l e n t s o l u t i o n \underset{―}{\underset{⏟}{⋚}}