Question-155183

Question Number 155183 by mathdanisur last updated on 26/Sep/21

Answered by aleks041103 last updated on 26/Sep/21

1 - {t a n}^{2} \frac{x}{2^{k}} = \frac{{c o s}^{2} \frac{x}{2^{k}} - {s i n}^{2} \frac{x}{2^{k}}}{{c o s}^{2} \frac{x}{2^{k}}} = \frac{c o s \frac{x}{2^{k - 1}}}{{c o s}^{2} \frac{x}{2^{k}}}

\underset{k = 1}{\prod^{n}} c o s \frac{x}{2^{k - 1}} = c o s x c o s \frac{x}{2} \dots c o s \frac{x}{2^{n - 1}} =

= \frac{1}{s i n \frac{x}{2^{n - 1}}} s i n \frac{x}{2^{n - 1}} c o s \frac{x}{2^{n - 1}} c o s \frac{x}{2^{n - 2}} \dots c o s \frac{x}{2} c o s x =

= \frac{1}{s i n \frac{x}{2^{n - 1}}} \frac{1}{2} s i n \frac{x}{2^{n - 2}} c o s \frac{x}{2^{n - 2}} \dots c o s \frac{x}{2} c o s x =

= \frac{1}{2^{2} s i n \frac{x}{2^{n - 1}}} s i n \frac{x}{2^{n - 3}} \underset{k = 0}{\prod^{n - 3}} c o s \frac{x}{2^{k}} = \dots =

= \frac{1}{2^{m} s i n \frac{x}{2^{n - 1}}} s i n \frac{x}{2^{n - 1 - m}} \underset{k = 0}{\prod^{n - 1 - m}} c o s \frac{x}{2^{k}} =

= \frac{1}{2^{n - 1} s i n \frac{x}{2^{n - 1}}} s i n x c o s x = \frac{s i n (2 x)}{2^{n} s i n (\frac{x}{2^{n - 1}})}

\underset{k = 1}{\prod^{n}} c o s \frac{x}{2^{k}} = \underset{k = 1}{\prod^{n}} c o s \frac{x / 2}{2^{k - 1}} =

= \frac{s i n (2 (x / 2))}{2^{n} s i n (\frac{x / 2}{2^{n - 1}})} = \frac{s i n x}{2^{n} s i n \frac{x}{2^{n}}}

\Rightarrow \underset{k = 1}{\prod^{n}} \frac{c o s \frac{x}{2^{k - 1}}}{{c o s}^{2} \frac{x}{2^{k}}} = \frac{(\underset{k = 1}{\prod^{n}} c o s \frac{x}{2^{k - 1}})}{{(\underset{k = 1}{\prod^{n}} c o s \frac{x}{2^{k}})}^{2}} =

= \frac{\frac{s i n (2 x)}{2^{n} s i n (\frac{x}{2^{n - 1}})}}{{(\frac{s i n x}{2^{n} s i n \frac{x}{2^{n}}})}^{2}} = \frac{s i n (2 x) 2^{2 n} {s i n}^{2} \frac{x}{2^{n}}}{2^{n} s i n \frac{2 x}{2^{n}} {s i n}^{2} x} =

= \frac{2 s i n x c o s x 2^{n} {s i n}^{2} \frac{x}{2^{n}}}{2 s i n \frac{x}{2^{n}} c o s \frac{x}{2^{n}} {s i n}^{2} x} = \frac{2^{n} c o s x s i n \frac{x}{2^{n}}}{c o s \frac{x}{2^{n}} s i n x} =

= 2^{n} \frac{t a n (\frac{x}{2^{n}})}{t a n (x)}

\Rightarrow Ω = \underset{x \to 0}{l i m} \frac{x}{t a n (x)} \underset{n \to \infty}{l i m} \frac{t a n (x / 2^{n})}{x / 2^{n}} = 1

Commented by aleks041103 last updated on 26/Sep/21

T h e r e i s a n e v e n e a s i e r m e t h o d

t a n (2 x) = \frac{2 t a n (x)}{1 - {t a n}^{2} (x)} \Rightarrow 1 - {t a n}^{2} x = 2 \frac{t a n (x)}{t a n (2 x)}

\Rightarrow \underset{k = 1}{\prod^{n}} (1 - {t a n}^{2} \frac{x}{2^{k}}) = 2^{n} \underset{k = 1}{\prod^{n}} \frac{t a n (\frac{x}{2^{k}})}{t a n (\frac{x}{2^{k - 1}})}

t e l e s c o p i c s u m :

\Rightarrow \underset{k = 1}{\prod^{n}} (1 - {t a n}^{2} \frac{x}{2^{k}}) = \frac{2^{n} t a n (2^{- n} x)}{t a n (x)}

\dots

\Rightarrow Ω = 1

Commented by mathdanisur last updated on 26/Sep/21

Very nice \boldsymbol S er, thank you

Very nice S er, thank you