Limit of product of cosines: $\lim\limits_{n \to \infty} \cos(\frac{x}{2})\cos(\frac{x}{4})\cos(\frac{x}{6})\cdots\cos(\frac{x}{2^n})$

Asked Nov 14 '13 at 14:27

Active Nov 14 '13 at 14:33

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We've got this problem $$ y= \lim_{n \to \infty} \cos\left(\frac{x}{2}\right)\cos\left(\frac{x}{4}\right)\cdots\cos\left(\frac{x}{2^n}\right)$$ I have got stuck in this problem of evaluating the limits. What I was thinking was to convert $$\cos\left(\frac{x}{2^n}\right)$$ somehow to $\sin$ and thus term might fully get dissolved for a shorter expression and we can evaluate value of limit.

edited Nov 14 '13 at 14:33

Lord_Farin

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asked Nov 14 '13 at 14:27

Tesla

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Let $y_n$ be the $n$-th partial product. What is $y_n\cdot \sin\left(\frac{x}{2^n}\right)$? – Daniel Fischer Nov 14 '13 at 14:29
As you can see, MathJax also works in titles -- and we make good use of it. Also, if you use \cos in place of cos, it'll be typeset in a nicer font: $\cos$ vs. $cos$. – Lord_Farin Nov 14 '13 at 14:34

Limit of product of cosines: $\lim\limits_{n \to \infty} \cos(\frac{x}{2})\cos(\frac{x}{4})\cos(\frac{x}{6})\cdots\cos(\frac{x}{2^n})$

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