Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This limit seemed quite unusual to me as there aren't any intermediate forms or series expansions which are generally used in limits. Stuck on this for a while now .Here's how it goes :

$$ \lim \limits_{n \to \infty} \left[\cos\left(x \over 2\right)\cos\left(x \over 4\right) \cos\left(x \over 8\right)\ \cdots\ \cos\left(x \over 2^{n}\right)\right] $$

Any Help is Welcome. ThankYou.

share|improve this question
1  
@GitGud : Man you're like my official editor now. Thanks again :) –  Simar Jul 30 '13 at 23:52
    
Is there anything that needs clearing out that you unaccepted? –  Pedro Tamaroff Jul 31 '13 at 0:10
    
Just wanted more answers.I'll do that in a few hours. The only thing that I couldn't get was that why does sin x/2^n come outside the product when cos was inside ? –  Simar Jul 31 '13 at 0:20
    
Though your answer seems correct, just a silly doubt it is. Please clarify if you can. Thanks :) –  Simar Jul 31 '13 at 0:21
    
Ah, fair enough. =) Is it clearer now? –  Pedro Tamaroff Jul 31 '13 at 0:23

3 Answers 3

up vote 12 down vote accepted

Hint $$\begin{align}{\sin x}&=2^1\sin\frac x 2 \cos\frac x2\\{}\\\sin x& =2^2\sin \frac x4\cos\frac x 4\cos \frac x 2\\{}\\\sin x& =2^3\sin \frac x8\cos \frac x8\cos\frac x 4\cos \frac x 2\\{}\\\cdots\;&=\hspace{2cm }\cdots\end{align} $$

One further hint

$$\sin x = {2^n}\sin \frac{x}{{{2^n}}}\prod\limits_{k = 1}^n {\cos \frac{x}{{{2^k}}}} $$

You'll need $\dfrac{\sin x}x\to 1$ as $x\to 0$.

Final spoiler:

$$\mathop {\lim }\limits_{n \to \infty } \prod\limits_{k = 1}^n {\cos \frac{x}{{{2^k}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sin x}}{x}{\left( {\frac{{\sin {2^{ - n}}x}}{{{2^{ - n}}x}}} \right)^{ - 1}} = \frac{{\sin x}}{x}$$

share|improve this answer
    
Got that.But how do you evaluate the product of cosine series? –  Simar Jul 30 '13 at 23:45
    
@Simar I have posted the full solution. –  Pedro Tamaroff Jul 30 '13 at 23:47

Setting $$ u_n(x)=\cos\frac{x}{2}\cdot\cos\frac{x}{4}\cdot\cdots\cos\frac{x}{2^n}, $$ we have $$ v_n(x):=u_n(x)\cdot\sin\frac{x}{2^n}=u_{n-1}(x)\cdot\cos\frac{x}{2^n}\cdot\sin\frac{x}{2^n}=\frac12u_{n-1}(x)\cdot\sin\frac{x}{2^{n-1}}=\frac12v_{n-1}(x). $$ It follows that $$ v_n(x)=\frac{1}{2^{n-1}}v_1(x)=\frac{1}{2^{n-1}}\cdot\cos\frac{x}{2}\cdot\sin\frac{x}{2}=\frac{1}{2^n}\sin x. $$ Hence, provided $\sin\frac{x}{2^n} \ne 0$, we have $$ u_n(x)=\frac{\sin x}{2^n\sin\frac{x}{2^n}}. $$ Thus $$ \lim_{n\to\infty}u_n(x)=\lim_{\xi\to0}\frac{\sin x}{x}\cdot\left(\frac{\sin\xi}{\xi}\right)^{-1}=\frac{\sin x}{x}. $$

share|improve this answer

Put $z=e^{i x/2^n}.$ Then the product becomes $$\prod_{k=1}^n \cos\left(\frac{x}{2^k}\right) = \frac{1}{2^n} \prod_{k=0}^{n-1} \left(z^{2^k}+z^{-2^k}\right) = \frac{1}{2^n} z^{-\sum_{k=0}^{n-1} 2^k} \prod_{k=0}^{n-1} \left(z^{2^{k+1}}+1\right)\\ = \frac{1}{2^n} z^{1-2^n} \frac{1}{z+1} \prod_{k=0}^n \left(z^{2^k}+1\right).$$ Now the product in this last formula is easily seen to be the generating function of the positive integers less than $2^{n+1}-1$ (consider the binary representation of an integer $q$ from this interval), so that we may continue with $$\frac{1}{2^n} z^{1-2^n} \frac{1}{z+1} \left(1+z+z^2+\cdots+z^{2^{n+1}-1}\right) = \frac{1}{2^n} z^{1-2^n} \frac{1}{z+1} \frac{1-z^{2^{n+1}}}{1-z}\\ = \frac{1}{2^n} \frac{1}{z+1} \frac{z^{1-2^n}-z^{1+2^n}}{1-z} = \frac{1}{2^n} \frac{z}{z+1} \frac{z^{-2^n}-z^{2^n}}{1-z} = \frac{1}{2^n} z \frac{z^{-2^n}-z^{2^n}}{1-z^2} \\ = \frac{1}{2^n} \frac{z^{-2^n}-z^{2^n}}{1/z-z} = \frac{1}{2^n} \frac{\sin(x)}{\sin(x/2^n)} = \frac{x/2^n}{\sin(x/2^n)} \frac{\sin(x)}{x}.$$ Finally recall that $\sin(x) \sim x$ in a neighborhood of zero, so that the limit of the first term is one, resulting in a final answer of $$\frac{\sin(x)}{x}.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.