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$$\lim_{n\to \infty } \,\cos (1) \cos \left(\frac{1}{2}\right) \cos \left(\frac{1}{4}\right)\cdots \cos \left(\frac{1}{2^n}\right)$$

How would you evaluate this limit? Is it just equivalent to $$\prod_{n=0}^\infty {\cos \left( \frac{1}{2^{n}} \right)}$$

Mithlesh Upadhyay
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1110101001
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1 Answers1

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Multiply first by $2^{n+1}\sin(1/2^n)$ and collapse it.

  • I don't quite follow. How does multiplying by that allow you to simplify it? – 1110101001 Jan 05 '14 at 19:11
  • Use $2\cos(x)\sin(x)=\sin(2x)$. The $\sin$, the $\cos$ and the $2$ collapse in another $\sin$ (which you can use for the next cosinus). –  Jan 05 '14 at 19:13
  • But doesn't multiplying by that change the value of the limit? – 1110101001 Jan 05 '14 at 19:15
  • Yes, you multiply and divide by it. The multiply one you use to collapse it. The divide part remains to compute the limit. –  Jan 05 '14 at 19:19
  • When you collapse it do you end up with 2*sin(1)? – 1110101001 Jan 05 '14 at 19:26
  • I think the $\cos(1)$ collapses too with the $\sin(1)$. So we get $\sin(2)$. But divided by what we multiplyed by (divided by $2^{n+1}\sin(1/2^{n+1})$). –  Jan 05 '14 at 19:29
  • Got it -- the answer is just sin(2), right? At the beginning I got why you multiplied by sin(1/2^n), but why did you use 2^n+1 instead of just 2^n? – 1110101001 Jan 05 '14 at 19:33
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    Because $1=2^{0}, 2=2^1, ..., 2^{n}=2^n$. So, then numbers $0,1,...,n$ are $n+1$ numbers. –  Jan 05 '14 at 19:34
  • Don't worry about the chat thing. Your questions might be the next guy's questions, so it is not bad if they can see the answers here. –  Jan 05 '14 at 19:36
  • So is sin(2) the right answer? Because when I evaluated with mathematica it gives sin(1)*cos(1) – 1110101001 Jan 05 '14 at 22:11
  • Check the cancelling of the $2$'s. Notice that $\sin(1)\cos(1)=2\sin(1)\cos(1)/2=\sin(2)/2$. –  Jan 05 '14 at 22:14
  • I keep getting sin(2) when I collapse the products... I should be getting sin(2)/2, right? Perhaps you could explain the method to collapse it? – 1110101001 Jan 05 '14 at 22:29
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    I think you might be computing wrong the limit. Notice that $\lim 2^{n+1}\sin(1/2^n)=2\lim \frac{\sin(1/2^n)}{1/2^n}=2$. –  Jan 05 '14 at 22:30
  • Ah, you were right -- I forgot about the n+1 in the limit. Thanks! – 1110101001 Jan 05 '14 at 22:37