2
$\begingroup$

Suppose $U_n = \prod_{k=2}^n \cos\left(\frac{\pi}{2^k}\right)$ and $V_n = U_n\cdot\cos\left(\frac{\pi}{2^n}\right)$.

How can I prove that $U_n$ and $V_n$ are adjacent?

What is the limit of $U_n$ and $v_n$?

Note: $W_n = U_n\cdot\sin\left(\frac{\pi}{2^n}\right)$ and $W_n$ is geometrical.

$\endgroup$
  • $\begingroup$ What does "adjacent series" mean? $\endgroup$ – DonAntonio Nov 5 '12 at 13:38
1
$\begingroup$

By the double angle formula, we have that

$$\sin x=2\sin \frac x 2 \cos \frac x 2 $$

Using the double angle formula once more

$$\sin x=2\cdot 2 \sin \frac x 4\cos \frac x 4 \cos \frac x 2 $$

You should realize that by induction,

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

This means that, for $x\neq 0$.

$$\frac{\sin x}x=\frac{2^n}x \sin \frac x {2^n} \prod_{k=0}^n \cos \frac x {2^k} $$

$$\frac{\sin x}{x}= \frac{\sin \dfrac x {2^n}}{\dfrac x {2^n}} \prod_{k=0}^n \cos \frac x {2^k} $$

Now let $x=\pi /2$. We get

$$\frac{{\sin \frac{\pi }{2}}}{{\frac{\pi }{2}}} = \frac{{\sin \frac{\pi }{{{2^{n + 1}}}}}}{{\frac{\pi }{{{2^{n + 1}}}}}}\prod\limits_{k = 0}^n {\cos } \frac{\pi }{{{2^{k + 1}}}}$$

or

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

$\endgroup$
  • $\begingroup$ so the limit is \frac{2}{\pi} $\endgroup$ – pourjour Nov 5 '12 at 13:46
  • $\begingroup$ and how did you conclude that \sin x=2^n \sin \frac x {2^n} \prod_{k=1}^n \cos \frac x {2^k}, I know you can use induction to prove but is there any easy way without wasting time looking for this formula $\endgroup$ – pourjour Nov 5 '12 at 13:48
  • $\begingroup$ @pourjour You aren't "wasting time". Induction is the most sensible way I can think about now. Just use the double angle formula to reduce the $n+1$ case to $n$ and you're done. And use dollar signs to get the LaTeX to render. $\endgroup$ – Pedro Tamaroff Nov 5 '12 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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