Is there a convenient way to write Viète's formula $\displaystyle \frac2\pi= \frac{\sqrt2}2\cdot \frac{\sqrt{2+\sqrt2}}2\cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2\cdots$ using sigma and/or pi notation

a) without recursive notation;

b) even with recursive notation?

As many ways as possible would be appreciated. Thanks in advance!


The RHS can be rewritten as $$\prod_{k=2}^{\infty}\cos \left(\frac{\pi}{2^k}\right)$$

Maybe there's no recurrence relation since $\pi$ is a transcendental number.

  • $\begingroup$ So we can write $\pi$ as 2$\displaystyle \prod_{k=2}^{\infty}\sec\left(\frac{1}{2^{k}\pi}\right)$? $\endgroup$ – Gabriel Jan 14 '15 at 4:02
  • $\begingroup$ Yes, I think there's no problem with it. $\endgroup$ – Tien Kha Pham Jan 14 '15 at 4:05

To reach the final result, you can follow these detailed steps:

  1. Write the double-angle formula $$ \sin x = 2 \sin\frac x2 \cos\frac x2 $$ and repeat using that formula so that we can get $$ \sin x = 2^n \sin\frac{x}{2^n} \prod_{i = 1}^{n} \cos\frac{x}{2^i} \quad (n \in \mathbb{Z}^*). \tag{$*$} $$

  2. Divide both sides of $(*)$ by $x$ (given $x > 0$), then take the limit of both sides of the equation when $n \to \infty$. Note that the change of value of $n$ has no effects on LHS, so we get $$ \frac{\sin x}{x} = \lim_{n \to \infty} \Big( \frac{\sin(x / 2^n)}{x / 2^n} \prod_{i = 1}^{n} \cos\frac{x}{2^i} \Big) = \prod_{i = 1}^{\infty} \cos\frac{x}{2^i}. $$ This is the result Tien Kha Pham provided directly. (Note that $\prod_{i = 1}^{\infty} \cos(x / 2^i)$ makes sense only if the corresponding limit $\lim_{n \to \infty} \prod_{i = 1}^{n} \cos(x / 2^i)$ exists.)

  3. To relate this result to the form of Viète's formula you mentioned in the question, substituting $x = \pi / 2$ with formulas $$ \cos \Big(\frac{\pi}{4}\Big) = \frac{\sqrt{2}}{2} \quad\text{and}\quad \cos \Big(\frac{x}{2^k}\Big) = \frac 12 \sqrt{2 + 2\cos \Big(\frac{x}{2^{k - 1}}\Big)} \quad (k \ge 2), $$ we finally conclude with $$ \frac{\pi}{2} = \prod_{k = 2}^{\infty} a_k, $$ where $a_2 = \sqrt{2} / 2$, $a_3 = \sqrt{2 + 2a_2} / 2 = \sqrt{2 + \sqrt{2}} / 2$ and $a_{k + 1} = \sqrt{2 + 2a_k} / 2$ for $k \ge 3$.

Once there was a minor mistake in the answer made by Tien Kha Pham. The typo no longer existed. But another consequent mistake, in the comment to that answer, will remain. Therefore I keep the following words visible only as a reminder for those who may get confused by the inconsistency between formulas.

Just a correction to the answer of Tien Kha Pham (since I don't have enough reputation to add a comment): it should be $$ \frac{2}{\pi} = \prod_{k = 2}^{\infty} \cos \Big( \frac{\pi}{2^k} \Big), $$ of which the $\pi$ appears in numerator, not denominator.

  • 1
    $\begingroup$ I raise my hat for your polite way of correcting an error in a given answer. Some people just run the down vote button as if they were opening a treasure box and don't event care to comment. Your attitude as shown in this post is a great example to be followed by others. $\endgroup$ – NoChance Jan 16 '16 at 21:38

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