MHB Find Infinite Product: $\sqrt{\frac{1}{2}}$

[Saitama]

Find :
$$ \sqrt{ \frac{1}{2}}\sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}}\sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} +\frac{1}{2}\sqrt{\frac{1}{2}}}}}
\cdots \infty $$

 

[magneto1]

Find :
$$ \sqrt{ \frac{1}{2}}\sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}}\sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} +\frac{1}{2}\sqrt{\frac{1}{2}}}}}
\cdots \infty $$

Spoiler

Rewrite this as a sequence $a_n$ defined by $a_0 = \sqrt{\frac{1}{2}}$, and
$a_{n+1} = \sqrt{\frac{1}{2} + \frac{1}{2} a_{n}}$, for $n >0$. We are tasked to find
$\prod_{n=0}^\infty a_n$.

Note that the recurrence can be written as $a_n = 2a_{n+1}^2 - 1$. This looks like the
double angle formula for cosines, $\cos(2x) = 2\cos^2(x) - 1$. In fact,
we can rewrite the whole product as $a_0 = \cos(\frac{\pi}{4})$, and that
$a_k = \cos(\frac{\pi}{4 \cdot 2^k})$ for all $k$.

We can apply Viete's formula:
\[
\frac{\sin(x)}{x} = \cos (\tfrac{x}{2}) \cdot \cos (\tfrac{x}{4}) \cdot \cos (\tfrac{x}{8}) \cdots
\]

The product is therefore $a_0 \frac{\sin(\pi/4)}{\frac{\pi}{4}} = \sqrt{\frac{1}{2}} \sqrt{\frac{1}{2}} \frac{4}{\pi} = \frac{2}{\pi}$.

 

[Saitama]

Spoiler

Rewrite this as a sequence $a_n$ defined by $a_0 = \sqrt{\frac{1}{2}}$, and
$a_{n+1} = \sqrt{\frac{1}{2} + \frac{1}{2} a_{n}}$, for $n >0$. We are tasked to find
$\prod_{n=0}^\infty a_n$.

Note that the recurrence can be written as $a_n = 2a_{n+1}^2 - 1$. This looks like the
double angle formula for cosines, $\cos(2x) = 2\cos^2(x) - 1$. In fact,
we can rewrite the whole product as $a_0 = \cos(\frac{\pi}{4})$, and that
$a_k = \cos(\frac{\pi}{4 \cdot 2^k})$ for all $k$.

We can apply Viete's formula:
\[
\frac{\sin(x)}{x} = \cos (\tfrac{x}{2}) \cdot \cos (\tfrac{x}{4}) \cdot \cos (\tfrac{x}{8}) \cdots
\]

The product is therefore $a_0 \frac{\sin(\pi/4)}{\frac{\pi}{4}} = \sqrt{\frac{1}{2}} \sqrt{\frac{1}{2}} \frac{4}{\pi} = \frac{2}{\pi}$.

Perfect!