Wallis this product?

Quick Solution:

If you know Euler's product for $\frac{\sin(x)}{x}$ this problem is simply a manner of plugging in $\frac{\pi}{3}$ :

$\frac{\sin(x)}{x} = \prod_{n=1}^{\infty} \left(1-\frac{x^{2}}{n^{2}\pi^{2}}\right) \Rightarrow \frac{\sin(\frac{\pi}{3})}{\frac{\pi}{3}} = \prod_{n=1}^{\infty} \left(1-\frac{1^{2}}{(3n)^{2}}\right) = \prod_{n=1}^{\infty} \left(\frac{3n-1}{3n} \cdot \frac{3n+1}{3n} \right)$

The final product is the multiplicative inverse of what we want to find. This means that the product we want to evaluate is actually equal to $\frac{\frac{\pi}{3}}{\sin(\frac{\pi}{3})} = \frac{2\pi}{3\sqrt{3}}$ This makes our answer $2+3+3 = \boxed{8}$

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Motivated Answer:

(This is not a complete solution!)

Say you didn't know Euler's product. How might you have come up with it from scratch?

The $\sin$ function has a Taylor expansion, which turns the sine function into a polynomial. As you know, polynomial functions contain roots where the function is zero. We will construct an infinite product of sine based on its roots;

$f(x) = c \prod_{k=1}^{n} \left( 1-\frac{x}{r_{k}} \right)$

Where $c$ is a constant and $r_{k} \in \mathbb{C}$ are the (possibly) complex roots. The constant $c$ is necessary since the polynomials $f(x) = 4x^{2}-4$ and $g(x) = x^{2}-1$ are different despite having the same roots. Note that this form of expressing a polynomial has the issue that $x=0$ can never be a root. This can easily be resolved by putting a factor of $x$ in front of the product.

You can show that the roots of $\sin(x)$ are purely real by simply expressing it as a complex exponential and looking for where it can be equal to zero:

$\sin(z) = \frac{e^{iz}-e^{-iz}}{2i} \ : \ z \in \mathbb{C}$

With this in mind, we can see that the roots of $\sin(x)$ are all of the form $k\pi \ : \ k \in \mathbb{Z}$ . This means that the expression for $sin(x)$ becomes

$\sin(x) = c \cdot x \cdot \prod_{k \in \mathbb{Z} \ k \neq 0} \left(1-\frac{x}{k\pi}\right)$

There is an x term out front since we pulled out the $k=0$ term from the product. Also notice that for each term of the form $1-\frac{x}{a}$ , there is also a term of the form $1+\frac{x}{a}$ and so we can further simplify the product using difference of squares:

$\sin(x) = c \cdot x \cdot \prod_{k = 1}^{\infty} \left(1-\frac{x^{2}}{k^{2}\pi^{2}}\right)$

This is getting close to how we want it, now we just need to evaluate $c$ . To do this, we need to recognize that that constant term represents how much we "stretch" the function in the $y$ direction. There are in principle many ways to evaluate $c$ , so long as you can make sure that some value $x_{0}$ which is not a root of our polynomial gives the correct value of $\sin(x_{0})$ .

Unfortunately, $x_{0} = 0$ is not going to work, since it is a root of our polynomial. (if you try this process for cosine though you'll find that you're in luck!) Any other values of $x_{0}$ leave a hard to evaluate infinite product as well. Fortunately, another way to think of stretching/shrinking in the y-direction is with derivatives! We can make sure that our value of $c$ is correct by ensuring that the derivative of our product at $x_{0} = 0$ matches that of $\sin'(x)$ at $x=0$ . This can be easily evaluates to $\sin'(0) = 1$ .

Now, we need to think about how to actually take the derivative of our crazy product. We can use the product rule, but a much easier way is to think about how to get any order term in the expansion. To get the $x^{1}$ term, note that if we ever take any of the $\frac{x^{2}}{k^{2}\pi^{2}}$ terms, we already have missed the mark. The exponent on that term will be greater than $1$ since its already at $2$ and can't go back down. Hence, the only contribution to the first order term is $cx$ .

The rest of the product is extremely complicated, but thankfully we know that it will be of at least third order and so the derivatives of all those terms at $x=0$ will vanish. We can then say that the derivative of our product at $x=0$ is $c$ , and so $c=1$ .

With this in mind, we arrive at our expression for $\sin(x)$ , or, simply by dividing out the leading $x$ , our expression for $\frac{\sin(x)}{x}$ :

$\frac{\sin(x)}{x} = \prod_{k=1}^{\infty} \left(1-\frac{x^{2}}{k^{2}\pi^{2}}\right)$

And we can follow from here with the quick proof.

Note that this isn't actually a formal nor valid proof. I haven't showed that the product converged, nor that we can guarantee that the functions were equal at any points other than at the roots and the derivative at $0$ . Since I haven't taken an analysis course yet, I'm curious at to what exactly there is left to show in order to make this a valid derivation/proof of Euler's product. Any feedback would be apreciated.

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As an added challenge, you can use the above product to evaluate $\zeta(2)$ where $\zeta(z)$ is the Riemann Zeta function

Relevant wiki: Gamma Function

$\begin{aligned} P & = \lim_{m \to \infty} \prod_{n=1}^m \frac {3n}{3n-1}\cdot \frac {3n}{3n+1} & \small \color{#3D99F6} \text{Divide up and down by }9 \\ & = \lim_{m \to \infty} \prod_{n=1}^m \frac {\color{#3D99F6} n}{\color{#D61F06}n-\frac 13}\cdot \frac {\color{#3D99F6} n}{\color{#D61F06}n+\frac 13} & \small \color{#3D99F6} \text{Gamma function }\Gamma (n+1) = n! \\ & = \lim_{m \to \infty} \frac {\color{#3D99F6} \Gamma (m+1)^2}{{\color{#D61F06}\frac {\Gamma \left(m+\frac 23\right)}{\Gamma \left(\frac 23\right)}}\cdot \color{#D61F06}\frac {\Gamma \left(m+\frac 43\right)}{\Gamma \left(\frac 43\right)}} & \small \color{#D61F06} \text{and }\Gamma (s+1) = s\Gamma(s) \\ & = \Gamma \left(\frac 23\right) \color{#D61F06} \Gamma \left(\frac 43\right) \\ & = \Gamma \left(\frac 23\right) \cdot \color{#D61F06} \frac 13 \Gamma \left(\frac 13\right) \\ & = \frac 13 \color{#3D99F6} \Gamma \left(\frac 13\right) \Gamma \left(1- \frac 13\right) & \small \color{#3D99F6} \text{also }\Gamma (z)\Gamma(1-z) = \frac \pi{\sin (\pi z)} \\ & = \frac \pi {3\sin \frac \pi 3} = \frac {2\pi}{3\sqrt 3} \end{aligned}$

Therefore, $a+b+c = 2+3+3 = \boxed 8$ .