# Euler Sine Product - #2 (answered)

## September 02, 2021 | 4 minutes, 6 seconds

# Introduction

There is a fairly well-known formula for sine, originally derived by Euler, but not rigourously. It is as follows:

\[ \frac{\sin(x)}{x} = \prod_{n=1}^{\infty}{\left(1-\frac{x^2}{n^2\pi^2}\right)}\]

I have recently attempted to derive this myself. Curiously, I've encountered the same issue that has been raised recently with Euler's derivation.

# Attempt

Let \(\sin(x)\) be defined by its roots;

\[ \sin(x) \equiv \begin{cases} \vdots & \vdots \\ 0 & x=2\pi \\ 0 & x=\pi \\ 0 & x=0 \\ 0 & x=-\pi \\ 0 & x=-2\pi \\ \vdots & \vdots \\ \star & \star \end{cases}\]

Already this is a nonstandard or unexpected derivation of the sine function; from here, it would be appropriate for once to assume that \(\sin(x)\equiv 0\) for the values we've given. However, let us continue.

\[ \begin{align} \sin(x) &\equiv \begin{cases} \vdots & \vdots \\ 0 & x=2\pi \\ 0 & x=\pi \\ 0 & x=0 \\ 0 & x=-\pi \\ 0 & x=-2\pi \\ \vdots & \vdots \\ \star & \star \end{cases}\\ &= x\begin{cases} \vdots & \vdots \\ 0 & x=2\pi \\ 0 & x=\pi \\ 0 & x=-\pi \\ 0 & x=-2\pi \\ \vdots & \vdots \\ \star & \star \\ \end{cases}\\ &= x\begin{cases} \vdots & \vdots \\ 0 & x^2=(2\pi)^2 \\ 0 & x^2=\pi^2 \\ \star & \star \\ \end{cases}\\ \end{align}\]

From here, one might be inclined to write:

\[ x\prod_{n=1}^{\infty}{\left(x^2-(n\pi)^2\right)}\begin{cases} \star & \star \\ \end{cases}=xr(x)\prod_{n=1}^{\infty}{\left(x^2-(n\pi)^2\right)}\]

This \(r(x)\) is important!! Everything that we *haven't* defined, potential properties missed by the inherent infinite product of roots is defined in this \(r(x)\). And hence the question becomes what on earth is \(r(x)\)? Firstly, it's not going to be simple in this case (it will in all likelihood also be an infinite product to match the other one). Secondly, we know that the given product alone does not converge at \(0\), which is bad. However, we *know* that this form we derived is not what we want (because of a simple manipulation issue). Instead, to make our lives simpler, let's rewrite our equation from before as:

\[ \sin(x)\equiv x\begin{cases} \vdots & \vdots \\ 0 & 1-\frac{x^2}{(2\pi)^2}=0 \\ 0 & 1-\frac{x^2}{\pi^2}=0 \\ \star & \star \\ \end{cases}=xr(x)\prod_{n=1}^{\infty}{\left(1-\frac{x^2}{n^2\pi^2}\right)}\]

This \(r(x)\) is different from the one we derived before! However, we *still* don't know what it is.

# Problems

We have that \(\sin(x)=xr(x)\prod_{n=1}^{\infty}{\left(1-\frac{x^2}{n^2\pi^2}\right)}\) and we want to figure out \(r(x)\).

From the oddness of \(\sin(x),\ x\) and evenness of the product, we know that \(r(x)\) is definitely even; \(r(x)=r(-x)\) which is an easy property to show. We also know that \(r(0)=1\) - we can find this by differentiating \(\sin(x)\) and the respective formula, or we can find it via \(\lim_{x\to 0}{\frac{\sin(x)}{x}}\).

This means that \(r(x)\) can be virtually anything. And yet evaluating the product at non-sine roots is incredibly difficult.

I've also tried using the property \(\sin(x)''=-\sin(x)\); this yielded the equivalent of \(\sin(x)=\sin(x)\) after way too much work with differential equations (not worth it).

So how can we find \(r(x)\)? We definitely know it is \(1\) for all \(x\), so perhaps I should restate:

How do we show that, for all \(x\in\mathbb{R}\), \(r(x)=r(0)\)? This becomes an incredibly difficult functional equation.

# Solution

After *a lot* of trial and error, I resolved to try and find the value of that nasty infinite product,

\[ p(x)=\prod_{n=1}^{\infty}{\left(1-\frac{x^2}{(n\pi)^2}\right)}\]

I did so by differentiating using an infinite product rule type idea (induction gurus go wild on the product rule for derivatives):

\[ p'(x)=\sum_{n=1}^{\infty}{-\frac{2x}{(n\pi)^2}\prod_{k=1\\k\neq n}^{\infty}{\left(1-\frac{x^2}{(n\pi)^2}\right)}}\]

Then simplifying a little bit:

\[ p'(x)=p(x)\sum_{n=1}^{\infty}{\frac{2x}{x^2-(n\pi)^2}}\]

This sum in and of itself has been problematic... but did find this....

And we're left with \(\frac{p'(x)}{p(x)}=-\frac{1}{x}+\cot(x)\). After some more work \(p(x)=\frac{\sin(x)}{x}\). That in and of itself implifies the original claim... so... fun. But on the plus side we know that \(r(x)=1\) for all \(x\) :)

~Ally.