Whats the correct way to factorize $x \sin(\frac{\pi}{x})$?

Question

I am inspired by Euler's factorization of $sin(x)$ as an infinite product. I was trying to apply to the expression $x \sin(\frac{\pi}{x})$.

Let $f(x) = x \sin{\frac{\pi}{x}}$. My findings were as follows:

1. $f(x) = 0$ at infinitely many points in the closed interval $[-1,1]$
2. $f(x) = 0$, namely at $x = 0$, and $x = \dfrac{1}{n}, n \space \epsilon \space Z \space \& \space n \neq 0$
3. $\therefore$ we can write $f(x) = A (x - 0)(x-1)(x+1)(x-\dfrac{1}{2})(x+\dfrac{1}{2})...$

$$ \implies f(x) = A x \Pi_{n=1}^{n=\infty}\left(x^2-\dfrac{1}{n^2} \right)$$

We can solve for $A$ by putting $x=2$, we get $A = \frac{1}{\Pi\left(4 - \dfrac{1}{n^2}\right)}$

So we get $$f(x) = x\Pi\left(\dfrac{x^2 - \frac{1}{n^2}}{4 - \frac{1}{n^2}}\right)$$

Now if we expand $f(x)$ in terms of its taylor series, we get a polynomial in powers of $\frac{1}{x}$, where as on the LHS we get powers of $x$. I can't equate coefficients as a result of this approach.

Where did I do wrong?

Answer 1

I guess it is non-factorisable.. We can factorize the RHS. But we cannot proceed to step of equating coefficients of terms of similar powers

Findings:

1. the function isn't differentiable at $x = 0$
2. because of the above fact taylor series of the function is impossible.