Infinite Factorization Power Series of $\sin(x)$ My teacher has given us a rather long problem and the last part is stumping me. How would one go about factoring the power series of sin(x)? Where:
p(x) = x - $\frac{x^{3}}{3!}$ + $\frac{x^{5}}{5!}$ - $\frac{x^{7}}{7!}$ ...
Better expressed as:
p(x) = $\sum_1^\infty (-1)^{n+1} * \frac{x^{2n-1}}{(2n-1)!}$
The exact phrasing of the last question is:
Use the roots to give an infinite factorization of p(x).

Where in a previous step of the problem, I have already figured out the roots to be -$\pi$, 0, and $\pi$.

I show here the method used by Euler.


*

*Initial Identities:
Euler showed that
\begin{eqnarray}
 \frac{\sin z}{z} = \prod_{k=1}^{\infty} \left ( 1 - \frac{z}{k^2 \, \pi^2} \right )
\end{eqnarray}
He started by considering his formula
\begin{eqnarray}
 e^z = \lim_{n \to \infty} \left ( 1 + \frac{z}{n} \right )^n,
\end{eqnarray}
and the substitution of $z$ by $\mathrm{i} z$, so
\begin{eqnarray}
 e^{\mathrm{i} z} = \lim_{n \to \infty} 
  \left ( 1 + \frac{\mathrm{i} z}{n} \right )^n
\end{eqnarray}
Then from using this in the identity, 
\begin{eqnarray}
 \sin z = \frac{e^{i z} - e^{-i z}}{2 \mathrm{i}}
\end{eqnarray}
he found
\begin{eqnarray}
 \sin z =
 \lim_{n \to \infty} \frac{ \left ( 1 + \frac{iz}{n} \right )^n
  - \left ( 1 - \frac{i z}{n} \right )^n }{2 \mathrm{i}}.
\end{eqnarray}
The key idea of Euler here was the factorization of this
binomial as presented in the next section.


*

*Factorization using roots of unity
Assume $n= 2 p + 1$. The roots of the equation
\begin{eqnarray}
 T^n -1 = 0
\end{eqnarray}
are
\begin{eqnarray}
 T_r = e^{\frac{2 \pi r \mathrm{i}}{n}}, \quad (r=0, \pm 1, \pm2, \cdots
  \pm p ).
\end{eqnarray}
One of the roots is $T_0=1$, while the other roots come in conjugate
pairs, leading to the second degree polynomial
\begin{eqnarray}
 (T - e^{\frac{2 \, \pi \mathrm{i}}{n}}) 
 (T - e^{-\frac{2 \, \pi \mathrm{i}}{n}} )
 = T^2 - 2 \, T \cos \left ( \frac{2 r \pi}{n} \right ) + 1
 \quad r=1,2, \cdots p,
\end{eqnarray}
so the polynomial $T^n -1$ can be factored out as
\begin{eqnarray}
 T^n -1 = (T-1) \prod_{k=1}^{p} \left [ T^2 - 2 T \cos \left ( 
  \frac{2 k \pi}{n} \right ) + 1  \right ]   \quad    (1)  .
\end{eqnarray}
Next we make the substitution $T=X/Y$ and clear fractions to get
\begin{eqnarray}
 X^n - Y^n = (X-Y) \prod_{k=1}^{p} 
  \left [ X^2 - 2 X Y \cos \left (
   \frac{2 \, k \pi}{n} \right ) + Y^2 \right ]
\end{eqnarray}
We now we do one more substitution. This is
\begin{eqnarray}
 X = 1 + \frac{iz}{n}, \quad \quad Y = 1 - \frac{iz}{n}
\end{eqnarray}
and after defining
\begin{eqnarray}
 q_n(z) = \frac{ \left ( 1 + \frac{iz}{n} \right )^n
  - \left ( 1 - \frac{i z}{n} \right )^n }{2 \mathrm{i} z}.
\end{eqnarray}
and using $X-Y= 2 \mathrm{i} z/n$, we find
\begin{eqnarray}
\label{qnz}
 q_n(z) &=& \frac{1}{n} \prod_{k=1}^{p} \left [
  \left ( 1 + \frac{iz}{n} \right )^2 -
  2 \left ( 1 + \frac{iz}{n} \right ) \left ( 1 - \frac{iz}{n} \right ) 
  \cos \left ( \frac{2 k \pi}{n} \right ) +
  \left ( 1 - \frac{iz}{n} \right )^2  \right ] \nonumber \\
  &=&  \frac{1}{n} \prod_{k=1}^{p} \left (
   2  - \frac{2 z^2}{n^2} - 2 \left ( 1 + \frac{z^2}{n^2} \right )
    \cos \frac{2 k \pi}{n} \right ) \nonumber \\
  &=&  \frac{1}{n} \prod_{k=1}^{p} 2 \left (
   1  - \frac{ z^2}{n^2} -  \left ( 1 + \frac{z^2}{n^2} \right )
    \cos \frac{2 k \pi}{n} \right ) \nonumber \\
  &=&  \frac{1}{n} \prod_{k=1}^{p} 2 \left [
   \left ( 1  - \cos \frac{2 k \pi}{n}  \right )
   - \frac{z^2}{n^2} \left ( 1 + \cos \frac{2 k \pi}{n} \right ) \right ] \nonumber \\
  &=&  \frac{1}{n} \prod_{k=1}^{p} 2 
   \left ( 1  - \cos \frac{2 k \pi}{n}  \right ) \left [
    1  - \frac{z^2}{n^2} 
    \frac{ 1 + \cos \frac{2 k \pi}{n}} {1 - \cos \frac{2 k \pi}{n}}  \right ] \quad (2) 
\end{eqnarray}
Now, from equation (1)  we see that
\begin{eqnarray}
 \frac{T^n -1}{T-1} = \prod_{k=1}^{p} \left [ T^2 - 2 T \cos \left ( 
  \frac{2 k \pi}{n} \right ) + 1  \right ].
\end{eqnarray}
On the other hand we know the trivial identity
\begin{eqnarray}
 \frac{T^n -1}{T-1} = 1 + T + T^2 + \cdots + T^{n-1}
\end{eqnarray}
Therefore
\begin{eqnarray}
 \lim_{T \to 1}\frac{T^n -1}{T-1} = \prod_{k=1}^{p} \left [ 2 - 2  \cos \left ( 
  \frac{2 k \pi}{n} \right )   \right ] = n
\end{eqnarray}
so equation~\ref{qnz} simplifies to
\begin{eqnarray}
 q_n(z) &=& \prod_{k=1}^{p} \left (
    1 - \frac{z^2}{n^2} \;
    \frac{ 1 + \cos \frac{2 k \pi}{n}} {1 - \cos \frac{2 k \pi}{n^2}}  \right )  \quad (3) .
\end{eqnarray}
We know that
\begin{eqnarray}
 \lim_{n \to \infty} q_n(z) = \frac{\sin z}{z}.
\end{eqnarray}
We now should find the limit as $n \to \infty$ in equation (3).
Euler assumed that the limit could go inside the product sign. We have the following
calculus identities (easily derived from Taylor expansion of the cosine function)
\begin{eqnarray}
\label{identu0}
 \lim_{u \to 0} \frac{1 - \cos a u}{a^2 u^2} &=& \frac{1}{2}  \nonumber \\
 \lim_{u \to 0}  \left ( 1 + \cos a \, u \right ) &=& 2  .
\end{eqnarray}
By choosing $a= 2 k \pi$ and $u=1/n$ in the first identity above we find
\begin{eqnarray}
 \lim_{n \to \infty} n^2 \, \left ( 1 - \cos \frac{2 k \pi}{n} \right ) &=& 2 k^2 \pi^2.
\end{eqnarray}
so
\begin{eqnarray}
\label{qnz3}
 \lim_{n \to \infty }q_n(z) =  
 \prod_{k=1}^{p} \left ( 1 - \frac{2 z^2}{2 k^2 \pi^2} \right ) =
 \prod_{k=1}^{p} \left (  1 - \frac{z^2}{k^2 \pi^2}  \right ).
\end{eqnarray}
Therefore
\begin{eqnarray}
 \frac{\sin z}{z} = \prod_{k=1}^{\infty} \left (  1 - \frac{z^2}{k^2 \pi^2}  \right )  \quad (4).
\end{eqnarray}


*

*Discussion
The materialization of this idea took Euler him more  than 10 years of work. 
Euler thought about the infinite series as extensions of polynomials where
the roots can be used to factorize them. If this is true, then it is natural
to think that
\begin{eqnarray}
 \sin z = z \left ( 1- \frac{z}{\pi} \right ) \left ( 1 - \frac{z}{2 \pi} \right ) 
 \cdots  \left ( 1-\frac{z}{k \pi} \right ) \cdots \nonumber \\
   \left ( 1- \frac{z}{-\pi} \right ) \left ( 1 - \frac{z}{-2 \pi} \right ) 
    \cdots \left ( 1-\frac{z}{-k \pi} \right ) \cdots
\end{eqnarray}
where $\sin z$ vanishes at each $z= \pm k \, \pi$, $k=0,1, \cdots $. Then by grouping
each factor with that where its root has the opposite sign, he found
\begin{eqnarray}
 \sin z = z \left ( 1- \frac{z^2}{\pi^2} \right ) \left ( 1 - \frac{z^2}{2^2 \pi^2} \right ) 
 \cdots  \left ( 1-\frac{z^2}{k^2 \pi^2} \right ) \cdots \nonumber \\
\end{eqnarray}
which coincides with (4).
However he got many objections. For example


*

*How does he knows that the order of the factorization
does not matter in the infinite.

*Are those the only complex roots of $\sin z$?, 

*$\sin s/s$ and $e^s \sin s/s$ have the same roots, but clearly can not be factored
with the same formula.
Euler was encouraged by the fact that his formula checked out fine for some particular
values found before by Newton and Leibniz's. Also he evaluated the Zeta function
$\zeta(2)=\pi^2/6$ and matched with good accuracy.
The point of convergence is not addressed here. Is there a radius of convergence
(as in infinite series)?
In the proof presented here, still we can not guarantee that the limit can be interchanged
with the $\prod$ sign. That is, is the limit of the product equal to the product of the limits?
That this could be done was proved at a much later time by Weierstrass who generalized
Euler's result with his theorem, that entire functions can be represented by a product
involving their zeroes and in addition, every sequence tending to infinity has an
associated entire functions with zeroes at precisely the points of that sequence
(see Wikerstrass factorization theorem'').
Reference: https://books.google.com/books?id=CYyKTREGYd0C
