How to prove that $\sin x = x \, _{0}F_{1}(-;\frac{3}{2};-\frac{x^{2}}{4})$? How to prove that $\sin x = x \, _{0}F_{1}(-;\frac{3}{2};-\frac{x^{2}}{4})$?
where $_{0}F_{1}$ is the hypergeometric series?
 A: One may recall the definition of the generalized hypergeometric function
$$
{}_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;z): = \sum_{n=0}^\infty \frac{(a_1)_n\dots(a_p)_n}{(b_1)_n\dots(b_q)_n} \, \frac {z^n} {n!}
$$ which gives
$$
{}_0F_1\left(-;\frac{3}{2};-\frac{x^{2}}{4}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{\left(\frac32\right)_n} \, \frac {x^{2n}} {4^n \:n!}.
$$ Then one may check that
$$
\begin{align}
\frac1{\left(\frac32\right)_n\:4^n \:n!}&=\frac1{\left(\frac32\right)\left(\frac32+1\right)\left(\frac32+2\right)\cdots\left(\frac32+n-1\right)\:4^n \:n! }
\\\\&=\frac{2^n}{3\times 5\times 7 \cdots \times (2n+1)\:4^n \:n! }
\\\\&=\frac{2^n\times \color{blue}{2\times 4\times 6 \cdots \times (2n)} }{\color{blue}{2}\times3\times \color{blue}{4} \times5\times \color{blue}{6}\times7 \cdots \times \color{blue}{2n}\times (2n+1)\:4^n \:n! }
\\\\&=\frac{2^n\times 2^n\times n! }{(2n+1)!\:4^n \:n! }
\\\\&=\frac1{(2n+1)!}
\end{align}
$$ giving 

$$
x\:{}_0F_1\left(-;\frac{3}{2};-\frac{x^{2}}{4}\right) = \color{blue}{\sum_{n=0}^\infty (-1)^n\frac {x^{2n+1}} {(2n+1)!}}=\color{red}{\sin x}
$$

as announced.
A: The function $_0F_1(;a;z)$ is a solution to the differential equation
\begin{equation}\tag{1}
  z \frac{d^2}{dz^2}y(z)+a\frac{d}{dz}y(z)= y(z).
\end{equation}
Knowing this, consider the function $u(x)=x y(-x^2/4)$ and take its second derivative. We obtain
$$\tag{2} \frac{d^2}{dx^2}u(x)=x\left[\frac{x^2}{4}y''\left(\frac{x^2}{4}\right)-\frac{3}{2}y'\left(\frac{x^2}{4}\right)\right] $$
Now choose $y(z)=$$_0F_1(;3/2;z)$, so that it satisfies eq.(1) with $a=3/2$ and $z=-x^2/4$. Using it we obtain
$$ \frac{d^2}{dx^2}u(x)=-xy\left(\frac{x^2}{4}\right)=-u(x) \tag{3}.$$
The general solution to (3) is $u(x)=\alpha \sin(x)+\beta \cos(x).$
Since $u$ is odd $u(x)=-u(x)$, $\beta=0$. Finally,
$$ \alpha  = \lim_{x\to0} \frac{u(x)}{x} =\; _0F_1\left(;\frac{3}{2};0\right)=1$$
so
$$ u(x)= \sin(x) = x  \; _0F_1\left(;\frac{3}{2};-\frac{x^2}{4}\right)$$
