Another way of expanding Taylor series 
I have to show that if $$f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\ldots, $$ then prove that $$ f(x)=f(a)+2\left[ \frac{(x-a)}{2}f'\left(\frac{x+a}{2}\right)+  \frac{(x-a)^3}{3!\times 8}f'''\left(\frac{x+a}{2}\right)+ \frac{(x-a)^5}{32\times 5!}f^{(v)}\left(\frac{x+a}{2}\right)+\ldots\right] $$

I am unable to proceed. Any hint will be appriciated. 
 A: We  write the Taylor expansion stated in OPs first line as
\begin{align*}
f(x)=\sum_{j=0}^\infty\frac{(x-a)^j}{j!}f^{(j)}(a)\tag{1}
\end{align*}
and the other expansion as
\begin{align*}
f(x)=f(a)+2\sum_{j=0}^\infty \frac{1}{(2j+1)!}\left(\frac{x-a}{2}\right)^{2j+1}f^{(2j+1)}\left(\frac{x+a}{2}\right)\tag{2}
\end{align*}

Expansion at $\frac{x+a}{2}$ instead of $a$ in (1) results in
  \begin{align*}
f(x)&=\sum_{j=0}^\infty\frac{1}{j!}\left(x-\frac{x+a}{2}\right)^j f^{(j)}\left(\frac{x+a}{2}\right)\\
&=\sum_{j=0}^\infty\frac{1}{j!}\left(\frac{x-a}{2}\right)^j f^{(j)}\left(\frac{x+a}{2}\right)\tag{3}
\end{align*}
  Exchanging the role of $x$ and $a$ in (3) gives
  \begin{align*}
f(a)&=\sum_{j=0}^\infty\frac{1}{j!}\left(\frac{a-x}{2}\right)^j f^{(j)}\left(\frac{x+a}{2}\right)\\
&=\sum_{j=0}^\infty\frac{(-1)^j}{j!}\left(\frac{x-a}{2}\right)^j f^{(j)}\left(\frac{x+a}{2}\right)\tag{4}
\end{align*}
  subtracting (4) from (3) gives
  \begin{align*}
\color{blue}{f(x)-f(a)}&=\sum_{j=0}^\infty\frac{1-(-1)^j}{j!}\left(\frac{x-a}{2}\right)^j f^{(j)}\left(\frac{x+a}{2}\right)\\
&\color{blue}{=2\sum_{j=0}^\infty \frac{1}{(2j+1)!}\left(\frac{x-a}{2}\right)^{2j+1}f^{(2j+1)}\left(\frac{x+a}{2}\right)}
\end{align*}
  and the claim follows.

