So I gotta derive the Implicit Midpoint method using taylor series for an ODE $y'(t)=f(t,y(t))$ and show that it is symmetric and A-stable.
So taylor series
$y(t)=y(t_{n-1/2})+\frac{h_n}{2}y'(t_{n-1/2})+\frac{h_n^2}{8}y''(t_{n-1/2})+\frac{h_n^3}{48}y'''(t_{n-1/2})+...$
and have to derive
$y_n=y_{n-1}+h_nf(t_{n-1/2},\frac{1}{2}(y_n+y_{n-1}))$
My attempt and where I'm stuck.
Note here we consider $t_{n-1/2}=t_n-\frac{1}{2}h_n$
So shuffling around...
$y(t)-y(t_{n-1/2})=\frac{h_n}{2}y'(t_{n-1/2})+\frac{h_n^2}{8}y''(t_{n-1/2})+...$ $\implies y(t)-y(t_{n-1/2})=\frac{h_n}{2}y'(t_{n-1/2})+\frac{h_n^2}{8}y''(t_{n-1/2})+...\implies \frac{y(t)-y(t_{n-1/2})}{h_n}=\frac{1}{2}y'(t_{n-1/2})+\frac{h_n}{8}y''(t_{n-1/2})+...$
Also, $y(t_n)=y(t_{n-1/2})+\frac{h_n}{2}y'(t_{n-1/2})+\frac{h_n^2}{8}y''(t_{n-1/2})+\frac{h_n^3}{48}y'''(t_{n-1/2})+...$
and $y(t_{n-1})=y(t_{n-1/2})-\frac{h_n}{2}y'(t_{n-1/2})+\frac{h_n^2}{8}y''(t_{n-1/2})-\frac{h_n^3}{48}y'''(t_{n-1/2})+...$
Combining yields
$\frac{y(t_n)-y(t_{n-1})}{h_n}=y'(t_{n-1/2})+\frac{h_n^2}{24}y'''(t_{n-1/2})+O(h_n^4)$
Now, I'm having trouble putting this all together. Any pointers?
Here's link to derivation from wikipedia, but I don't think it's what I'm looking for. https://en.wikipedia.org/wiki/Midpoint_method