How to derive Adam Moulton 2 step implicit method using taylor expansion I have some confusion on the derivation of multistep method using Taylor expansions. For example, we want to derive the linear 2 step Simpson's rule: My professor first write down the scheme of an implicit multistep method as follows:
$y_{n+1} = a_1y_{n} +b_1hy^{'}_{n}+a_2y_{n-1}+b_2hy^{'}_{n-1}+b_0hy^{'}_{n+1}$
Then we expand around $y_n$ for the rest of the terms up to $O(h^6)$ respectively for $y_{n+1},y_{n-1},y^{'}_{n-1},y^{'}_{n+1}$, we then collect the like powers of $h$, and set the first 5 coefficients of $y,y^{'},y^{''},y^{'''},y^{''''}$ to $0$, then we solve and get the respective $a_1,a_2,b_1,b_2,b_0$ and substitue back into the scheme to get the Simpsons method $$y_{n+1} = y_{n-1}+\frac{h}{3}(y'_{n+1}+4y^{'}_n+y'_{n-1})$$
However I am very confused on how to use taylor expansion to derive another implicit 2 step method, say, the Adam Moulton 2 step: $$y_{n+1} = y_{n}+\frac{h}{12}(5y'_{n+1}+8y^{'}_n-y'_{n-1})$$
Please help me with this. I tried only taking the first 4 coefficients of the derivatives of $y$ and form 4 equations and get the answer, but I was just guessing, furthermore, since there are 5 unknowns in a 2 step implicit method, I should form 5 equations.
Also, I am aware on how to use integration to derive the above methods, but not taylor expansion
 A: I was stuck on something similar to this. I am not sure how to use the taylor expansion to derive the Adam Moulton 2 step implicit method, but you can use lagrange interpolation polynomial:
First, derive the $f(t_{i+1},w_{i+1})$ value using the Adams-Bashforth Two-step explicit method
Then, integrate the range interpolation polynomial of the 3 points: $f(t_{i+1},w_{i+1}),f(t_{i},w_{i}),f(t_{i-1},w_{i-1})$
eg:
\begin{equation}
\int^{t_{i+1}}_{t_i}\frac{(t - t_{i})(t - t_{i-1})}{(t_{i+1} - t_{i})(t_{i+1} - t_{i-1})}f_{i+1}+\frac{(t - t_{i+1})(t - t_{i-1})}{(t_{i+1} - t_{i+1})(t_{i+1} - t_{i-1})}f_{i}+\frac{(t - t_{i+1})(t - t_{i})}{(t_{i-1} - t_{i+1})(t_{i-1} - t_{i})}f_{i-1}dt
\end{equation}
where
\begin{equation}
\begin{aligned}
f_{i+1} &= f(t_{i+1},w_{i+1})\\
f_{i} &= f(t_{i},w_{i})\\
f_{i-1} &= f(t_{i-1},w_{i-1})
\end{aligned}
\end{equation}
The way this can be integrated, let:
\begin{equation}
\begin{aligned}
t_{i+1} &= t_i + h\\
f_{i} &= t_i\\
f_{i-1} &= t_i -h
\end{aligned}
\end{equation}
so that we get:
\begin{equation}
\int^{t_{i}+h}_{t_i}\frac{(t - t_{i})(t - (t_{i}-h))}{(t_{i}+h - t_{i})(t_{i}+h -( t_{i}-h))}f_{i+1}+\frac{(t - (t_{i}+h))(t - (t_{i}-h))}{(t_{i}+h - (t_{i}+h))(t_{i}+h - (t_{i}-h))}f_{i}+\frac{(t - (t_{i}+h))(t - t_{i})}{(t_{i}-h - (t_{i}+h))(t_{i}-h - t_{i})}f_{i-1}dt
\end{equation}
this should give you:
\begin{equation}
\frac{h}{12}[5f(t_{i+1},w_{i+1})+8f(t_{i},w_{i})-f(t_{i-1},w_{i-1})]
\end{equation}
and ultimately:
\begin{equation}
w_{i+1} = w_i + \frac{h}{12}[5f(t_{i+1},w_{i+1})+8f(t_{i},w_{i})-f(t_{i-1},w_{i-1})]
\end{equation}
To do the three-Step or Four-Step, simply just integrate the lagrange interpolation polynomial for the 4 or 5 points you will find.
