# Taylor expansion of $f$ in stability analysis of 2-step Adams-Bashforth method

Given the two-step Adams-Bashforth method $$u_{n+1} = u_n + \tfrac{h}{2}(3f_n - f_{n-1})$$ find its order.

Some notation: $t_n = t_0 + nh$ is the $n$-th node and $y_n = y(t_n)$; $f_n$ stands for $f(t_n,y_n)$ and $u_n$ is an approximation of $y_n$. Being $u^*_{n+1} = y_n + \tfrac{h}{2}(3f_n - f_{n-1})$, what one can say about the local truncament error is that

$$\tau_{n+1} = \frac{y_{n+1} - u^*_{n+1}}{h}$$

A first step in getting the LTE is expanding $y_{n+1}$ in $y_n + hy'_n + \tfrac{h^2}{2}y''_n + \tfrac{h^3}{6}y'''_n(\xi)$, however I can't properly explain to myself why

$$u^*_{n+1} = y_n + \tfrac{3}{2}hy'_n - \tfrac{h}{2}\left(y'_n - hy''_n + \tfrac{h^2}{2}y'''_n(\chi)\right)$$

How does one get there? What I'm asking is basically an explanation about how should I do the Taylor expansion of $3f_n - f_{n-1}$.

The order of the method is 2. The solution I'm trying to understand is the one proposed by my teacher.

I assume you are exploring the standard first order equation $\frac{dy}{dt}=f(t,y)$. Then

$$\frac{dy}{dt}\Big|_{t=t_k}=f(t_k,y_k)$$

for each $k$. Hence

$$f_n=y'_n$$

and

$$f_{n-1}=y'_{n-1}=y'_n-hy''_n+\frac{h^2}{2}y'''(\chi)$$

for some $\chi$ between $t_{n-1}$ and $t_n$. The expression for $f_{n-1}$ is determined by finding the Taylor series for $y'(t_n-h)$ centered at $t_n$.