<s>[INST] Find the **Local Truncation error** in the form of $ \ o(h^k) \ $ of the multi-step method Find the Local Truncation error in the form of $ \ o(h^k) \ $ of the multi-step method  $ 2u_{i+3} = −3u_{i+2} + 6u_{i+1} − u_i + 6hf(t_{i+2}, u_{i+2}) $. 
Also determine it is convergent or not. 
Answer:
The scheme is $ \ 2u_{i+3} = −3u_{i+2} + 6u_{i+1} − u_i + 6hf(t_{i+2}, u_{i+2}) $. 
We know that $ f(t_i,u_i )=\frac{u_{i+1}-u_i}{h} ,\ by \ Taylor \ series \ $
Then , $ \ 2u_{i+3} = −3u_{i+2} + 6u_{i+1} − u_i +6 u_{i+3}-6u_{i+2} $, 
or,$ \ 4u_{i+3}-9u_{i+2}+6u_{i+1}-u_i=0 $
or, $ 4(u_{i+3}-u_{i+2})-5(u_{i+2}-u_{i+1})+(u_{i+1}-u_i)=0 $
Dividing by $ h \ $ , 
$ 4 .\frac{u_{i+3}-u_{i+2}}{h}-5. \frac{u_{i+2}-u_{i+1}}{h}+\frac{u_{i+1}-u_i}{h}=0 $ , 
or, $ 4 u'''+o(h^4)-5 u'' +o(h^2)+u'+o(h)=0, $ 
So the Local Truncation error is  $ \ (h^4) \ $.
Am I right ? Is there any help ?
 [/INST] I'll assume this is approximating $y'(t) = f(t,y)$.  The method is 
$$
2y_{n+3} =-3 y_{n+2} + 6 y_{n+1} - y_n + 6 h f (t_{n+2}, y_{n+2})
$$
To find the (local) error, taylor expand each term:
\begin{align*}
2 y_{n+3} &= 2\left( y_n + 3h y_n ' + \frac{(3h)^2}{2} y_n '' + \frac{(3h)^3}{6} y_n ''' + \frac{(3h)^4}{24} +O(h^5)\right) \\
-3 y_{n+2} &= -3 \left( y_n + 2h y_n ' + \frac{(2h)^2}{2} y_n '' + \frac{(2h)^3}{6} y_n ''' + \frac{(2h)^4}{24} + O(h^5) \right) \\
6 y_{n+1} &= 6 \left( y_n + h y_n ' + \frac{h^2}{2} y_n '' + \frac{h^3}{6} y_n ''' + \frac{h^4}{24} + O(h^5) \right) \\
-y_n &= -y_n \\
6 h f(t_{n+2}, y_{n+2}) &= 6h \left( y_n' + 2h y_n '' + \frac{(2h)^2}{2} y_n ''' + \frac{(2h)^3}{6} y_n '''' + O(h^4) \right)
\end{align*}
and matching the coefficients of the scheme reveals truncation error $O(h^4)$.
To check if it converges, we look to the root condition (i.e. apply to the test equation $y'=f(t,y)=0$, and see if the scheme is zero-stable.
$$
2 y_{n+3} = -3 y_{n+2} + 6 y_{n+1} - y_n \implies 2 r^3 = -3 r^2 + 6r -1
$$
One of the roots is larger than 1 in absolute value, so by Dahlquist Equivalence Theorem, the scheme does not converge (since it doesn't satisfy the root condition)
</s>