The question is to prove that error order of backward euler method is $o(h)$
We know that in backward euler method $$f_i’=\frac{f_i-f_{i-1}}{h}$$
By using taylor seris we can get $$f(x)=f(x_i)+(x-x_i)f’(x_i)+\frac{(x-x_i)^2}{2!}f’’(x_i)+….$$
By putting $x=x_{i-1}$ i get :
$$\frac{f_i-f_{i-1}}{h}+f’i=\frac{h}{2!} f’’_i+\frac{h^2}{3!}f’’’_i+…$$
Which is not what i want to get
Cause the error is $$\frac{f_i-f_{i-1}}{h}-f’_i$$
Any help ?
I could prove that $o(h)$ is the order of error of forward Euler method by using $x=x_{i+1}$
But for the backward method it seems it doesn’t work
Best Answer
You also need to take into account that $x-x_i$ at $x=x_{i-1}$ has the value $-h$.
Check also the other signs, the Taylor terms should be alternating.