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
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.