Find convergence rank of a single-step scheme

Question

Hi i have a huge problem with finding convergence rank of the following method: $$x_{n+1}=h\cdot(\alpha\cdot f_{n}+(1-\alpha)f_{n+1})$$ where $\alpha\in(0,1)$ I will be more than glad for thorough explanation as i am pretty much hopeless in it. I though about something like that : $$x(t_{n+1})=h\cdot(\alpha\cdot x'(t_{n})+(1-\alpha)x'(t_{n+1}))$$ and somehow use Taylor? To bring it further?

Answer 1

Set $h=2s$, $t=t_n+s$ so that $x_n=x(t-s)$, $x_{n+1}=x(t+s)$ and use the Taylor expansions $$ x(t\pm s)=x(t)\pm x'(t)s+\frac12x''(t)s^2\pm\frac16x'''(t)s^3+…\\ x'(t\pm s)=x'(t)\pm x''(t)s+\frac12x'''(t)s^2\pm… $$ to combine the error term of the method.