# ODE Taylor series expansion using midpoint Euler method

I am trying to solve part (c) of the question I attached as a picture as preparation for an exam.

I can see how to get the result in the correct form using a taylor series expansion, however I don't understand where $$t+h/2$$ has come from in each occurrence. Can anyone explain?

Question

## 1 Answer

The (implicit) midpoint method $$y_{k+1}=y_k+hf(x+\tfrac12h,\tfrac12(y_{k+1}+y_k))$$ is the combination of an implicit and explicit Euler half-step. Both half steps are easiest to analyze from the midpoint $$m=\tfrac12(y_{k+1}+y_k)$$ on. $$m=y_k+\tfrac12hf(x+\tfrac12h,m)\\ y_{k+1}=m+\tfrac12hf(x+\tfrac12h,m)$$

Also, as the method is symmetric about the midpoint, it makes sense to have the Taylor expansions also about the midpoint, as (anti-)symmetric terms will combine and may simplify (to zero).

Note that for an exact solution $$m=\tfrac12(y(x_{k+1})+y(x_k))=y(x_k+\tfrac12h)+O(h^2),$$ so that the change from the mean value $$m$$ to the value $$y(x_k+\tfrac12h)$$ at the midpoint in the method formula only adds another $$O(h^3)$$ term.

• I suppose that makes sense, but what is still confusing me is how part (b) could be used to make part (c) since the $f_t$ term would remain the same before and after substituting part b into the Taylor expansion of $f(x(s),s)$ – Questioner May 15 at 11:59