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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

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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.

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  • $\begingroup$ 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 $f(x(s),s)$ $\endgroup$ – Questioner May 15 at 11:59

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