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Find the order of the method $$y_{n+1}=y_n+hf(t_n+(1-\theta )h,\theta y_n+(1-\theta)y_{n+1}) \ \ \ \ \ (\star)$$ for $\theta \in [0,1]$.

My attempt:

We know for $\theta=0 $, we have that $(\star)$ is Euler's method "backward" which is of order one. If $\theta=1$, $(\star)$ is Euler's method "Forward" and of order one. If $\theta= \frac{1}{2}$, then we have the midpoint rule which is of order 2. I do not know how to start showing the order of $(\star)$ in the otherwise case, so I really appreciate any kinf of help with this.

Ahmed
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  • You have defined $y_{n+1}$ in terms of $y_{n+1}$. Perhaps the latter $y_{n+1}$ should instead be a $y_{n-1}$? – Math1000 Jan 25 '20 at 00:36
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    No it is $n+1$, I just copy it from my book – Ahmed Jan 25 '20 at 00:40
  • Do you not see a problem in defining $y_{n+1}$ in terms of $y_{n+1}$? If one does not know the value of $y_{n+1}$, how can one possibly hope to compute the value of $y_{n+1}$ from the equation you have given? – Math1000 Jan 25 '20 at 00:48
  • Look at the similar questions https://math.stackexchange.com/q/1486008/115115 and https://math.stackexchange.com/q/3381289/115115. Also https://math.stackexchange.com/q/1213614/115115 for an implemented example. These theta methods get the Heun method as second order method, while this results in the midpoint method. – Lutz Lehmann Jan 25 '20 at 07:40
  • @Math1000 : Implicit Runge-Kutta methods and multi-step methods are a well-established topic. They are used for "stiff" ODE, which are non-linear and/or containing different time scales. – Lutz Lehmann Jan 25 '20 at 07:54
  • You should somewhere in a central place use the interpolation error formula $$ y(t_n+(1−θ)h)-[θy(t_n)+(1−θ)y(t_{n+1})] = \frac{θ(1−θ)h^2}2y''(t_n+\tau h). $$ – Lutz Lehmann Jan 25 '20 at 08:09

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