# Local truncation error from Taylor/Variable timestep

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maistral
I am not sure where I should be posting this; I am perfectly aware this is Taylor expansion calculus, but considering the fact that I am working with numerical ODE solutions, I posted this here.

In the paper: Implementation of an Adaptive BDF2 Formula and Comparison with the MATLAB Ode15s (Celaya, et al); I have found this formula:

It's supposed to be a local truncation error formula which compares the numerical solution yn+2 to that of a Taylor expansion y(tn+2). While I can set the values of h to be equal with each other (thus retrieving the constant step version), I have no idea how to derive this term if I want to bring it to higher orders (say, I want a fourth-order solution which would den require the calculation of a fifth derivative term with variable stepsize).

TLDR: Does anyone know how this thing has been made? I have been scouring the internet for a while about this and I seem to have no information as to how this thing appeared (I mean, looking at the four papers I have now, this just magically appeared).

Thanks!

Homework Helper
2022 Award

I found https://student.cs.uwaterloo.ca/~cs370/notes/LTE.pdf which explain how to calculate LTEs and give the forwards Euler and Crank-Nicholson as examples.

Thus your starting point is the advancing formula (27) in which you set $y_{n+1} = y(t_{n+1})$, $y_n = y(t_n)$ and $f_{n+2} = y'(t_{n+2})$. The result you are looking for gives the error in terms of $y^{(3)}(t_{n+2})$, so you will then need to use these Taylor expansions: \begin{align*} y(t_{n+1}) &= \sum_{k=0}^\infty \frac{(t_{n+1} - t_{n+2})^k}{k!} y^{(k)}(t_{n+2}) \\ y(t_n) &= \sum_{k=0}^\infty \frac{(t_{n} - t_{n+2})^k}{k!} y^{(k)}(t_{n+2}) \end{align*} I will say that starting from the constant-step version (26) I obtained $$y_{n+2} - y(t_{n+2}) = \frac 29 h^3y^{(3)}(t_{n+2}) + O(h^4)$$ so either I have gone astray or the paper contains an error.

hutchphd
maistral

I found https://student.cs.uwaterloo.ca/~cs370/notes/LTE.pdf which explain how to calculate LTEs and give the forwards Euler and Crank-Nicholson as examples.

Thus your starting point is the advancing formula (27) in which you set $y_{n+1} = y(t_{n+1})$, $y_n = y(t_n)$ and $f_{n+2} = y'(t_{n+2})$. The result you are looking for gives the error in terms of $y^{(3)}(t_{n+2})$, so you will then need to use these Taylor expansions: \begin{align*} y(t_{n+1}) &= \sum_{k=0}^\infty \frac{(t_{n+1} - t_{n+2})^k}{k!} y^{(k)}(t_{n+2}) \\ y(t_n) &= \sum_{k=0}^\infty \frac{(t_{n} - t_{n+2})^k}{k!} y^{(k)}(t_{n+2}) \end{align*} I will say that starting from the constant-step version (26) I obtained $$y_{n+2} - y(t_{n+2}) = \frac 29 h^3y^{(3)}(t_{n+2}) + O(h^4)$$ so either I have gone astray or the paper contains an error.