I am working on deriving the local truncation error (LTE) for what I think is a two-step Adams-Bashforth-Moulton IVP scheme by taylor series expansion. The given scheme is as follows.

$y_{n+1}^* = y_n + \frac{h}{2}[3f(y_n)-f(y_{n-1})]$

$y_{n+1} = y_n + \frac{h}{2}[f(y_n)-f(y_{n+1}^*)]$

My intuitive understanding of the process of deriving the LTE for a numerical scheme is to expand an "exact" solution as far as necessary to cancel out all of the terms in the approximation such that $LTE = y_{n+1} - \hat{y}_{n+1}$ where the LTE will have some asymptotic behavior $O(h^{p+1})$, where $p$ is the order of accuracy. This means I simply need to get the numerical scheme into a format where it can cancel out with the Taylor Expansion of an exact solution. I read around about how to approach this problem with a predictor-corrector scheme like the one I was given and was unsatisfied with answers given for similar schemes, such as the modified euler, and wanted to see if my intuition of combining the predictor-corrector method into one equation and then subtracting that from the "exact" value would result in a simplified derivation of the LTE.

Trying it out, the math seemed to work but I'm unsure if it's correct. My question is, for deriving the LTE of numerical schemes such as this one, is it really as simple as getting the temporal scheme into a form that is suitable for cancellation by Taylor Series Expansion to an "exact" solution? This makes sense to me as the order of accuracy will be left by the highest taylor expansion term, the remainder, that is unable to cancel. It would be lovely if someone could confirm/deny this intuition so I might change my approach to deriving LTEs for numerical schemes.


1 Answer 1


Yeah, your intuition seems correct. You simply need to expand using the Taylor form and then find the difference. That should give you the LTE.

You can do something similar to this question : Determine the local truncation error of the following method

  • $\begingroup$ Thanks or confirming my intuition and linking to a detailed mathematical derivation of an LTE. $\endgroup$
    – VdZ
    Aug 16, 2020 at 1:44
  • $\begingroup$ @VdZ can you please select it as the right answer if you found it useful? Thanks $\endgroup$
    – User2345
    Aug 17, 2020 at 13:06

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