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Suppose I wish to derive a numerical integrator for the first-order ODE $$x'(t)=F(x(t)).$$ By differentiating both sides of the expression in $t$, I can write a second-order relation also satisfied by $x(t)$: $$x''(t)=\frac{d}{dt}\left(x'(t)\right)=\frac{d}{dt}F(x(t))=DF(x(t))\cdot x'(t)=DF(x(t))\cdot F(x(t)),$$ by the chain rule, where $DF$ denotes the Jacobian of $F$.

Using Taylor series, I could write $$x(t+h)=x(t)+F(x(t))\cdot h + \frac12 DF(x(t))\cdot F(x(t))\cdot h^2 + O(h^3).$$ If we omit the $O(h^3)$ term, we get quadratic-error explicit time integrator for our ODE.

Does this time integration strategy where we know derivatives of $F(\cdot)$ have a name? What are its drawbacks?

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Not too surprisingly, it is referred to as a "Taylor series method." The drawback is that, in general, it may not be easy to compute the derivatives of $F$, particularly as you add more terms to the Taylor series to improve the accuracy.

But the method can be quite attractive if $F$ has a specific form, say a polynomial in $t$.

Other approaches to implementation of the method include using automatic differentiation methods or symbolic calculus systems to calculate the higher-order derivatives.

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    $\begingroup$ Thanks! This is indeed the Google search term I was missing ;-) . I can't find many documents that take this method seriously --- it seems to be among the zoo of relatively unused numerical integrators. I can't tell if it's just because of the cost of computing DF, or if the integrator itself isn't useful. Maybe now that automatic differentiation is more common (e.g. using libraries like Pytorch) we should revisit... $\endgroup$
    – Justin Solomon
    Commented Jul 8 at 21:00
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    $\begingroup$ In many real applications, $F$ is very complicated. Computing its derivatives is not always easy. Worse, $F$ is in many cases no differentiable -- say if a part of the device you're modeling switches on or off at specific times or based on conditions on $x$. $\endgroup$
    – Wolfgang Bangerth
    Commented Jul 9 at 3:25
  • $\begingroup$ This makes sense! I work on a lot of graphics problems where $F$ is likely to be fairly clean/differentiable --- maybe I'll play with these! $\endgroup$
    – Justin Solomon
    Commented Jul 9 at 15:11
  • $\begingroup$ One other potential pitfall is that, although using derivative information can lead to a higher order scheme, the error coefficient may grow quite large and the higher-order scheme may require smaller step sizes than the lower order scheme to hit some desired error tolerance. $\endgroup$
    – whpowell96
    Commented Jul 11 at 16:25

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