Infinite stepsize integrating a function

[mforets]

For the context see this discussion. In short, integrating a function such as

@taylorize function bball!(t, x, dx)
    dx[1] = x[2]
    dx[2] = -one(x[1]) # this one is constant
    return dx
end

with a taylor series in time whose order is larger than two (the exact solution has degree two), then TaylorIntegration.stepsize! returns a stepsize which is Inf.

This is maybe correct? Since one gets the exact solution for all t? So i don't know the expected action point here, but then some error happens down the line, in TaylorModels.validated_integ.

[lbenet]

Thanks @mforets for opening this issue!

Indeed, the returned computed local solution is actually exact, which means that if the order is 4 or larges, the returned time step will be Inf. What we could do is check for such a Inf time step, and only in that case, use the last non-zero coefficient of the series to compute a finite step size.

cc @PerezHz

[PerezHz]

Perhaps we use could use TaylorSeries' findlast? For example, here, instead of doing ord = x.order, we could do ord = findlast(x)

[lbenet]

Good idea! Could you test it?

[lbenet]

Closing; solved by #73