Re: Taylor Series Reversion error: 'quotient' by 'zero'

It seems that revert needs a term in t within the taylor serie.
For eq, the t coefficient is 0, so the error quotient by zero appears.
I have tried this to have a linear term, but don't know if the result is
correct
ser2:subst(u^(1/3),t,ser);
sol:revert(ser2,u);
subst(t^3,u,sol);
i get
\[\frac{{{\ensuremath{\pi} }^{3}} {{t}^{9}}}{2240}\mathop{+}\frac{3
\ensuremath{\pi}  {{t}^{3}}}{4}\]

Le sam. 15 août 2026 à 10:45, Justin Jensen <[email protected]> a
écrit :

> I'm new to Maxima and Computer Algebra Systems in general. So far Maxima
> seems to be exactly what I need right now.
>
> As part of a side project I'm trying to find the inverse of a function. It
> has no closed-form function, at least in the general case, so I'm
> approximating it with a Taylor series. Note that the series has
> coefficients of 0 for the 0th, 1st, and 2nd terms. When I try to do a
> reversion of series using the `revert` command, it fails and gives the
> error: "'quotient' by 'zero'" (see below). How do I resolve this? Once upon
> a time, I used Mathematica to find the inverse of a very similar series and
> it worked fine. How do I do this with Maxmia?
>
> (%i1) eq: (2*t-sin(2*t))/%pi;
> eq (2*t-sin(2*t))/%pi
> (%i9) ser: taylor(eq,t,0,9);
> ser
> (4*t^3)/(3*%pi)-(4*t^5)/(15*%pi)+(8*t^7)/(315*%pi)-(4*t^9)/(2835*%pi)+...
> (%i10) revert(ser,t);
> (%o10)
> revert(-((4*t^9)/(2835*%pi))+(8*t^7)/(315*%pi)-(4*t^5)/(15*%pi)+(4*t^3)/(3*%pi),t)
> (%i11) load("revert")$
> (%i61) revert(ser,t);
> `quotient' by `zero'
>
>  -- an error. To debug this try: debugmode(true);
>
>
> Thanks in advance,
>
> --
> Justin
> _______________________________________________
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> [email protected]
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>

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