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

I'm afraid revert() won't work in this case, because the function in 
question has no ordinary Taylor-series reversion. The error message you 
get is legit, since f'(0) = 0.

Mathematica does something more sophisticated, like a Puiseux power 
series expansion, which is beyond the capabilities of revert().

You might be able to do something like this by hand:

load(revert)$
eq:(2*t-sin(2*t))/%pi$
h:ratsimp(eq/t^3);
g:taylor(t*h^(1/3),t,0,7);
block([r:expand(revert(g,t))],ev(r,t=t^(1/3)));

I believe this reproduces a result consistent with that produced by 
Mathematica.


Viktor


On 2026-08-14 17:06, Justin Jensen wrote:
> 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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