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Bill Allombert on Mon, 05 May 2025 15:21:30 +0200
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Re: reversing a series modulo
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- To: pari-users@pari.math.u-bordeaux.fr
- Subject: Re: reversing a series modulo
- From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>
- Date: Mon, 5 May 2025 15:21:27 +0200
- Delivery-date: Mon, 05 May 2025 15:21:30 +0200
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On Sat, May 03, 2025 at 02:38:33PM -0400, Max Alekseyev wrote:
> Hello,
>
> Reversing a series and taking it modulo 5 works, but not the opposite way
> around. Why?
>
> ? f = x + x^2 + 4*x^6 + 4*x^7 + x^11 + x^12 + 4*x^16 + 4*x^17 + O(x^21)
> %1 = x + x^2 + 4*x^6 + 4*x^7 + x^11 + x^12 + 4*x^16 + 4*x^17 + O(x^21)
> ? serreverse(f) * Mod(1,5)
> %2 = Mod(1, 5)*x + Mod(4, 5)*x^2 + Mod(2, 5)*x^3 + Mod(4, 5)*x^5 + Mod(4,
> 5)*x^6 + Mod(2, 5)*x^8 + Mod(3, 5)*x^10 + Mod(3, 5)*x^11 + Mod(4, 5)*x^12 +
> Mod(4, 5)*x^17 + Mod(3, 5)*x^18 + O(x^21)
> ? serreverse(f * Mod(1,5))
> *** at top-level: serreverse(f*Mod(1,5))
> *** ^----------------------
> *** serreverse: impossible inverse in Fl_inv: Mod(0, 5).
The formula used by PARI is
g=intformal(1/subst(f',x,g))
which involves division by small primes (when integrating)
What you can do is
? serreverse(f*(1+O(5^2)))*Mod(1,5)
%34 = Mod(1, 5)*x + Mod(4, 5)*x^2 + Mod(2, 5)*x^3 + Mod(4, 5)*x^5 + Mod(4, 5)*x^6 + Mod(2, 5)*x^8 + Mod(3, 5)*x^10 + Mod(3, 5)*x^11 + Mod(4, 5)*x^12 + Mod(4, 5)*x^17 + Mod(3, 5)*x^18 + O(x^21)
Cheers,
Bill