ASKSAGE: Sage Q&A Forum - Individual question feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Sat, 05 May 2018 16:35:30 -0500How to solve a differential equation in polynomial ring (modulo polynomial)https://ask.sagemath.org/question/42252/how-to-solve-a-differential-equation-in-polynomial-ring-modulo-polynomial/I'm working in an arbitrary field F=GF(p^m,'a',modulus=f) where f is a irreducible polynomial of degree m in F_p.
I then took an irreducible polynomial h of degree k in PolynomialRing(F).
Now i am trying to solve the differential equation for phi with degree(phi)= d with d < k fixed
phi * ksi = phi' mod h, where ksi is a fixed polynomial with degree h-1
I tried solving this the regular way by defining phi as a function and using desolve(), but i don't see how to implement the mod h part and how to demand phi to have degree d.
Any help would be greatly appreciated!Thu, 03 May 2018 09:04:36 -0500https://ask.sagemath.org/question/42252/how-to-solve-a-differential-equation-in-polynomial-ring-modulo-polynomial/Comment by Necoroyals for <p>I'm working in an arbitrary field F=GF(p^m,'a',modulus=f) where f is a irreducible polynomial of degree m in F_p.
I then took an irreducible polynomial h of degree k in PolynomialRing(F).
Now i am trying to solve the differential equation for phi with degree(phi)= d with d < k fixed</p>
<p>phi * ksi = phi' mod h, where ksi is a fixed polynomial with degree h-1</p>
<p>I tried solving this the regular way by defining phi as a function and using desolve(), but i don't see how to implement the mod h part and how to demand phi to have degree d.</p>
<p>Any help would be greatly appreciated!</p>
https://ask.sagemath.org/question/42252/how-to-solve-a-differential-equation-in-polynomial-ring-modulo-polynomial/?comment=42261#post-id-42261f=x^13+x^4+x^3+x+1,
p=2,
m=13,
F=GF(p^m, name="a", modulus=f),
h is a random irreducible polynomial of degree 8 in the polynomial ring of GF,
d is the degree of phi, it should be less than or equal to 4Sat, 05 May 2018 05:56:36 -0500https://ask.sagemath.org/question/42252/how-to-solve-a-differential-equation-in-polynomial-ring-modulo-polynomial/?comment=42261#post-id-42261Comment by tmonteil for <p>I'm working in an arbitrary field F=GF(p^m,'a',modulus=f) where f is a irreducible polynomial of degree m in F_p.
I then took an irreducible polynomial h of degree k in PolynomialRing(F).
Now i am trying to solve the differential equation for phi with degree(phi)= d with d < k fixed</p>
<p>phi * ksi = phi' mod h, where ksi is a fixed polynomial with degree h-1</p>
<p>I tried solving this the regular way by defining phi as a function and using desolve(), but i don't see how to implement the mod h part and how to demand phi to have degree d.</p>
<p>Any help would be greatly appreciated!</p>
https://ask.sagemath.org/question/42252/how-to-solve-a-differential-equation-in-polynomial-ring-modulo-polynomial/?comment=42260#post-id-42260Could you please provide the explicit construction of `p, m, F, f, h, k, d` ? Then we have something to start with.Sat, 05 May 2018 05:21:32 -0500https://ask.sagemath.org/question/42252/how-to-solve-a-differential-equation-in-polynomial-ring-modulo-polynomial/?comment=42260#post-id-42260Answer by tmonteil for <p>I'm working in an arbitrary field F=GF(p^m,'a',modulus=f) where f is a irreducible polynomial of degree m in F_p.
I then took an irreducible polynomial h of degree k in PolynomialRing(F).
Now i am trying to solve the differential equation for phi with degree(phi)= d with d < k fixed</p>
<p>phi * ksi = phi' mod h, where ksi is a fixed polynomial with degree h-1</p>
<p>I tried solving this the regular way by defining phi as a function and using desolve(), but i don't see how to implement the mod h part and how to demand phi to have degree d.</p>
<p>Any help would be greatly appreciated!</p>
https://ask.sagemath.org/question/42252/how-to-solve-a-differential-equation-in-polynomial-ring-modulo-polynomial/?answer=42262#post-id-42262What is wrong is the following argument ?
The degree of $\phi-\phi'$ is d, which is smaller than the degree of h since $d<h$
Hence, $\phi-\phi' = 0 \mod h$ is equivalent to $\phi-\phi' = 0$. Hence $\phi = 0$ is the only solution to the equation.Sat, 05 May 2018 15:38:38 -0500https://ask.sagemath.org/question/42252/how-to-solve-a-differential-equation-in-polynomial-ring-modulo-polynomial/?answer=42262#post-id-42262Comment by Necoroyals for <p>What is wrong is the following argument ?</p>
<p>The degree of $\phi-\phi'$ is d, which is smaller than the degree of h since $d<h$ <="" p="">
</p><p>Hence, $\phi-\phi' = 0 \mod h$ is equivalent to $\phi-\phi' = 0$. Hence $\phi = 0$ is the only solution to the equation.</p>
https://ask.sagemath.org/question/42252/how-to-solve-a-differential-equation-in-polynomial-ring-modulo-polynomial/?comment=42263#post-id-42263Sorry i checked again and it should be phi * ksi = phi' mod h, where ksi is a fixed polynomial with degree h-1, terribly sorry about the confusion....Sat, 05 May 2018 16:35:30 -0500https://ask.sagemath.org/question/42252/how-to-solve-a-differential-equation-in-polynomial-ring-modulo-polynomial/?comment=42263#post-id-42263