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Derivative in infinite polynomial ring

asked 2020-09-05 18:32:15 +0200

mathstudent gravatar image

I am defining my ring as R.<x>=InfinitePolynomialRing(QQ), and this should give me ring with variables x[1],x[2],... etc. right? Now I want to differentiate a polynomial with respect to x[1] variable. So I defined f=x[1]^3 (for example). I am trying f.derivative(x[1]) but that does not work. It shows " 'typeerror': argument 'var' has incorrect type (expected sage.rings.polynomial.multi_polynomial_libsingular.MPolynomial_libsingular, got InfinitePolynomial_dense)." Can someone please explain what is wrong and what I should do to fix it?

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answered 2020-09-05 20:53:08 +0200

Emmanuel Charpentier gravatar image

A nice one...


sage: R.<x>=InfinitePolynomialRing(QQ)
sage: f=R(x[1]^3)

Note that :

sage: [u.parent() for u in f.variables()]
[Multivariate Polynomial Ring in x_2, x_1, x_0 over Rational Field]

(yes, I played with R a bit...) contrasting with:

sage: x[1].parent()
Infinite polynomial ring in x over Rational Field

Hence the error you noticed. However :

sage: x[1] in f.variables()

It is there ; you just have to find it :

sage: f.variables().index(x[1])

Hence the (awkward) :

sage: f.derivative(f.variables()[f.variables().index(x[1])])

A better way would be to cast x[1] to the proper class. Finding which isn't intuitive...

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Thanks a lot.

mathstudent gravatar imagemathstudent ( 2020-09-06 10:42:42 +0200 )edit

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Asked: 2020-09-05 18:32:15 +0200

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Last updated: Sep 05 '20