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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

Can this nonconformity of "Multivariate Polynomial Ring" and "Infinite polynomial ring" be brought into conformity so that operations (for example the derivative) can be executed intuitively?

Thrash gravatar imageThrash ( 2022-11-19 17:12:31 +0200 )edit

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

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Last updated: Sep 05 '20