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Finding Schur coefficient of a concrete polynomial

asked 2021-06-08 15:34:54 +0200

convergency gravatar image

updated 2021-06-09 00:46:31 +0200

slelievre gravatar image

By concrete polynomial, I mean something like

$$ (x1+x2) (x1+x3) (x2+x3) $$

What I tried didn't work here:

sage: R.<x0,x1,x2> = PolynomialRing(QQ, 3)
sage: Sym = SymmetricFunctions(QQ)
sage: s = SymmetricFunctions(QQ).s()
sage: f = x0^2*x1 + x0*x1^2 + x0^2*x2 + 2*x0*x1*x2 + x1^2*x2 + x0*x2^2 + x1*x2^2
sage: s(f)

It gives a long error message.

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answered 2021-06-09 07:44:33 +0200

rburing gravatar image

I don't know why this conversion isn't implemented. But you can use the from_polynomial method instead:

sage: s.from_polynomial(f)
s[2, 1]
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Asked: 2021-06-08 15:34:54 +0200

Seen: 58 times

Last updated: Jun 09