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12 votes
Accepted

Resultant of linear combinations of Chebyshev polynomials of the second kind

Since $U_n + t U_{n-1}$ is of degree $n$ and $\sum_{k=0}^{n-1} U_k$ is of degree $n-1$ with leading coefficient $2^{n-1}$, the resultant factors as $$ 2^{n(n-1)} (-1)^{n(n-1)} \prod_{j=1}^{n-1} (U_n(...
Terry Tao's user avatar
  • 114k
12 votes
Accepted

Resultant of $f(x)$ and $f(-x)$

We have $\mathrm{Res}(f(x),f(-x))=2^n a_n P(\alpha)^2$, where $P(\alpha)=\prod_{1\leq i<j\leq n}(\alpha_i+\alpha_j)$. By e.g. the case $d=2$ of Exercise 7.30 in Enumerative Combinatorics, vol. 2, ...
Richard Stanley's user avatar
12 votes

Splitting the Resultant, as when the Determinant becomes the square of the Pfaffian

I think it is better to homogenize and view $P$, $Q$ as binary forms in $\mathbb{C}[X_1,X_2]$. An invertible $2\times 2$ matrix $g$ acts on the variables $x_i$ by $(gx)_i=\sum_{j=1,2}g_{ij}x_j$ and on ...
Abdelmalek Abdesselam's user avatar
11 votes
Accepted

Determinant is to Pfaffian as resultant is to what?

Pfaffian resultant formulas are obtained in Resultants and Chow forms via Exterior Syzygies (2001), where the polynomials are represented by coordinates on a Grassmanian manifold.
Carlo Beenakker's user avatar
11 votes
Accepted

What is the essence of the constant factor in the standard definitions of the discriminant?

As Robert said, if you want everything to work in $\mathbb Z[f_0,\ldots,f_m]$, you need that factor. I'll also mention that your polynomial indexing is messed up, you probably meant the sum to go from ...
Joe Silverman's user avatar
10 votes

What is the essence of the constant factor in the standard definitions of the discriminant?

The factor $f_0^{2m-2}$ makes the discriminant a polynomial in the coefficients $f_0, \ldots, f_m$.
Robert Israel's user avatar
9 votes
Accepted

Polynomials that share at least one root

The locus of real polynomials $p(x)$ sharing a root with $P(x)$ is the union of the hyperplanes $H_\alpha : p(\alpha)=0$, where $\alpha$ runs over the roots of $P(x)$. This is an arrangement of ...
François Brunault's user avatar
6 votes
Accepted

Has vol. 3A of Cullis's "Matrices and Determinoids" been scanned and vol. 3B been archived?

Q1: Volume 3 part 1 (a.k.a. volume 3A) has been digitized and reissued as a paperback by Cambridge UP, see Amazon. The digital version is online in the HathiTrust digital library, but with limited "...
Carlo Beenakker's user avatar
4 votes

What is the essence of the constant factor in the standard definitions of the discriminant?

Perhaps you will find the following helpful, which expands a bit on Robert Israel's answer: Write $f(x)=f_0x^m+\ldots+f_m=f_0\cdot(x-\alpha_1)\cdots(x-\alpha_m).$ Multiplying out the RHS we get $f(x)...
Chris McDaniel's user avatar
3 votes
Accepted

Defining polynomial of compositum of splitting fields

For monic polynomials $f,g$ define $f*g=\mathrm{Res}_y(f(y),g(x-y))$. This is an associative and commutative operation on univariate monic polynomials with neutral element $x$. If $f=\prod_{i=1}^m(x-\...
Alexei Entin's user avatar
3 votes
Accepted

Library/Database of parametric polynomial systems

There is a database of polynomial systems, which comes with PHCpack by Jan Verschelde.
Max Alekseyev's user avatar
3 votes
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Simple zeroes of complex polynomial $f(\cdot,a)$: condition on $P(a)=\operatorname{Res}_z(f,f')$

No, it's not true, as is shown by the polynomial $f(z,a):=(z-a)(z-1)^2$.
Loïc Teyssier's user avatar
3 votes
Accepted

A variation on Abhyankar–Moh–Suzuki theorem

The answer is no. You can simply choose $$f=t(t^2-t+1)$$ $$g=t(t^2+1)$$ These satisfy all your conditions, but $\mathbb{C}[f,g]\subsetneq \mathbb{C}[t]$. Let us check this: $(1)-(2)$: $f=tf_1$ and $g=...
Jérémy Blanc's user avatar
3 votes

Polynomial defined recursively by a resultant

This question is close to a central one in Rational Trigonometry, which is to determine the relations between successive quadrances between a cyclic list of points on a line. Here quadrance refers to ...
Norman Wildberger's user avatar
2 votes

What is the essence of the constant factor in the standard definitions of the discriminant?

A trivial observation, which was not pointed out so far: the non-normalized discriminant is 'better' in that it comes 'closer' to being a 'homomorphism' $P[x]\rightarrow P$, though it 'still' is not ...
Peter Heinig's user avatar
  • 6,051
2 votes

Polynomials that share at least one root

I tried to make a 3D image for $P(x)=x^3+3 x^2-2 x-1$. The set consists of three planes, each tangent to the discriminant surface. But it became too visually complex, partly because the discriminant ...
2 votes

Using computer algebra to check if a family of algebras are pair-wise non-isomorphic

For this and other Macaulay2-related questions, I highly recommend the Macaulay2 google group. In general, there are some "exploratory" techniques (i.e. not quite a proof) which are more ...
tim's user avatar
  • 396
2 votes
Accepted

Polynomial function defined recursively by a resultant - is it well defined?

Notice that $$ \let\eps\varepsilon q(x_1,x_2,x_3) =-\bigl(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}\bigr) \bigl(-\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}\bigr) \bigl(\sqrt{x_1}-\sqrt{x_2}+\sqrt{x_3}\bigr) \bigl(-...
Ilya Bogdanov's user avatar
2 votes
Accepted

Dimension of the set of singular hypersurfaces

Given a smooth projective variety $X\subset\mathbb{P}^{N-1}$, let $I(X)\subset X\times(\mathbb{P}^{N-1})^*$ denote the locus of pairs $(x,H)$ where $x\in H$ and $T_x(X)\subset T_x(H)$; here we are ...
Kapil's user avatar
  • 1,566
1 vote
Accepted

Conditions for resultants of bivariate quadratics to be perfect squares

At very least it is necessary that each of the following three numbers is a square: $$a_{02}^2 b_{20}^2-a_{02} a_{11} b_{11} b_{20}-2 a_{02} a_{20} b_{02} b_{20}+a_{02} a_{20} b_{11}^2+a_{11}^2 b_{02} ...
Max Alekseyev's user avatar
1 vote

Application of Resultant in Computer Algebra

Perhaps this is not the best forum to ask such a question; nevertheless I can provide some answer. You may like the paper Computing resultant matrices with Macaulay2 and Maple by Busé, which not only ...
Jose Brox's user avatar
  • 2,992
1 vote

A variation on Abhyankar–Moh–Suzuki theorem

Since you (OP) seem to be interested in Lüroth’s theorem, you might want to look at a "constructive" proof that I just wrote up from Schinzel's Selected topics on polynomials. It seems using ...
pinaki's user avatar
  • 5,339
1 vote
Accepted

A variation on $k(x^2,x^3)=k(x)$

Such $F, G$ do exist, unless I missed a criterion. One possibility is to take $F(T)$ a polynomial such that $F(h)$ is never zero, such as $T^3 + x^2$, and then $G(T) = x F(T)$. Then you can use the ...
Will Sawin's user avatar
  • 148k
1 vote
Accepted

Using computer algebra to check if a family of algebras are pair-wise non-isomorphic

There's a practical computer-algebra way to determine whether (a) they're all isomorphic up to finitely many exceptions, or (b) there "(bounded finite)-to-one" non-isomorphic, i.e., there ...
YCor's user avatar
  • 63.9k
1 vote

Resultants for compactly represented product form polynomials?

Too long for a comment so I am posting this as an answer. It would help if you said precisely what notion of resultant you are using and which variables are being eliminated. There are at least two, ...
Abdelmalek Abdesselam's user avatar
1 vote

Efficient algorithm to compute resultants of sparse polynomials?

See the oevre of I. Emiris: Emiris, Ioannis Z.; Pan, Victor Y., Improved algorithms for computing determinants and resultants, J. Complexity 21, No. 1, 43-71 (2005). ZBL1101.68981. Jeronimo, ...
Igor Rivin's user avatar
  • 96.4k

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