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20

In order to avoid definiteness problems, I will consider $a$, $b$ and $c$ as independent variables in the field of rational fractions $\mathbb{C}(a,b,c)$, and I will seek some kind of a parametric representation of the solutions $(a,b,c)$ of the proposed equation as rational functions from $\mathbb{C}(u,v)$ where $u$ and $v$ are independent variables. ...


11

If $x_1,x_2,x_3$ are the roots of $x^3+2x^2+3x+4=0$ then $$x^3+2x^2+3x+4 = (x-x_1)(x-x_2)(x-x_3) $$ $$= x^3 - (x_1 + x_2 + x_3)x^2 + (x_1 x_2 + x_1 x_3 + x_2 x_3)x - x_1 x_2 x_3 = x^3 - e_1 x^2 + e_2 x - e_3.$$ So $e_1 = -2$, $e_2 = 3$ and $e_3 = -4$. Now the trick is to express the power sums $x_1^3 + x_2^3 = x_3^3$ and $x_1^4 + x_2^4 = x_3^4$ in terms of ...


7

Probably a simple general formula for your matrix $C_n=(C_{\lambda,\mu})_{\lambda,\mu\in\mathcal P_n}$, where $\mathcal P_n$ denotes the partitions of $n$ (ordered in decreasing lexicographic ordering) does not exist. However a number of things can be said, notably your above guesses can be confirmed. One thing that your examples suggest but which is false ...


6

Okay, I'll answer your question on the cross ratio: First of all, recall its definition $$\lambda = \lambda(z) = [z,z_2,z_3,z_4] = \frac{z-z_3}{z - z_4} : \frac{z_2 - z_3}{z_2 - z_4}$$ which is the unique Möbius transformation $z \mapsto \lambda(z)$ sending the ordered triple $(z_2,z_3,z_4)$ to $(1,0,\infty)$—it is a general and easy fact that ...


4

Not a general answer, but in this particular case you can write $$ f(x_1,x_2,x_3,x_4)= \frac{(x_1-x_3)(x_2-x_4)}{(x_1-x_4)(x_2-x_3)} $$ which makes it easier to see the symmetries.


4

The claim is true. I find it convenient to state the problem in a "Boolean hypercube" form . The variables will be the $\binom{2d}{d}$ strings of length $2d$ over the alphabet $\{L,R\}$ with exactly $d$ Rs. For any string $x\in\{L,R\}^{2d}$ with exactly $d+1$ $R$'s, let $\partial(x)$ denote the sum of the variables labelled by strings obtained by ...


4

Ones first impression is that this is a set of three quadratic equations. So if we would proceed by eliminating two of the three variables, we would get a higher order equation that presumably can not be solved. However, in this case the functions are such that we can side-step these complications. With a nice trick we can work with linear equations! x*y - ...


3

Note that $|e_n|$ is dominated by $\left(\sum_j|x_j|\right)^n/(n!)$, so yes, everything like what you wrote converges and pretty fast.


3

You have four symbols $x_1,x_2,x_3$ and $x_4$. If you have two different permutations which fix your function, then they will fix it if applied consecutively. The identity fixes the function, as does the inverse of any permutation which fixes it. The permutations fixing the function therefore form a group. Since there are four objects being permuted, the ...


3

First of all notice that $a$ and $b$ appear "symmetrically": if you exchange $a$ with $b$, the system doesn't change. In these cases, setting $$x=a+b,\quad y=ab$$ usually helps. Indeed, you can note now that the left hand side of the second equation is just $x/y$, while $$ a^3+b^3 = x^3 -3xy $$ The first equation already tells you what $x$ is; so the ...


2

We have $$\begin{align} a+b+c & = 3 \\ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} &= \frac{5}{12} \\ a^3 + b^3 + c^3 & = 45 \\ \end{align} $$ The second equation tells us that $\frac{ab+bc+ca}{abc} = \frac{5}{12} $, or that $12(ab+bc+ca) = 5(abc)$. Using $(a+b+c)^3 - a^3 - b^3 - c^3 = 3(a+b)(b+c)(c+a) = 3(3-a)(3-b)(3-c)$, we get that $27-45 = ...


2

Beside the very nice answer by Marc, I found this Introduction to Symmetric Functions by Mike Zabrocki: $$ e_\mu = \sum_\lambda B_{\lambda \mu} m_\lambda\tag{2.28} $$ where $B_{\lambda \mu}$ is the number of matricies with entries in $\{0, 1\}$ whose column sum is $\mu$ and row sum is equal to $\lambda$. $$\dots$$ $B_{\lambda \mu}$is the number of ...


2

The pairwise sums of the roots are $2a_1, 2a_2,$ and $(a_1+a_2)\pm(b_1 \pm b_2)i$, so the right-hand side is the product of these six factors and an extra minus sign. So your expression could be written as $-\prod_{i<j}(r_i + r_j)$. So, since the right-hand side is symmetric in the four roots, it's automatic it can be written as a polynomial expression ...


2

In order to see the "correct" behavior of the Schur polynomials and to be able to compare different formulas for the Schur polynomials, the number of variables you need is equal to the total number of boxes in the partition (the value $d$, in the Wikipedia notation), not just the number of (nonzero) parts in the partition. Try taking $n=d=4$ in both of your ...


2

It is natural to try to manipulate symmetric polynomials as one does polynomials, representing them as a sum of monomials, and to take advantage of the symmetry by representing a whole orbit of monomials only once, as a minimal symmetric polynomial$~m_\lambda$. Also one can now leave the number of indeterminates unspecified, and thus effectively manipulate ...


2

A function $f$ is radially symmetric if $f(Ax) = f(x)$ for all orthogonal $A$. Thus, if $f$ and $g$ are radially symmetric, and the convolution exists, we have for any $A\in O(n)$ $$\begin{align} (f\ast g)(Ax) &= \int_{\mathbb{R}^n} f(y) g(Ax-y)\,dy\\ &= \int_{\mathbb{R}^n} f(y) g(A(x-A^T y))\,dy\\ &= \int_{\mathbb{R}^n} f(y) g(x-A^T ...


2

I'm not sure if this is exactly what you're looking for, but I recently worked on a paper regarding just this question (http://arxiv.org/abs/1403.8139). We were generalizing Tokuyama's deformation formula (http://projecteuclid.org/euclid.jmsj/1230129805) for the Schur polynomial, so we used Gelfand-Tsetlin patterns rather than Young tableaux. Tokuyama's ...


2

Do this $$f(x) = 2(x^2 + 4x) + 5 = 2(x^2 + 4x + 4) + 5 - 8 = 2(x + 2)^2 - 3.$$ Note that $f$ is insensitive to the sign of $x + 2$ so $f$ is symmetric about the line $x = -2$


2

Doing $(a,b,c)→(a^2,b^2,c^2)$ two times is clearly the easier way. Let $A=a+b+c,B=ab+bc+ca,C=abc$. Consider $$\begin{align}&(x^{1/2}-a)(x^{1/2}-b)(x^{1/2}-c)\times (x^{1/2}+a)(x^{1/2}+b)(x^{1/2}+c) = \\&(x^{3/2}-Ax+Bx^{1/2}-C)\times (x^{3/2}+Ax+Bx^{1/2}+C) = \\&(x^{3/2}+Bx^{1/2})^2 - (Ax+C)^2 = \\&x^3+2Bx^2+B^2x-A^2x^2-2ACx-C^2\end{align}$$ ...


2

This solution is similar to what you are saying, but not phrased in your language of implicit function theorem. Consider the polynomial $$ \sum_{i=1}^n \left[ x_i ^m \prod_{j < k, j\neq i, k\neq i} (x_j-x_k)\right] $$ If $x_i = x_j$, then you should verify that the polynomial is equal to 0 (many terms with $x_i-x_j)$ are zero, and then ensure that the ...


2

I'll suppose $R$ is an integral domain, and therefore contained in an algebraically closed field $K$ (one can do without this assumption, but it simplifies the argument). Suppose $p(s_1,\ldots,s_n)=0$ but $p$ is not the zero polynomial; then certainly there exist constants $a_1,\ldots,a_n\in K$ such that $p(a_1,\ldots,a_n)\neq0$. Now consider $$ ...


2

I think what you need is Newton's identities, in particular the section about their application to the roots of a polynomial.


1

Let $A \subset \mathbb{R}^2$ be a set of points given by equation $f(x,y) = 0$, then for any invertible linear transformation $T : \mathbb{R}^2 \to \mathbb{R}^2$ set $T(A)$ is given by $(f \circ T^{-1})(x,y) = 0$. This is because $A = \{(x,y) \mid f(x,y) = 0\}$, and \begin{align} T(A) &= \Big\{(x',y')\ \Big|\ (x',y') = T(x,y) \land f(x,y) = 0\Big\} \\ ...


1

When we reflect a point $(x_0,y_0)$ across the line $y=-x,$ our new $x$-coordinate will be $-y_0$ and our new $y$- coordinate will be $-x_0.$ Consequently, our general transformation in this case is to make $x\mapsto-y,$ $y\mapsto-x$. What about the more general case of reflecting about a line $\ell$ through the origin, though? Well, first, see where the ...


1

Well you essentially have $g(x)=f(-x)$ and hence by the chain rule, $g'(x)=-f'(-x)$. I suspect this is enough for you to be able to determine any extensions, etc. yourself.


1

I do not quite understand what you are trying to do. You said you want to define a kernel on $\boldsymbol{x} \sim \mathcal{N}(\boldsymbol{x}_i, \Sigma_i)$ and $\boldsymbol{y} \sim \mathcal{N}(\boldsymbol{x}_j, \Sigma_j)$. Why does the distribution of your points $\{ \boldsymbol{x}, \boldsymbol{y} \}$ matter ? In general, you do not need any distribution ...


1

I doubt it. Let's look at $m_{(1,0,0,0,0)}$. You want the solutions of $$a+b+c+d+e=0$$ with all variables on the unit circle. It will be hard enough to find a formula for that special case, much less for the general case. Note that $m_{(1,0)}$ is $a+b=0$ which is solved by $a=e^{it}$, $b=e^{i(t+\pi)}$. Then $m_{(1,0,0)}$ is $a+b+c=0$, which forces $a,b,c$ ...


1

You consider the map $c\colon (x_1,\ldots,x_n)\mapsto (a_0,\ldots,a_{n-1})$ where $x^n+a_{n-1}x^{n-1}+\ldots+a_0=(x-x_1)\cdots(x-x_n)$. For any point $\mathbf x=(x_1,\ldots,x_n)$ with all entries different(!), you can find an inverse map $c^{-1}$ from a neighbourhood of $c(\mathbf x)$ to a neighbourhood of $\mathbf x$. Of course you can combine this with ...


1

One way is the following: $$\begin{align}(x-a^4)(x-b^4)(x-c^4)&=x^3-\left(\sum\limits_{cyc} t^4\right)x^2+\left(\sum\limits_{cyc} \left(\frac{1}{t}\right)^4\right)x-1\tag 1\\ &=x^3-\frac12\left(\left(\sum\limits_{cyc} t^3\right)-1\right)x^2 +\frac12\left(\sum\limits_{cyc} \left(\frac{1}{t}\right)^2\right)x-1\tag2\\ ...


1

Using the power of the Dandelin-Gräffe iteration. For any polynomial $p$ with roots $z_i$, the polynomial $p^{(1)}(x)$ defined via $$ p^{(1)}(x^2)=p(x)·p(-x) $$ has the same degree as $p$ and roots $z_i^2$. Repeating this process, the polynomial $p^{(2)}(x)$ defined via $$ p^{(2)}(x^2)=p^{(1)}(x)·p^{(1)}(-x) $$ has still the same degree as $p$ and roots ...



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