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19

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 ...

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 ...

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.

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

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

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 $$... 2 We have$$\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} $$The second equation tells us that \frac{ab+bc+ca}{abc} = \frac{5}{12} , or that 12(ab+bc+ca) = 5(abc). Using (a+b+c)^3 - a^3 - b^3 - c^3 = 3(a+b)(b+c)(c+a) = 3(3-a)(3-b)(3-c), we get that 27-45 = ... 2 This solution is similar to what you are saying, but not phrased in your language of implicit function theorem. Consider the polynomial$$ \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 think what you need is Newton's identities, in particular the section about their application to the roots of a polynomial. 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 ... 1 You can definitely get a computer algebra system to do this for fixed d: It's just linear algebra! Notice that, if you can prove this for |I|=2d, then it follows for all larger I. Specifically, if you can show that 123 \cdots d|(d+1)(d+2) \cdots (2d) = (d+1)(d+2) \cdots (2d) | 12 \cdots d, then it follows for any 2d elements by replacing ... 1 According to Galois theory, since S \subset S(h) \subset F, S(h) is F^H, the field of elements of F fixed by some subgroup H of S_n. Since h is only fixed by the identity automorphism, H = \{id \}, and S(h) =F. In more detail : Let P be the minimal polynomial of h over S and let \sigma be in S_n, so that \sigma(h) = x_{i_1} + ... 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 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 There is no nice way. This is, however, the theory of symmetrization which attempts to show that for some extremal problems the solutions have certain symmetry. It goes back at least to Steiner symmetrization of 19th century. Much is known, but many open problems remain, and there is no universal theorem with which one could hammer problems without looking ... 1 It seems like the real question is what one can say about (the computation of) the coordinate functions for the basis of monomial symmetric functions; this does not involve a scalar product for which the monomial symmetric functions are orthonormal. (Of course once one knows the coordinate functions, the scalar product making the monomial symmetric functions ... 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 A solution is a quadratic Bezier curve (a parabola), with control points (0, 0), (p, 1-p) and (1, 1). The initial slope of this curve is \frac{dy}{dx}=\frac{1-p}{p}=\tan(a+45°). UPDATE: a more direct approach to the parabola. Rotate the axis by 45° clockwise to make the axis of the parabola vertical. The equation is v=\lambda u(\sqrt2-u), with ... 1 I doubt it. Let's look at m_{(1,0,0,0,0)}. You want the solutions ofa+b+c+d+e=0$$with all variables on the unit circle. It will be hard enough to find a formula for that special case, much less for the general case. Note that m_{(1,0)} is a+b=0 which is solved by a=e^{it}, b=e^{i(t+\pi)}. Then m_{(1,0,0)} is a+b+c=0, which forces a,b,c ... 1 The transpositions s_i = (i \; i+1) \in S_n, (1 \le i < n) generate and satisfy the Coxeter relations$$ \begin{align} s_i^2 &= 1 \\ s_i s_j &= s_j s_i \quad &\text{if } |i - j| > 1 \\ s_i s_j s_i &= s_j s_i s_j \quad &\text{if } |i - j| = 1 \end{align}  In other words, any permutation $\tau \in S_n$ can be written (not ...

1

It's totally false. Take any constant function defined on $X \times X$.

1

There are numerous counterexamples, such as $f(x,y)=xy$. In fact, any symmetry function of satisfies that $f(x,y)=f(y,x)$ for every $(x,y)$ in the domain.

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