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21

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. ...


12

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 ...


8

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 ...


6

The trick it's just to add and subtract $1$ from each fraction and then to factorize: $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a+b+c}{b+c}-1+\frac{a+b+c}{c+a}-1+\frac{a+b+c}{a+b}-1=(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})-3=7\cdot 0.7 -3=1.9$$


6

$$\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{a+c}$$ $$=\frac{7-(a+b)}{a+b}+\frac{7-(b+c)}{b+c}+\frac{7-(a+c)}{a+c}$$ $$=\frac{7}{a+b}+\frac{7}{b+c}+\frac{7}{a+c}-3$$ $$=7\cdot 0.7-3=1.9$$


4

Simplify it to $$ f(x)={{5a^2+6ax+9x^2}\over {a+3x}}=\frac{4a^2}{a+3x} + {{a^2+6ax+9x^2}\over {a+3x}}=\frac{4a^2}{a+3x} + {{(a+3x)^2}\over {a+3x}}\\ =\frac{4a^2}{a+3x} + (a+3x) $$ and substitute $y=a+3x$ then $$ f(y)=\frac{4a^2}{y}+y $$ which has the following property: $$ f(-y)=-\frac{4a^2}{y} -y = -f(y) $$ which means that $f(y)$ is an odd function, point ...


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 - ...


4

One possibility is to compute $S$-polynomials from the given polynomials, namely $$ 25x + 9y + 16z - 12=0, \\ 24yz + 16z^2 - 32z + 5 =0. $$ Then we can eliminate $x$ and $y$ and substitute into the three original equations. We obtain $$ (64z^3 - 80z^2 - 52z + 5)(4z - 1)=0. $$ Although we have all roots real here, we see that the only positive real solution ...


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

Take $$\|(x_1,x_2)\|^2 = x_1^2+x_1x_2+x_2^2.$$ This relation defines a norm on $\Bbb R^2$. Clearly, $$\|(x_1,x_2)\|=\|(x_2,x_1)\|\ne \|(x_1,-x_2)\|.$$ edit Write this norm in terms of a matrix $A=\frac 12 \begin{pmatrix}2&1\\1&2\end{pmatrix}$ and scalar product: $$\|\vec x\|^2 = \vec x^TA\vec x = (A\vec x,\vec x).$$ The matrix $A$ is positive ...


3

From the first equation we have $$y=\frac6x.$$ Plugging this into the second equation yields $$x^{\tfrac6x}+\left(\frac6x\right)^x=17.$$ Call this expression on the left hand side $f(x)$. We want to solve the equation $f(x)=17$. It is obvious that $x=2$ and $x=3$ solve this equation. It remains to show that there are no other solutions. To see this, first ...


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

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 ...


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

eliminating $y,z$ we get for $x$: $$125 x^4-60 x^3-58 x^2+18 x=1$$


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

Multiply the first equation by $5x$, the second by $3y$, the third by $4z$. Then add all three up. This gives you $25x + 9y + 16z = 12$. Now this is clearly a useful intermediate result. For example, you can substitute it into the original equations to eliminate one variable. This yields six new relations: [1] $4/x -25x - 21y = 8$; [2] $3/x + 25x + 28z = ...


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

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 ...


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

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

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

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

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 think your assumption is right. In the Introduction to Symmetric Functions by Mike Zabrocki there is a nice table on page 18, that relates all kinds of symmmetric functions. There $m_\mu$ is related to $e_\lambda$ by the same coefficients $G_{\lambda\mu}$ as $f_\mu$ and $h_\lambda$ are, but he leave[s] it as an exercise to determine some sort of formula ...


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 ...



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