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I want to write the function $$ F_N=\sum_{i=1}^N\sqrt{x_i} $$ in terms of the $N$ elementary symmetric polynomials of the $N$ positive variables $x_1,\dots,x_N$.

The $N=1$ case is trivial, as we have $F_1=\sqrt{x_1}$ so $$F_1=\sqrt{s_1}$$

The $N=2$ case is simple, as we have $F_2^2=(\sqrt{x_1}+\sqrt{x_2})^2=x_1+x_2+2\sqrt{x_1x_2}$, which means that $$F_2=\sqrt{s_1+2\sqrt{s_2}}$$

The $N=3$ case is not so simple, but after a bit one finds $$F_3=\sqrt{s_1+2\sqrt{s_2+2\sqrt{s_3}F_3}}$$ so one can at least solve for $F_3$.

There is kind of a pattern evolving here. However, I cannot proceed beyond $N=3$... Any insights?

EDIT: I'm not sure this is helpful, but as $F_N$ is an eigenfunction of the operator $2\sum_ix_i\frac{\partial}{\partial x_i}$, one can easily find that $$F_N=\sqrt{s_1}G_N\left(\frac{s_2}{s_1^2},\frac{s_3}{s_1^3},\dots,\frac{s_N}{s_1^N}\right)$$ where $G_N$ is a function of $N-1$ arguments.

EDIT 2: Actually, $F_N$ is an eigenfunction of the operators $$\frac{2 (-1)^{n-1}}{(1/2)_{n-1}}\sum_ix_i^n\frac{\partial^n}{\partial x_i^n}$$ for every positive integer $n$, where $(a)_n$ is the Pochhammer symbol. I wonder if one can use this fact to further reduce the structure of $F_N$.

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  • $\begingroup$ your definition of $F$ seems lacking. How does $F_2$ have nested square roots? Also what is $G_N$. $\endgroup$ – Alex R. Aug 26 '14 at 1:43
  • $\begingroup$ I edited the text, is it clearer now? $G_N$ is a function of those $N-1$ variables. $\endgroup$ – Ziofil Aug 26 '14 at 2:33
  • $\begingroup$ I assume this question stems from working on math.stackexchange.com/questions/720114 ? I like this version better, but I'm not sure I can make anymore progress on it than on the other one. $\endgroup$ – David E Speyer Aug 26 '14 at 15:34
  • $\begingroup$ Ah, I had not seen that one! Although here I'm not restricting the $x_i$'s to be integers, they can be any positive number. $\endgroup$ – Ziofil Aug 26 '14 at 16:03
  • $\begingroup$ @DavidSpeyer, for N=2 using the fact that $-4(x_1^2\partial_{x_1^2}+x_2^2\partial_{x_2^2})(\sqrt{s_1}G_2)=\sqrt{s_1}G_2$ I can write a differential equation for $G_2$, which I solve and I obtain $G_2(z)=\sqrt{1+2\sqrt{z}}$, which is exactly what is should be. It seems plausible that by using the higher-order operators one could obtain higher-order differential equations for $G_N$ and keep reducing the number of arguments until one finds a final solution. This is highly not trivial though. $\endgroup$ – Ziofil Aug 27 '14 at 19:36

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