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correct typo: $p \leq 0$ (not $p \geq 0$) for $n=5$
Noam D. Elkies
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The conjecture for $n=4$ is correct. For general $n$ and any $x_1,\ldots,x_n$, $$ f(p) := \sum_{i=1}^{n}\dfrac{x^p_{i}}{\prod_{j\neq i}(x_{i}-x_{j})} $$ is positive for $p > n-2$ and switches sign at $p=0,1,2,\ldots,n-2$ and nowhere else. For example:
If $n=5$ then $f(p) \geq 0$ iff $p \geq 3$ or $1 \leq p \leq 2$ (or $p \leq 0$);
if $n=6$ then $f(p) \geq 0$ for $p \in [0,1]$, $[2,3]$, or $[4,\infty)$;
if $n=7$ then $f(p) \geq 0$ for $p \in (-\infty,0]$, $[1,2]$, $[3,4]$, or $[5,\infty)$;
etc.

The key fact is the following exponential analogue of "Descartes' law of sines":

Proposition. Let $x_1,\ldots,x_n$ be pairwise distinct positive real numbers, and $c_i$ any real numbers not all zero. Then the function $f(p) := \sum_{i=1}^n c_i^{\phantom.} x_i^p$ has at most $n-1$ real zeros, counted with multiplicity.

Proof: Induction on $n$. For $n=1$ the function $f$ is a nonzero multiple of $x_1^p$, which is never zero; this establishes the base case. Now let $n>1$, and assume we have proved the case $n-1$ of the Proposition. Let $f_1(p) = x_n^{-p} f(p) = \sum_{i=1}^n c_i (x_i/x_n)^p$, which is of the same form as $f$ with $x_n=1$ and has the same zeros and multiplicities. Then the inductive hypothesis applies to the derivative $$ f'_1(p) = \sum_{i=1}^{n-1} (c_i \log (x_i/x_n)) \, (x_i/x_n)^p; $$ so $f'_1$ has at most $n-2$ zeros, counted with multiplicity. By Rolle's Theorem it follows that $f_1$ has at most $n-1$ zeros, counted with multiplicity. This completes the induction step and the proof. QED

It remains to check that our $f$ vanishes at $p=0,1,\ldots,n-2$; then we can use our Proposition to deduce that each of these $n-1$ zeros is simple and there are no others. This can be done by recognizing $f(p)$ as the sum of the residues of the differential $x^p \, dx \left/ \prod_{j=1}^n (x-x_j) \right.$ (which is regular at $x=\infty$ for integers $p \leq n-2$). Alternatively we can relate $f(p)$ to a Schur-Vandermonde determinant with a repeated row.

Noam D. Elkies
  • 79.9k
  • 15
  • 281
  • 376