3
$\begingroup$

How can I prove that for real numbers $x_k$ and $y_k$,

$x_1^k+x_2^k+ \dots + x_n^k=y_1^k+y_2^k+ \dots + y_n^k$ $(k=1,2,\dots,n)$,

where $x_1\le x_2 \le \dots \le x_n$ and $ y_1\le y_2 \le \dots \le y_n$

implies $x_k=y_k (k=1,2,\dots,n)$?

I considered the polynomial with roots $x_k$, $(t-x_1)(t-x_2)\dots(t-x_n) $and its coefficients, but could not proceed.

$\endgroup$
  • $\begingroup$ Are the numbers supposed to be nonnegative? Otherwise $1^2=(-1)^2$ is a counterexample. $\endgroup$ – Jose Brox Jun 19 '18 at 14:17
  • $\begingroup$ If you have a single number, the only possible value for $k$ is $1$. $\endgroup$ – José Carlos Santos Jun 19 '18 at 14:20
  • $\begingroup$ @JoséCarlosSantos Oh, I see now that the equation is meant for all $k$ up to $n$. $\endgroup$ – Jose Brox Jun 19 '18 at 14:26
2
$\begingroup$

It follows from Newton's identities and from your hypothesis that, if $\sigma_1,\ldots,\sigma_n$ are the elementary symmetric polynomials, then$$(\forall k\in\{1,\ldots,n\}):\sigma_k(y_1,\ldots,x_n)=\sigma_k(y_1,\ldots,y_n).$$Therefore$$(X-x_1)\ldots(X-x_n)=(X-y_1)\ldots(X-y_n),$$and this means that $\{x_1,\ldots,x_n\}=\{y_1,\ldots,y_n\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.