I am currently stuck with the following question:

Let $q$ be a polynomial of degree $n+1$ with distinct positive zeros $x_0, ... , x_n$. Find a polynomial $p \in P_n$ that satisfies the functional equation $q(x)p(x) + q(-x)p(-x) = 1$ for all $x$ in $\Re$. Is such a polynomial unique?

Would appreciate your kind explanations! Thank you!

  • 1
    $\begingroup$ You can solve the problem by looking at the smallest (or largest coefficient first) and then working your way up (or down). For example, by setting $x=0$ you get $p(0) = 1/2q(0)$. So you can try to work the coefficients of $p$ one by one. For an abstract argument, consider $a_i$ to be the coefficients of $p$ (you have $n+1$ of them). Then $F(x) = q(x)p(x)$ has degree $2n+1$. When looking at $F(x) + F(-x)$ all monomials of odd degree cancel. There remains $n+1$ even degree monomials. So $n+1$ linear constraints and $n+1$ variables. There is a solution (use distinct roots for uniqueness). $\endgroup$ – ARG May 9 at 10:55

Nice exercise, though I doubt that it can be considered as a research question. Put $y=x^2$, and write $q(x)= a(y)+xb(y)$. I claim that the polynomials $a$ and $b$ are coprime: if a non-constant polynomial $f$ divides $a$ and $b$, then $f(x^2)$ divides $q(x)$, but that implies that some of the roots $x_i$ of $q$ are negative. Note also that since the roots are positive $y$ does not divide $a$. Therefore there exist polynomials $c,d$ such that $a(y)c(y)+yb(y)d(y)=\dfrac{1}{2} $. Put $p(x)=c(x^2)+xd(x^2)$; then $q(x)p(x)=\dfrac{1}{2}+ x g(x^2) $, with $g= ad+bc$, so $q$ answers the question.

As usual you can replace $c(y),d(y)$ by $c(y)-yb(y)e(y), d(y)+ a(y)e(y)$ for some polynomial $e$, so the solution is far from unique.

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  • $\begingroup$ Thank you! I have understood! $\endgroup$ – user157299 May 6 at 14:08
  • $\begingroup$ Out of curiosity, are we able to use Lagrange interpolation formula for this question? @abx $\endgroup$ – user157299 May 6 at 14:29
  • $\begingroup$ I don't think so. For one thing, there are many solutions for $p$, with different zeroes. $\endgroup$ – abx May 7 at 15:20

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