Prove that $f_n$ has $n$ distinct roots which are symmetric about $0$

Question

Given $f_0(x)\equiv 1$ and define $f_n$ recursively via $$f_{n+1}(x)=xf_n(x)-f'_n(x).$$ Prove that $f_n$ is a polynomial of degree $n$. Further prove that $f_n$ has $n$ distinct real roots which are symmetric about $0$ (by symmetric we mean that if $a$ is a root of $f_n$, then so is $-a$).

Work so far: Proving that $f_n$ is a degree $n$ polynomial is rather easy, this can be done by induction. The difficult part comes from the second part. To show that $f_n$ has $n$distinct roots, I first multiply $e^{-x^2/2}$ on both sides of the recursive equation, and this yields $$e^{-\frac{x^2}{2}}f_{n+1}(x)=(e^{-\frac{x^2}{2}}f_n(x))'.$$ Put $g_n(x):=e^{-\frac{x^2}{2}}f_n(x)$, we see that $g_n$ has a formular $$g_n(x)=\frac{d^n}{dx^n}g_0(x)=\frac{d^n}{dx^n}e^{-\frac{x^2}{2}}.$$ Thus the question now is to prove that $g_n$ has $n$ distinct real roots which are symmetric.

To show the roots are symmetric, we may show that the function is symmetric(either about the $y-$axis or the origion). What I really have trouble on is to prove $n$ distinct real roots. I'm not sure if this approach really works. Any suggestions?

Answer 1

[Hint: Show that $f_n$ is a polynomial with terms either all in odd powers or even powers; i.e., either odd or even.]

From that I guess you can probably see the symmetry.

To show that there are indeed $n$ roots for $f_n$, we use the induction again:

1. Base Case: trivial.
2. Inductive Step: Suppose $n=k$ the claim is true. Need: $xf_k(x) = f'_k(x)$ happens exactly $k+1$ times. Consider $g_k(x) = f_k(x) e^{-\frac{x^2}{2}}$. Observation: (i) $g'_k(x) = \left( -xf_k\left( x \right) +f'_k(x)\right )e^{-\frac{x^2}{2}} = 0$ if and only if $xf_k(x) = f'_k(x)$; (ii) the number of roots of $g_k$ is exacty $k$. Now, $g_k$ tends to $0$ as $x \rightarrow \pm \infty$. In between the two "furthest" roots of $g_k$ (or, $f_k$), $g'_k = 0$ happens exactly $k-1$ times. How about the two infinite intervals outside? ;) What can you conclude by invoking its continuity and observing its limit?

Answer 2

A "generic" way to do this (for arbitrary sequences of orthogonal polynomials, not just the Hermite family whose Rodrigues formula you've described) is to

• derive the recurrence relation,
• prove the Christoffel–Darboux formula from the recurrence relation, and
• use the Christoffel–Darboux relation and Sturm chain considerations to show that the roots of adjacent orthogonal polynomials interlace.

See Sec. 4.2, 4.3 and Thm. 4.10 in https://cel.archives-ouvertes.fr/cel-00661847/document