The roots of Hermite polynomials are all real?

The Hermite polynomials are defined as $$H_n(x)=(-1)^n e^{x^2}\dfrac{d^n}{dx^n}e^{-x^2}.$$

How does one prove that all the roots of the Hermite polynomial are real?

Solution 1:

By induction, $H_n$ is a polynomial of degree $n$. Its roots are the zeros of $u_n(x) = \frac{d^n}{dx^n} e^{-x^2}$. But by Rolle's theorem, between any two zeros of a differentiable function there is a zero of its derivative. The same is true between a zero and $+\infty$ or $-\infty$ for a function that goes to $0$ at $\pm \infty$.

Solution 2:

The high-minded, linear algebraic route involves deriving the recursion relation

$$\hat{H}_{n+1}(x)=x\hat{H}_n(x)-\frac{n}{2}\hat{H}_{n-1}(x)$$

for the monic Hermite polynomial $\hat{H}_n(x)=2^{-n}H_n(x)$ (that is, the polynomial normalized to have unit leading coefficient), and from this derive the symmetric tridiagonal Jacobi matrix

$$\begin{pmatrix}0&\sqrt{\frac12}&&&\\\sqrt{\frac12}&0&\sqrt{\frac22}&&\\&\sqrt{\frac22}&\ddots&\ddots&\\&&\ddots&\ddots&\sqrt{\frac{n-1}2}\\&&&\sqrt{\frac{n-1}2}&0\end{pmatrix}$$

whose characteristic polynomial is $\hat{H}_n(x)$. Show that the eigenvalues of a symmetric matrix are all real, and you're done.