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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?

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up vote 16 down vote accepted

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 $\pm \infty$.

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This is a great answer! – rotten Feb 2 '12 at 6:48
@Robert Israel I will complete the answer in the following:Assume the result is right when k=n-1, that is $u_{n-1}(x)$ has n-1 real roots.Then consider when k=n, by Rolle's theorem, there are n-2 real roots which satisfies u_n(x)=0, name them $$a_1<a_1<...<a_{n-2}$$ Then, notice that ±∞ are also the roots.Use Rolle's theorem again, there are one real root between −∞ and $a_1$, one real root between $a_{n-2}$ and +∞. So we have already found n real roots. Given H_n is a polynomial of degree n, these are all roots of H_n. – 89085731 Feb 2 '12 at 7:06
Right. [comments must be at least 15 characters in length] – Robert Israel Feb 2 '12 at 7:07
This is also one route to show that the Hermite polynomials form a Sturm sequence. – J. M. Feb 3 '12 at 4:06

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


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


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

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