Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

up vote 12 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 $\pm \infty$.

share|improve this answer
    
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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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