Hermite polynomials recurrence relation

Question

Hermite polynomials $H_n (x)$ can be obtained using the recurrence relation $$H_{n+1} (x)=2xH_n (x)-2nH_{n-1} (x).$$ To prove this, I started by calculating the first derivative of the Hermite's Rodrigues formula $H_n (x)=(-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2 } $. The process goes like this: $$ \frac{d}{dx}H_n (x)=(-1)^n 2xe^{x^2}\frac{d^n}{dx^n} e^{-x^2 }+(-1)^n e^{x^2} \frac{d^{n+1}}{dx^{n+1}}e^{-x^2 } $$ Rearranging the terms in the previous equation produces $$H_{n+1} (x)=2xH_n (x)-\frac{d}{dx}H_n (x)$$ or $$H_{n+1} (x)=2xH_n (x)-H_n'(x).$$

That's it. Im stuck here. I don't know how to show that $H_n' (x)=2nH_{n-1} (x)$. Can you help me?

@FelixMarin He has to prove the last recurrence to prove the first recurrence...so he cannot compare... — alexjo, Nov 27 '13 at 11:36

alexjo · Accepted Answer · 2013-11-27T10:51:33.940

21

Let’s write $D = \frac{\operatorname{d}}{\operatorname{d}x}$, and take the derivative of $H_n$ $$ \begin{aligned}D H_n &= D \left( e^{x^2} D^n e^{-x^2} \right) \\ &= 2 x e^{x^2} D^n e^{-x^2} + e^{x^2} D^n \left( - 2 x e^{-x^2} \right) \\ &= 2 x e^{x^2} D^n e^{-x^2} + e^{x^2} \sum_{k=0}^n \binom{n}{k} D^k (-2 x) D^{n-k} e^{-x^2} \\ & \left(\text{note that $D^k(-2x)=0$ for $k\ge 2$, so $\sum_{k=0}^n=\sum_{k=0}^1$}\right)\\ &= 2 x e^{x^2} D^n e^{-x^2} + e^{x^2} \sum_{k=0}^1 \binom{n}{k} D^k (-2 x) D^{n-k} e^{-x^2}\\ &= {{2 x e^{x^2} D^n e^{-x^2} }}+ e^{x^2} \left(\underbrace{\binom{n}{0}}_1\underbrace{D^0(-2x)}_{-2x}D^{n}e^{-x^2}+\underbrace{\binom{n}{1}}_n\underbrace{D^1(-2x)}_{-2}D^{n-1}e^{-x^2} \right) \\ &= \color{red}{{2 x e^{x^2} D^n e^{-x^2} }}+ e^{x^2} \left(\color{red}{-{2 x D^{n} e^{-x^2} }}- 2 n D^{n-1} e^{-x^2} \right) \end{aligned} $$ So we have the rather simple end result $$\frac{\operatorname{d}}{\operatorname{d}x} H_n(x)=2 n H_{n-1}(x)$$

edited Nov 27 '13 at 10:51

answered Nov 26 '13 at 14:00

alexjo

14,976

I didnt get this portion of the above proof youve given me...∑k=0n(nk)Dk(−2x)Dn−ke−x2 – Glenio Rosario Nov 26 '13 at 14:14
1

It's the General Leibniz rule $(fg)^{(n)} = \sum_{k = 0}^n {n\choose k}f^{(k)}g^{(n-k)}$ or with the $D$ operator $D^{(n)}(fg) = \sum_{k = 0}^n {n\choose k}D^{(k)}fD^{(n-k)}g$ – alexjo Nov 26 '13 at 14:20
When i tried to solve it, i obtained −2xDne−x2−2xDn−1e−x2. How did 2n appeared there? – Glenio Rosario Nov 27 '13 at 08:52
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It comes from the binomial $\sum_{k=0}^1 \binom{n}{k} D^k (-2 x) D^{n-k} e^{-x^2}=\underbrace{\binom{n}{0}}1\underbrace{D^0(-2x)}{-2x}D^ne^{-x^2}+ \underbrace{\binom{n}{1}}{n}\underbrace{D^1(-2x)}{-2}D^{n-1} e^{-x^2}$ – alexjo Nov 27 '13 at 09:19
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Aren't you missing the prefactor $(-1)^{n}$ of $H_n$? – ViktorStein May 27 '21 at 16:24

score 4 · Answer 2 · answered Feb 20 '22 at 14:30

$H_n'(x) = 2n H_{n-1}(x)$ can be proved by induction. The base case $H_1'(x) = 2H_0(x)$ is straightforward to verify. Now, assume $H_k'(x) = 2k H_{k-1}(x)$ and we will show $H_{k+1}'(x) = 2(k+1) H_{k}(x)$. Taking derivatives on both sides of the recurrence relation $H_{k+1}(x) = 2xH_k(x)-H_k'(x)$, one has $$ H_{k+1}'(x) = 2H_k(x) + 2xH_k'(x) - H_{k}''(x). $$ Using the inductive hypothesis $H_k'(x) = 2k H_{k-1}(x)$, we have \begin{aligned} H_{k+1}'(x) &= 2H_k(x) + 4kxH_{k-1}(x) - 2kH_{k-1}'(x) \\ & = 2H_k(x) + 2k[2xH_{k-1}(x) - H_{k-1}'(x)] \\ &= 2H_k(x) + 2kH_k(x) \\ &= 2(k+1)H_k(x). \end{aligned}

Hermite polynomials recurrence relation

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