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Hermite Polynomials' Recursive Relations 📂Functions

Hermite Polynomials' Recursive Relations

Theorem

The Hermite polynomials satisfy the following recursive relation.

Hn′(x)=2nHn−1(x)Hn+1(x)=2xHn(x)−2nHn−1(x)=2xHn(x)−Hn′(x) \begin{align} H_{n}^{\prime}(x) &= 2nH_{n-1}(x) \\ H_{n+1}(x) &= 2xH_{n}(x)-2nH_{n-1}(x) \\ &= 2xH_{n}(x)-H_{n}^{\prime}(x) \nonumber \end{align}

Proof

(1)(1)

Solving using the Generating Function

Generating function of the Hermite polynomials

Φ(x,t)=e2xt−t2=∑n=0∞Hn(x)tnn! \Phi (x,t) = e^{2xt-t^{2}}=\sum \limits _{n=0}^{\infty} H_{n}(x)\frac{t^{n}}{n!}

Differentiating the generating function of the Hermite polynomials gives,

2te2xt−t2=∑n=0∞Hn′(x)tnn! 2te^{2xt-t^{2}} = \sum \limits _{n=0}^{\infty}H_{n}^{\prime}(x)\frac{t^{n}}{n!}

Then, the left side, by the definition of the generating function, is,

2t∑n=0∞Hn(x)tnn!=2te2xt−t2=∑n=0∞Hn′(x)tnn! 2t\sum \limits _{n=0}^{\infty} H_{n}(x)\frac{t^{n}}{n!}=2te^{2xt-t^{2}}=\sum \limits _{n=0}^{\infty}H_{n}^{\prime}(x)\frac{t^{n}}{n!}

Rearranging gives,

∑n=0∞Hn′(x)tnn!=∑n=0∞2tHn(x)tnn!=∑n=0∞2(n+1)Hn(x)tn+1(n+1)! \begin{align*} \sum \limits _{n=0}^{\infty}H_{n}^{\prime}(x)\frac{t^{n}}{n!} &= \sum \limits _{n=0}^{\infty} 2tH_{n}(x)\frac{t^{n}}{n!} \\ &= \sum \limits _{n=0}^{\infty}2(n+1)H_{n}(x)\frac{t^{n+1}}{(n+1)!} \end{align*}

Comparing the coefficients of tnn!\dfrac{t^{n}}{n!} in both sides gives,

Hn′(x)=2nHn−1(x) H_{n}^{\prime}(x)=2nH_{n-1}(x)

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Solving using the Differential Operator

Hermite polynomials

Hn(x)=(−1)nex2dndxne−x2 H_{n}(x)=(-1)^{n}e^{x^{2}}\frac{ d ^{n}}{ dx^{n} }e^{-x^{2}}

Let’s denote the differential operator as D=ddxD = \dfrac{d }{dx}. Differentiating the Hermite polynomial once gives,

DHn(x)=D[(−1)nex2Dne−x2]=(−1)n2xex2Dne−x2+(−1)nex2Dn+1e−x2=(−1)n2xex2Dne−x2+(−1)nex2Dn[(−2x)e−x2] \begin{align*} DH_{n}(x) &= D\left[ (-1)^{n}e^{x^{2}}D^{n}e^{-x^{2}} \right] \\ &= (-1)^{n}2x e^{x^{2}}D^{n}e^{-x^{2}}+(-1)^{n}e^{x^{2}}D^{n+1}e^{-x^{2}} \\ &= (-1)^{n}2x e^{x^{2}}D^{n}e^{-x^{2}}+(-1)^{n}e^{x^{2}}D^{n}\left[ (-2x)e^{-x^{2}}\right] \end{align*}

Applying the Leibniz rule to the second term gives DHn(x)=(−1)n2xex2Dne−x2+(−1)nex2∑k=0nn!(n−k)!k!Dk(−2x)Dn−ke−x2=(−1)n2xex2Dne−x2+(−1)nex2∑k=01n!(n−k)!k!Dk(−2x)Dn−ke−x2=(−1)n2xex2Dne−x2+(−1)nex2[−2xDne−x2−2nDn−1e−x2]=2n(−1)n+1ex2Dn−1e−x2=2n(−1)n−1ex2Dn−1e−x2=2nHn(x) \begin{align*} DH_{n}(x) &= (-1)^{n}2x e^{x^{2}}D^{n}e^{-x^{2}}+(-1)^{n}e^{x^{2}}\sum \limits _{k=0}^{n}\frac{n!}{(n-k)!k!}D^{k}(-2x)D^{n-k}e^{-x^{2}} \\ &= (-1)^{n}2x e^{x^{2}}D^{n}e^{-x^{2}}+(-1)^{n}e^{x^{2}}\sum \limits _{k=0}^{1}\frac{n!}{(n-k)!k!}D^{k}(-2x)D^{n-k}e^{-x^{2}} \\ &= (-1)^{n}2x e^{x^{2}}D^{n}e^{-x^{2}}+(-1)^{n}e^{x^{2}}\left[ -2xD^{n}e^{-x^{2}}-2nD^{n-1}e^{-x^{2}} \right] \\ &= 2n(-1)^{n+1}e^{x^{2}}D^{n-1}e^{-x^{2}} \\ &= 2n(-1)^{n-1}e^{x^{2}}D^{n-1}e^{-x^{2}} \\ &=2n H_{n}(x) \end{align*} The second equality is valid when k≥2k\ge 2 is Dk(−2x)=0D^{k}(-2x)=0.

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(2)(2)

The proof follows the same methodology as for (1)(1). Differentiating the generating function for tt gives,

(2x−2t)∑n=0∞Hn(x)tnn!=(2x−2t)e2xt−t2=∑n=0∞Hn(x)tn−1(n−1)! (2x-2t)\sum \limits _{n=0}^{\infty}H_{n}(x)\frac{t^{n}}{n!}=(2x-2t)e^{2xt-t^{2}}=\sum \limits _{n=0}^{\infty}H_{n}(x)\frac{t^{n-1}}{(n-1)!}

Reorganizing gives,

∑n=0∞Hn(x)tn−1(n−1)!=∑n=0∞[2xHn(x)−2tHn(x)]tnn! \sum \limits _{n=0}^{\infty}H_{n}(x)\frac{t^{n-1}}{(n-1)!}=\sum \limits _{n=0}^{\infty}\left[2xH_{n}(x)-2tH_{n}(x)\right]\frac{t^{n}}{n!}

Comparing the coefficients of tnn!\dfrac{t^{n}}{n!} on both sides gives,

Hn+1(x)=2xHn(x)−2tHn(x) H_{n+1}(x)=2xH_{n}(x)-2tH_{n}(x)

Furthermore, by (1)(1) the following is true,

Hn+1(x)=2xHn(x)−Hn′(x) H_{n+1}(x)=2xH_{n}(x)-H_{n}^{\prime}(x)

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