Deriving Hermite polynomial derivative recurrence relation straight from differential equation.

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Starting with the equation $$ y''(x) - 2x y'(x) +2n y(x) = 0, $$ you can differentiate: $$ y'''-2y'-2xy''+2ny' = 0 \\ (y')''-2x(y')'+2(n-1)(y')=0. $$ In other words, $f=y'$ satisfies $$ f''-2xf'+2(n-1)f=0, $$ which is the Hermite equation with $2n$ replaced by $2n-1$. There are two linearly independent solutions of the Hermite equation, but only one solution is a polynomial. So you are correct that $H_n'=\alpha_n H_{n-1}$ must hold for some constant $\alpha_n$. The constant $\alpha_n$ depends on normalization, which means that $\alpha_n$ is not uniquely determined by the equation; that's why you're stuck at that point.

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Arturo don Juan
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Arturo don Juan

Updated on May 30, 2020

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  • Arturo don Juan
    Arturo don Juan over 3 years

    I want to derive the derivative recurrence relation for the Hermite polynomials straight from the Hermite differential equation. That is, I want to go from left to right in the following sequence without going through the generating function, Rodrigues formula, or any other representation of the Hermite polynomials.

    $$H''_n(x)-2xH'_n(x)+2nH_n(x)=0 \longrightarrow H'_n=2nH_{n-1}$$

    Every derivation of the above identity that I've seen has been from either the generating function or Rodrigues formula. I essentially would like to find the simplest method for getting from the Hermite differential equation to that recurrence relation.

    Sorry if this sounds/looks trivial. I've been stuck on it for hours. My first instinct was to differentiate the equation, which gives

    $$H'''_n(x)-2xH''_n(x)+2(n-1)H'_n(x)=0$$

    From this, restricting ourselves to the finite-polynomial Hermite polynomials, I would say that $H'_n=\alpha H_{n-1}$, where $\alpha$ is some constant. However, I don't know how to then deduce that $\alpha=2n$. I've tried further differentiation, integration, substitution, putting it in Sturm-Liouville form, etc.

    • D.A.N.
      D.A.N. over 7 years
      It has to come from normalization right? These equations can only specify the hermite polynomials to within a multiplicative constant.
    • Arturo don Juan
      Arturo don Juan over 7 years
      @D.A.N. Yeah I think you're right. I believe that particular normalization is what defines the Hermite polynomials (as opposed to any set of orthogonal solutions to the Hermite differential equation). I know for sure that the Rodrigues formula for the Hermite polynomial is defined by that normalization, so I'd guess its the same for most other representations.