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Derive Rodrigues’ formula for Laguerre polynomials $$ L_n(x)=\frac{e^x}{n!}.\frac{d^n}{dx^n}(x^ne^{-x}) $$

The Rodrigues’ formula for Hermite polynomials can be obtained by taking $n^{th}$ order partial derivatives of its generatig function $$ g(x,t)=\sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}=1+tH_1(x)+\frac{t^2}{2!}H_2(x)+\cdots\cdots\cdots+\frac{t^n}{n!}H_n(x)+\cdots\cdots\cdots\\ \frac{\partial^n}{\partial t^n}\Big(e^{2xt-t^2}\Big)=H_n(x)+\frac{(n+1)n(n-1)\cdots2}{(n+1)!}tH_{n+1}(x)+\cdots\\ H_n(x)=\Bigg[\frac{\partial^n}{\partial t^n}\Big(e^{2xt-t^2}\Big)\Bigg]_{t=0}=e^{x^2}\Bigg[\frac{\partial^n}{\partial t^n}\Big(e^{-(x-t)^2}\Big)\Bigg]_{t=0}\\ =(-1)^ne^{x^2}\Bigg[\frac{\partial^n}{\partial x^n}\Big(e^{-(x-t)^2}\Big)\Bigg]_{t=0}=(-1)^ne^{x^2}\Bigg[\frac{\partial^n}{\partial x^n}\Big(e^{-x^2}\Big)\Bigg] $$

Generating function for laguerre polynomials is $g(x,t)=\dfrac{e^{-\frac{xt}{1-t}}}{1-t}=\sum_{n=0}^\infty L_n(x) t^n$

I do not think the same technique applies in the case of Laguerre polynomials. So how do I derive that for Laguerre polynomials ?

GNUSupporter 8964民主女神 地下教會
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Sooraj S
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2 Answers2

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Note that the Taylor-Maclaurin formula cannot be used directly (as in the case of Hermite polynomials) as there is some $n$ dependency inside the $n^{th}$ derivative. How can we get around this ? ... Let use a coefficient extrator \begin{eqnarray*} x^n= [u^0]: \frac{u^{-n}}{1-xu}. \end{eqnarray*} We have \begin{eqnarray*} \sum_{n=0}^{\infty} L_n(x) t^n &=&[u^0]: e^x \sum_{n=0}^{\infty} \left(\frac{t}{u} \right)^n \frac{1}{n!} \frac{d^n}{dx^n} \frac{e^{-x}}{1-xu} \\ &=&[u^0]: e^x \frac{e^{-(x+t/u)}}{1-u(x+t/u)} \\ &=& \frac{1}{1-t} [u^0]: \frac{e^{-t/u}}{1-\frac{ux}{1-t}}. \\ \end{eqnarray*} Expand these two functions and observe that the central term is ... what we want \begin{eqnarray*} \left( \sum_{i=0}^{\infty} \frac{(-t/u)^i}{i!} \right) \left( \sum_{j=0}^{\infty} \left( \frac{ux}{1-t} \right)^j \right) = \cdots+ \sum_{k=0}^{\infty} \frac{1}{k!} \left( \frac{-xt}{1-t} \right)^k + \cdots \end{eqnarray*} Now reverse engineer all this ... and your result follows.

Donald Splutterwit
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  • Could you please elaborate a little more as I have difficulty in following your approach. The coefficient corresponds to $u^0$ in the expansion of $\dfrac{u^{-n}}{1-xu}$ is $\frac{1}{n!}x^n$, right ? – Sooraj S Dec 23 '20 at 06:46
  • The coefficient of $u^0$ in \begin{eqnarray} u^{-n} \left( 1+xu+x^2u^2+\cdots \color{red}{x^n} u^n +\cdots \right) \end{eqnarray} would be $x^n$. (not sure where the $1/n!$ came from ?) The coefficient of $u^0$ in $u^{-n} e^{xu}$ would be $x^n/n!$ – Donald Splutterwit Dec 23 '20 at 12:37
  • ohh thats my mistake, its an infinite GP right, I got confused by thinking about binomial expansion of the expression. – Sooraj S Dec 23 '20 at 14:23
  • I also have trouble understanding how the expression $[u^0]:e^x\sum_{n=0}^\infty\big(\frac{t}{u}\big)^n\frac{e^x}{n!}\frac{d^n}{dx^n}\frac{e^{-x}}{1-xu}$ becomes $[u^0]:e^x\frac{e^{-(x+t/u)}}{1-u(x+t/u)}$, rest seems fine, could you please clarify on that ? – Sooraj S Dec 23 '20 at 14:39
  • This is the Taylor-Maclaurin Series \begin{eqnarray} f(x+a)=f(x)+af'(x)+\frac{a^2}{2!}f''(x)+ \frac{a^3}{3!} f^{(3)}(x)+\cdots \end{eqnarray} see https://en.wikipedia.org/wiki/Taylor_series – Donald Splutterwit Dec 23 '20 at 14:43
  • Thanks. I just had hard time identifying the Taylor series in the summation form. – Sooraj S Dec 23 '20 at 20:54
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If you are allowed to use the explicit representation for Laguerre polynomials, $$ L_n(x) =\sum_{k=0}^n \binom{n}{k} \frac{(-x)^k}{k!} \quad ,$$ then the result follows easily from the Leibniz differentiation rule for products $$ (f(x) \ g(x) )^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x) .$$

Refer to the Rodrigues formula you want to prove. Let $f(x)=x^n$ and $g(x)=e^{-x}.$ Then $f^{(k)}(x) = x^{n-k}\ n!/(n-k)! $ and $g^{(n-k)}(x) = (-1)^{n-k}e^{-x}.$ Now clean it up and redo the sum in reverse order, i.e., $k \to n-k,$ and you'll get the top formula.

user321120
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