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I am trying to show that the Laguerre differential equation, given in my homework problem as

$xL''_n(x)+(1−x)L'_n(x)+ nL_n(x) = 0$,

is indeed solved by the Laguerre polynomials in their closed sum form:

$L_n(x)=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} x^k$

I've tried substituting the sum into the equation but I haven't been able to resolve it:

$xL''_n(x)+(1−x)L'_n(x)+ nL_n(x)$
$ = x \sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!}k(k-1) x^{k-2} + (1-x)\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!}k x^{k-1} +n\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} x^k$
$=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} [xk(k-1)x^{k-2}+(1-x)kx^{k-1}+nx^k]$
$=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} [(k^2-k)x^{k-1}+kx^{k-1}-kx^k+nx^k]$
$=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} [k^2 x^{k-1}+(n-k)x^k]$
$=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} [x^{k-1}(k^2+(n-k)x)]$

What am I doing wrong? How would I show that this equals zero? Or is this the completely wrong approach? I've seen various similar questions here but they always either work with a different version of the differential equation, or with recurrence relations - which I am hesitant to use, since my homework problem has not introduced the recurrence properties of the Laguerre polynomials so I'd have to derive and prove them before being allowed to use it in my proof.

Beyond the closed sum form I am also given the Rodrigues form of the polynomials:

${\displaystyle L_{n}(x)={\frac {\mathrm {e} ^{x}}{n!}}{\frac {\mathrm {d} ^{n}}{\mathrm {d} x^{n}}}{\bigg (}x^{n}\mathrm {e} ^{-x}{\bigg )}}$

But substituting those into the differential equation only leads me to terms with $L_{n+1}$ and $L_{n+2}$.

Can anybody tell me how to approach this? Is there any way to prove that the polynomials in either of these two forms satisfy the above differential equation? Or is there no way around the recurrence relations and shall I prove these first, even though they were mentioned neither in my lecture nor the homework problem?

Trang Nguyen
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2 Answers2

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We have $$x{{L}_{n}}''\left( x \right)+\left( 1-x \right){{L}_{n}}'\left( x \right)+n{{L}_{n}}\left( x \right)=0$$ where $${{L}_{n}}\left( x \right)=\sum\limits_{k=0}^{n}{\left( \begin{matrix} n \\ k \\ \end{matrix} \right)\frac{{{\left( -1 \right)}^{k}}}{k!}{{x}^{k}}}$$

As you have done, substitute to get $$\sum\limits_{k=0}^{n}{\left( \begin{matrix} n \\ k \\ \end{matrix} \right)\frac{{{\left( -1 \right)}^{k}}k\left( k-1 \right)}{k!}{{x}^{k-1}}}+\left( 1-x \right)\sum\limits_{k=0}^{n}{\left( \begin{matrix} n \\ k \\ \end{matrix} \right)\frac{{{\left( -1 \right)}^{k}}k}{k!}{{x}^{k-1}}}+n\sum\limits_{k=0}^{n}{\left( \begin{matrix} n \\ k \\ \end{matrix} \right)\frac{{{\left( -1 \right)}^{k}}}{k!}{{x}^{k}}}=0$$

Then cancel terms and collect $$\sum\limits_{k=0}^{n}{\left( \begin{matrix} n \\ k \\ \end{matrix} \right)\frac{{{\left( -1 \right)}^{k}}{{k}^{2}}}{k!}{{x}^{k-1}}}+\sum\limits_{k=0}^{n}{\left( \begin{matrix} n \\ k \\ \end{matrix} \right)\left( n-k \right)\frac{{{\left( -1 \right)}^{k}}}{k!}{{x}^{k}}}=0$$

Note that the first term in the first sum is zero so re-index, and collect terms $$\sum\limits_{k=0}^{n}{\left\{ \left( \begin{matrix} n \\ k+1 \\ \end{matrix} \right)\frac{{{\left( k+1 \right)}^{2}}}{\left( k+1 \right)!}-\left( \begin{matrix} n \\ k \\ \end{matrix} \right)\left( n-k \right)\frac{1}{k!} \right\}{{\left( -1 \right)}^{k}}{{x}^{k}}}=0$$ It’s very easy using the definition of the binomial coefficients to show $$\left( \begin{matrix} n \\ k+1 \\ \end{matrix} \right)\left( k+1 \right)-\left( \begin{matrix} n \\ k \\ \end{matrix} \right)\left( n-k \right)=0$$

mathstackuser12
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1

I shouldn't do your homework for you, but I'll verify it the fast way, which your teacher appears to hint at: the Sheffer sequence. See wikipedia.

The point is it's much easier to use operator (Heaviside) calculus, and the tasteful Rodrigues representation $$ L_{n}(x)={\frac { e ^{x}}{n!}} \partial^n ( e ^{-x} x^n )= {\frac { (\partial-1)^n } {n!}}~~ x^n. $$ The equality holds by virtue of (NB operators acting on everything to their right), $$ \partial ~~ e^{-x} f(x) = e^{-x} (\partial-1) f(x) ~, $$ hence,
$$ e^x \partial ~ e^{-x} = \partial-1, ~~~\leadsto ~~~ e^x \partial^n ~ e^{-x} = (\partial-1)^n~, $$ by intercalating $e^{-x} e^x=1$s among powers of the gradient.

Now, likewise, easily show the identity (recall the commutator of quantum mechanics!) $$ \bbox[yellow]{x(\partial-1)^k = (\partial-1)^k x - k(\partial-1)^{k-1}}. $$

We then, for $n>1$, plug this representation into Laguerre's second order linear equation, multiplied by $(n-1)!$, $$ {1\over n} x (\partial -1)^n (\partial^2 x^n -\partial x^n) + {1\over n} (\partial -1)^n (\partial x^n + n x^n )\\ = x (\partial -1)^{n+1} x^{n-1} + (\partial -1)^n (x^{n-1}+ x^n) \\ = (\partial -1)^{n+1}(x x^{n-1}) -(n+1) (\partial -1)^{n} x^{n-1} + (\partial -1)^n (x^{n-1}+ x^n) \\ = (\partial -1)^{n} \Bigl((\partial -1) x^n -(n+1) x^{n-1} +x^{n-1}+ x^n\Bigr )=0. $$ In the penultimate line, we used the above identity for $k=n+1$.

So, the Rodrigues expression satisfies the Laguerre equation, alright. Heaviside calculus is something effective you should be able to use, anyway.

Cosmas Zachos
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