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I'm trying to show that Kummer's equation can be solved by deriving

$$xL_{n}^{''}(x)+(1-x)L_{n}^{'}(x)+nL_{n}(x)=0$$

from the Laguerre polynomials:

$$g(x,t)=\frac{e^{-xt/(1-t)}}{1-t}=\sum_{n=0}^{\infty}L_{n}(x)\frac{t^n}{n!},$$

where $0<t<1.$

I'm having trouble working this one out.

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1 Answer 1

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The first equation (or sequence of equations) follows from the second equation because $$ x g''(x,t)+(1-x)g'(x,t)+t\frac{\partial g(x,t)}{\partial t} = 0$$ where the prime indicates $\partial/\partial x$. You may just replace $g$ by $L_n$; in the last term, $t\cdot \partial/ \partial t$ picks the factor of $n$: just look how this operator acts on $t^n$.

The identity I wrote above is straightforward to prove. $$-\frac{e^{-\frac{t x}{1-t}} t (1-x)}{(1-t)^2}+\frac{e^{-\frac{t x}{1-t}} t^2 x}{(1-t)^3}+t \left(\frac{e^{-\frac{t x}{1-t}}}{(1-t)^2}+\frac{e^{-\frac{t x}{1-t}} \left(-\frac{x}{1-t}-\frac{t x}{(1-t)^2}\right)}{1-t}\right) = 0$$

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Why can you replace $g$ by $L_n$? –  Alex Dec 2 '12 at 19:25
1  
Dear Alex, the equation for $L_n$ is nothing else than a coefficient in front of $t^n$ in the Taylor expansion of the equation for $g$ with respect to $t$. You just compare the terms term-by-term. Please just expand my equation for $g$ as a Taylor expansion in $t$ and don't ask any more questions. –  Luboš Motl Dec 2 '12 at 19:41

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