I trying to figure out whether a simpler form of this series exists.

$$\sum_{i=0}^{n-2}\frac{L_{i+1}(-x)-L_{i}(-x)}{i+2}\left(\sum_{k=0}^{n-2-i} L_k(x)\right)$$

$L_n(x)$ is the $n$th Laguerre polynomial.

Wikipedia offers several useful recurrence relations in terms of the associated Laguerre polynomials. Two of which show that $$L_n^{(\alpha+1)}{x}=\sum_{i=0}^n{L_i^{(\alpha)}(x)}$$ and $$L_n^{(\alpha)}(x)=L_n^{(\alpha+1)}(x)-L_{n-1}^{(\alpha+1)}(x)$$ These allow us to simplify the expression a bit to $$\sum_{i=0}^{n-2}\frac{L_{i+1}^{(-1)}(-x)L_{n-2-i}^{(1)}(x)}{i+2}$$

Wikipedia also gives us the relation $$L_n^{(\alpha+\beta+1)}(x+y)=\sum_{i=0}^nL_{i}^{(\alpha)}(x)L_{n-i}^{(\beta)}(y)$$ Removing the denominator, our expression would completely reduce to $$\sum_{i=0}^{n-2}L_{i+1}^{(-1)}(-x)L_{n-2-i}^{(1)}(x)=L_{n-2}^{(1)}(0)=n-1$$ This is where I'm stuck. How might we evaluate it considering the denominator in the sum? Or is there some other way we can simplify the expression?


Let we set $N=n-2$. The generating function of Laguerre's polynomials is: $$ \sum_{n\geq 0}L_n(x)\,t^n = \frac{1}{1-t}\,\exp\left(-\frac{tx}{1-t}\right)\tag{1} $$ hence: $$ \sum_{n\geq 0}\left(\sum_{k=0}^n L_k(x)\right)\,t^n = \frac{1}{(1-t)^2}\,\exp\left(-\frac{tx}{1-t}\right)\tag{2} $$ $$ \sum_{n\geq 0}L_n(-x)\,t^n = \frac{1}{1-t}\,\exp\left(\frac{tx}{1-t}\right)\tag{3} $$ $$ \sum_{n\geq 0}\left(L_{n+1}(-x)-L_{n}(-x)\right)t^{n+1} = t\exp\left(\frac{tx}{1-t}\right)-t(1+x)\tag{4} $$ Now we may replace $t$ with $tu$ in $(4)$ and integrate both sides over $(0,1)$ with respect to $u$ to get: $$ \sum_{n\geq 0}\frac{L_{n+1}(-x)-L_n(-x)}{n+2}\,t^{n} \\= \frac{2-4t+2t^2-t^2x^2-t^2 x^3}{2t^2 x^2}+\frac{(1-t)(tx+t-1)}{t^2 x^2}\,\exp\left(\frac{tx}{1-t}\right)\tag{5}$$ and now: $$ \sum_{i=0}^{N}\frac{L_{i+1}(-x)-L_i(-x)}{i+2}\left(\sum_{k=0}^{N-i}L_k(x)\right)\tag{6a}$$ is the coefficient of $x^N$ in the product between the RHS of $(2)$ and the RHS of $(5)$.
It can be computed through $(1)$ with a bit of patience.

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  • $\begingroup$ So is 6c the final answer? My calculations don't agree. $\endgroup$ – tyobrien Jun 17 '16 at 16:38
  • $\begingroup$ @TyO'Brien: if I didn't mess up with shifts, the final answer should be $-L_{n+2}(x)$, exactly. If you show your computations, it is easier for me to spot the mistake in the above lines, if there is. Anyway, the above argument should be easy to follow and fix, if necessary. $\endgroup$ – Jack D'Aurizio Jun 17 '16 at 16:58
  • $\begingroup$ View functions 3,4, and 5 on desmos.com/calculator/x5bp91l3ub $\endgroup$ – tyobrien Jun 17 '16 at 17:15
  • $\begingroup$ @TyO'Brien: by "your computations" I meant what do you think is wrong in the above lines, plots are not very useful to spot the mistake, they just suggest there is one. $\endgroup$ – Jack D'Aurizio Jun 17 '16 at 17:22
  • $\begingroup$ @TyO'Brien: mistake spotted and fixed. $\endgroup$ – Jack D'Aurizio Jun 17 '16 at 17:56

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