calculusorthogonal-polynomialspolynomialsrecurrence-relationsspecial functions
We knew that, the generalized Laguerre polynomials $L_n^{(-1/2)} $ are related to the Hermite polynomials $H_{2n}$ by:
$$ H_{2n}(x) = (-1)^n 2^{2n} n! \, L_n^{(-1/2)} (x^2).$$
I look for a formula connecting the generalized Laguerre polynomials $L_n^{(-1)} (x)$ and the Hermite polynomials $H_{n}$.
Thank you in advance
Very roughly speaking: Laguerre polynomials look like the family of trigonometric functions ($\sin nx$ or $\cos nx$) in the region where the weight is concentrated. That is, they have moderate size and slowly increasing oscillation with $n$. Outside of this region, they look pretty much like any random collection of polynomials.
When $\alpha=0$, the weight $e^{-x}$ is concentrated near $0$. On the graph below the weight is shown in black, and the first five Laguerre polynomials $L_n$ are in various colors. You can see that they look like $1-\sin nx$ near $0$, and then wildly diverge when the weight becomes small.
For comparison, I did the same with $\alpha=2$. The polynomials have civilized appearance where the weight is not too small, roughly in the interval $[1,4]$. They wildly diverge both to the left and to the right.
do these polynomials still converge using the Least Square approximant
When $\alpha=0$, the Laguerre polynomial form on orthonormal basis of $L^2([0,\infty),e^{-x})$ (reference here). Therefore, for any function $f\in L^2([0,\infty),e^{-x})$ we have $\sum c_n L_n\to f$ in the norm, where $c_n=\langle f,L_n\rangle$.
When $\alpha>0$, the Laguerre polynomial form on orthogonal (but not normalized) basis* of $L^2([0,\infty),x^\alpha e^{-x})$. Therefore, for any function $f\in L^2([0,\infty),x^\alpha e^{-x})$ we have $\sum c_n L_n^{(\alpha)}\to f$ in the norm, where $c_n=\langle f,L_n^{(\alpha)}\rangle / \|L_n^{(\alpha)}\|$. (The value of $\|L_n^{(\alpha)}\|$ is given on Wikipedia).
(*) Disclaimer: I don't have a reference for the the completeness of $\{L_n^{(\alpha)}\}$, but I just put a bounty on unanswered question On the completeness of the generalized Laguerre polynomials in the hope someone does. Or you can consult the books G. Polya Orthogonal polynomials and G. Sansone Orthogonal functions, in case they treat $L_n^{(\alpha)}$.
Well, using the falling factorial $$ (z)_n=\prod_{k=0}^{n-1} (z-k) $$ you can write $$ L_n^{(\alpha)}(x)=\frac1{n!}\sum_{k=0}^n\frac{(-n)_k (k+\alpha+1)_{n-k}}{k!}x^k $$ which is valid for all $\alpha\in\mathbb{R}$ and all nonnegative integer $n$.
Best Answer
Now bear in mind, I've never used Laguerre polynomials before so be warned. Here's my attempt, using some recurrnce relations described on Wikipedia. Maybe there's simpler, and better, ways of doing it than mine.
https://en.wikipedia.org/wiki/Laguerre_polynomials#Recurrence_relations
$$ \begin{align} L_n^{(\alpha+\beta+1)}(x+y)&= \sum_{i=0}^n L_{i}^{(\alpha)}(x)L_{n-i}^{(\beta)}(y) \\ \Rightarrow L_n^{(1/2)}(x)&= \sum_{i=0}^n L_{i}^{(-1/2)}(x)L_{n-i}^{(0)}(0) \\ \Rightarrow L_n^{(1)}(x)&= \sum_{i=0}^n L_{i}^{(1/2)}(x/2)L_{n-i}^{(-1/2)}(x/2) \\ L_n^{(\alpha)}(x) &= L_n^{(\alpha+1)}(x) - L_{n-1}^{(\alpha+1)}(x) \\ \Rightarrow L_n^{(0)}(x)&= L_n^{(1)}(x) - L_{n-1}^{(1)}(x) \\ \Rightarrow L_n^{(-1)}(x)&= L_n^{(0)}(x) - L_{n-1}^{(0)}(x) \\ \Rightarrow L_n^{(-1)}(x) &= \left(L_n^{(1)}(x) - L_{n-1}^{(1)}(x) \right) - \left(L_{n-1}^{(1)}(x) - L_{n-2}^{(1)}(x) \right) \\ &= L_n^{(1)}(x) - 2L_{n-1}^{(1)}(x)+ L_{n-2}^{(1)}(x) \\ &=\small{ \left[\sum_{i=0}^n L_{i}^{(1/2)}(x/2)L_{n-i}^{(-1/2)}(x/2)\right] -2\left[\sum_{i=0}^{n-1} L_{i}^{(1/2)}(x/2)L_{n-1-i}^{(-1/2)}(x/2)\right]+ \left[\sum_{i=0}^{n-2} L_{i}^{(1/2)}(x/2)L_{n-2-i}^{(-1/2)}(x/2)\right]} \\ &=\small{L_{n}^{(1/2)}(x/2)L_{0}^{(-1/2)}(x/2)+L_{n-1}^{(1/2)}(x/2)L_{1}^{(-1/2)}(x/2)-2L_{n-1}^{(1/2)}(x/2)L_{0}^{(-1/2)}(x/2)+\sum_{i=0}^{n-2} L_{i}^{(1/2)}(x/2)\left[L_{n-i}^{(-1/2)}(x/2)-2L_{n-1-i}^{(-1/2)}(x/2)+L_{n-2-i}^{(-1/2)}(x/2)\right]} \\ &=\small{L_{n}^{(1/2)}(x/2)+L_{n-1}^{(1/2)}(x/2)\left(\frac{1-x}{2}-2\right)+\ldots} \\ &=\small{\left[\sum_{j=0}^n L_{j}^{(-1/2)}(x)\right]+\left[\sum_{j=0}^{n-1} L_{j}^{(-1/2)}(x)\right]\left(\frac{-3-x}{2}\right)+\sum_{i=0}^{n-2}\left(\left[\sum_{j=0}^n L_{j}^{(-1/2)}(x)\right]\left[L_{n-i}^{(-1/2)}(x/2)-2L_{n-1-i}^{(-1/2)}(x/2)+L_{n-2-i}^{(-1/2)}(x/2)\right]\right)} \end{align}$$
So now you can substitute in your relation $H_{2n}(x) = (-1)^n 2^{2n} n! \, L_n^{(-1/2)} (x^2)$, and you're done, albeit with a rather convoluted formula.