Laguerre polynomial question

Question

Can someone help me with this

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

The author said that we should just expand it but I don't understand how and what $L_{n}$ is equal to.

Since there is $\frac{t^{n}}{n!}$ I guess the exponential function should be expanded but I don't know what to do with $\frac{1}{1-t}$ neither what is the full expression for the $L_{n}$. Thank you for your help, :)

Answer 1

Using the following definition of Laguerre polynomials \begin{eqnarray*} L_{n}=\sum_{k=0}^{k=n}\frac{(-1)^{k}}{k!} {n\choose k} x^{k} \end{eqnarray*} Now we should have ... (See https://en.wikipedia.org/wiki/Laguerre_polynomials#Recursive_definition,_closed_form,_and_generating_function) \begin{eqnarray*} \sum_{n=0}^{\infty}L_{n}(x)\frac{t^{n}}{\color{red}{1}} &= & \sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{(-1)^{k}}{k!} {n\choose k} x^{k} t^{n} \\ &= & \sum_{k=0}^{\infty} \sum_{n=k}^{\infty }\frac{(-1)^{k}}{k!} {n\choose k} x^{k} t^{n}\\ &= & \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} x^{k}\sum_{n=k}^{\infty } {n\choose k} t^{n}\\ &= & \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} x^{k} \frac{t^{k}}{(1-t)^{k+1}} \\ &= & \frac{1}{1-t} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} \left( \frac{xt}{1-t} \right)^{k} \\ &= & \frac{e^{-\frac{xt}{1-t}}}{1-t}. \end{eqnarray*}

Answer 2

$L_n^{(0)}(x)$ which is also written $L_n(x)$ is $L_n(x) = \sum_{k=0}^n (-1)^k {n \choose k}\frac{x^k}{k!}$. The expression in the question is an exponential generating function. Try to show that coefficients of like powers of $t$ on the left and right sides of the equation are equal. You can multiply the right side by 1 - t.

The equation in the question is incorrect. the correct generating function is not an exponential one as noted in red. Please see the Digital Library of Mathematical Functions, equation 18.12.13. I suggest you try some other examples first.