Laguerre polynomial 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, :)
 A: 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*}
A: $L_n^{(0)}(x)$ 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.
