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Laguerre polynomials has the recurrence relation

$$(n+1) L_{n+1} (x)=(2n+1-x) L_n (x)-nL_{n-1} (x)$$

In proving this, I differentiate the generating function of laguerre polynomials

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

The process goes like this:

\begin{align}

\frac{\partial g}{\partial t}

&=\frac{(1-t) D_t \left(e^{-\frac{xt}{1-t}}\right)-e^{-\frac{xt}{1-t}} D_t \left(1-t\right) }{(1-t)^2}\\

&=e^{-\frac{xt}{1-t}} \frac{1-t-x}{(1-t)^3}

\end{align}

$$(1-t)^2 \frac{\partial g}{\partial t}=(1-t-x)g \tag 1$$

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solving the right-hand side of eq. (1), we have:

\begin{align}

(1-t-x)g

&=(1-t-x) \sum_{n=0}^\infty L_n (x)t^n\\

&=L_0 (x)-xL_0 (x)+\sum_{n=1}^\infty(L_n (x)-L_{n-1} (x)-xL_n (x) ) t^n

\end{align}

also, Solving the left-hand side of eq. (1), we have:

$$

\begin{align}

(1-t)^2 \sum_{n=0}^\infty nL_n (x) t^{n-1}&=(1-2t+t^2)\sum_{n=0}^\infty nL_n (x) t^{n-1}\\

&=\sum_{n=0}^\infty nL_n (x)(t^{n-1}-2t^n+t^{n+1})\\

&=L_1 (x)+\sum_{n=1}^\infty t^n [(n+1) L_{n+1}(x)-2nL_n (x)+(n-1)L_{n-1} (x) ].

\end{align}

$$

My question is: How did the derivative of the generating function with respect to $t$ in the left hand side equal to $\sum_{n=0}^\infty nL_n (x) t^{n-1}$

that is,

$$

\frac{\partial g}{\partial t}=\sum_{n=0}^\infty nL_n (x) t^{n-1}

$$

Thanks! ðŸ™‚

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