I'm having some difficulty understanding the solution to a particular Laguerre expansion.
The problem reads "Expand the term $ e^{-x}$ as a Laguerre expansion, noting the orthogonality of $$ < f|g> = \int_{0}^{\infty}f^*(x)g(x)e^{-x}dx $$
I know the first 4 polynomials $L_{n}(x)$, which states that $$ e^{-x} = a_{0}L_{0}(x) + a_{1}L_{1}(x) + a_{2}L_{2}(x) + a_{3}L_{3}(x) +...$$ $$ \rightarrow e^{-x} = \sum a_{n}L_{n}(x)$$
Here's the problem. The SOLUTION states that the coefficients are obtained by integrating $e^{-2x}$ with the polynomials: $$ a_{i}=\int_{0}^{\infty}L_{n}(x)e^{-2x}dx$$
Where did this integral come from? I can't make any sense of why the exponential is squared. Even if you let $g(x)=f(x)$ the entire term $$<f|f> =\int_{0}^{\infty}|f(x)|^{2}*e^{-x}dx$$ which equals $\int e^{-3x}$. We haven't covered polynomials expansions yet in this class so I doubt I'm expected to know detailed resolutions to them. This came out of a chapter on vector spaces.
I can solve the rest of the problem on my own, I just cannot understand why coefficients are solved with the above integral.
You can show from the orthogonality of the Laguerres that each coefficient in the expansion of a function $f(x)$ is
$$a_i = \frac{<f|L_i>}{<L_i|L_i>} = \int_0^{\infty} dx \, f(x) L_i(x) e^{-x}$$
given that the Laguerres are properly normalized. Note that the inner product space here requires the weighting function $e^{-x}$ in the integral definition. In your case, $f(x) = e^{-x}$, so that
$$a_i = \int_0^{\infty} dx \, L_i(x) \, e^{-2 x} $$
That's all there is to it.