If I'm going to construct a second degree Laguerre least squares approximation to the function $f(x)=\exp(x)$ over the right open interval $[0,\infty)$, will I get such a polynomial?
This is because I have tried to evaluate the involved integrals but they all does not converge in the given interval.Thus I couldn't find the respective coefficients in the second degree Laguerre approximation.
Any hints on solving this.
It is not astonishing. $exp:x \rightarrow e^x$ cannot have a Laguerre expansion.
For a function $f$ to have such an expansion, it must be square integrable on $(0,\infty)$ with respect to measure $e^{-x}dx$, i.e., be such that the integral:
$$\int_0^{\infty}f(x)^2 e^{-x}dx$$
is convergent, which is not the case for $f=exp$.
Nevertheless, everything is not lost. Here is a little trick: instead of $e^x$, take the function defined by $f(x)=e^{x/3}$. Its square is integrable: $\int_0^{\infty}e^{2x/3}e^{-x}dx=\int_0^{\infty}e^{-x/3}dx$ is convergent. Then take the order-2 Laguerre decomposition:
$$\tag{1}e^{x/3} \approx a_0L_0(x)+a_1L_1(x)+... \text{with} \ \ a_k=\int_0^{\infty}e^{x/3}L_k(x)e^{-x}dx$$
and set $X=x/3$. Let us write (1) explicitly:
$$e^{x/3}\approx\left(\frac{3}{2}\right)1+\left(\frac{-3}{4}\right)(1-x)+\left(\frac{3}{8}\right)\frac{1}{2}(2-4x+x^2)$$
Expanding and grouping:
$$e^{x/3}\approx\frac{9}{8}+\frac{3}{16}x^2$$
The curves corresponding to the LHS and RHS resp. are plotted below and show a good agreement, at least on $(0,3)$. Thus, setting $x/3=X$, the curves of
$$y=e^X \ \ \ \text{and} \ \ \ y=\frac{9}{8}+\frac{27}{16}X^2$$
which will display a good agreement... on $(0,1)$ (because it not more than a "shrinking" of $x$ coordinates).