Q:

Laguerre polynomials and Gram Schmidt

Last two days I was trying to solve the following problem But I couldn't. 
It's a problem (#5.2.2) from Mathematical Methods for Physicists by George B. Arfken and Hans J. Weber, 7th Edition. 
I tried to look for the solution in the Solution Manual here but it just doesn't explain the solution completely and has nothing mentioned but the actual answers.  I'm looking for the complete procedure that gets us to answers. 
Thanks. 

Here's the problem:
Apply the Gram-Schmidt procedure to form the first three Laguerre polynomials:
${u_{n} (x) = x^n ,\qquad  n=0,1,2,...,\qquad 0\le x< \infty, \qquad   w(x)=e^{-x}}$
The conventional normalization is:
$$\int_0^\infty L_m(x)L_n(x)e^{-x}\,dx=\delta_{mn}.$$
The answers are:
$${L_0=1,\qquad L_1=(1-x),\qquad L_2=\frac{2-4x+x^2}{2} }$$

A:

We apply the Gram Schmidt orthogonalization procedure to
$$ P_1(x)=c \qquad P_2(x)=ax \qquad P_3(x)=bx^2$$
1) Take $P_1$ and normalize it 
$$ ||P_1(x)||^2=\int_0^{\infty}c^2e^{-x}=c^2 $$ 
So $c=1 $ and $L_0=1 $
2) Take $P_2$ and subtract to it its projection on $L_0$, then normalize the result
$$P_{2|L_0} = \int_0^{\infty} (ax)e^{-x}dx =a   \\P_2-P_{2|L_0}=a(x-1) \\
||P_2-P_{2|L_0}||^2 = \int_0^{\infty} a^2(x-1)^2e^{-x}=a^2$$
Now we are free to choose either $ a=1$ or $a=-1 $ , to remain consistent with your notation we choose the latter and get
$$L_1=1-x$$
3)Take $P_3$ and subtract to it its projection on $L_0$ and on $L_1$ then normalize it and get $L_2$ with very analogous procedure

