gram-schmidtlaguerre-polynomialsorthogonal-polynomialspolynomials
Consider the set of functions $u(x)=x^n,\,\,$ with $n=0,1,2, \dots$.
Use the Gram-Schmidt procedure to construct the first 3 orthogonal polynomials of:
$$\text{Laguerre:} \;\;\;\;L_n(x),\;\;\;\; \text{Range:}\;\; 0 \le x < \infty, \;\;\;\; \text{Weight:}\;\; w(x)=e^{-x}.$$
For these polynomials the normalization is
$$\int_0^\infty [L_n(x)]^2 e^{-x} dx \,=\, 1.$$
Suppose $e_k$ are orthonormal, then $\|f-\sum_k \alpha_k e_k \|^2 = \|f\|^2 -2 \sum_k \alpha_k \langle f, e_k \rangle + \sum_k \alpha_k^2$.
To find the best approximation, we minimise the above over the $\alpha_k$. This is a convex quadratic separable problem, and the solution is given by just differentiating with respect to the $\alpha_k$. This gives $\alpha_k = \langle f, e_k \rangle$.
Hence the best approximation is $\sum_k \langle f, e_k \rangle e_k$.
In the above problem, you are given $g_1(x) = 1, g_2(x) = x, g_3(x) = x^3$. Use Gram Schmidt to orthonormalise the $g_k$ to get $e_k$. Then the coefficients $\alpha_k$ can be computed as above.
Here are the orthonormalised functions: $e_1(x) = {1 \over \sqrt{2}}$, $e_2(x) = \sqrt{3 \over 2} x$, $e_3(x) = \sqrt{7 \over 8}(5 x^3-3x)$.
Then for $f(x) = x^5$ we have $\langle f, e_1 \rangle = 0$, $\langle f, e_2 \rangle = { \sqrt{6} \over 7}$, $\langle f, e_3 \rangle = {\sqrt{32} \over 9 \sqrt{7}}$.
Grinding through the details gives the best approximation $\sum_k \langle f, e_k \rangle e_k = -{5 \over 21} x + {10 \over 9} x^3$.
The polynomials obtained from (1) are the normalized Legendre polynomials. For all three cases, it is possible to calculate them by hand, but (especially for (3)) the calculation can be quite cumbersome. So, I've used the following Mathematica code for the Gram-Schmidt (including normalization) procedure:
Clear[f, g, ip];
ip[f_, g_] := Integrate[f*g*1, {x, -1, 1}];
proj[g_, f_] := g*ip[f, g]/ip[g, g];
normalized[f_] := f/Sqrt[ip[f, f]];
gs[fs_List] := Module[{u}, normalized /@
Table[u[k] = fs[[k]] - Sum[proj[u[j], fs[[k]]], {j, 1, k - 1}],
{k, 1, Length[fs]}]];
lps = Expand[gs[Table[x^k, {k, 0, 4}]]]
This is the code for the first part of the question - if you want to do it for (2) and (3), just replace the part in the second line after f * g with the appropriate weight function and the correct interval.
Best Answer
Let's call the first 3 orthogonal polynomials we are looking for $v_0(x), v_1(x), v_2(x)$. We are going to solve for $v_0(x)$ by starting with $u_0(x) = 1$ as your first polynomial. You can find the normalization constant $N$ by solving:
$\int^\infty_0 v_0(x)^2 e^{-x}dx = \frac{1}{N^2} $
Your normalized polynomial $v_0(x)$ will be $Nu_0(x)$. In this case, you will find that N=1, or that your function $v_0(x) = 1$ is already normalized.
Then continue with $u_1(x) = x$. You have to make it orthogonal with your first orthogonal polynomial $v_0(x)$ by evaluating the integral:
$\int^\infty_0 u_1(x)v_0(x) e^{-x}dx$
You have to subtract the result of this integral from $u_1(x)$ to get your first orthogonal polynomial. The integral turns out to be -1 (solved by integration by parts), and so you get $v_1 = x - 1$. You can find that the normalization constant is 1 by doing the integration
$\int^\infty_0 v_1(x)^2 e^{-x}dx = \frac{1}{N^2} $
Finally,to find $v_2(x)$, we start with $u_2(x) = x^2$ and make it orthogonal to both of our polynomials $v_0(x)$ and $v_1(x)$. This time we have to subtract the following two integrals from $u_2(x)$:
$\int^\infty_0 u_2(x)v_0(x) e^{-x}dx$
$\int^\infty_0 u_2(x)v_1(x) e^{-x}dx$.
This should lead to $v_2(x) = x^2 - 4x + 2$. Do the usual normalization integral, and this time you will find that $N=\frac{1}{2}$.
You can check your answers here: https://en.wikipedia.org/wiki/Laguerre_polynomials#The_first_few_polynomials (Note that our $v_1(x)$ has a minus sign as compared to theirs. That is fine, since a minus sign doesn't change the fact the orthnormality of the polynomial)