# Orthogonal polynomials and Gram Schmidt

How can we use the Gram Schmidt procedure to calculate $L_0,L_1, L_2, L_3$, where ${L_0(x), L_1(x), L_2(x), L_3(x)}$ is an orthogonal set of polynomials on $(0, \infty)$ w.r.t. the weight function $w(x) = e^{-x}$ and $L_0(x)=1$.

Can someone illustrate how to calculate $L_2$ and $L_3$? So I get the idea? I get some integrals but they seem to lead me not far.

Also, now with $L_0,L_1, L_2, L_3$, how can we compute the least squares polynomial of degree 1, 2, and 3 on the interval $(0,\infty)$ for the weight function $w(x) = e^{-x}$ for $f(x) = e^{-2x}$?

Thanks.

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You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. –  Zev Chonoles Feb 10 '13 at 2:07

Just follow the Gram-Schmidt algorithm outlined there, where your inner product is the weighted inner product $$\langle u,v\rangle_w = \int_a^b u(x)v(x)w(x)\,dx.$$ Here, $w(x)$ is called the weight function.

In your context, it sounds like you want to generate the first four Laguerre polynomials, $\{L_0(x),L_1(x),L_2(x),L_3(x)\}$, by applying the Gram-Schmidt algorithm to the standard monomials $\{1,x,x^2,x^3\}$. The resulting Laguerre polynomials will form an orthogonal (or orthonormal if you include the normalization step in the Gram-Schmidt algorithm) family on $0<x<\infty$ with respect to the weight function $w(x)=e^{-x}$.

So, following the algorithm linked above (including the normalization) and using the weighted inner product above, you get (using the notation in the link):

\begin{align} u_0(x)&=1,\\ u_1(x)&=x-{\langle x,1\rangle_w\over \langle 1,1\rangle_w}\cdot 1=x-{\int_0^\infty x\cdot 1\,e^{-x}\,dx\over \int_0^\infty 1^2\,e^{-x}\,dx}=x-1,\\ u_2(x)&=x^2-{\langle x^2,1\rangle_w\over \langle 1,1\rangle_w}\cdot 1-{\langle x^2,x-1\rangle_w\over \langle x-1,x-1\rangle_w}\cdot (x-1)\\ &=x^2-{\int_0^\infty x^2\cdot 1\,e^{-x}\,dx\over \int_0^\infty 1^2\,e^{-x}\,dx}\cdot 1-{\int_0^\infty x^2\cdot (x-1)\,e^{-x}\,dx\over \int_0^\infty (x-1)^2\,e^{-x}\,dx}\cdot (x-1)\\ &=x^2-4x+2,\\ u_3(x)&=x^3-{\langle x^3,1\rangle_w\over \langle 1,1\rangle_w}\cdot 1-{\langle x^3,x-1\rangle_w\over \langle x-1,x-1\rangle_w}\cdot (x-1)-{\langle x^3,u_2\rangle_w\over \langle u_2,u_2\rangle_w}\cdot (x^2-4x+2)\\ &=x^3-9x^2+18x-6. \end{align}

To get the $L_k$, you just need to divide each $u_k$ above by its (weighted) norm: \begin{align} L_0(x)&={u_0\over \|u_0\|_w}={u_0\over \langle u_0,u_0\rangle_w^{1/2}}=1,\\ L_1(x)&={u_1\over \|u_1\|_w}={u_1\over \langle u_1,u_1\rangle_w^{1/2}}=x-1,\\ L_2(x)&={u_2\over \|u_2\|_w}={u_2\over \langle u_2,u_2\rangle_w^{1/2}}={x^2\over 2}-2x+1,\\ L_3(x)&={u_3\over \|u_3\|_w}={u_3\over \langle u_3,u_3\rangle_w^{1/2}}={x^3\over 6}-{3\over 2}x^2+3x-1. \end{align} and you get the first four Laguerre polynomials: $$\left\{1,x-1,\frac{x^2}{2}-2 x+1,\frac{x^3}{6}-\frac{3 x^2}{2}+3 x-1\right\}.$$

Since these form an orthonormal family, you can expand $f(x)=e^{-2x}$ as follows:

$$f(x)=\sum_{n=0}^k c_n L_n(x), \qquad c_n=\langle f,L_n\rangle_w=\int_0^\infty f(x)L_n(x)w(x)\,dx,$$ where $k=0,1,2,3$ indicates (as you call it) the "degree" of the approximation, i.e., an expansion in terms of the first $k$ Laguerre polynomials.

For example, \begin{align} c_0&=\langle e^{-2x},1\rangle_w=\int_0^\infty e^{-2x}\cdot 1\cdot e^{-x}\,dx=1/3,\\ c_1&=\langle e^{-2x},x-1\rangle_w=\int_0^\infty e^{-2x}\cdot (x-1)\cdot e^{-x}\,dx=-2/9,\\ c_2&=\langle e^{-2x},\frac{x^2}{2}-2 x+1\rangle_w=\int_0^\infty e^{-2x}\cdot \left(\frac{x^2}{2}-2 x+1\right)\cdot e^{-x}\,dx=4/27,\\ c_3&=\langle e^{-2x},\frac{x^3}{6}-\frac{3 x^2}{2}+3 x-1\rangle_w=\int_0^\infty e^{-2x}\cdot \left(\frac{x^3}{6}-\frac{3 x^2}{2}+3 x-1\right)\cdot e^{-x}\,dx=-8/81. \end{align}

Here are the $k=0,\dots,3$ approximations of $f(x)=e^{-2x}$:

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This makes sense, but can you show how to compute L2(x) for example in the first part? I am not following it. –  Buddy Holly Feb 10 '13 at 5:31
I edited to add those calculations. Hope that helps. –  JohnD Feb 10 '13 at 23:18
thanks so much. Can you also provide an example calculation for the least squares polynomial for the other part of the question? –  Buddy Holly Feb 10 '13 at 23:22
I added those to my explanation. –  JohnD Feb 10 '13 at 23:30
Thanks for your help! –  Buddy Holly Feb 10 '13 at 23:33