Theledge.net

Sunday, March 25. 2007

An example of what I've been doing recently:

Posted by Matthew Groeninger in A "Matt" Story, Math at 22:15

Since I haven't really had time to post, I figured I would take a quick opportunity to round out the end of the month with a post that illustrates the kind of thing I’ve been doing to ~~waste my time~~ keep me busy.

Here goes:

If our inner product is defined as $\left\langle p,q \right\rangle = \int_{0}^{\infty}e^{-x}p\left(x\right)q\left(x\right) dx$ and our basis is defined as $\left\{1,x,x^2,x^3\right\}$ our Gram-Schmidt process is as follows:

Obviously,

$\begin{center}\begin{math}b_1 = \frac{1}{\left|1\right|} = 1\end{math}\end{center}$

Once we have found a normalized polynomial in our basis we can generate an orthogonal polynomial.

$\begin{center}\begin{math}w_2 = v_2 - \left\langle v2,b_1 \right\rangle b_1 = x - \int_{0}^{\infty}e^{-x}\left(x\right)\left(1\right) dx = x - \left(\left(-1-x)\right)e^{-x}\Big |^{0}_{\infty}\right)\end{math}\end{center}$

$\begin{center}\begin{math}w_2 = x - 1\end{math}\end{center}$

Once we have an orthogonal polynomial we need to normalize it.

$\begin{center}\begin{math}b_2 = \frac{x - 1}{\left|x-1\right|} = -x+1\end{math}\end{center}$

Again, now having two normalized, orthogonal polynomials in our basis, we can generate another orthogonal polynomial.

$\begin{center}\begin{math}w_3 = v_3 - \left\langle v3,b_2 \right\rangle b_2 - \left\langle v3,b_1 \right\rangle b_1 = x^2 - \left(\int_{0}^{\infty}e^{-x}\left(x^2\right)\left(-x+1\right)dx\right) \left(-x+1\right)-\int_{0}^{\infty}e^{-x}\left(x^2\right) dx \end{math}\end{center}$

$\begin{center}\begin{math}= x^2 - \left(-6+2)\right)\left(-1-x)\right)-2\end{math}\end{center}$

$\begin{center}\begin{math}w_3 = x^2-4x+2\end{math}\end{center}$

Again we need to normalize the polynomial:

$\begin{center}\begin{math}b_3 = \frac{x - 1}{\left|x^2-4x+2\right|} = \frac{x^2-4x+2}{2}\end{math}\end{center}$

Since we now have three orthonormal polynomials we can complete our basis:

$\begin{center}\begin{math}w_4 = v_4 - \left\langle v4,b_3 \right\rangle b_3 - \left\langle v4,b_2 \right\rangle b_2 - \left\langle v4,b_1 \right\rangle b_1\end{math}\end{center}$

$\begin{center}\begin{math}= x^3 - \left(\int_{0}^{\infty}e^{-x}\left(x^3\right)\left(\frac{x^2-4x+2}{2}\right)\right) dx \left(\frac{x^2-4x+2}{2}\right) - \left(\int_{0}^{\infty}e^{-x}\left(x^3\right)\left(-x+1\right) dx\right) \left(-x+1\right) -\int_{0}^{\infty}e^{-x}\left(x^3\right) dx\end{math}\end{center}$

$\begin{center}\begin{math}= x^3 - \left(-18\right)\left(\frac{x^2}{2}\right) - \left(18\right)\left(2x\right) - 18 - 18x + 18 -6\end{math}\end{center}$

$\begin{center}\begin{math}w_4 = x^3 - 9x^2 + 18x -6\end{math}\end{center}$

$\begin{center}\begin{math}b_4 = \frac{x^3 - 9x^2 + 18x -6}{\left|x^3 - 9x^2 + 18x -6\right|} = \frac{x^3 - 9x^2 + 18x -6}{-6} = \frac{-x^3 + 9x^2 - 18x + 6}{6}\end{math}\end{center}$

Therefore our new, orthonormal basis is:

$\begin{center}\begin{math}\left\{1,-x+1,\frac{1}{2}\left(x^2-4x+2\right),\frac{1}{6}\left(-x^3 + 9x^2 - 18x + 6\right)\right\}\end{math}\end{center}$

Several things to note:

This is not what I am learning. This is something I learned a year and a half ago. This is something I am doing in order to learn other things.
This was essentially part “a” of a five part problem.
The assignment had five problems.
We were given a week to finish it.
These are the first couple of Laguerre polynomials. We used them to approximate a function. It was a terrible approximation.
I’m not bitching. This is actually the class I am doing well in and that I enjoy.

Addendum to anyone who finds this by searching to solve a homework problem: I have no idea if this is right. I would not suggest using it.

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Defined tags for this entry: math, school

Thursday, January 6. 2005

Classes!

Posted by Matthew Groeninger in News at 19:26

Classes start Monday!!! Light load this semester:

Matrix Methods
Intermediate Microeconomics
Economic Statistics
Introduction to Analysis

Plus, I might have a lab class that deals with nonlinear systems and chaos...
We will play with Chua's circuit, which means I'll learn a little EE, but that's ok! *grin*

It seems interesting... We'll have to see...

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Defined tags for this entry: school

Wednesday, December 22. 2004

Grades!

Posted by Matthew Groeninger in News at 09:50

Diff/Eq	A-
MacroTheory	A
Math Econ	A
Probability	B

So, I traded an A in diffeq to get the B in Probability.. and it was so worth it...

I felt like I was failing probability the whole semester... For some reason, I just couldn't do it... I understood it all, I just couldn't apply it...

Amazingly frustrating...

Anyways, I'm all done and stuff, and now I can relax...

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Defined tags for this entry: school

Saturday, December 18. 2004

So...

Posted by Matthew Groeninger in Just words.. at 01:37

So, I switched the site over to Serendipity.... no really good reason, other than I was tired of looking at drupal and knowing it was just overhead, cause I couldn't do what I wanted too...

I have some things I need to work on... I'd really like an amazon plugin, but I think that would work best as an input filter....

And I'd love a blogroll system, but I'm not going to use another site for it... (I'll have to cast around and see if I can find a roll-your-own method....)

And I'm tired right now...

But I got a B in probability!!!! Whohoa!!!

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Defined tags for this entry: blog, php, s9y, school, serendipity

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