February « 2009 « jkwiens.com

Archive for February, 2009

Getting Things Done

February 28th, 2009

Lately, I have been reading a book called “Getting Things Done” by David Allen. It is an awesome book so far and I would recommend it to anyone. The basic idea of the book is to implement a system that gets your projects and TODO lists out of your head and into a more dependable system. The analogy David Allen makes is that your brain is like RAM in your computer. RAM is not meant for long term storage and there is only a limited supply of it. What you need is another storage device (like a harddrive). This way you can free up RAM for more important tasks than your mental TODO list. This will, in the end, make you more productive and less stress. The problem that I noticed with myself is that I have a lousy memory. This is why I could never be a biologist. This means that my TODO list doesn’t make me stress because I’m totally preoccupied with the task at hand. However, this also means that I don’t always work on the items that I SHOULD be working on at any given moment.

This leads me to my second point. Once I understood the system, I needed a place to put my system. The natural place for me to put it is on the web. I would be able to access “my list” from work, home, and on my iphone. This means that I could access it practically anywhere. Additionally, I am planning on keeping a notepad and pen with me at all times, so that I can write down projects and tasks as I think about them. As David Allen said, you never remember that you need toilet paper when you are in the grocery store. Anyways, I digress. The problem that I found was finding a good webapp that implemented the GTD system. The webapp that showed the most promise was Vitalist. However, you need to pay if you want to track more than five projects. The system is way to simple to justify paying $10 a month for it. This lead me to a post on LifeHacker which explains how to implement the GTD system using Google Notes. This system is a lot more feasible. It is not glamorous, but it gets the job done. The big thing that I don’t like about it is that Google hasn’t made a version of this webapp for the iPhone. You can still view the site in Safari as a regular webpage, but it is tricky to use. Anyways, until a better system comes along (that is free), I will stick with this system.

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How To Survive A Fall From 35,000 Feet

February 22nd, 2009

A couple of my friends finally posted a video that they helped create on youtube. The film title is called “How To Survive A Fall From 35,000 Feet”. I originally saw the film over a year ago and absolutely loved it. However, I didn’t know that they put it up on youtube until today. The absolutely amazing part of the film is that they did all the editing and filming in two days during the final audition rounds for Spielberg’s reality show “On The Lot”.

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The Wronskian and Generalized Solutions

February 14th, 2009

Assume we have a linear differential equation that has the following form:

(1)

$\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + \ldots +p_{n-1}(t)\frac{dy}{dt} +p_n(t)y = 0$

If $y_1, ~ y_2, ~y_3, ~\ldots,~ y_{n-1}, ~y_n$ are solutions to (1), we will then define the Wronskian as:

(2)

$W(y_1,\ldots,y_n) = \left | \begin{array}{cccc} y_1 & y_2 & \cdots & y_n \\y_1^\prime & y_2^\prime& \cdots & y_n^\prime \\ \vdots & \vdots & ~ & \vdots \\y_1^{(n-1)} & y_2^{(n-1)} & \cdots & y_n^{(n-1)} \end{array} \right |$

If the functions $p_1, ~ p_2, ~ \ldots , ~ p_{n-1},~ p_n$ are continuous on the open interval I, if the functions $y_1, ~ y_2, ~y_3, ~\ldots,~ y_{n-1}, ~y_n$ are solutions of (1), and if $W(y_1,\ldots,y_n)(t) \neq 0$ for at least one point in I, then every solution of (1) can be expressed in the following form:

(3)

$y = c_1y_1 + c_2y_2+ \ldots + c_{n-1}y_{n-1} + c_n y_n$

As an additional challenge, I would like to see a proof of this theorem. I haven’t attempted to prove this theorem and have no idea how complicated the proof is for it.

Comments(0) - Differential Equations

Homogeneous 2nd Order DE

February 11th, 2009

I am going to keep moving through my whirlwind introduction to differential equations. If you find yourself bored, I apologize. These questions, at least at the beginning, will be quite simple. However, if you are bored, you can take 20 seconds out of your day and solve it. The problem should at least be as entertaining as a Sudoku.

Today we are going to look at how to solve a Linear Homogeneous Second Order Differential Equation with constant coefficients. The equation will have the form

(1)

$ay^{\prime\prime} + b y^{\prime} + cy = 0$

where a, b, and c are constants.

We are going to look at the example problem

(2)

$6y^{\prime\prime} -5 y^{\prime} + y = 0$

with the boundary conditions

(3)

$y(0) = 4, ~~ y^{\prime}(0)=3$

First, let us assume that $e^{\lambda t}$ is a solution. If the problem has the same form as equation (1), this is a safe assumption. Now, since $y = e^{\lambda t}$ , $y^\prime =\lambda e^{\lambda t}$ , and $y^{\prime\prime} =\lambda^2 e^{\lambda t}$ , equation (2) will become

(4)

$(6\lambda^2 -5\lambda+1)e^{\lambda t} = 0$

Therefore, because $e^{\lambda t} != 0$ , we know

(5)

$6\lambda^2 -5\lambda+1 = 0$

This is known as the characteristic equation. The values of $\lambda$ that satisfy this equation are $\frac{1}{2}$ and $\frac{1}{3}$ . Please note that the roots that satisfy the characteristic equation are real numbers and are not equal to each other. If the numbers are complex or have repeated roots, the next few steps will be different and will be solved in a future post.

Because the roots are real and different, the following equations will be solutions to (2).

(6)

$y = e^{\frac{t}{2}}$

(7)

$y = e^{\frac{t}{3}}$

However, because (2) is linear, we know that the superposition of equations (6) and (7) will be a solution. Therefore, the generalized solution would be

(8)

$y = \lambda_1 e^{\frac{t}{2}} + \lambda_2 e^{\frac{t}{3}}$