Day 25 in Mat229

Last time

Next time

Today:

Announcements
- Next time we move on to Sequences and Series -- a bit of a change in mindset from what we've been doing. We'll start with Vol 2, Sec 5.1: Sequences.
- Your quiz on Thursday will be over numerical integration.
Today: we'll be working with a few more improper integrals.
- Last time we had a look at Example 3.49, for several reasons:
  - To talk about probability density functions (non-negative functions whose integrals over the entire real line are 1),
  - to talk about units (and how we can use them to sanity check a calculation), and
  - to correct an error in the exercise.
  I also noted the importance of the \(dx\) in an integral, where \(x\) has units: \(dx\) has the same units.
  We explored what is known as the probability density function (pdf) of an exponential distribution, \[ \int_{0}^{+ \infty} k e^{-k x} d x = 1 \]
  The mean of an exponential distribution with rate parameter \(k\) is \(\frac{1}{k}\). We were able to show that with a calculation:
  The average of any function of \(x\), \(f(x)\), is obtained by multiplying \(f(x)\) times the probability density, and integrating over the domain where the integrand is non-zero: \[ \overline{f(x)} = \int_{0}^{+ \infty} f(x) k e^{-k x} d x \] Since we wanted the average of \(x\), this becomes \[ \overline{x} = \int_{0}^{+ \infty} x k e^{-k x} d x \] which is indeed \(\frac{1}{k}\). Therefore, if the average time between accidents is 3 months, \(k=\frac{1}{3}\).
  So the probability of a time between accidents of 8 months or more (and this explains why we let the upper limit go to infinity -- anything bigger than 8) is \[ \int_{8}^{+ \infty} \frac{1}{3} e^{-\frac{x}{3}} d x \approx .0077 \]
- Then we had a look at Example 3.51. It illustrates that improper integrals still behave like ordinary integrals in a lot of ways. We can break them up, for example: \[ \int_{-\infty}^{\infty} x e^x d x = \int_{-\infty}^{0} x e^x d x + \int_{0}^{\infty} x e^x d x \]
  We can quickly reach the conclusion that \(I\) is divergent, because in one case (on the positive side) the integrand is headed for infinity as \(x \rightarrow \pm \infty\).
- Let's get to our our improper integral handout for some practical work. (Here's the Mathematica version if you wish.)
  - #1, 2bd, 3a
- In particular, we want to talk about comparisons. for that, I have an additional handout (and here's the Mathematica version)
Links:
- Wolfram Alpha
- Mathematica on-line is an option, if you are at a computer without Mathematica installed. I've tried it and it worked pretty well!

Website maintained by Andy Long. Comments appreciated.