Day 34 in MAT229

• We're in week 10.

  I hope that you enjoyed some break in your "Spring Break". This is that time when I personally need a little revitalization; and a week off usually does it. I'm not sure that a "long weekend" is going to have the same effect.

• I'm noticing a dramatic falling off. Maybe it's discouragement; maybe it's zoom fatigue; maybe it's stress, or feeling overworked; maybe it's the beautiful spring weather.

  Time to power through, people.

  I do have to get the midterm grades in by tomorrow. I'm always one of the last, because I'm trying to get every grade possible in so that I can give my most honest assessment. So you'll have those by tomorrow morning (Tuesday). I try to err on the realistic side....

• I'm issuing a general IMath/Lab/Hw amnesty. Any assignment that's closed on you can be re-opened. Just let me know which you want to work on. It's important to practice.

• I've graded your lab 8 (for those who submitted -- fewer and fewer -- only 13 submitted lab 9; I appreciate you all taking it easy on the grader, but I'd actually rather see more work.).

  Here's a key.

  A few comments:

  1. Integration by parts leads to one part which is indeterminate, so we expect you to try L'Hopital on it.
  2. There's some issues with terminology on this and successive problems.

     There are two kinds of convergence/divergence at work here. One is convergence of integrals, and we compare integrals to integrals. In order for an integral to converge, it will be necessary for an integrand to converge, too.

     When comparing integrals, each result only has one "working side". For now, assume all functions are positive. If $f(x)\ge g(x)$ everywhere, and \[ A=\int_a^b f(x) dx \]

     converges, then so must \[ B=\int_a^b g(x) dx \]

     But if $A$ diverges, then we can't say anything about $B$ -- since $B$ may be small enough to converge (or it may diverge; we don't know yet!).

     On the other hand, if $f(x)\le g(x)$ everywhere, and \[ A=\int_a^b f(x) dx \]

     diverges, then so must \[ B=\int_a^b g(x) dx \]

     But if $A$ converges, then we can't say anything about $B$ -- since $B$ may be small enough to converge (or it may diverge; we don't know yet!).

  3. Convergent, by comparison.
  4. Can't tell!
  5. Pretty straight forward integral. I'm sorry I did say "700 pounds" when I should have said "700 kg"! But that shouldn't have messed up your calculations.

     However I put that example in there so that you could test your work. Having a known value and solution is sometimes helpful, and encouraging if you're wondering if you're doing something correctly....

• I also graded your Weekly assignment 6 (key).
  1. Most split $x\left(e^{-2x}+e^{-3x}\right)$ into two integrals: keep it as one, it's a little faster when doing your integration by parts: \[ \int x\left(e^{-2x}+e^{-3x}\right) dx \] Let $f(x)=x$, $g'(x)=(e^{-2x}+e^{-3x})$; and away you go.

  2. Make sure that you change the limits of integration appropriately. your $u$ integral has limits $-\infty$ to 1.

     One of the interesting things I wanted to point out here is that we can turn an integral with an infinite domain (the $u$ integral) into one with a finite domain (the $x$ integral); and we know how to use the midpoint rule to approximate the finite domained one.

     So this is a trick to allow us to compute numerical approximations of the infinite domained improper integrals: turn them into finite-domained ones, via some transformation (in this case $u=\ln(x)$).

  3. The work problem involved some unit conversion at the end, which was a little tricky for folks. I asked you to convert to exponential functions with base $e$ because this is calculus class.

     A lot of folks struggled with that request. I simply wanted you to write \[ p(x)=10^{stuff} \] as \[ p(x)=\left(e^{\ln(10)}\right)^{stuff}=e^{\ln(10)stuff} \]

     In calculus class we prefer to use base-e....

Now we do something cruel and unusual: we've just introduced you to these improper integrals, and now we're going to use them as tools to study infinite series.

So you're kind of confronting two somewhat unfamiliar animals simultaneously. What ties them together is this idea of running off to infinity, and trying to figure out how to successfully add up an infinite amount of things.

Great thought experiments, but perhaps intellectually troubling. That's perhaps why calculus is difficult: there are lots of intellectually troubling things going on. You don't want to turn it into abstract nonsense, however. If you have any questions, please pop by for an office hour!

1. Integral Test (.nb)
2. Integral Test (.pdf)
3. Long talks you through the Integral Test.

• Nothing to mention.