Day 35 in MAT229

• We're in week 10.

  Today's lab continues with sequences and series, which is our current theme.

• Midterm grades were submitted. Some of you are perhaps worried, and may need to make an appointment to talk to me about your progress. Please reach out!

  If you can't make the office hours, just email and I'll find some time to talk things over.

• 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. There is the convergence of functions, and there is the convergence of integrals.

     In order for an improper integral (with a limit of integration of $\infty$, say) to converge, the integrand must converge (asymptote) to 0.

     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 \[ A=\int_a^\infty f(x) dx \]

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

     because \[ 0 \le \int_a^\infty g(x) dx \le \int_a^\infty f(x) dx \]

     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 \[ A=\int_a^\infty f(x) dx \]

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

     because \[ 0 \le \int_a^\infty f(x) dx \le \int_a^\infty g(x) dx \]

     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!).

• 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:

     \[ \int_{0}^{e}x (\ln(x))^2 \ dx = \int_{-\infty}^{1}u^2 e^{2u} \ du \]

     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....

Make sure that you understand the difference:

1. A sequence is an infinite, ordered list of numbers: $\left\{a_k\right\}_{k=1}^{\infty}$

   We can think of it as a function $a(x)$, but with integers for arguments, and real numbers for values: \[ a: \mathbb{N} \rightarrow \mathbb{R} \]

   It might be better simply to write it as $a(n)$, but mathematicians have adopted the alternative notation $a_n$, perhaps to make clear that this function has a domain restricted to integers.

2. A series is an infinite sum of terms in a sequence. You can think of it as the analogue of an improper integral: \[ \sum_{k=1}^{\infty}a_k \]

   And this is one of the reasons that we are not too surprised when we find ourselves comparing series and improper integrals.

• Nothing to mention.