18.01 Section, November 23, 2015
Section leader: Eva Belmont (ebelmont@mit.edu, E18-401B)
.
Improper integrals, series convergence via integrals, probability densities
Review
R∞
• Dominated
convergence
for
integrals:
if
a g(x) dx converges and |f (x)| ≤ g(x) in the interval [a, ∞),
R∞
then a f (x) dx converges.
R∞
R∞
• And the reverse: if a |g(x)| dx diverges and f (x) ≥ |g(x)| for x in [a, ∞), then a f (x) dx diverges.
• Integral convergence criterion for series: if f is a decreasing function with f (x) ≥ 0, then:
R∞
• More
R ∞ precisely, if convergence holds then the value of the series is between a f (x) dx and f (a) +
a f (x) dx.
R∞
• Probability density function: p(x) ≥ 0 such that ∞ p(x) dx = 1.
x2
• Normal distribution: p(x) = s√12π e− 2s2
R∞
• Expectation: E = −∞ x p(x) dx
R∞
• Variance: −∞ (x − E)2 p(x) dx
√
• Standard deviation: Variance
Problems
1. Show whether
R∞
3
ln(x)
x
dx converges:
(a) directly;
(b) using dominated convergence for integrals.
What about
R1
0
ln(x)
x
dx?
Use whatever method you like.
1
2. (a) Give an upper bound for the value of
P∞
1
n=1 n3 .
P
1
(b) Get a different upper bound by writing the series as 1 + 213 + · · · + 613 + ∞
n=7 n3 and using an
integral to estimate the latter infinite sum. Draw a picture of what you’re calculating.
(c) Do the same for a lower bound.
(d) Is the estimate in (c) within 0.01 of
(e) Is 1 +
1
23
+ ··· +
1
63
within 0.02 of
3. For what values of a does
Rb
1
0 xa
P∞
1
n=1 n3 ?
P∞
1
n=1 n3 ?
dx converge?
4. Write an integral expressing the standard deviation of the probability density function
(
1
sin(x) if 0 < x < π
p(x) = 2
0
otherwise.
5. Bonus question: Show that the integral convergence criterion for series doesn’t work, in general,
if f isn’t required to be a decreasing function.
So, you have to come up with a specific f that satisfies the conditions except for not being decreasing, and show that
R
P
f converges and f doesn’t, or vice versa.
2