29,875 reputation
273163
bio website math.pugetsound.edu/~mspivey
location Tacoma, WA
age 41
visits member for 3 years, 10 months
seen 50 mins ago

I am a math professor at the University of Puget Sound. My background is in operations research, and I teach typical OR courses such as optimization, modeling, and probability, as well as calculus, statistics, and differential equations.

My math blog, A Narrow Margin, includes (among other things) discussion of some of my favorite posts - of mine and of others - from math.SE.


Jul
27
revised Famous uses of the inclusion-exclusion principle?
fixed broken link
Jul
27
comment Famous uses of the inclusion-exclusion principle?
@alancalvitti: Thanks. I'll fix it.
Jul
22
awarded  Nice Answer
Jul
20
awarded  Good Answer
Jul
20
awarded  Nice Answer
Jul
16
awarded  Nice Answer
Jul
12
awarded  Necromancer
Jul
1
awarded  Announcer
Jun
18
awarded  Good Answer
Jun
15
awarded  Announcer
Jun
7
comment Why is $1^{\infty}$ considered to be an indeterminate form
@ThisIsNotAnId: Now, suppose you don't fix the base on your expression and you allow it to vary instead. What happens to the value of $\lim_{n \to \infty} f(n)^{g(n)}$, where $f(n) \to 1$ and $g(n) \to \infty$? Does it go to $\infty$, or to 0, or to something, like 1, in between 0 and $\infty$? Because 1 is a boundary case for the base of the exponential expression the answer turns out to depend on the relative rates at which $f(n)$ goes to 1 and $g(n)$ goes to $\infty$. My original post expands on that idea.
Jun
7
comment Why is $1^{\infty}$ considered to be an indeterminate form
@ThisIsNotAnId: I'm not sure that the comment box is an adequate place to address your question, but I'll give it a shot. The number 1 is special in the $1^{\infty}$ form because it is a boundary case. Take any fixed number slightly smaller than 1 and raise it to an arbitrarily large power, and you're heading toward 0. Fix 1, and take it to an arbitrarily large power, and you still have 1. Take any fixed number slightly larger than 1 to an arbitrarily large power, and you're heading toward $\infty$. The number 2 as the base does not represent such a boundary case.
Jun
1
awarded  Revival
May
29
comment Beautiful identity: $\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$
@Marc: Done! And thanks for bringing this post back to my attention. There was an index error in it.
May
29
revised Beautiful identity: $\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$
fixed couple of math typos
May
23
comment Exponential Generating Functions For Derangements
+1. Quite nice!
May
23
comment Exponential Generating Functions For Derangements
@martycohen: Well, that depends on what you already know about the derangement numbers. But it is easy to get from the recurrence Brian uses to the one I use. See, for example, this blog post.
May
19
awarded  Good Question
May
18
awarded  Revival
May
17
awarded  Announcer