29,655 reputation
271161
bio website math.pugetsound.edu/~mspivey
location Tacoma, WA
age 41
visits member for 3 years, 9 months
seen 4 hours 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
16
awarded  Nice Answer
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12
awarded  Necromancer
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awarded  Announcer
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awarded  Good Answer
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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.
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awarded  Good Question
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awarded  Revival
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awarded  Announcer
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awarded  Caucus
Apr
29
awarded  Nice Answer
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awarded  Necromancer
Apr
9
comment Ways to evaluate $\int \sec \theta \, d \theta$
@MichaelHardy: Thanks for the notification. I look forward to seeing it.
Apr
8
awarded  Notable Question