Mathematical physicist and mentor to mathematically talented youth. Talent is that which bridges the gap between what can be taught and what must be learned.
Aug 27, 2020 • 42 tweets • 8 min read
Believe me, @graciegcunning, you sounded smart (indeed, brilliant) before, and you sound even smarter here. Your questions are ancient, deep, and fascinating. And it's vastly more important to ask good questions than to know all the answers. It's much, much harder, too.
The first thing to know about mathematics is that it is really two things: a body of knowledge, and an artistic practice.
Most people focus on the body of knowledge, and it's easy to see why. Mathematical knowledge is very powerful, and its logical justification is very clear.
Nov 26, 2019 • 19 tweets • 3 min read
Contemporary researchers in the mathematical sciences enjoy countless advantages over their counterparts of a century ago. A vastly larger and better-funded community now has convenient access to the literature, and to communication channels that make global collaboration easy.
General conditions, too, overwhelmingly favor contemporary researchers. Rates of extreme poverty have plummeted. Life expectancies are longer by two decades. No mathematician or physicist today dies of tuberculosis, or of peritonitis following a ruptured appendix.
Sep 17, 2019 • 4 tweets • 2 min read
@stevenstrogatz An analogy between musical and poetic rhythm may help. 6/8 and 3/4 both refer to rhythms in which each "line" of a musical "poem" has six "syllables." But in the former case, the the stress is as in "elegant discipline," whereas in the latter, it's as in "nothing really matters."
@stevenstrogatz To a poet, the distinction is between six syllables as two dactyls (which is what you get with 6/8) and six syllables as three trochees (which is what you get with 3/4.) 6/8 therefore refers to a duple rhythm, with two stressed beats per line, and 3/4 to a triple one, with three.
Mar 3, 2019 • 14 tweets • 4 min read
@jamestanton God wants us to write our polynomials like this:
with their coefficients multiplied by entries drawn from Pascal's triangle. Makes everything nicer.
In particular, when the coefficients a, b, c, ... are in geometric progression (i.e., b = ar, c = ar^2, etc.) then so are these polynomials. Indeed, the nth polynomial is then just
a(x + r)^n.
So in this case. all the roots coincide: all n of them are equal to -r = -b/a.