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> What is it like to understand advanced mathematics? It's really nice! Get to understand a lot of stuff. Some of the advanced math is really powerful for a
by graycat 10y ago
> What is it like to understand advanced mathematics?
It's really nice! Get to understand
a lot of stuff.
Some of the advanced math is really
powerful for applications.
And, can look back at the math
saw in physics where wondered if the
physics profs really understood the
math and conclude, right, they
didn't or at least not very well.
E.g., recently I saw a physics lecture
where they explained more than once
that a quantum mechanics wave
function was differentiable
and "also continuous" as if
there was some question, doubt,
choice, or chance otherwise. Of COURSE
it is continuous! Every differentiable
function is continuous!
There's a lot more on why it's
nice to understand the advanced
math!
- smaddox 10y ago> Every differentiable function is continuous! Under who's definition? The Heaviside step function is discontinuous, but it's derivative is usually considered to be the Kronecker delta function. Both of these are used extensively in physics and engineering.
- zodiac 10y agoDid you mean Dirac delta instead of Kronecker delta?
- smaddox 10y agoYes, my mistake. Kronecker is used often in quantum mechanics, but when working in Hilbert space rather than real space.
- witty_username 10y ago> If f is differentiable at a point x0, then f must also be continuous at x0. https://en.wikipedia.org/wiki/Differentiable_function#Differentiability_and_continuity https://en.wikipedia.org/wiki/Differentiable_function#Differ...
- Smaug123 10y agoThe Dirac delta function is not a function but a distribution. This is just the product of sloppy notation by the physicists.
- Klockan3 10y agoNo, it is just that physicists means something different than mathematicians when they use the word "function". Just assume that they always are distributions that are evaluated with dirac deltas and everything makes sense.
- graycat 10y ago> Under who's definition W. Rudin, Principles of Mathematical Analysis. W. Fleming, Functions of Several Variables. For the set of real numbers R, a function f: R --> R x in R, d in R, function f is differentiable at x with value d provided for h in R lim_{h --> 0} (f(x + h) - f(x))/h = d In this case we write f'(x) = df(x)/dx = d In that case, in particular, lim_{h --> 0} (f(x + h) - f(x)) = 0 so that f is continuous at x. Differentiation is important in physics, e.g., in Newton's second law.
- Klockan3 10y agoThey most likely said "The wave function is differentiable and its derivative is continuous", which isn't obvious. If you think that it sounded strange you should have asked during the lecture and they would have explained. Now instead you just assumed that you knew better and made a fool of yourself... Also this isn't advanced mathematics, this is something most who are interested learns in high-school and I can assure you that most physics professors are aware of this fact.
- graycat 10y ago> They most likely said No, they said what I wrote they said. It was at YouTube, from MIT, from a course in quantum mechanics. Uh, I do understand this stuff: E.g., given a positive integer n, the set of real numbers R, the usual topology for R^n, a set C a subset of R^n closed in that topology, there exists a function f: R^n --> R so that for x in R^n, f(x) is 0 for x in C, f(x) > 0 for x not in C, and f infinitely differentiable. E.g., for n = 1, the result applies for set C a Cantor set or a Cantor set of positive measure. For n > 2, the result applies for a sample path of Brownian motion or the Mandelbrot set. When I was a grad student, I discovered and proved this result. Later I published the result in JOTA.
- selimthegrim 10y agoA bigger problem is the difference between Hermitian and self adjoint, a lot of physics professors don't understand that. Or for example, this famous dialogue between Geoffrey Chew and Arthur Wightman from the 1960s bootstrap fad: Wightman asked Chew: why assume from the start that the S-matrix was analytic? Why not try to derive it from simpler principles? Chew replied that "everything in physics is smooth". Wightman asked about smooth functions that aren't analytic. Chew thought a moment and replied that there weren't any.
- selimthegrim 10y agoBy the way, the reason they said that was no doubt to motivate the students to choose correct boundary conditions for some Sturm-Liouville problem (as the previous poster sort of alluded to)