jadagul:

antisquark replied to your post: Oh god. Grading a problem on my calc test, and I…

Congrats, you invented a “trick question” ;)

Sure, but I try really hard not to do that (and tell my students so). Trick questions belong in homework, not on timed tests.

Hmm, I dunno. On the one hand I understand that timed tests produce stress and so forth, on the other hand, does it mean all questions in tests have to be trivial? What is the purpose of the test anyway? I don’t have a strong opinion on this (also I feel liked my perspective is “privileged” since I finished most of my math tests well ahead of time). I can say that, when wearing the hat of an employer, I want candidates that can solve non-trivial questions and I have to test for it myself since a university degree is not much of an indication (then again, my recruiting experience is for pretty elitist groups).

light-rook:

evolution-is-just-a-theorem:

Physicist followers: physic’s textbooks for mathematicians recommendations?

I have a copy of Quantum Theory for Mathematicians, which is pretty good, but I’d like more books. Difficulty of the math isn’t relevant.

@bartlebyshop, @light-rook, @ anyone else who feels like jumping in.

Deligne is the standard recommendation, and also really good as far as I can tell and have heard. I never got around to doing a principled study of it (hurts me to admit it, but it’s a little too mathy for my tastes), and probably “all” you really need (it’s quite long).

cc @nostalgebraist, @mark-gently and I think @alexyar is technically in a mathematical physics group, but I’m pretty sure she hates physics.

I second this recommendation (note that there are two volumes).

Somewhat off topic (but reminded to me by that book), I once spent quite a bit of time translating the theory of harmonic superspaces to mathematical language (i.e. describing them as Grassmanian-like objects in supertwistor space instead of the mess of indexes used by physicists) and the result was quite beautiful, but unfortunately I haven’t published it anywhere.

jadagul:

diffractor:

Learning about eigenvalues for the first time…

Is producing some interesting reactions in my mind, since I don’t understand all the applications so far, so lots of elegant facts about them are not obvious at all and just look miraculous.

So when I come across tidbits like “the sum of the eigenvalues is equal to the trace of the matrix” and “the product of the eigenvalues is equal to the determinant of the matrix”

I’m just like

You eventually start thinking about this the other way. The determinant is the product of the eigenvalues. It happens that, given a matrix, there’s this really annoying formula for calculating the determinant without figuring out what the eigenvalues are explicitly.

(There are a number of advantages to this approach. The big one is probably that it still makes sense if you’re thinking about “an operator” and not actually thinking about a matrix.)

I don’t think this is the ideal way to think about the determinant, in terms of “Platonic mathematical beauty.” The ideal way to think about the determinant, is via alternating n-forms.

Given a vector space V over a field K, a multilinear form or rank r is a function \alpha: V^r -> K which is linear in every argument when fixing all other arguments. A form is call alternating when substituting two equal vectors to any positions yields zero:

\alpha(v_1, v_2 … v_i, w, v_{i+1} … v_j, w, v_{j+1} … v_{r-1}, v_r) = 0

An alternating form is always antisymmetric, i.e. switching any two arguments flips the sign of the result. For K of characteristic other than 2, any antisymmetric form is alternating. For K of characteristic 2 (i.e. 1+1=0 in K), antisymmetric forms and symmetric forms are the same thing while alternating forms are a proper subspace of either (for V of non-zero dimension).

Alternating forms of any given rank obviously form a vector space of their own. For V of finite dimension n, the space of alternating forms of rank n is always 1-dimensional. 

Now, consider a linear operator A: V->V. This operator induces a linear operator B on multilinear forms defined by

(B\alpha)(v_1, v_2 … v_r) = \alpha(A v_1, A v_2 … A v_r)

On the space of alternating forms of rank n, the operator B has to be multiplication by a scalar since this space is 1-dimensional. This scalar is the determinant of A. 

eka-mark:

If we ever build a hypercomputer, algorithmically incompressible strings will suddenly become very interesting! :)

I think that a hypercomputer is not only physically impossible but logically incoherent. What does it mean for an object in the physical world to be a hypercomputer? We don’t have “direct” access to the properties of physical objects: our only access if through testable models. These models have to be computable since our mind is computable*. In other words, if someone handed you a black box claiming it is a halting oracle, there would be no way for you to verify it is indeed a halting oracle. Moreover, there would be no way for you to verify the box is anything incomputable.

* That is, any precisely defined function that you can evaluate using your mind is computable, even if you could think for infinite time and had a perfectly reliable infinite external memory module. Some might argue that since you actually have neither infinite time nor infinite space for your mental computations, we should embrace the more radical position of ultrafinitism. I think this is debatable.

diffractor:

Learning about eigenvalues for the first time…

Is producing some interesting reactions in my mind, since I don’t understand all the applications so far, so lots of elegant facts about them are not obvious at all and just look miraculous.

So when I come across tidbits like “the sum of the eigenvalues is equal to the trace of the matrix” and “the product of the eigenvalues is equal to the determinant of the matrix”

I’m just like

To see that the trace is the sum of the e-values and the det is product of the e-values, it’s enough to understand why

1. Trace and det are invariant under changes of basis.

2. All matrices have a basis in which they have the Jordan normal form. In fact, most matrices (i.e. all but a set of zero measure) have a basis in which they are diagonal (all of this is true over algebraically closed fields, e.g. complex numbers).

The above two facts explain the “miracles” since in Jordan normal form checking these facts is trivial. Now, let’s see how to understand 1 & 2.

1. Algebraically, these are trivial. Namely, det(PAP^{-1}) = det P det A det P^{-1} = det P det A (det P)^{-1} = det A. Also tr(PAP^{-1}) = tr(AP^{-1}P) = tr A. All you need is the formula for changing bases (which follows more or less straight from the definitions) and basic properties of det and tr (although it takes some juggling to show that det is multiplicative). Geometrically, you can see it by observing that applying A to any n-dimensional shape multiplies its volume by |det A|, whereas the sign of det A says whether A preserves orientation (for real matrices; you can in principle derive invariance in the complex case from invariance in the real case by analytic continuation; alternatively, over any field det A is just the action of A on the 1-dimensional space of alternating n-forms). tr A is a bit more tricky but you can see it’s actually the derivative of det: d/dt (det A(t)) = (det A) (tr dA/dt).

2. Proving Jordan normal form is a bit technical so let me just do some hand waving to explain why most matrices are diagonalizable. Introducing an indeterminate x, we can compute the polynomial p(x) = det (A - xI) (characteristic polynomial of A). For A of size nxn, this is a polynomial of degree n. The e-values are just the roots of this polynomial since det(A - lambda I) = 0 iff A - lambda I is degenerate matrix iff it has a kernel i.e. exists non-zero v s.t. (A - lambda I)v = 0 but this is the definition of an e-value / e-vector. Now, most polynomials of degree n have n different roots. This means you have n e-vectors with different e-values and it’s not hard to see they have to be linearly independent*, so they form a basis. In this basis, A becomes diagonal.

* Assume they are dependent. Then, Sum_i c_i v_i = 0. Applying A, Sum_i c_i lambda_i v_i = 0. Multiplying the first equation by lambda_1 and subtracting, we conclude that a subset of size n-1 is linearly dependent. Continuing by induction we arrive at a contradiction.

jadagul:

I think I broke my number theory class today.

They already were losing it on Wednesday because I described the concept of infinity as “an infamous brainfuck.” They didn’t stop laughing for like two or three minutes. (Apparently it’s intrinsically funny when your professor says a curse word in class?)

But today I was discussing the prime number theorem, which says that the number of prime numbers less than x is asymptotic to x/log(x). And of course I had to clarify that that was a natural log, not a log base ten.

And my class was outraged by this, why would I do that, why not just write ln, everyone uses log for log base 10. (Despite my total inability to find anyone who agrees they actually use log for log base 10. Or, really, at all. If you do, please chime in and share!

But at any rate this conversation went on for a while and I wanted to move on, so I just said I was going to do this. And they should just remember that all my logs were natural.

It’s a good thing this was like three minutes from the end of class, because I was worried a couple of them were going to pass out.

I use log for log base 2 and ln for log base e. In computer science at least, it seems like a useful convention.

diffractor:

Mathblr.

If we know a function is one-to-one and continuous, does that imply it is either monotonically increasing or monotonically decreasing?

Yes. Assume to the contrary that a < b < c but f(a) < f(b) > f© (the case f(a) > f(b) < f© is analogous). Choose any y s.t. y < f(b) but y > max(f(a),f(b)). By the intermediate value theorem there is x1 in (a,b) s.t. f(x1) = y and also x2 in (b,c) s.t. f(x2) = y so we got a contradiction with the one-to-one assumption.