# Maths

Discussion in 'General Advice' started by Imoyram, Nov 1, 2015.

1. ### SnitchanonWhat's a mod to a nonbeliever.

Logs in base 2 show up a bit in computing, but in my experience it's pretty much all base e unless something odd's going on. And then people go out of their way to say "this is a log in base [X]" to avoid confusion.

2. ### ExohedronDoesn't like words

If you do any relativity, you're going to run into the signature convention issue: is it (+---) or (-+++)? I have heard this described as a math versus physics issue, but also as an east coast versus west coast issue.

4. ### ExohedronDoesn't like words

Okay, so Poisson brackets. The primary use in physics is to talk about time evolution. You have your Hamiltonian function H, and then to describe how some physical quantity q changes you compute {q, H}, which, according to Hamilton's laws of motion, is the time derivative of q. In particular, Hamilton's laws of motion, which look different for position and momentum, are unified as df/dt = {f, H} for f being either position or momentum (those should be partial derivatives).

So it gives relations between observable quantities. Notably, if two quantities are conserved, like momentum and energy, then their Poisson bracket is also conserved, and you can determine conserved quantities by their bracket with the Hamiltonian. This is easier than actually computing said quantities over all physically-valid trajectories, as finding said trajectories usually involves solving differential equations, while this involves just checking if various derivatives cancel.

Also the Poisson bracket can be deformed to give quantum mechanics.

I wouldn't say that it is an inner product for two reasons: it outputs functions, not scalars, and it is asymmetric, {f, g} = -{g, f}. If anything, it's more akin to the cross product than the dot product, in the technical sense that it is a Lie bracket.

Last edited: Jan 18, 2016
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5. ### EulersBidentitye^i*[bi] + 1

Anyone explain to me a sentence? It's algebra.

Corollary (iii): If |G| is prime then G is a cyclic group.
[Proof]
From the latter corollary, observe that any group of prime order is unique up to isomorphisms.

What does this mean by "unique up to isomorphisms"?

6. ### ExohedronDoesn't like words

So mathematicians have a thing where they're very careful to distinguish between things that are the same and things that are simply equivalent in some sense. It's part annoying and part fascinating and has led to some very interesting mathematical and philosophical questions about what it means to be equal.

Anyway, the statement means that if G and H are both of order p where p is prime, then G is isomorphic to H. "Unique" without any qualifiers would mean that G would have to be equal to H, but "unique up to isomorphism" means that G and H don't have to be the same, merely isomorphic. The "up to isomorphism" bit means that we're allowed to use isomorphisms to get from one to the other, rather than them having to be equal to start with.
Contrast with G having order, say, 6, where there are at least two groups with order 6 that are not isomorphic. So if G and H are both order 6, that doesn't mean that G is isomorphic to H.

You'll see similar statements in other fields of mathematics, like "up to homeomorphism" in topology or "up to isometry" in geometry, and the idea is the same. G isn't equal to H, but we have some from map G to H that preserves everything we like so that G has the same shape, same structure as H, and since G and H have everything in common, we don't distinguish between them and just say "unique".

Last edited: Feb 13, 2016
7. ### EulersBidentitye^i*[bi] + 1

Hey math folks, I'm 5/8ths of the way through my exam period with exam #6 tomorrow. I (along with everyone else in my year, it seems?) really misjudged the difficulty of the exam & now I'm cramming material which I wasn't expecting I'd need to know. Also, I'm stuck, which is why I'm here. Anyone feel like helping out with some (relatively elementary) complex analysis/PDEs? Right now I'm dragging myself thru separation of variables to solve a BVP and failing to understand how the coefficients of the separated ODEs are calculated.

8. ### ExohedronDoesn't like words

Ah, PDEs. I hated those.
So what kind of issues are you having?
We start with some equation D(x, y, u(x,y)) = 0 for some function u of x and y and some differential operator D.
You should be getting something like u(x, y) = a(x)b(y), which then by some magic becomes D_1(x, a(x))/a(x) = D_2(y, b(y))/b(y), right? And then since the two sides depend on different variables, we get that both have to be constant, and we're left with D_1(x,a(x)) = ka(x), D_2(y, b(y)) = kb(y) for some constant k. Now we're left with hopefully solvable ODEs. Is this where you're having trouble?

9. ### EulersBidentitye^i*[bi] + 1

Yeah, exactly. I've got a general solution in for u as a product of solutions in x,y, and I don't know what to do with it. I've substituted in values from the boundary conditions, and I'm still just sitting here like a lemon. One sec and I'll take pics of where I'm at. I also have a written solution for this problem but I can't follow it.

10. ### esotericPrognosticatorstill really excited about kobolds tbqh

I'm in advanced precalc and what we (and our graphing calculators) call log base e is "natural log," abbreviated "ln(x)." is that not standard?

I get the impression that log base 10 is generally pretty useless, but we've been using it just in the context of exponential variables, because log(a^x) = xlog(a), which is pretty helpful when you're, like, trying to find the equation of an exponential function. I'm pretty sure log base e would also work in that context? also we did a little bit of work messing around with log bases in the context of, like, 5^x=8, find x without taking the log of both sides (usually in a more complex context, obviously), 'cause then x is just log base 5 of 8. but yeah I think those are just, like, novelty applications.

11. ### EulersBidentitye^i*[bi] + 1

Tyvm as always. My exams which required logs and Poisson brackets went okay.

13. ### ExohedronDoesn't like words

Are you sure you have the problem correct? Because from what I can tell, those boundary conditions don't work. In particular, at (0, 0), we get that u(x,0) is supposed to look like sin(3x), which has derivative 3 in the x direction, but we're told that ux(0, t) = 0. Now, we're given that u(x, 0) = sin(3x) is only supposed to hold when x > 0, but it does mean that we aren't going to get a solution that extends to the boundary, at least not with a continuous derivative.

14. ### EulersBidentitye^i*[bi] + 1

Straight from the horse's mouth, I'm afraid. But I can find a different problem to work through.

15. ### ExohedronDoesn't like words

Maybe find a different one.

Anyway, here's how I would solve the problem, ignoring the fact that this isn't going to give a solution to the problem as stated.

We have our general case of

u(x, t) = ax + b + Σμ not 0 e2t/3(Aμcos(μx)+Bμsin(μx))

Note: since μ2 = (-μ)2, cos(μx) = cos(-μx) and sin(μx) = -sin(-μx), we can combine terms and look only at positive μ

u(x, t) = ax + b + Σμ > 0 e2t/3(Pμcos(μx)+Qμsin(μx))

Let's look at our first boundary condition: ux(0, t) = 0. That gives us

ux(0, t) = a + Σμ > 0 e2t/3μQμ = 0

Note that since the functions ekt are linearly independent for different t, we need that the coefficient of e2t/3 to vanish. In other words, μQμ = 0 for each μ. But since all of these μ are nonzero, we can divide out and get Qμ = 0.
Also a = 0.

Next we have u(π, t) = 0. That gives us

u(π, t) = b + Σμ > 0 e2t/3Pμcos(μπ) =0

Again we need the coefficients of e2t/3 to vanish, so we get that Pμcos(μπ) =0 . This has two solutions:
μ = (2n - 1)/2π for positive integer n, or Pμ = 0. Also b = 0.

So we're left with

u(x, t) = Σμ = (2n - 1)/2π e2t/3Pμcos(μx)

At this point you would try to sub in to the third condition and fail.

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16. ### ExohedronDoesn't like words

General tip: Don't write that two expressions are equal to each other if you know their shared value. It doesn't tell you anything new and obscures the fact that you know what their value. So for instance, writing f'(0) = f(π) will only lead you in circles, because both expressions are complicated enough on their own.

Also remember which families of functions are linearly independent. e to the stuff forms a linearly independent family, as do trig functions. So if you get a sum of things that add up to 0, see if you can then say that the coefficients have to be 0.

Last edited: May 26, 2016
17. ### EulersBidentitye^i*[bi] + 1

Is this working on the assumption that μ is an integer?
That part is given in the provided solution for this problem, but it's fairly unclear why. I was mostly hoping that if I kept writing, the meaning would reveal itself...

Honestly, I think I should probably cut my losses and go practise answering questions that I at least vaguely understand.

18. ### ExohedronDoesn't like words

No, we're not assuming μ is an integer. Generally you can't assume integers when doing analysis except maybe when something is a dimension.

And yeah, equation (4) is bad pedagogy if you ask me. They're not trying to say anything meaningful, they're just trying to combine two equations into one equation, which leads to confusion.

19. ### EulersBidentitye^i*[bi] + 1

I did a dumb: was trying to sub x=pi instead of x=0 into ux, couldn't figure how your expression had simplified that way.

20. ### EulersBidentitye^i*[bi] + 1

Yep, cutting losses and hoping I pick up enough marks on the first half of the question to pass this damn exam. I appreciate the help, @Exohedron!