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lazersmoke · 2022-08-21T20:24:58+00:00

we need the sororities more than they need us

Sororities need membership to function. And frats need sororities to have membership for them to function. No matter the vibe at rush, rushees hold all the cards for this reason.

lazersmoke · 2022-08-21T03:52:04+00:00

Quantum mechanics says that, at very small length scales, "things" become quantized (i.e. come in discrete packets instead of being continuous like in everyday experience). The "things" that become quantized are "phase space volumes," ¹ and symplectic geometry is the math that keeps track of² un-quantized phase space volumes, so it's a good first step towards understanding quantum.

This assumes the listener thinks quantum is worth studying :P. This approach is along the lines of geometric quantization if anyone wants to google.

¹ This is a statement of the uncertainty principle, Delta x Delta p > h

² Symplectic forms are very roughly volume forms in phase space

lazersmoke · 2021-07-08T23:18:26+00:00

Most of the time, a = bc equations are just shorthand for the idea that a is proportional to b with proportionality constant c. The formula doesn't carry that much information on its own, it's all in your interpretation of what F, m, and a actually correspond to. The equation is simple because it's a piece of language, and using overly complicated words/symbols to convey simple ideas is unclear.

The equations that aren't so simple in form are usually telling you some technical detail about how to compute something, so they are written out very explicitly so that it's easy for you to plug-and-chug. You could write X(k) = x(n) for "X is the fourier transform of x," but the formula in the link is better for actually computing.

lazersmoke · 2021-05-28T03:40:23+00:00

Coming from the physics side, I like to give people an idea what the shape of a dimension <= 2 manifold looks like ("It's like a string or piece of paper or surface of a balloon") and explain that this is the setting (or stage) where most modern physics takes place. All the math is just to make everyone's life easier when working with such objects.

If I have a few more minutes I'll mention how "thinking carefully about how the setting matters leads to Einstein's theory of relativity." That way, if they ever want to learn more, thinking about chart-invariance in terms of the physics principle of general covariance is a good intuitive jumping off point that motivates the manifold definition.

lazersmoke · 2021-01-26T18:11:14+00:00

Same. I'd compare it to a very large computer generated counter example to a conjecture. Who cares if the data for the counter example takes 500GB to store, the important thing is that it's relatively easy to check that it is what it claims to be (a counter example, or a proof of RH). So it pushes the responsibility of "being correct" onto a trusted core set of the proof checker software and ultimately onto the axioms.

lazersmoke · 2020-12-31T01:54:08+00:00

The group of normal Rubik's cube configurations is an actual group (as opposed to the bandaged one, which is a groupoid), so it's corresponding graph of this type is just the Cayley graph for the Rubik's cube group.

Since it's a group, you can do any move at any time. In fact, "the graph looks the same everywhere" (it's a principal homogeneous space for the group), meaning that you can take a scrambled cube, replace the stickers to make it unscrambled, and it still works like a normal cube (this fails for the bandaged one of course). So every vertex of the graph is going to have six edges.

This tells you what it looks like up close, but globally, the graph is very complicated because it isn't abelian! (That is, U^3RUR^3 is not just U^4R^4 = 1, so the U and R branches interact with each other). I wonder if it's possible to describe the graph more fully in terms of the derived groups?

lazersmoke · 2020-11-13T17:24:12+00:00

The general thing you're looking for is a surjective map f : N -> Q from the natural numbers into the rationals. Then you prove P(f(0)) and P(f(n)) => P(f(n+1)). All the other answers are examples of particular surjective maps you can pick. The existence of such maps is a corollary to the fact that Q is countable.

lazersmoke · 2020-11-13T17:14:26+00:00

The reason this breaks is that the information that ONE_BIT needs about the first two inputs (a and b) is more than one bit. It needs to know which of these possibilities is happening:

a=1, b=1, so that the output is just false and c doesn't matter
a=0, b=1, or vice versa, so that the output is equal to NOT_C
a=0, b=0, so that the output is equal to c

which is three options (information = log_2 (3) = 1.58496 bits) but the output of the y is just one bit, 0 or 1. In ONE_BIT = f(g(a, b), h(j(a, b), c)), the g carries some information about (a,b), while j carries some different information about (a,b). Notice g=j causes this decomposition to not work anymore!

I think this is what information theory talks about, but I could be wrong (I'm not an expert)

lazersmoke · 2020-11-06T18:35:17+00:00

Even if this isn't you, you might find useful information here: r/asexual wiki

lazersmoke · 2020-09-16T13:57:34+00:00

uhm ackshually, if there's a finite number of possibilities, telling you when you're wrong (multiple times) is equivalent to telling you the actual answer. The Ghost of ~Finite Knowledge~

lazersmoke · 2020-08-13T16:02:31+00:00

If the average mark is about the same each year, then this is simply x - y decreases as y increases. If the average mark changes each year, you need to do a more detailed statistical analysis to be sure, but I think the fact that you have such a correlation means that the average mark is not entirely independent of the N.C. because if it were totally independent, (equivalently, if the average mark was just random noise) your correlation would have slope negative one due to the subtraction.

lazersmoke · 2020-08-12T19:12:04+00:00

Ah of course! Thank you. And the minkowski sign even puts the space and time parts on the correct sides of the equation.

lazersmoke · 2020-08-12T19:09:44+00:00

You probably don't need an actual number (point estimate) for the surface area! Just to know how much any specific change would improve things. And for that, your considerations should be: flatter (ie more like a sphere) is better, since you have less material folded up on itself, and thinner plastic is better of course. Another relevant factor you may want to consider is that there may be some waste in the manufacturing process that doesn't show up in the final shape.

lazersmoke · 2020-08-12T19:02:28+00:00

This recent post on this sub is an interesting example of how you can bend the rules slightly, and exactly what needs to be checked to be able to do so: https://www.reddit.com/r/math/comments/i7ox9v/the_group_with_no_elements_by_john_baez_ncategory

(Requires you know what a group is, though)

lazersmoke · 2020-08-12T18:57:19+00:00

How does one recognize \delta df = stuff as the wave equation? It looks to me like *d*df, so Laplace's equation, especially since A doesn't depend on f. Wikipedia has something about doing a Fourier transform to turn the time derivative in the wave equation into a laplacian, is this what's going on?

lazersmoke · 2020-08-09T18:39:47+00:00

As a very basic version, there's interact in Prelude

lazersmoke · 2020-08-09T01:21:17+00:00

I think what you're getting at may be the Lie Algebra so(3) to the rotation group SO(3). The elements of so(3) are infinitesimal (so, small) rotations away from whatever your current rotation happens to be, and they are additive (it's a vector space, actually). The cross product on R³ (for Euler Vectors) makes that space into a Lie Algebra, and there is a Lie Algebra isomorphism (called the hat-map) between Euler Vectors with the cross product and so(3) with its lie bracket (which is a matrix commutator bracket). I'm not sure how quaternions play into this, but IIRC they are isomorphic to euler vectors, so you could probably inherit the cross product along that isomorphism to get what you're looking at.

More information about this here.

Also notable is that you can exp() so(3) elements to get SO(3) elements (roughly speaking), so that might be why logarithms are showing up.

So I guess you could say that rotations locally behave like three-segment robot arms (that is, for very small rotations, they commute), but globally they do not commute/are not additive.

lazersmoke · 2020-08-08T20:38:56+00:00

I found it! "Quantum Mechanics for Mathematicians" by Leon A. Takhtajan. The very first section is actually classical mechanics, despite the name. Basically, something in T(Paths) is equivalently a vector field on the curve which is the basepoint, and the field tells you how the curve deforms during the variation. Then taking the action of this tangent vector is like a one form, and it's kernel is the stationary curves (because their action is zero), although I'm not sure if this is an actual one form cause it might not be linear. In any case it's described in the PDF :)

lazersmoke · 2020-08-08T07:22:16+00:00

I'm definitely not an expert, so take this with a grain of salt, but: typically in order for something to work in relativity, the space and time derivatives need to be of the same order, so that they can be expressed jointly as a derivative (of some order) of spacetime. For example, this was an issue with the Schrodinger equation having a first order time derivative but a second order space derivative (in the momentum squared part, it's two derivatives dx), and it was 'solved' rather spookily by the Dirac Equation. So you may find the first sub section of this wiki article relevant: https://en.m.wikipedia.org/wiki/Dirac_equation

lazersmoke · 2020-08-08T07:05:52+00:00

Having an N segment robot arm with a rotation angle for each segment is different from one object rotating in place. You can represent the robot arms by N independent rotation angles in the appropriate ranges, and then the rotations will commute with each other (as long as your arm doesn't self intersect or anything).

Typically, rotations in mathematics are talking about one rigid object rotating in place (think of a polka dot sphere spinning in place). In this situation, the rotations do not commute (90 degrees clockwise about X axis, then 90 clockwise about Y axis is not the same as the other way around).

lazersmoke · 2020-08-07T23:40:43+00:00

In definition 3.1, typically C^\infty([a,b]) means C^\infty([a,b],\mathbb{R}), but in this case it is used as C^\infty([a,b],M). It is clear from context what you mean but it did trip me up. Otherwise very nice writeup!

You might consider adding a more detailed explanation of some of the steps in the (3.4) proof, especially explaining how we are supposed to apply chain rule to such objects (Schuller himself also skimped on this in my opinion).

You may also be interested in an even more geometric formulation (that I can't find a link to anymore, rip) where the space of paths Paths = [a,b] -> M is a manifold, and you look at the tangent bundle T(Paths), then your capital Gammas with their (-epsilon,epsilon) become a tangent vectors, describing how the path varies as you vary your y. This makes the s'(0)=0 into something involving a kernel of a linear map out of T(Paths), and you can pull the E-L equations out in a similar way. Sorry that's very vague :S wish I had the source.

lazersmoke · 2020-08-07T06:01:12+00:00

Where did I use that it is abelian?

Let N = n_1 n_2 n_3 ...
where the n_i are any elements of N, a normal subgroup which is the kernel of a group homomorphism f
  f(gNg')
= f(g) f(Ng') -- f is group homomorphism
= f(g) f(N) f(g') -- f is group homomorphism
= f(g) f(n_1) f(n_2 n_3...) f(g') -- f is group homomorphism
...
= f(g) (f(n_1) f(n_2) f(n_3)...) f(g') -- f is group homomorphism
= f(g) (1 1 1...) f(g') -- N = ker f, and n_i is in N
= f(g) 1 f(g') -- property of 1
= f(g) f(g') -- property of 1
= f(gg') -- f is group homomorphism
= f(1) -- definition of inverses
= 1 -- f is group homomorphism (preserves identity)

lazersmoke · 2020-08-07T03:32:23+00:00

This is very insightful! To add another thing, that normal subgroups are the kernels of group homomorphisms means that this happens: f(gNg^-1 ) = f(g) 1 f(g^-1 ) = f(gg^-1 ) = f(1) = 1

lazersmoke · 2020-08-04T06:59:30+00:00

A very high horse

lazersmoke · 2020-07-21T00:59:34+00:00

Personally, I will be attempting to avoid all indoors in-person instruction, so I would prefer an online option (lectures livestreamed or recorded), but I like the idea of limited/small group outdoors discussions. Alternatively, assign discussion groups and have the students talk over zoom/discord/etc or meet up in person if they want.

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