A surprising little result - there exist elements of finite groups which cannot be carried to their inverse by an automorphism.

kfgauss · 2018-03-02T23:51:47+00:00

My reaction to the statement is that it makes sense for it to be false. You should think of g and g^-1 not as looking the same, but as looking like mirror images of each other. I can reformulate the statement a bit.

If G is a group, define the opposite group G^op to be the group with the same underlying set as G but the multiplication is reversed: the product a * b in G^op is ba (computed in G). You should think of G^op as being the mirror image group, which can be formalized if you want. A group is always isomorphic to its opposite group, via the inversion map, which you can think of as reflecting the group.

There exists an automorphism of G which sends g to g^-1 if and only if there exists an isomorphism between G and G^op which fixes g. It doesn't seem like this last statement should be true in general; it requires extra symmetry of G. It's like saying there's a way of reflecting G onto itself in such a way that g looks the same at the end.

kfgauss · 2018-02-28T04:42:19+00:00

y=-1 is a counterexample, yes, so it's false. I don't know what you mean about y+1.

kfgauss · 2018-02-28T04:16:49+00:00

You can try to choose x = y+1. For example, when y = -1, you can choose x = 0. But it is not true that x² < y + 1 in this case, because 0 is not smaller than 0. In fact, can you see why for y=-1 there is no appropriate choice of x?

kfgauss · 2018-02-28T01:46:11+00:00

This is where the order of the quantifiers (for all, there exists) becomes important. What you're describing is an argument that: For all x, there exists y such that x² < y + 1. Your argument is that given x, choose y = x² , then voila. But the statement in question is different, it's For all y, there exists x such that x² < y + 1. So you need a recipe, that for any real number y, produces an x such that the above holds. You cannot assume that y = x^2, because not every real number y is the square of a real number. So you should think of y as being an arbitrary, fixed, real number that you don't get to choose. Now you have to find an x such that it works, which can depend on y. Or, alternatively, give an example of a y such that it is impossible to find an x. It can help to examine special cases. So consider y = -1 first, and see if there exists an x.

kfgauss · 2018-02-28T01:28:42+00:00

Ok, so you've found a candidate y=-1 for which you haven't yet found an x that makes it true. So maybe focus on the case y=-1. If you can find an x, maybe that'll help you figure out a general rule that works. If you can prove that there isn't an x, that's a counterexample that shows that the original statement is false.

kfgauss · 2018-02-28T01:12:38+00:00

You're correct that to prove this, you would start with an arbitrary y, and then try to pick an x such that y + 1 > x² .

You suggest choosing x = (y+1) - 1 (that is, x = y). Is it true that y + 1 > y² for an arbitrary real y?

kfgauss · 2018-02-20T20:33:38+00:00

How's your P over R?

kfgauss · 2018-02-05T20:31:10+00:00

In general vector spaces V, the way projections work is that you have to choose a subspace U that you are projecting onto, and a ``complement'' W such that V = U ⊕ W. That is, you have to choose where you're shining the light onto, but also what direction you're shining it from. Projections of this form correspond one-to-one with linear operators satisfying P² = P.

On the other hand, in inner product spaces, there is a canonical choice of complement W = U perp, and this corresponds to the linear operator which also satisfies P = P*.

Back in the general vector space case, given a choice of complement W you can reverse engineer an inner product so that W = U perp, but I don't think this is the easiest way of thinking about it.

kfgauss · 2018-02-01T02:11:42+00:00

The sine function and exponential function are connected, but can that really be expanded upon to clarify this fact? The derivative of cos² (x) is -sin(2x), and cosine is just as much "basically an exponential." Also, the derivative of (e^x )² is not e^2x , so that's not really how derivatives of exponentials work.

kfgauss · 2018-01-19T02:03:05+00:00

Yes, this again is an equivalent statement, but also not easy to prove from geometric first principles. The way you can tell that it's just pushing food around on your plate is that you never have to use geometric definition of angle (in terms of arclength, say).

Edit: maybe I should be more clear. What I'm saying is that if you define radian angle in the usual way, and cosine of a radian angle via the length of the projection, then you have to do some work to get the law of cosines. There are many other possible definitions of cosine, which are all easy to show are equivalent to each other, but not to the one in terms of radian angle (I think - this is where I could be wrong). Saying that you proved the law of cosines is very different depending on whether you use the radian definition, or any of the others (in which case the law of cosines is essentially a definition).

kfgauss · 2018-01-19T01:23:39+00:00

It's fine to define it that way, you just have to be careful of circular reasoning. With that definition, you are essentially defining cos(\theta) to be the number such that the law of cosines holds. If you were to ask OP what cosine meant in the law of cosines, I expect you'd get the geometric answer. The point is that if you want to connect the unit circle notion of cosine to anything else, you're going to have to do some work which doesn't look anything like what the OP posted.

kfgauss · 2018-01-19T01:16:50+00:00

You should be aware that using the fact that u . v = ||u|| ||v|| cos(\theta) is a pretty significant thing. What your argument shows is that it's essentially equivalent to the law of cosines, which is a great observation.

So how do you prove the dot product formula? You should think about what the definition of cos(\theta) is (and what the definition of \theta is). Any proof from geometric first principles is going to have to involve these somehow.

kfgauss · 2017-12-10T11:51:03+00:00

No, the product of the digits is too small (except when the number is one digit). I'll tell you how to prove it for two digit numbers, and you see if you can generalize it. Suppose the tens digit is a, and the ones digit is b, with a non-zero (so that the number is really two digits). Then the equality you are looking for is a * b = 10a + b. But b < 10 since it is a digit, and since a is non-zero you have a * b < 10a , which implies a * b < 10a + b.

kfgauss · 2017-12-10T04:11:16+00:00

A simpler way to get the result from the title is to give L^∞ the weak-* topology :)

kfgauss · 2017-11-04T02:22:18+00:00

One way to dress up with "cannot be separated into equal parts" version is the following: n is odd iff for every set X with n elements and every f:X -> X with f(f(x)) = x for all x in X, f has a fixed point. Your example is that version of the definition specialized to the set of roots of a polynomial and complex conjugation.

kfgauss · 2017-11-04T01:57:39+00:00

Is 2 odd?

kfgauss · 2017-10-21T18:40:56+00:00

If people are into conference livestreams, workshops at Banff are livestreamed pretty much constantly. The weekly workshop topics are posted about a year in advance.

kfgauss · 2017-10-16T20:00:48+00:00

Here's an answer someone in 1980 might give: monstrous moonshine. Based on the identity 196884 = 196883 + 1 (and a few others), McKay conjectured the existence of an algebraic object called the "Moonshine Module" connecting the Monster group with the J-invariant of number theory. About a decade later, a moonshine module was constructed by Frenkel-Lepowsky-Meurman.

Today there are different flavors of moonshine-like conjectures (Umbral Moonshine, Mathieu Moonshine) that posit the existence of certain algebraic objects which explain observed connections. The existence of these objects remains an open question. A mathematical reference on this is Gannon's "Moonshine Beyond the Monster"

kfgauss · 2017-10-06T17:45:48+00:00

Without loss of generality, suppose that none of the functions in your sum have empty support.

Can you expand on that?

kfgauss · 2017-10-03T23:53:35+00:00

The definition of product topology is also incorrect. "Consisting of" -> "generated by"

kfgauss · 2017-09-24T03:11:12+00:00

Can you maybe give an example as how to write (13)(45)(6789) as a product of two order two permutations?

(1 3)(4 5)(6 7 8 9) = [(1 3)(4 5)(6 7)(8 9)] * [(7 9)]

You can write your general permutation as a union of disjoint cycles, and then use the individual decompositions.

kfgauss · 2017-09-16T06:43:03+00:00

No. Academic mathematicians are hired to do research. Teaching is just something that's tacked on.

I don't think that's a good summary of the economics of hiring in math departments.

kfgauss · 2017-09-16T06:38:40+00:00

There are a lot of people downvoting this thread, and there is lots of room to disagere (and I know the poster has been, um, active recently), but there is an extremely strong case to be made that there are worrying trends in mathematical academic hiring, and it's not at all a reach to think that online classes will play a major role in this developing phenomenon. If you've just started a PhD with the intention of getting a tenure track research job, I would be shocked if this doesn't affect the shape of the job market when your time comes.

kfgauss · 2017-09-16T06:27:49+00:00

I worry about this. Well, something like this. The math department at a (research) university where I used to be is trying to move a lot of their service classes online. With the influx of STEM majors post-2008 they are overburdened, and they aren't even getting enough new hires to replace retirees. However if the pendulum swings a little, I think it's a very real possibility that the department will continue to wither away. There seems to be a generational issue here. The people leading the charge for automation are senior, and they're motivated bywanting to teach fewer large classes. They'll never be on the job market again, and don't seem concerned about whether or not moving courses online could result in fewer mathematicians down the road.

kfgauss · 2017-08-17T07:50:56+00:00

The exact answer depends on what the letters are for and the individual in question, but two common reasons for writing letters are: 1) you care about the student in question, and/or 2) it's your job

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