12:07 AM
let $\chi_1, \dots, \chi_n$ be the irreducible characters and let $\chi$ be the character of the regular representation. We want to compute $\sum_{i=1}^n \chi_i(1)$ and show that this is equal to $\frac{1}{|G|}\sum_{g \in G}\chi(g^2)$ (that's the number of elements such that $g^2=1$, as we explicitly know what $\chi$ looks like.)
Now write $\chi=\sum_{i=1}^n \chi_i(1)\chi_i$ (this follows from the decomosition of the regular representation), then we have $\frac{1}{|G|}\sum_{g \in G} \chi(g^2)=\frac{1}{|G|} \sum_{g \in G} \sum_{i=1}^n \chi_i(1)\chi_i(g^2)$ interchange the sums, this is equal
that was a neat problem
I hope you know Frobenius-Schur indicators

Yeah it's cute
That's why I posted it :)

though your statement is off by one
you need to count $1$ as well, but that doesn't have order two

Ah, that's true
Anyways I'm off to sleep
Bye

12:46 AM
@anakhro In case you didn't figure it out yet, start with this: $P(y)=y$ iff $y\in \text{im}(P)$.

looks like Federer could end his career with his knee surgery
:(

1:11 AM
@TedShifrin that one is trivial to prove. :(

i
hi

hi mr. hawk

I got a stupid algebra question. If x > 0, and x^2 = a^2, why can't we have x = -a where a< 0?

What makes you think you can't? @Hawk

it is a derivative of this closed question math.stackexchange.com/questions/2838005/…. There are other equations, but they don't matter in this case.
all the answers insist that the sign of x>0 forces a>0, I can't just can't see it

1:26 AM
From $x>0$ alone, one can deduce that $x=-y$ for some $y<0$, @Hawk.

so how are they insisting a > 0?

Oh, and hello everyone :)

start from first principles

Consider $x=1, a=-1$. Then $x^2=a^2$ while $x=-a$.

r^2 = s^2 if and only if r = s or r = -s.
for all real numbers r and s

1:32 AM
okay...but isn't tis in agreement with what I wrote?

what you wrote has restrictions on the signs of r and s

what are you trying to tell me? If x > 0, x = a or x = -a, if x = -a (a <0), the original map isn't one to one

@anakhro But think of it in terms of lengths: Both $P$ and $Q$ decrease the length of the vector unless the vector lies in their respective images. That does it.

That's the whole $\|Px\| \leq \|x\|$ shindig.
So $\|P(Q(x))\| \leqq \|Q(x)\| \leqq \|x\|$

I ran into an instance of that inequality in an intro physics course this week, amusingly

1:46 AM
In quantum mechanics, I guess.

Nope

w h a t

@anakhro and equality holds iff ...

If you drop an object, it accelerates downward at g=9.8 m/s^2. If you allow a cart to slide down an inclined ramp, then (effectively) you project the acceleration vector onto the direction down the ramp
And this acceleration is always smaller than g

yup ...

1:49 AM
Iff $x\in im(Q)$

for the first (second) one

If you take the coordinates to be relative to the ground rather than the ramp, then you further project the acceleration down the ramp onto its x and y components

x and y components are overrated

Which means you get a quite small acceleration in the vertical coordinate, assuming the angle is small
Precisely because it’s the projection of a projection

something to do with complementary angles

1:52 AM
Hot dog I see it.
Ted, do you mind if I call you Teddy from now on.
As a declaration of my admiration

You also see this kind of inequality in QM, to be sure

Definitely Cauchy-Schwarz in QM ...

i had in mind $|Px|\leq|x|$, to be clear
In which case it translates in QM to “you’re never more likely to get some measurement result than any measurement result” :P
Main use for Cauchy-Schwarz in QM off the top of my head is to prove a version of the Heisenberg uncertainty principle

2:08 AM
@Semiclassical nice insertion of the quantifier: "measurement" :P

yeah, 'get something' was a bit too loose

did you point out the importance of the statement to the intro students?

to clarify: the QM statement wasn't for intro students
I did point out the geometry of the lab setup, though
(not as |Px|<|x|, but the idea that projecting a vector always makes it smaller)

good enough
are you using the rubber band experiment, yet?

the experiment worked out quite well, at least qualitatively. (i've yet to see a good way to make that lab quantitative)

2:14 AM
coolio

my favorite part of that lab has to be the following: Suppose you arrange for the rubber band to be supporting most of the weight, with the scale supporting almost one of it. Then when the rubber band is heated and contracts, it'll actually lift the weight off the scale.
not by much: best I've gotten is about a centimeter, and I might be overestimating that
but still fun to watch.

I want some inequality like $(a+b)^2 \leq 2(a^2 + b^2)$...
It looks familiar but I can't put my finger on it.

$2(a^2+b^2)-(a+b)^2=a^2-2ab+b^2=(a-b)^2$

One of these days I will ask you guys harder questions, I promise.
looks off into the distance.
@Semiclassical thanks
You must understand, I would come up with an endearing nickname for you much like "Teddy" if I could.

2:28 AM
"Classy"?

Okay, let's not go that far, @skullpatrol

Photobucket flashbacks

SemiSavage :P

3:09 AM
@AkivaWeinberger can you send me that facebook news article again about the DiffEQs?

4 hours later…
6:45 AM
@Akiva DogAteMy: Have you heard the Brentano Quartet? They're in residence at Yale. I just heard them in concert tonight.

4 hours later…
10:54 AM
0

This was a Scientist job advertisement of famous space Institution. Third column describes the job responsibility. They may conduct skill test. How do a person develop these skills? Can you suggest How does a person prepare for this skill test? Please suggest textbooks.

@Mathgeek Looks very general the description in the image you posted. I would not know how to prepare for such a job unless they wrote somehting more specific (in the description).

sir, I need to study the basics of these mathematical modelling
@MatsGranvik

yes I hear you. My experience is mainly in janitoring so I don't know how to help.

I think I need to learn the fundamentals of Mathematical modelling using Matlab.

Do you know Mathematica?

11:03 AM

Or Scilab, which is a Matlab clone.

How do I learn the basics of mathematical modelling?
can you suggest some textbook?
this poster was of the ISRO.
Indian space research organisation.

Why does wolframalpha give -2 for 10 mod -3 and -10 mod 3? Shouldn't it be +/- 1?
https://www.wolframalpha.com/input/?i=10+mod-3
https://www.wolframalpha.com/input/?i=-10+mod+3

Microsoft Office Excel: =MOD(10,-3)
gives -2 as well
clock wise arithmetic
one way or the other
left or right
Mod[10, Range[20] - 10]
{-8, -6, -4, -2, 0, -2, -2, 0, 0, Indeterminate, 0, 0, 1, 2, 0, 4, 3, 2, 1, 0}
Mod[Range[20] - 10, -3]
{0, -2, -1, 0, -2, -1, 0, -2, -1, 0, -2, -1, 0, -2, -1, 0, -2, -1, 0, -2}
which is periodic, and which would explain why
Conjecture: There is a distribution that consists of the truncated bounded Dirichlet series that converges in the limit to the inverse error function.
nnn = 5;
g1 = Table[
T = Tuples[
Table[Table[If[k == 1, nn, n], {n, -(k - 1), k - 1}], {k, 1, nn}]];
ListLinePlot[
Sort[Table[
Total[T[[n]]/Range[Length[T[[n]]]]] - nn, {n, 1,
Length[T]}]]], {nn, 1, nnn}]
Why is that?

11:42 AM
Is it true that |sqrt(z)|^2 (the squared modulus of the principal square root of a complex number) always equals |z|?

11:53 AM
Or more generally, take a function f and assume that sqrt(f(z)) is analytic in a certain symmetric strip around the real axis (so f does not vanish and sqrt(f(z) has no branch points in this region). Is it true that |sqrt(f(z))|^2 = |f(z)| ?

@courge9 yes
@courge9 yes

So that's a "yes" to both questions? :-D
Then I just have to convince myself from the definition that my intuition is right

12:58 PM
@TedShifrin

3 hours later…
4:06 PM
Let $\mathrm{dim}$ denote the Lebesgue covering dimension. What are reasonable assumptions on a space $X$ that guarantee $Y\subseteq X\implies\mathrm{dim}(Y)\leq\mathrm{dim}(X)$?

4:24 PM
Hi chat! How can I easily see that these two are equivalent (linear algebra) $$T\alpha_i=\sum_{j=1}^n A_{ji}\alpha_j,\quad 1\leq i\leq n\iff \sum_{j=1}^n(\delta_{ij}T-A_{ji}I)\alpha_j=0,\quad 1\leq i\leq n.$$

4:45 PM
@Lukas there's something that's not quite clear to me, idk if it's obvious or not, but in the source I'm reading the author claims that two forms are independent just by looking at the first two coefficients of the q-expansions; why is this the case?

5:21 PM
Because dependent form have equal first coefficients in the q-expansions @Edward
(I have no idea what I'm talking about)

rofl actually I think it's quite a dumb reason that I missed
if there's a linear dependence then one is a constant times the other

Does that imply that the coefficients are also multiples by the same constant?

guess so

Because then if the first two pairs of coefficients have different ratios you should be done I guess?

5:25 PM
But I don't even know what kind of forms you're talking about (or what a q-expansion is) so don't trust me :P

modular forms lol
q expansion is just a fourier expansion

5:44 PM
Hello guys! I have this statement abut Calculus: Given $f(x)=3/x^2$, let $A=(5,0),B=(x,0),C=(x,f(x))$ with $5<x<20$ be points on $\Bbb{R}^2$. Find $x$ such that the area of the triangle $ABC$ is max
My question is: The area of the triangle should be $A(x)=(x-5)\cdot3/x^2\cdot1/2$ right? Thank you!

6:09 PM
yes

4 hours later…
9:49 PM
@StupidQuestionsInc I've never seen something quite like that that I recall. Well, $\sum_j \delta_{ij}T\alpha_j = T\alpha_i$, and $A_{ji}I\alpha_j = A_{ji}\alpha_j$, right?

Hi @Ted

10:09 PM
Hi, demonic @Alessandro.

how do you attach a video here
it only says you can upload an image

I do not know that videos can be uploaded.

oh one time Igunummeg uploaded one i think, oh well..
oh I can just do a link

Hello
@geocalc33 I can't open the link so I doubt it is a fun read.

@JackOhara google it if you really want to read it. It is fun though, at least for me
homeowmorphism.com
the article is called: Intro to Mapping Class Groups, Teichmuller Spaces and Hyperbolic Surfaces.

10:26 PM
@geocalc33 Thanks!
now we can have an opinion about it since the link works

Yeah I have a few questions about the information so far

10:51 PM
HEllo, I'm a little new to stack exchange, is this an ok forum to ask specific questions?

Of course. Whether you get a good answer depends on who's around. :)

Awesome!
So here is my question: I am studying mathematical statistics, and I am currently struggling with a question about linear functionals. It looks so simple and straightforward that I know there must be a trick, but I'm having trouble seeing what the conceptual undercurrent of the question is
here goes:
Let X1,X2,...,Xn ∼ F. Let θ = F(t) where t is a fixed number.
(a) Is θ a linear functional? Why/why not?

I don't understand your "let" sentence.

It's just the premise of the question

It's not a sentence.

10:55 PM
As in, X_1, X_2 etc, are a set of random variables

and whats the - F?

a distribution

So is there a sentence you can write so this makes sense?

Oh, that is supposed to be a "follows a distribution" symbol

Oh.
If $t$ is a fixed number, $\theta$ is a fixed number. Why should this be a linear functional? $F$ is not a linear function on the real numbers, so I don't see how this could possibly be.

10:58 PM
Thank you. I want to see if I understand what you are saying:
Since F(t) is a CDF, I apologize if I did not make that clear, then F(t), with t as a fixed number, is also a fixed number, and as a fixed number, it cannot be a linear functional?

Right. A linear functional is a (real-valued) linear function on a vector space.

And this doesn't have anything to do with a statistical functional, as something that takes a function and delivers a number?

For example, integration $\int_a^b\,dt$ is a linear functional on the space of continuous functions on $[a,b]$.
What is the context you have that makes you think $\theta$ should be a linear functional?

Only context is in the problem itself...

It's just a plain old real-valued function. If $F$ was varying, you could possibly interpret it as evaluation at $t$ on the vector space of functions, but CDF's don't form a vector space.