What is the difference between a characteristic value of a linear functional and a characteristic value of a linear opearator.

@NV-US: I've never heard of the first one.

Ok. Thank u

Have you really seen that?

No, sorry, was confused

I have a question, typing......

Heya @Ali!!

Given a linear operator T on the space of all cts. real valued functions given by, T(f) = integral (from 0 to x) f(t)dt. Show that T has no characteristic values. I solved T(f) = pf and got, f(x) = Ae^(x/p), where A is constant

How to proceed now?

Sorry about, no latex. I am on my phone, it hangs

@TedShifrin

OK, so let's see. You get $pf'(x)=f(x)$, so $f'(x)/f(x) = 1/p$.

I agree with your solution.

Hmm ... So I'm confused.

How can i show that T has no char. values?

It seems like you showed every $p\ne 0$ is a characteristic value.

For all i know, the char. function i got, does belong in the space for p not equal to 0

Ohhhhh ...

Yeah, i think so too

But we're being silly.

What is your $A$?

Constant

Is there some value of $x$ where you know what $T(f)(x)$ must be?

I dont understand

Look at $T(f)$.

Do you know its value, for sure, at some value of $x$?

Like a constant value?

Yes.

Look at $\int_0^x f(t)dt$.

f(t)=5

T(x)=5x

f(t)=t gives T(f)=x^2/2

No, no. We don't know anything about $f(x)$. No matter what continuous function $f$ you look at, what value of $x$ can you tell me the value of $\int_0^x f(t)dt$?

At x=0

BINGO!!!

And the value is .... ?

0

So now tell me what you know about your constant $A$?

Thinking

A is 0?

Yup. Now can you answer the question? :)

Hi @Ted

Oh oh, @Balarka, you're breaking your sleep cycle again? :)

I have to go to school tomorrow so probably I'll go to bed in half an hour

Wow, you actually go to school?

Heh

BTW, with regard to that $S^{\omega + 1}$ question, I realized that if you glue two contractible things along something contractible, van Kampen isn't very interesting. :P

@TedShifrin hello Ted :)

I said hi cuz it said you were here earlier :)

Hey @Ted, @Ali, and @Balarka!

Sure, $\pi_1$ is not hard to calculate :P

Hi Demonark

Hi @BalarkaSen

Gluing two contractible things along something contractible actually gives you a contractible object though

Right :)

So T has no char values because we cannot find a non zero function such that T(f) = pf

Perfect, @NV-US.

Hi @Ali

Thank you for the help :)

You're most welcome.

But how did u get that we were doing something wrong earlier?

We didn't do anything wrong. We just didn't figure out $A$ ... And I saw it instantaneously.

You have to use some initial value information when you have it!

Ok, i will remember that

@Ted What does it mean to take connection of tensors? I suspect $\nabla_X T(X_1, \cdots, X_n) = -X T(X_1, \cdots, X_n) + \sum_i T(X_1, \cdots, \nabla_X X_i, \cdots, X_n)$

Good :)

Oh also hi @Daminark

@Balarka: A connection on vector fields induces a connection on tensors of type $(k,\ell)$.

It is a derivation that commutes with contraction.

For (k, 0) tensors is it by the formula I wrote?

Ah

If you consider a covering of X with included finite intersections then you can make this into a category O with morphisms the usual inclusions. Then all van kampen says is the fundamental groupoid of X is the colimit of fundamental groupoids of objects of O

i don't really care about that point of view

Are you talking to me or to Ali?

That was to Ali

probably me... :(

If $T$ is an $(n,0)$ tensor then $T(X_1,\dots,X_n)$ is a scalar function and it's just differentiation in the direction of $X$. But it equals the sum when you use the derivation property. So your formula isn't right.

Categorical language is a sure way for me to ignore you, Ali, but that's OK. :P

Particularly today. I had a tooth extracted, and I'm bleeding and in pain :P

@TedShifrin Hm

I thought there had to be a minus sign somewhere because metric connections satisfy $\nabla g = 0$

@Ted sorry, I hope you feel better soon

Give me 6-8 months, Demonark.

The issue is that what you wrote is ambiguous, Balarka. You mean $(\nabla_X T)(X_1,\dots,X_n)$. :P

Wait is it gonna take that long? Damn

@Ted Oh, I thought it was apparent?

It wasn't to me until you bitched.

Sorry lol

@TedShifrin is torture standard procedure now?

yelloha

Yeah you're right

Demonark: A bunch of time to heal, then implant, then a few months, then crown ...

So eating will be interesting for a while, even after a few days :P

@TedShifrin I gotta prove that GL_2 (Z/2) is isomorphic to S_3

Also I hope you get better.

I dont need help but just tips

Thanks, Balarka.

I have the 6 elements and studiying them by order

Just explicitly write out $GL_2(\Bbb Z_2)$, Kasmir. Write down the 6 matrices.

Ohhh.

-.-

Tooth problems are the absolute worst. Unfortunately I don't brush so maybe my teeth will fall off in my twenties

That's really bad news, Balarka, plus you stink.

I just need some tips :D

Ehm well

Make the multiplication table, Kasmir.

the order of them are 1, 2, 3

hint how many groups of order 6

He doesn't know that yet, @Ali, I bet.

They're trying to get him to write down the isomorphism explicitly, I bet.

2 of order 3 how to distingusih them in a fast way ?

I have such an exercise in my algebra book!

you got a book ?

on algebra too ?

Yup.

@Ted "Stink" is the mainstream counterculture style that comes after "punk"

OMGGGGGGG

I embrace it

where can I buy it?

LOL, Balarka.

Ted pls give me link :D

type in Ted Shifrin anywhere

You can find it linked on my university homepage, which is linked on the homepage here

I thought you did multivarible calculus and geometry only =p

you told me this in fact i think :D

Abstract geometry: An algebraic approach

LOL, Demonark. Right.

Okay searching now :D

Smacking Demonark: A Geometric Approach

Actually, my very first coauthor complained that I was too much of an algebraist because I loved to use differential forms. ... We never wrote a second paper together :P

amazon.com/Abstract-Algebra-Geometric-Theodore-Shifrin/dp/…

this ?

Anyhow, @Kasmir, who cares? The multiplication tables will match up regardless. You just have to do it.

@TedShifrin hahah

Yes, yes. Stop making it look like I am running a business here!

HAHAHA

Lol I want to see that author talk to Nori...

omg I really did not know you have a book on algebra

That is the best news :D

It has a bunch of projective geometry in the last chapter, which makes it a bit unique.

becaue you explain things in a good way

Well, I do groups way after everything, so it won't match your course.

I did integers, polynomials, commutative rings first because they're easier than groups.

So, as I say, it won't match your course.

It is good anyway to pick few chapters

we gonan do rings and polys anyway

what did you mean by they will match ?

there are 3 elements of order 2

I think normal subgroups and quotient groups are harder than ideals and quotient rings, but I guess it's readable without the earlier chapters.

so how do I know whitch is mapped to which

You just need to make the multiplication tables consistent.

You don't want to think just about order.

Yes how the elements "multiplication" do to each one

but I ll do it my way, and then tell me if there is a better systematic way ( if i did not find it )

=P

Where one element of order 2 goes and where one element of order 3 goes will determine everything, Kasmir. Check it out.

okay =p

@Balarka: I think you're off by a minus sign with your formula!

@TedShifrin Actually maybe the minus should have been on the second term instead of the first.

Right.

I was writing it all out and confuzzling myself with that. Thanks

Can anyone think of ANY function g such that $\lim_{x \rightarrow 1} \frac{g(x)}{x-1} = 4$ but $\lim_{x \rightarrow 1} g(x) \neq 0$?

Tried a bunch of things but I keep getting = 0

ANY function is fine, need not be continuous

@Wildcard math.stackexchange.com/questions/2420799/…

@Wildcard for the record, even though intuitively the process is 'smooth' (you could balance the lines out by slowly moving them) I think with larger line (or plane) counts you start running into the problem where the sides of the square/cube are very large compared to the combinatorial explosion of tiny regions you get in the center of the cube where all lines (or planes) overlap

@IrregularUser it isn't possible, x-1 tends to 0 so if g(x) doesn't tend to 0, the quotient blows up

@Wildcard $L(n) = \frac{1}{2}n(n+1) + 1$, $P(n) = \frac{1}{6}n^3 + \frac{5}{6}n + 1$

@Wildcard even stronger statement that I can't disprove is the following

given a set of lines on a square, can you smoothly translate and/or rotate the lines such that all regions have equal area, without ever reducing or increasing the amount of regions?

@Daminark Ah, thanks for that

@orlp I think that is the key.

@orlp I fixed the definition. Got interrupted but now it's done.

@Wildcard it's definitely a stronger statement though

Let me try again....

although I have to say I have 0 knowledge in advanced geometry proofs

@Wildcard \binom{a}{b} $ = \binom{a}{b}$

@orlp $L(n)={n\choose0}+{n\choose1}+{n\choose2}$

@orlp $P(n)={n\choose0}+{n\choose1}+{n\choose2}+{n\choose3}$

Is it safe to assume that the maximum number of regions in $d$ dimensions using $n$ cuts of $d-1$ dimensional dividers is $\displaystyle R_d(n) = \sum_{k=0}^d \binom{n}{k}$?

(there must be a proper term for '$d-1$ dimensional dividers' that form regions, right?)

@anon hi

hi

am trying to show that GL_2 (Z/2) is isomorphic to S_3

using table ( we did not do fancy formulas yet )

Is there a systematic way of doing this?

I mean a way that is clever ( not trial and error)

I found the order of the elements

you can interpret the vector space as {0,a,b,c}, where adding a letter to itself gives 0 and adding two distinct letters gives the third. every permutation of {a,b,c} (fixing 0) is then a linear map (preserves this addition operation).

can you give an example of such method?

I mean how it works

I just did

if you interpret the addition the way I said to, the result becomes obvious

but if you're in an intro class, this problem is not too much to do by simply working out everything

@orlp I haven't worked out why the pattern works the way it does.

I didn't work out that pretty closed form on my own, unfortunately.

@anon am in intro class =p

@Wildcard have you looked at the relevant OEIS sequences and their references?

The exercise in the book is to work out what the recurrence is, not to put it in closed form. I did that successfully. When I checked my answer, the neat patterned closed form was given.

I can see that the recurrence holds that:

@anon that is a fancy way :D thanks

But if I list the two tables

how much info can I get to find what element maps to what

$R_d(n)=R_d(n-1)+R_{d-1}(n-1)$ for all $d$ and all $n$, and that $R_d(0)=1$ for all $d$.

I have three elements of order 2 , and two elements of order 3

do you know about group actions?

nope =p

@orlp So all that's left is to prove that the recurrence I've defined above is equivalent to this summation and you're all set.

We only had the definition of isomorphism

I need to find the map phi

after that I can work out 1-1 and onto

I think we have to do it the hard way on this exercice to appriate the fanciar methods

@orlp But my question is really about the specific case of cutting a cube with 4 planes. That's the first interesting 3 dimensional case.

you don't know what the correspondence between the matrices and the permutations is?

exactly

label the three nonzero vectors 1,2,3. then for each matrix, figure out how it permutes those three vectors.

@Wildcard I don't know why you're jumping straight into cubes

try solving squares & lines first

(lines and points is trivially true)

@orlp If a square can be cut in 7 equal area portions by 3 straight lines then I think a cube can probably be cut into 15 equal volume portions with 4 planes.

@orlp because it's what I thought of first. ;)

@anon hmm can you please explain in more detail? what vectors?

If it is possible, I'd like a puzzle consisting of 15 pieces of different colors, all the same volume, that can be assembled into a cube.

@KasmirKhaan "label the three nonzero vectors"

@orlp yes, hmmm.

do you consider (123) a vector?

no. do you know what GL(2,Z/2Z) means?

yes its the general linear group of 2x2 matrices in mod 2

2x2 matrices with entries from Z/2Z. what do we do with 2x2 matrices? we apply them to column vectors. column vectors with entries from Z/2Z in this case.

yes I have written all the 6 such matrices

if we write down row vectors, the finite vector space (Z/2Z)^2 has four elements: {(0,0), (0,1), (1,0) and (1,1)}

okat

okay*

the three nonzero vectors are (0,1), (1,0), (1,1)

okay so far so good

call them a=(0,1), b=(1,0), c=(1,1)

okay

for each matrix, figure out how the corresponding matrix transformation permutes a,b,c

for example, the matrix with top row (0,1) and bottom row (1,0) swaps coordinates, so it permutes a and b while fixing c, so the permutation in one-line notation is (ab)(c).

genious :D

@anon Thanks alot ! :D first time i see something like this =P

mmhmm

@anon are you a proffesor of math olympiades?

no

Okay =p

Ill keep working using this method and see where things go :D

@anon the top row (1 1) and bottom row (0 1 ) , does this correspond to (bc) ?

no, that matrix fixes b=(1,0)^T doesn't it?

do we interpet them as colums or rows?

the default is to apply a matrix to a column vector from the left

problem is writing columns in the chatroom is annoying and I don't know if you have chatjax

Hmm i see that you got a = ( 0,1 ) , i put it as (1,0)

as a colum , that was the mistake

sure you can do that

okay

using a = (1 ,0 ) , b = (0,1) , c = (1, 1 ) as colums

the matrix first row (1,1) second row ( 0,1)

then yeah (a)(bc)

exactly a stayed fixe

fixed*

and c went to b

make sense that b went to c :D

okay one more try to see if I got the hang of this

first row ( 1, 1) second row (1,0) is (abc)

@anon sorry if am annoying you , but found this method quite cool =p

yep. it's a group action.

...are you going to write the Putnam again? @anon

?

@anon did you mean that when we study the chapter group actions we will learn such methods? :D

what you did is use a group action

All right =p I dont know what that is yet so =p

it sounds like a fun topic

whats up guys!

hi-yo

how's it going sir!

yo-hi

since the order of both groups are the same , to show isomorphism , is enuf to show injective right?

not bad

I'm so excited for my first classes tomorrow!

feels like time has slowed down

cool

what's the schedule?

linear algebra is fun

loved that class

neat

feel free to ask about physics

I'm back at attempting to integrate 4x/(x^4+1)

I have the U substitution, but, you guys told me to do partial fraction inside of u substitution and I am confuzzled

Because now I need to solve for A and B, but, I have DU and a U

to clarify what we had earlier

you can do that integration in two ways. one way is to do partial fractions immediately, and integrate each of the resulting fractions

the other is to do a u-substitution first, and only then do partial fractions and integrate the resulting fractions

Ok, I'll would like to figure out both, but, the second one is currently stumping me.

@semi

omg i keep doing that :(

I will certainly ask about physics

I love our math chats.

heh, cool

you really helped me study for calc :3

@10Replies kk. do you know what u-sub you want to do?

U = x^2

having a bit of calc knowledge should help with physics.

then I get du/(u^2 + 1)

semi

if we have 2 groups of same order

here you have to take physics and calc together

injective hom is iso right?

since physics is calc based

i mean, kinematics is basically "so here's what having constant second derivative implies"

@Semiclassical.

yeah

@KasmirKhaan not sure, tbh.

well the first assignment is all calc it looks like

some derivative stuff

right

@Semiclassical okay thanks anyway =p

If you want an interesting thing to ask your teacher, ask them what the integral of position means

it's when you do forces that it stops being just calculus implications

@10Replies should still be a factor of 2 overall. (the u-sub gets rid of a factor of 2, but you start with a factor of 4)

not to mention energy :^)

Hopefully nobody tries to help me answer this.

because I want to try on my own first

so you've got $\dfrac{2}{1+u^2}du$

and haven't even had first class

but this is a sampling of my assignment

imgur.com/a/kcW6G

"oh man that's so easy the trick is [REDACTED] and the answer is 42."