Mathematics - 2016-05-01 (page 2 of 3)

fadelm0

4:00 PM

you have a point. i just wasn't confident enough with the question

i don't even know the right way to ask it

Mike Miller

Stsrt from the beginning

fadelm0

i'll just ask the way i can, and you tell me if it's not clear

Mats Granvik

@fadelm0 In that case it is some times better to not ask. For me it usually kills the creative process.

fadelm0

@MatsGranvik I'm just working on something, and I need some insight from math people, since I'm not a math person myself

i don't need specific solutions, just a push in the right direction

user 1618033

@MatsGranvik What would you say about this integral? $$\int_{-\infty}^{\infty} \frac{\cos(a x)}{x} \ dx$$

Mats Granvik

4:06 PM

@fadelm0 I am not a math person either.

fadelm0

and I love the chat rooms on SE

Mats Granvik

@user1618033 It is a good one.

fadelm0

@MatsGranvik well, it's a math chat room...

anyway

user 1618033

@MatsGranvik :D

(just consider the Cauchy Principal Value combined with the idea of odd function and you're done)

fadelm0

how do you write a function that adds the same number to itself with increment to it's exponent?

Mats Granvik

4:09 PM

"Finishing a book is just like you took a child out in the back yard and shot it."
Truman Capote.

Do you mean like this:
x = 1/2
x = x^(x + 1)
x = x^(x + 1)
x = x^(x + 1)

..

...

user 1618033

For the other version, $$\int_0^{\infty} \frac{\log(\cos^2(a x))}{b^2-x^2} \ dx=\infty$$

fadelm0

@MatsGranvik i mean something like:
x = 1
y = a^(x - 1)
x = 2
y = a^(x - 1) + a^(x - 2)
x = 3
y = a^(x - 1) + a^(x - 2) + a^(x - 3)
...

I just didn't know how to explain it

And I have no idea of how to even START thinking about a solution

Balarka Sen

4:29 PM

Google geometric series. A variant of that answers your question.

Mats Granvik

In[920]:= Clear[x, y, a]
Reduce[y == a^(x - 1) + a^(x - 2) + a^(x - 3), a]
During evaluation of In[920]:= Reduce::nsmet: This system cannot be solved with the methods available to Reduce. >>

Out[921]= Reduce[y == a^(-3 + x) + a^(-2 + x) + a^(-1 + x), a]

user 1618033

My fingers hurt while checking the table of integrals and nothing interesting enough.

Adeek

4:48 PM

hey @BalarkaSen

Balarka Sen

Hi

Adeek

I am solving the following question

Determine whether there exists a short exact sequence $ 0 \rightarrow Z_4 \rightarrow Z_8 \oplus Z_2 \rightarrow Z_4 \rightarrow 0$

I solved this one

Balarka Sen

ok

Adeek

but I am thinking now of the following

fadelm0

@BalarkaSen Thanks, I think that put me in the right direction!

Adeek

4:49 PM

Determine which abelian group A fit into a short exact sequence $0 \rightarrow Z_{p^m} \rightarrow A \rightarrow Z_{p^n} \rightarrow 0$

Balarka Sen

Mhm?

JeSuis

Hello, I have the following set $S=\{(x,y,z)\in\Bbb{R}^3: xyz=1\}$, I already proved that the set is a two dimensional submanifold, now I would like to find the connected component and to prove they're isomorphic to $\Bbb{R}^2$. I am not sure how can I proceed, it's not "difficult" to imagine that there is $4$ component connected "for" the four integer solution.

Agawa001

@user1618033 ur fingers ? or ur finger ?

Adeek

I am kinda stuck on this one

Balarka Sen

Use classification of finitely generated abelian groups.

JeSuis

4:52 PM

@BalarkaSen can you look at my question?

Adeek

oh

what is that theorem @BalarkaSen ?

JeSuis

@Adeek en.wikipedia.org/wiki/Finitely_generated_abelian_group

Balarka Sen

@JeSuis How would you prove that for $xy = 1$ in $\Bbb R^2$?

I.e., how would you show that the connected components of that are homeomorphic to R^1?

Mike Miller

Preferably you should think about the classification of finite abelian groups :)

user 1618033

@Agawa001 My fingers, I use more.

Adeek

4:56 PM

I didn't see it before ok so according to wikipedia

Agawa001

@user1618033 you didnt understand

Adeek

$Z_{p^m} = Z_p \oplus Z_p .... $ m-times ?

Balarka Sen

I defer further questions to @MikeMiller if he's willing to help more. I am not, because I am busy right now.

Adeek

and same thing for $Z_{p^n}$?

user 1618033

@Agawa001 Yes, I did understood.

Balarka Sen

4:57 PM

No that's very false and not what the theorem says.

Anyways, I'm off. @JeSuis In case you need more hint on proving that connected components are homeo to R^2, think about projection.

JeSuis

@BalarkaSen hum I will think, thank you!

Agawa001

@user1618033 ur fingerz when typing, ur finger when writing

user 1618033

@Agawa001 lol, my fingers (hurting) while scrolling :-)

Semiclassical

5:24 PM

@user1618033 G&R: so useful, yet such a pain to find exactly what you're looking for

that's not a strike against its organization, though, just that it covers a lot of stuff

user 1618033

@Semiclassical Agree. Sometimes it is hard to find what you're looking for, indeed.

Semiclassical

finding elliptic integral representations of definite integrals is a real pain, for example, just due to how many variations they have to cover

of course, the main reason i end up having to use G&R for that is because Mathematica just kind've sucks at simplifying elliptic integrals

user 1618033

@Semiclassical Yeah, I know.

Agawa001

@user1618033 what are u searching ?

have u already thought to code ur investigation process ?

user 1618033

@Agawa001 The craziest integrals you might possibly imagine and that appear in my calculations. Working on a new integral family and not sure if these have some nice closed-forms. Trying to find anything similar to them.

@Agawa001 Unfortunately, no.

Agawa001

5:39 PM

@user1618033 everything has a closed form, whereas the difference resides in how long it takes to reach this closed form

depends, intelligence, foresight, reachability, available-time, speed(that is optimised since we have computers now, god bless computers)

user 1618033

@Agawa001 Well, yeah, and if things go incredibly bad, define new functions, define new constants, and things will look like that in the end, but ...

For instance, could zeta(3) be expressed in some known constants? No one knows it.

Agawa001

it reminds me of some formula i failed to reach any closed form of it, i thought over when i was in the bus, i found a nice link between it and another form then deduced its closed expression

Adeek

@BalarkaSen here

Balarka Sen

am now

user 1618033

5:56 PM

@Agawa001 The calculation of integrals, series and limits may be a very tough environment. I don't say in vain that it takes many years of extremely hard work to get reasonable results.

Adeek

so

@BalarkaSen according to classification theorem we have $A = Z_{p^a} \oplus Z_{p^b}$ with a + b = n + m

right ?

Balarka Sen

Yes.

Well, at least, those are the candidates which fit in the short exact sequence.

Adeek

yeah

so is that all of them ?

Balarka Sen

That's for you to figure out.

Mike Miller

6:12 PM

@Karim: So the middle term has order $p^{n+m}$. You seem to have concluded that it is generated by two elements. How? (Just saying that needs proof.)

Mike Miller

6:28 PM

@Semiclassical What did you say?!

Jessy Cat

Anybody else having trouble posting questions? I keep getting timed out.

Semiclassical

horrible, terrible things. i am ashamed.

Jessy Cat

Shame! ding ding Shame! ding ding

I'm also having trouble connecting to chat, but only on my laptop. I'm going to try restarting it... I'm on my iPhone now

Adeek

6:45 PM

how can I guarantee that $(Z_{p^a} \oplus Z_{p^b}) / H$ isomorphic to $Z_{p^n}$?

@BalarkaSen

Balarka Sen

What's $H$?

Adeek

the image generated by basis element of $Z_{p^m}$

Balarka Sen

By what map?

Jessy Cat

Worked :)

Adeek

I got in order to embed $Z_{p^m}$ into $Z_{p^a} \oplus Z_{p^b}$ we must have $p^m \leq p^{max\ {a,b}}$

Jessy Cat

6:47 PM

0

Q: Residue of $1/(\sin(1/z))$ defined at $z=0$? Trying to derive Laurent Series of $\csc (1/z)$ to find it.

This question is related to this one. I was able to figure out on my own that the residue of $\displaystyle \sin \left(\frac{1}{z} \right)$ is defined at $z=0$ by finding the Laurent Series of $\displaystyle \sin \left( \frac{1}{z}\right)$ centered at $z=0$, and noting that the coefficient of th...

Adeek

I map generator of $Z_{p^m}$ onto an element of order $p^m$ in A

@BalarkaSen

fadelm0

why is it that there's more perceivable difference between two numbers with the same ratio when they are small, than when they are large?

has this phenomenon been studied?

Balarka Sen

Which element? There are lots of elements of order $p^m$.

fadelm0

i'm not talking about actual difference, just perceivable difference

Balarka Sen

Your question isn't making sense to me, @Adeek.

Tobias Kildetoft

6:49 PM

@fadelm0 that does not sound like a mathematical property

fadelm0

@TobiasKildetoft yeah, i know. it's not. but I'm trying to study why people usually can't imagine the magnitude of difference between say 10^27 and 10^26, while the difference between 10 and 100 is easily imagined

i'm wondering whether that is a property of positional notation

Adeek

7:18 PM

hi @BalarkaSen here?

Balarka Sen

Yep.

Adeek

I don't understand so I got in order to embed $Z_p^m$ onto A we must have $p^m \leq p^{max\{a,b\}}$

but now I need to define the map so that $(Z_{p^a} \oplus Z_{p^b}) /H ~= Z_{p^n}$

where H is the image of the basis element of $Z_{p^m}$

I don't know how to proceed further

@BalarkaSen any hints ?

Mambo

@robjohn

Hi

I have a problem in extending a module action

Hello @ThomasRot

Do you any idea about Hochschild cohomology of Banach algebras?

Balarka Sen

@Adeek Sorry, I was away. What you said is right, you need $m \leq \text{max}(a, b)$ to start with.

I don't know how to give a hint without revealing the whole solution. You just need to write down a map.

How did you solve the $\Bbb Z_8 \times \Bbb Z_2$ thing? Same idea.

Adeek

I just mapped 1 to (2,1) so 2 maps to (4,0) and 3 maps to (6,1)

Denote that image by M. The powers of coset are of order 4

so (Z_8 (+) Z_2) / M ~= Z_4

Balarka Sen

7:32 PM

So use the same idea.

I don't know why generalizing that should be too hard.

Adeek

alright

1 sec I will go eat and do it

Adeek

7:55 PM

back

what about if we define the first map as $1 \mapsto (x_1,1)$

@BalarkaSen

Cbjork

8:25 PM

$\displaystyle\sum_{k=0}^\infty \dfrac{\cos(k t)}{k!} =e^{\cos (t)} \cos (\sin (t))$

How do I show this?

user 1618033

@Cbjork consider Euler's formula ... $e^{ix}=\cos(x)+i\sin(x)$

Cbjork

Ok I'll do some more work with it

s.harp

8:39 PM

@Cbjork $\cos(kt)=\rm{Re}(e^{ikt})$ if $t$ is real. So $\sum_k \frac{\cos(kt)}{k!}=\rm{Re}\left(\sum_k \frac{(e^{it})^k}{k!}\right)=\rm{Re}(\exp(e^{it}))$

Cbjork

I see!

thank you very much

Balarka Sen

@Adeek What's $x_1$?

Adeek

$x_1 \in Z_{p^a}$

Balarka Sen

An arbitrary element?

Adeek

yeah

I wanted to show if we have H defined that way then G/H is cyclic

Balarka Sen

8:49 PM

Then of course that's not going to work. $\Bbb Z_4 \to \Bbb Z_8 \times \Bbb Z_2$ given by $1 \mapsto (4, 1)$ is not even injective.

Adeek

yeah but wait

but I was not done yet

but wait I started out with $Z_{p^a} \oplus Z_{p^b}$ wlog suppose that a | b so in this case your counter example doesn't work

Balarka Sen

Huh? Why are you assuming $a | b$?

You should either give me a worked out solution, ask a reasonable question, or think about it carefully, @Adeek. I have work to do, I can't always point out why your things do not work.

Mapping things to arbitrary things never going to work out. Never.

I could take $x_1= 0$.

Adeek

I see

I could say $x_1 \neq 0$

Balarka Sen

Still false. THINK.

Ari Nubar Boyacıoğlu

Heyy, can anyone explain what generalized eigenvector is?

anon

9:06 PM

@AriNubarBoyacıoğlu v is a generalized eigenvector of A, with eigenvalue lambda, if A^k v = lambda^k v ... for k large enough.

Hi @anon

hi

What's up?

thinking I can construct ternary cross products from binary ones using composition algebras

everything's a go 'cept it's not obvious it's linear in the one argument

Balarka Sen

Cross product where? In R^4?

anon

9:09 PM

binary in R^3 and R^7, ternary in R^4 and R^8

the ternary one in R^4 is forced by the properties of the cross product, so I want to construct it in from quaternion operations in order to generalize it to constructing the ternary cross product on R^8 using octonions.

Balarka Sen

Mhm. So how do you propose to construct it?

Adeek

How is (1,0) + H isn't cyclic ? With H is subgroup generated by (x_1,1) with x_1 neq 0 ?

Balarka Sen

Ah. From quaternions.

Adeek

@BalarkaSen

anon

also trying to understand triality

Adeek

9:11 PM

even if $x_1 = 0$ then it will be cyclic as well

$(0,1)$ generates all of $Z_{p^{b}}$

Balarka Sen

@Adeek Dude, even before that, the map $\Bbb Z/p^m \to \Bbb Z/p^a \times \Bbb Z/p^b$ given by $1 \mapsto (x_1, 1)$ need not even be injective for arbitrary $x_1 \neq 0$.

Mike Miller

Ramanujan movie was pretty good

anon

it's out?

Balarka Sen

yeah @anon

Adeek

I am not considering defining the map yet

i am just seeing that if we define the subgroup H the quotient of it then it is cyclic

then I will define the map that makes it work

Balarka Sen

9:13 PM

What's the point if the first map is not injective? You're looking at short exact sequences $0\to \Bbb Z/p^m \to A \to \Bbb Z/p^n \to 0$.

$\Bbb Z/p^m \to A$ has to be injective

Adeek

I know

Balarka Sen

I don't know what you have in mind then.

But I don't want to discuss about this anymore, partially because I am thinking about something else right now.

Adeek

ok

Balarka Sen

Try asking someone else.

Adeek

@anon or @MikeMiller can you help me with this problem ?

I think I am kinda stuck

I am almost done with it

Determine which abelian group A fit into a short exact sequence $0 \rightarrow Z_{p^m} \rightarrow A \rightarrow Z_{p^n} \rightarrow 0$

so far I got

that we must have $A$ isomorphic to $Z_{p^a} \oplus Z_{p^b}$ with a + b = n + m

and $p^m \leq p^{max\{a,b\}}$

Now I need the property that $(Z_{p^a} \oplus Z_{p^b}) / Im(Z_{p^m} \rightarrow A) ~= Z_{p^n}$

Balarka Sen

9:38 PM

@KarlKronenfeld What are you deciding today?

Oh, I should know what the answer is going to be.

"Trying to decide what to decide".

;)

Karl Kronenfeld

I like the "trying to" phrase. I am not even deciding. lol

Semiclassical

"trying to make a decision about what to be indecisive about"

Karl Kronenfeld

Putting in my best effort here.

Balarka Sen

@KarlKronenfeld That's new.

Adeek

9:55 PM

hey @BalarkaSen I think I solved it

Balarka Sen

ok

Adeek

we must have the following condition $|Z_{p^a} \oplus Z_{p^b} / H| = |Z_p^n|$ so we must have $p^{a + b - m} = p^n$

if we choose $p^m \leq p^{max\{a,b\}}$ and $p^{a + b - m} = p^n$ along with defining the first map which send $1$ to an element of order $p^m$ in A then that should work

right ?

Balarka Sen

sounds good

Anthony

10:21 PM

@BalarkaSen you around?

Balarka Sen

@Anthony Yes I am.

Anthony

lol

Adeek

I don't think I am completely correct

Anthony

Could you help me with my set theory hw? I'm trying to show that assuming the axiom of choice, for every nonempty family T of nonempty sets, there is a function function $F:\cup T \rightarrow T$ such that $x\in F(x)$ for all $x\in\cup T$. I feel like it should be really simple, but I just keep running in circles. :/

Balarka Sen

That's too hard for me. Ask someone else, maybe.

Anthony

10:24 PM

Alrrrrrrighty.

Thanks.

Balarka Sen

Glad to be of no help.

Anthony

:D

Forever Mozart

@Anthony that is sort of obvious

Anthony

Care to elaborate? I think I'm just running on fumes here.

Forever Mozart

Axiom of choice says that given any collection $T$ of sets, there is a new set consisting of exactly one element from each set in the collection

Anthony

10:27 PM

Yeah.

Forever Mozart

So it is a mapping $T\to \bigcup T$

all you have to do is reverse it

Anthony

Oh man, this is going to hurt lol

I was trying to do something dumb with powersets.

Let me look at it again, thanks.

Forever Mozart

lets say $f:T\to \bigcup T$ is the choice function

define $F:\bigcup T \to T$ by $F(x)=T$ iff $f(T)=x$.

Anthony

Should one of those $f$'s be a $F$?

Forever Mozart

yes fixed now

Anthony

10:32 PM

I'm not sure I understand what you're trying to say.

First, do you mean $T$ in the second line, or do you mean $T'\in T$?

Second, I don't see how this approach works, given that the function needs to be defined on all of $\bigcup T$, and the choice function will only give us a function on a subset.

Forever Mozart

$F:\bigcup T \to T$ by $F(x)=t$ iff $f(t)=x$

where $t\in T$

Anthony

Sure, but then as I said about this won't define a function unless $f$ was onto.

Forever Mozart

Oh ok that is easy to fix

For each $x\in \bigcup T$ pick $F(x)\in T$ such that $x\in F(x)$.

easier than before

Anthony

I suppose that's the heart of it, but this feels highly non-rigorous lol

I'll try to write that up... Thanks.

Forever Mozart

well it depends on your specific definition of AC

Anthony

10:38 PM

I'm using:
For any set X of nonempty sets, there exists a choice function f defined on X.

Forever Mozart

ok so what should be the collection of sets

Anthony

Initially we said $T$, but this is too small of a collection of sets.

Forever Mozart

how about

Anthony

Ugh I can't think.

Forever Mozart

for each $x\in \bigcup T$ define $T_x=\{t\in T:x\in T\}$.

Then consider the collection $\{T_x:x\in \bigcup T\}$

There is a choice function on this set

Then define $F$ in the natural way

Anthony

10:43 PM

D'oh. That sounds golden.

Thanks.

Forever Mozart

that might work

Anthony

I think it does, I'll make sure.

Forever Mozart

bah sorry I mixed my T's again, should have been $T_x=\{t\in T:x\in t\}$

but yeah it works

I've been using AC so much that I thought your question was obvious

Anthony

Yeah, I know, and it is obvious

Thanks

<3

Balarka Sen

10:59 PM

Hi @AkivaWeinberger

Jessy Cat

Hello :)

0

Q: Finding residues at a point $a$ given function values there

I am faced with the following problem: (a) Find $\displaystyle res_{a} \frac{\varphi(z)}{(z-a)^{n}}$ where $\phi$ is a given function analytic at $a$, $\varphi(a) \neq 0$, and $n$ is a positive integer. (b) Suppose $a$ is a simple pole of $f$, and let $\displaystyle res_{a} f = A$. Find...

complex-analysis proof-verification residue-calculus

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