Mathematics - 2013-03-07 (page 2 of 2)

user58512

19:08

is the outer automorphism group of GL_3(2) the same as line-point duality of fano plane?

Dominic Michaelis

19:30

hi

user58512

hi

skullpatrol

hi

Arkamis

hi

skullpatrol

$=(hi)^4$

Charlie

e^hi

skullpatrol

19:38

$\pi^{hi}$

$\pi^{hi}+e^{hi}=Charlie-Skull's$ last theorem.

Dominic Michaelis

$hi^{ack(g_{64}, g_{64})}$

smackcrane

hello all

skullpatrol

hello

user58512

hello

Charlie

Hello you

anon

19:47

@user58512 I have made some good progress on your question (reducing it to checking some a handful of things), but I have to leave to go to the bank, lunch etc. and then I have work again in a couple hours. I intend to answer it tonight if no one else does.

user58512

which one?

anon

the one with the so-called elliot configuration

user58512

oh great!

why so-called??

anon

because I cannot find any references for it!

user58512

mm me neither.. I did google it and found nothing

ok have a good day!

smackcrane

19:52

I don't suppose anyone knows much complex analysis ... I have a pretty basic question: what is the difference between analytic and holomorphic? I'm beginning to develop an intuition for thinking about analytic functions, and I'd like to know if I need to change anything to think about holomorphic functions

Arkamis

@smackcrane There is none

They are two different words for the same thing.

smackcrane

@Arkamis that's just what I was hoping. Thanks!

skullpatrol

synonyms

Arkamis

Holomorphic means "whole shape" basically, meaning that the shape of the function is "whole" in the sense that it doesn't contain any bad parts.

user58512

I don't quite agree @Arkamis

Arkamis

19:56

Analytic in the complex sense links the concept of differentiability to analytic real-valued functions; however, a conceptual linkage is all that exists (analyticity of real vs. complex valued functions is mathematically different, although conceptually similar)

user58512

for functions C -> C then the fact analytic and holomorphic are the same is a theorem

Dominic Michaelis

wait wait wait

user58512

but you can call a function R -> R analytic, whereas it doesn't make sense to call it holomorphic

Dominic Michaelis

what definition of analytic are you using ?

user58512

infinitely differentiable

and holomorphic just means that it's complex and differentiable once

Dominic Michaelis

19:58

than there is a big big uge big difference :D

Arkamis

Not really. Because in complex analysis they are the same.

Dominic Michaelis

for example the real valued function

user58512

(the theorem is that is if you can differentiate a function once, f' has a primitive, so it's infinitely differentiable)

Dominic Michaelis

$e^{-\frac{1}{x^2}}$

Arkamis

@DominicMichaelis Complex analysis

user58512

19:59

that's a really good example: the notions differ in the real case

Arkamis

Did you guys even read my second statement?

user58512

I guess you picked that up from the integral domain question? :)

Dominic Michaelis

is as a function from $\mathbb{R}\rightarrow \mathbb{R}$ a infinetly often differentiable

Arkamis

"Analytic in the complex sense links the concept of differentiability to analytic real-valued functions; however, a conceptual linkage is all that exists (analyticity of real vs. complex valued functions is mathematically different, although conceptually similar)"

smackcrane

holomorphic isn't even a thing in the 'real' case is it?

Arkamis

20:00

Not really

Dominic Michaelis

@user58512 nope that was the key for a problem i faced for a long time

Arkamis

No one talks about "holomorphic" real-valued functions like that.

user58512

@smackcrane, that's right

Arkamis

And in complex analysis, that there is no distinction, to me, means that the existence of the less-general term is unjustified

Dominic Michaelis

@arkamis the only real valued holomorphic functions are constants

Arkamis

20:02

It just adds terminology for no appreciable gain, except to math historians, or someone reading a book written before, say, 1960.

Dominic Michaelis

i have to go eating excuse me

user58512

it's important to be able to say holomorphic in complex analysis

when you start using real valued functions along with your complex ones

user2041143

Hello :)

Arkamis

@user58512 I think at that point, the nomenclature of function spaces deals quite adeptly at clarifying any conflict.

Dominic Michaelis

i am back

Arkamis

20:06

Holomorphic remains a stupid word in my opinion. I suppose I find it's greatest benefit is to provide a conceptual linkage to meromorphic functions

smackcrane

what's the conceptual linkage there? is the intuition very similar?

Dominic Michaelis

in my analysis class we definied analytic functions as functions from $\mathbb{R} \to \mathbb{R}$ which taylors series converge to the function

Arkamis

@smackcrane A function meromorphic in a region is analytic in that region except at countably many points, which are isolated poles.

Yes, it's countably many

user2041143

Anyone good with multiplication?

Dominic Michaelis

meromorphic 2 peridocal functions are much more interesteing than holomorhic ones :D

@user204 for sure mathematica

user2041143

20:10

Do you mind checking my problem ?

Dominic Michaelis

i can say it only after i saw it :)

will you show it to me ?

user2041143

0

Q: Find the multipliers of given numbers whose sum doesn't exceed the result (explained)

Ok. I have to find the multipliers of given numbers whose sum (of multipliers) must NOT be smaller than ANY of the result by multiplication of the number with the multiplier. For example, if given numbers are: 15.5, 4.60, 10.5, 5.80 So, we multiply the first number by numbers with the following...

complex-multiplication

Dominic Michaelis

your tag is completly wrong :)

sry i don't get what you want to do

Arkamis

20:42

user2041143

Sorry, I thought I explained it well.. which part you didn't understood?

Arkamis

@user2041143 Basically the part where you're pulling random numbers out of thin air

user2041143

Well, I put random numbers

So the sum of these numbers is not bigger by any of the results

Arkamis

What you're asking is not clear at all.

You have a set of numbers, you want to generate a second set of numbers so that multiplying the first set by the second set, and taking the sum, results in something larger than the second set?

Or you want to find a set of numbers such that their sum is smaller than any multiple?

Why not just take 1 1 1 1?

user2041143

So the sum of these numbers is not bigger by any of the results

Well..

No!

I don't want to do that.

Arkamis

20:57

It's not clear what you want to do. Also your example is exactly the opposite of what you state that you want.

I suggest spending some time writing the question much more clearly.

user2041143

I have set of numbers (first set), which I want to find set of numbers (second set), when we multiple them, the sum of the SECOND SET, is NOT lower than ANY result by multiplication of any number of set 1 with set 2

Arkamis

So if your first set is $\{a,b,c,d\}$, you want to find $\{e,f,g,h\}$ such that $e+f+g+h > x$, where $x$ is any product of a single element from the first set, and a single element from the second set?

user2041143

For the example I have given, the sum of second integers is 740 and is lower than any result by multiplication of first set and second set.

What do you mean by $\{\}$ ?

Arkamis

it's set notation.

If you don't use ChatJax, it should look like {a,b,c,d} on your screen

There are infinitely many such sets of numbers.

user2041143

ok well

yes, the first set is in my example {a,b,c,d}

and i want to find second set {e,f,g,h)

Arkamis

21:02

There are infinitely many such second sets.

Argon

@robjohn I conjectured the following integral. Could you suggest ways that I may be able to (dis)prove it? $$\int_{-\infty}^{\infty}\frac{dx}{\cosh^{2n+1}x}\stackrel{?}{=}\pi\frac{\binom{2n}{n}}{4‌^{n}}\qquad\forall x\in\mathbb{Z}^{+}$$

user2041143

Akamis, I need to find only one.

Where e+f+g+h > a*e OR b*f OR c*g OR d*h.

Arkamis

Let's say $d$ is your biggest number in the first set.

what about a*f?

user2041143

Nope. a*f not possible.

Arkamis

ok

user2041143

21:03

Not allowed

Only a*e, b*f, c*g or d*h

Each number in first set must have only ONE number assigned to it.

My English is not my native language. :)

Nimza

Good evening!

user58512

hello

Arkamis

Ok, well, assume d is your largest number

Nimza

Help me please, is it possible to say something about random variables $X$ and $Y$ if their moment-generating functions satisfy $M_X(t)$ = $M_Y(t^{\alpha})$ for some $\alpha > 0$ ?

user2041143

Ok.

d is my largest number, then?

Arkamis

21:06

hang on -- dealing with something else

user2041143

Alright !

Dominic Michaelis

huhu

could someone explain why asaf would deduct at least half of the points for my solution math.stackexchange.com/questions/323969/…

kiamlaluno

@RegDwighт Geesh... I am not Calabrian; I am Lombard. The fact I know some Calabrian words doesn't make me Calabrian.

Next, you will say I am German, just because I know some German words.

Dominic Michaelis

I am German :D

user2041143

21:21

Any update? :)

Arkamis

21:32

@user2041143 There is no general solution.

But to show that is pretty complicated.

user2041143

Any idea how to solve it?

Fomulas, theorems, or whatever to use to help me

Arkamis

You can look at it a couple of ways

The easiest way is to see that a, b, c can each be written as multiples of d:

a = m_1 d, b = m_2 d, c = m_3 d

Likewise, e,f,g can all be written as multiples of h

e = n_1 h, f = n_2 h, g = n_3 h

So, the sum of your second set is (n_1+n_2+n_3+1)h

your possible products are: ae = m_1 d n_1 h, bf = m_2 d n_2 h, cg = m_3 d n_3 h, dh = dh

So then you have conditions, depending on which product is the biggest:

ae < dh, for example, implies m_1 d n_1 h < dh implies m_1 n_1 < 1

Dominic Michaelis

lol i finally unterstand that mapsto stands for maps to

Arkamis

But this may or may not be true, depending on your initial sets

Some sets may have infinite solutions. Some may have none at all.

Equivalently, you can look at it as a vector problem

user2041143

What do you mean m_1 d

What is m_1

Arkamis

21:38

m subscript 1

it's a multiple

so if, for example a = 2 and d = 11, then a = m_1 * 11, and hence m_1 = 2/11

user2041143

oh.

let me check

Arkamis

If you write y as a vector, then the sum of the elements in y is y DOT 1, where 1 represents [1 1 1 1]

Then, using the dot product identity, we have y DOT 1 = 2|y|cos t, where t is the angle between y and 1.

Then, each multiple can be written as a dot product of vectors: ae = y DOT [a 0 0 0] = a|y|cos t_1

Since you want y DOT 1 > ae = y DOT [a 0 0 0], we have 2 cos t > a cos t_1

and then equivalently for bf, cg, dh

user2041143

erm

Are you sure this would work?

Arkamis

No, I'm not sure it will work. I'm sure it's correct.

user2041143

Going this way.

Arkamis

21:42

Existence/nonexistence of solutions depends on your initial set.

user2041143

We are going to get the LOWEST possible values?

Arkamis

Look, if your initial set is [1 1 1 1], then I can pick [2 2 2 2].

user2041143

Ok

Then?

Arkamis

If your initial set is [21312 98123 211 32090] then it might not have a solution. I don't know.

user2041143

I understand that.

So, can you give me an example

With numbers... instead of going theory

Arkamis

21:44

I just did.

[1 1 1 1] leads to choosing [2 2 2 2]. the max element-wise product is 2. The sum of the multiples is 8.

I'd actually say if your largest number is less than 4, then it's easy.

If it's larger than 4, it probably doesn't have a solution.

Assuming positive entries.

user2041143

Yes, but how do you get the [2 2 2 2]?

Arkamis

I randomly picked numbers that would obviously work.

user2041143

If my largest number is less than 4, it's easy then?

I don't think so.

What would you do if the numbers were 2.20, 3.25 and 3.20?

All numbers are <4

And yes, only positive entries are possible, no negatives

Arkamis

I thought you always had sets of 4 numbers

Then it should be easy if the numbers are less than n, where n is your set of numbers.

So, 2.2, 2.5, 2.8, then you can just pick [3 3 3]

user2041143

No, the numbers can be 4, 5, 9 etc.

So, for 2.2, 2.5 and 2.8, if we take 3,3,3

Then we get, 3*2.2=6,6, then 3*2.5=7,5 and 3*2.8=8,4. So, 3+3+3=9, which is bigger than all results.. it should be smaller by 6,6 and 7,5 and 8,4.

Arkamis

21:50

You said you wanted it larger.

user2041143

the sum of second set of numbers

Arkamis

"the sum of the SECOND SET, is NOT lower than ANY result"

user2041143

Nooo.. I didn't mean that, sorry... bad English.

I didn't explain it

Very well

I'm sorry about that.

Arkamis

Back to my original point. Explain it better. I need to leave now, though.

user2041143

Ah, alright then.

Maybe we can finish this later, if we meet together again!

Arkamis

21:51

But if you want it smaller, than the converse holds. If your smallest number is larger than the size of the set, then it's also easy

user2041143

Have a nice day though.

Arkamis

for example: 4 5 6

user2041143

ok

Arkamis

Then just choose 1 1 1

user2041143

And get what?

Arkamis

21:52

1+1+1 = 3 < 4*1, 5*1, 6*1

user2041143

You're correct, but

That would work for big numbers only?

What if the numbers are 1.7, 3.5 and 4.9

Arkamis

Yep.

user2041143

My point is to make it work with smaller numbers, from 1.5 to 3-4

Arkamis

Then I'm afraid there is no general solution.

user2041143

Well I'm ready to take any steps.

By no general solution you mean no solution at all?

000

22:15

$test$

RegDwighт

@kiamlaluno haha, dream on. No you don't and no I won't.

000

@robjohn Your bookmarklet really stands the test of time. Awesome job.

robjohn

@000 Ever since I had to modify it for the lack of the AjaxComplete callback, I don't think I've changed it.

@000 Thanks :-)

000

@robjohn Have you participated in math competitions?

user2041143

Anyone online

000

22:31

@user2041143 Hi.

user2041143

Hi

000

How are you?

robjohn

@000 Once in high school when I was in 10th grade. I came in higher than the others in the school, but not well enough to warrant any state placement. There were no real opportunities after that.

user2041143

I'm fine, thanks.

Looking for solution of my problem

Impossible

000

@robjohn Ah. Would you have any general study tips?

@user2041143 Nothing is impossible.

robjohn

22:40

@000 are you going to be in a competition?

000

@robjohn Indeed. State-level.

I've been studying for three weeks.

robjohn

@000 find practice problems and don't try to force things.

000

@robjohn Be more specific with the latter if you do not care.

robjohn

@000 Forcing things creates unnecessary stress, which usually stifles creativity in my experience.

000

@robjohn Right. So, you are implying that the best studying would occur when I'm in the right mood for it.

robjohn

22:45

@000 I'm saying to relax and don't stress work problems, move on and don't get stuck

000

@robjohn I totally agree.

It's useless to focus on a single problem, especially in a timed testing environment. Often, I can't do a problem simply because my brain didn't fire properly when I first read it. Returning to it often allows me to do it.

user2041143

My problem is impossible! :)

Dominic Michaelis

mpfh today i won't cap again -.-

@robjohn you don't look well whats the matter?

robjohn

@DominicMichaelis I thought I would try out St Patrick's Day colors.

Dominic Michaelis

i could still get 100 points today, damn work

robjohn

22:59

@DominicMichaelis You only have an hour left...

Dominic Michaelis

i won't manage that one :(

Argon

@robjohn I think I found a neat way to prove $\zeta(2)=\frac{\pi^2}{6}$ with contour integration, though perhaps it is somehow flawed

Consider contour $C$, a rectangle with vertices at $0, R, R+i \pi$ and $i\pi$

Then $$\oint_C \frac{x}{e^x-1}\, dx = 0$$

Dominic Michaelis

yeah obviously

Argon

$$0 = \int_0^\infty \frac{x}{e^x-1}\, dx - \int_0^\infty \frac{x+i \pi }{-e^x-1}\, dx+\frac{\pi i}{2}\operatorname*{Res}_{z=i \pi}\left( \frac{z}{e^z-1}\right)$$

Right?

user58512

R=infinity?

Argon

23:12

Yes

user58512

I don't see what the second integral is

Charlie

Seems that robjohn is sick...

user58512

hm

Argon

@Charlie Ah

user58512

e^{x + ipi} = - e^x

Dominic Michaelis

23:12

@Charlie i said the same :D

Argon

@user58512 $$\int_0^\infty \frac{x}{e^x+1}\, dx$$

(taking real part)

Charlie

@DominicMichaelis hahahahaha

user58512

why is it up to infinity

Argon

@user58512 $R \to \infty$

user58512

$$0 = \int_0^\infty \frac{x}{e^x-1}\, dx - \color{blue}{\int_0^{i \pi} \frac{x+i \pi }{-e^x-1}\, dx}+\frac{\pi i}{2}\operatorname*{Res}_{z=i \pi}\left( \frac{z}{e^z-1}\right)$$

Argon

23:14

Oh, ya, my error

user58512

oh this is the R+ipi to ipi part

Argon

I forgot an integral!

$$\int_0^{\pi} \frac{-x}{e^{ix}-1}$$

Then take real part and integrate

The real part is $-\frac{x}{2}$

So the integral equals $-\frac{\pi^2}{4}$

Dominic Michaelis

wait wait wait

user2041143

Why integrals are so hard!

Dominic Michaelis

i got pi^2 /4 + 4 i log (2) for the integral

Argon

23:18

Now the residue is $0$...

so we have

user58512

you can throw away the imaginary parts I guess

they must all have something they cancel with

Dominic Michaelis

but i don't have a -

Argon

$$-\frac{\pi^2}{4}+\int_0^\infty \frac{x}{e^x-1}\, dx+\int_0^\infty \frac{x}{e^x+1}\, dx = 0$$

Ya, I left imaginary parts (they explode around $i \pi$)

Dominic Michaelis

never mind the imaginary part but i get $\frac{\pi^2}{4}$ not $-\frac{\pi^2}{4}$

@user2041143 if they are to hard you are to soft ;)

user2041143

I don't really understand them!

I know how to do them by table ones only.

But when we have to do shift/change

I am not sure what to take

Dominic Michaelis

23:21

integrating is an art

user58512

@Argon, how do you change those integrals into the sum?

Dominic Michaelis

i still want to know why i have no $-$

Argon

@DominicMichaelis $$-\frac{x}{e^{ix}-1} = -\frac{e^{-ix/2}}{e^{ix/2}-e^{-ix/2}} = -\frac{e^{-ix/2}}{2 i \sin(x/2)}$$ Taking real parts, we get that

Wait

user19161

@robjohn Nice green! You need more Vitamin C!

Argon

@DominicMichaelis Oh, I think I see

Did you change the integral from $\pi$ to $0$ to $0$ to $\pi$ by changing sign?

Dominic Michaelis

23:24

you have $- ( -\sin(x/2))$

Jonas Teuwen

My crack. My crack. It is the Antichrist.

Argon

@DominicMichaelis One negative from switching integration bounds, one from squaring $i$ from parametrization, one more from $\Re \exp(-ix/2)$

user19161

-1

Q: What is the purpose of defining the notion of inflection point?

What is the purpose of defining inflection point? I know that it is defined to be the point where the second derivative is zero and the second derivative sign changes. It have to have some purpose for pure math.

algebra-precalculus terminology graph math-history

Dominic Michaelis

oh i did calculate $$\int_0^\pi \frac{-x}{\exp(ix)-1} \, \mathrm{d}x= \frac{1}{4} \pi (\pi +4\pi i \log(2))$$

user19161

I am disturbed that a perfectly good question like this gets downvoted.

Argon

23:27

@DominicMichaelis This is true (perhaps, I ignored the complex part). But the integral was originally from $\pi$ to $0$, so to make it from $0$ to $\pi$, multiply by $-1$

So $-\frac{\pi^2}{4}$

Dominic Michaelis

ah ok sry i missed that one

user19161

Yet the answer gets three upvotes. Why do people downvote the question? Why? Why?

Argon

Then $$\frac{\pi^2}{4} = \int_0^\infty \frac{x}{e^x-1}\, dx+\int_0^\infty \frac{x}{e^x+1}\, dx$$

user58512

ok

Dominic Michaelis

throwing the integrals together and substituting right ?

Argon

23:29

It's not hard to show, by turning the integral into a sum, that the first integral is $\zeta(2)$ and the second is $ \frac{\zeta(2)}{2}$

So $\frac{3}{2}\zeta(2) = \frac{\pi^2}{4} \implies \zeta(2) = \frac{\pi^2}{6}$

Dominic Michaelis

thats a neat way

Argon

I don't know if it is properly justified, I just made it up :)

robjohn

$$
\begin{align}
\int_{-\infty}^\infty\frac{\mathrm{d}x}{\cosh^{2n+1}(x)}
&=\int_{-\infty}^\infty\frac{\mathrm{d}x}{(1+x^2)^{n+1}}\\
&=\int_{-\infty}^\infty\frac{\mathrm{d}x}{(x-i)^{n+1}(x+i)^{n+1}}\\
&=\frac1{(2i)^{n+1}}\int_{-\infty}^\infty\frac1{(x-i)^{n+1}}\frac1{\left(1+\frac{x-i}{2i}\right)^{n+1}}\mathrm{d}x\\
&=\frac1{(2i)^{n+1}}\int_{-\infty}^\infty\frac1{(x-i)^{n+1}}\sum_{k=0}^\infty\binom{-n-1}{k}\left(\frac{x-1}{2i}\right)^k\mathrm{d}x\\
&=\frac{2\pi i}{(2i)^{n+1}}\frac1{(2i)^n}\binom{-n-1}{n}\\

Argon

@robjohn WOW! Thanks, I always love your giant LaTeX blocks :)!

user19161

The green looks so nice I feel like changing into green.

user58512

23:32

but that adds to 3/2 zeta(2)?

Dominic Michaelis

@user585 yeah $\frac{1}{2}+ 1=\frac{3}{2}$

user58512

oh i get it

how do you turn the integrals into sums?

Peter Tamaroff

@robjohn Oh, noes. You got the flu?

user19161

@PeterTamaroff Wrong grammar Pedro.

robjohn

@PeterTamaroff nah... just trying some Irish for St P's Day

Dominic Michaelis

23:34

ok i go sleeping good night everyone

Argon

@user58512 Check out 1.6: de.wikibooks.org/wiki/…

user58512

ah

great

user19161

@Argon Wow you read German!

robjohn

@Argon what is your derivation of $\zeta(2)$?

Argon

Feb 11 at 21:35, by Jacob Black

@Argon You read German?

@robjohn Hehe, it is all above. I integrated $$\oint_C \frac{x\, dx}{e^x-1}$$ over (indented) rectangle contour $C$ with verticies at $0, R, R+i \pi$ and $i \pi$.

Then, after taking real parts of everything, we get $$\int_0^\infty \frac{x}{e^x-1}\, dx + \int_0^\infty \frac{x}{e^x+1}\, dx = \frac{\pi^2}{4}=\zeta(2)+\frac{\zeta(2)}{2}$$

(Letting $R \to\infty$, of course)

user19161

23:42

@Argon HAHAHAHAHA

robjohn

@Argon How do you handle the part from $0$ to $i\pi$?

Argon

@robjohn I parametrized the contour as $iz$ and found the real part of the integral ($\pm\frac{x}{2}$, where the sign depends on what side the integral is)

user19161

I am appalled that people feel the need to downvote complete solutions to homework problems.

user19161

I should write an essay on this and post on meta, but I am not in the mood.

robjohn

@Argon That works.

Argon

23:49

@robjohn NICE! Thanks a lot for everything, once again.

Bye! Get well soon.

robjohn

@Argon There is a paper with a lot of ways of computing $\zeta(2)$, compiled by Robin Chapman. Let me find it.

Argon

@robjohn I saw it, it was great

user19161

@robjohn The mod who vanished!

user19161

@Argon It's Fri here!

robjohn

@Argon This URL seems to be good.

Charlie

23:53

@peter there's someone in my uni that looks like you... It's scary...

user19161

@Charlie How can anyone look like Pedro?

Charlie

Looking

robjohn

@Argon None there use contour integration as far as I can see.

user19161

@robjohn You remind me of the green goblin in Spiderman.

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$Mathematics$ Mathematics