Mathematics - 2014-10-05 (page 4 of 5)

Alizter

16:00

@BalarkaSen I told you I am being daft

Balarka Sen

@Alizter Compute the residue using Laurent expansion of that stuff.

Alizter

@BalarkaSen that is the problem

I forget these

Nick

@Sarah: ~~Why is that funny?~~ I don't think I should know

Alizter

hence why I am practicing

gonna read up brb

Balarka Sen

Have fun

Sarah

16:01

@Nick shhhh

Alizter

ahhhh I left the book in my locker

:(

Nick

@Sarah: A shhhh to you too.

Sarah

@Alizter how can you forget laurent?

surely you learn that before residues?

Alizter

meh thats what I get for teaching myself

Nick

Note to self: Pay rent before residing

Sarah

16:08

@Alizter $-e^{i\pi/3}/9$

Alizter

@Sarah is that the res?

Sarah

no its the answer to life

Nick

@Sarah: 42?

$-e^{i\pi/3}/9 \approx -(0.055 + 0.1i) \neq 42$

Balarka Sen

blah

Mats Granvik

Note to self: Upgrade to Mathematica 10 before doing homework in linear algebra.

Nick

16:13

I suck your blah!

Muhuhaha'

Alizter

crap

contour integration is hard

Nick

lol

Balarka Sen

what are you integrating, @Alizter?

Nick

Note to @Mats : It's not the version of the boat that matters, it's the motion of the ocean.

Balarka Sen

@Nick you need to get rid of this chat and do some actual mathematics

Alizter

16:15

@BalarkaSen $\int_0^\infty \frac{x}{(x^3+1)^2}dx$

@BalarkaSen I tried contour

set it up nicely

Balarka Sen

Blah

Alizter

however I cant seem to get it to work

Balarka Sen

Just use partial fractions

Alizter

yeah im doing the old fashioned way now

Balarka Sen

I am not going to kill a fly with a cannon, thank you

Mats Granvik

16:17

@Nick I see. But I would have saved myself a lot of time if I had had Mathematica 10.

Alizter

not as good as using FLT in an olympiad

Nick

@BalarkaSen: I'm multitasking. Currently doing a few answers on main + Learning about Groups... It's too boring

Balarka Sen

i don't care olympiad

@Nick Groups, eh? OK, what is your understanding?

Explain them to me. I'll give a walkthrough to a better idea.

Sarah

@Nick why are groups boring then?

Nick

@BalarkaSen: I'm 2 pages into the book but sure. A group, in my words, is an algebraic structure which obeys some rules.

Balarka Sen

16:19

@Nick Have you ever heard of mapping? From set to set?

Nick

@BalarkaSen: Yes.

@Sarah: Too much blah. Too little Aargh.

Balarka Sen

@Nick OK, consider the set X = {1, 2, 3, ..., n}. Think about the maps X --> X.

What do they look like?

How many are there?

Nick

A spiderweb. Um, infinite?

Teddy

hi again

Balarka Sen

No, think carefully.

First consider what the maps from X --> X are.

Nick

16:22

I should go read the book.

Teddy

the unreduced suspension of a n-simplex is a (n+1)-simplex, is it correct?

Sarah

@Nick how many ways can an element be mapped to another element

for a single one

Balarka Sen

@Nick walkthroughs > books

@Sarah is finally helping.

Nick

@Sarah: $n$ ways

Sarah

so howmany ways for another one

Balarka Sen

16:23

Right, @Nick. Now 1 can be mapped to n elts. What about 2?

Nick

n-1 ways

Balarka Sen

Exactly!

Sarah

no

Nick

Oh

Balarka Sen

@Sarah no he is right

Nick

16:23

@Sarah: Are we talking about relations or functions?

Sarah

@BalarkaSen no

There are $n^n$ ways not $n!$ ways.

Balarka Sen

There are $n!$ permutations

Think about it carefully.

Sarah

you said how many maps from X --> X

$|X|^{|X|}=n^n$

Nick

Cool, I've reached higher algebra... I'm stuck at counting

Balarka Sen

Yes, those are all permutations

{1, 2} --> {1, 2} as f(1) = 1 and f(2) = 2 and f(1) = 2 and f(2) = 1. There are 2 maps, not 2^2 = 4

You fail. Fail hard.

Nick

16:26

2 mins ago, by Nick

@Sarah: Are we talking about relations or functions?

Mapping is a vague term.

Sarah

I count 4

Nick

@BalarkaSen: You describe mapping for functions

@Sarah: You are doing it for relations

Balarka Sen

@Sarah Enumerate.

Sarah

what about $f(1)=f(2)=1,2$

Balarka Sen

Huh?

Sarah

16:28

unless the mappings are injective in which case that is different

Balarka Sen

That's not even a map.

Oh right right

Nick

1 min ago, by Nick

Mapping is a vague term.

Balarka Sen

I was thinking of bijective maps all along

Sarah

Thank you.

Balarka Sen

@Sarah Thanks for the refresher

Nick

16:29

@Sarah is the winner!

Alizter

ding ding?

Nick

Well, atleast I learnt something :D

Balarka Sen

@Nick OK, bijective maps from X to X

Nick

There are many maps :D

Balarka Sen

There are |X|! of them, right?

Nick

16:30

@BalarkaSen: Yes

Balarka Sen

OK, consider a map f : X --> X and a map g : X --> X

Nick

ok

considered

Balarka Sen

is f composition g a (bijective) map from X --> X?

whenever i say map, it's bijective.

Sarah

Nick

Not necessarily

Sarah

16:31

Is that bijective?

Nick

@Sarah: haha, Dora :D

Gracias

Balarka Sen

@Nick Logic behind your contradiction of ideas?

Nick

Wait, did you mean f(g(x))?

Balarka Sen

mmhmm

Sarah

Composition of bijections is bijective

Nick

16:33

^ What sarah said, yeah

Balarka Sen

let him figure out that one @Sarah

@Nick OK, exercise : prove that composition is a group operation on the bijective maps from X to X

and thus X is a group under this operation, of order n!

it's pretty trivial but you need to get used to these proofs

what's the identity element in here, @Nick?

Nick

@BalarkaSen: ... I should go find that book now.

Balarka Sen

walkthroughs > books

$F : X \to X$ be bijective. You want a map $G : X \to X$ such that $F(G(x)) = G(F(x)) = F(x)$ for all $x \in X$

What do you think?

Nick

@BalarkaSen: Should I construct some F and G such that the above is satisfied? If so, there are many. lol

I'm horrible with proofs

Balarka Sen

you're given an F

Nick

16:38

Yay!

Balarka Sen

in fact F can be anything

you want a corresponding G

Nick

(I though F was my grade. lol)

Balarka Sen

bah. if you're going to joke along i am not gonna waste my time helping

Nick

Yeah, I should get a base on this before you give me a walkthrough.

Maybe some other time.

Balarka Sen

just google [transformation groups] and be done

leaves

Nick

16:41

Hopefully, I haven't upset him.

Balarka Sen

nah

i am just annoyed to see such an unmotivated student

Nick

@BalarkaSen: ... Well, ok. revisiting your question

Prove that composition is a group operation on the bijective maps from X to X
and thus X is a group under this operation, of order n!

Balarka Sen

i knew that was going to work

Yes, @Nick

Nick

what do you mean by order n!

Balarka Sen

@Nick ok, better notations. we have proved that there are |X|! bijective maps from X to X

Nick

16:44

ah

Sarah

Little britain is quite funny. I can't understand some of it though :P

Balarka Sen

denote the set of all the maps from X to itself as Map(X). any problem, @Nick?

Nick

@BalarkaSen: other than notation, logic and madness... nah

Balarka Sen

OK, so we are going to prove that Map(X) is a group under composition.

what are the group laws again?

Mike Miller

@BalarkaSen All maps? A group?

Balarka Sen

16:48

bijective maps.

Mike Miller

This is an important difference...

Nick

Closure, Associativity, (fill in the blank), and Commutivity if it's abelian

Balarka Sen

@MikeMiller every map is bijective. don't kid me.

@Nick it has an identity element and has inverses.

Nick

@BalarkaSen: No, isn't there one more

Balarka Sen

apart from the fact that there is a binary operation, no.

you have already mentioned closure

Nick

16:50

The composition of functions is always associative. So strike 1

Balarka Sen

mmhmm

Nick

f ∘ (g ∘ h) = (f ∘ g) ∘ h

What's the latex for ∘?

Balarka Sen

mmhmm

@Nick $\circ$, as \circ

Nick

Assuming f and g are invertible $(f \circ g)^{−1} = ( g^{−1} \circ f^{ −1})$

So, the inverse exists

strike 2

um, what's left

Balarka Sen

no how do you know inverse exists?

Nick

16:53

f and g, i assume bijective.

Mike Miller

@BalarkaSen His point is that the composition of invertible functions is invertible. Hence, closure.

Nick

composition of bijective functions is bijective.

Balarka Sen

ah

yes, @Nick

Nick

Hence closure

@BalarkaSen: btw, i figured how to prove the bijection thing.

Balarka Sen

what thing

Nick

16:55

1 min ago, by Nick

composition of bijective functions is bijective.

Balarka Sen

that's what you proved above

Nick

no, i mean through formal argument. Every composite of injections is also an injection. Every composite of surjections is also a surjection. Hence, QED

Mike Miller

no, you really did just prove it above :P You wrote down an inverse for the composition of invertible functions; hence their composition is invertible. But invertible = bijective.

Nick

ohk. If that's enough for you. :D

Balarka Sen

what @Mike said

Nick

16:59

ok, so composition is a non-abelian group operation. Yeah?

Balarka Sen

you haven't decided the inversion yet

not even the existence of identity

Nick

Doesn't inversion naturally give you an identity.

Balarka Sen

where have you even decided inversion?

user112495

If I have an infinite sequence of step function $\phi_k$ and the infinite series of $\phi_k(x)$ functions converges for all x in $[a, b]$, is this infinite series also a step function?

Balarka Sen

you have proved that composition of invertible functions are invertible not that every map in Map(X) can be inverted.

Nick

17:05

@BalarkaSen: hmm. $f^{-1} \circ f = \text{identity}$

Balarka Sen

that's the definition of inverse functions, yes. what is the identity element in Map(X)?

Nick

the identity can be whatever $x$ you want

Balarka Sen

heh?

Nick

$f^{-1}(f(x)) = x$

Balarka Sen

identity elt $e$ of a group $(G, \cdot)$ is defined as $e \cdot a = a \cdot e = a$ for all $a \in G$

Nick

17:06

ah

Balarka Sen

what is this in the case of $(G, \cdot) = (\mathbf{Map}(X), \circ)$?

user112495

@BalarkaSen Have you got any ideas about my question?

Balarka Sen

no, i don't care about analysis

:P

Nick

I'm missing that.

What is the identity?

f o e = f = e o f

Balarka Sen

@Nick identity is a function F in Map(X) such that F(f(x)) = f(F(x)) = f(x) for all x in X and all f in Map(X)

not?

Mike Miller

17:09

@BalarkaSen wat

Balarka Sen

wat @Nick

Nick

No, just saying the first thing that popped into mind. The identity should be some constant function

Balarka Sen

@MikeMiller just point out the typo instead of watting

you're close @Nick read what is in there carefully

Mike Miller

I'm doing other things, @BalarkaSen.

And you've since fixed it. Your RHS originally said $x$.

Balarka Sen

AND pointing out my typos @Mike :P

not that i don't want you to. i want you to be present during my excellent representation of groups.

Nick

17:13

@BalarkaSen: Nah, I'm confused on this part

Balarka Sen

then F(f(x)) = f(f(x)) which is not equal to f(x)

Mike Miller

@Nick You want $f(F(x)) = f(x)$, yes?

Nick

yes

Mike Miller

You know $f$ is injective...

Balarka Sen

Blah. Ok.

Mike Miller

17:15

One thing at a time, @BalarkaSen

Nick

F(x) = x

yeah

Balarka Sen

Yeah

That's it @Nick

Nick

... I feel like I've been led around in a $\circ$

Balarka Sen

Well that was rather quicker than my classmates...

LOL @Nick

@Nick so what next.

Nick

@BalarkaSen: Now, I should probably go do groups formally. Any tips before I go? (other than I should study)

G'night :D

Balarka Sen

17:18

You're not done yet

:P

Nick

No, my net times expired so yeah.

Balarka Sen

blah. OK. think about the inverse part while you sleep.

Nick

@BalarkaSen: We'll pick this up later. :D

@BalarkaSen: Yeah, i think doing things formally will help.

toodles

Balarka Sen

No it won't

Mike Miller

everything you've done is formally... :P

Balarka Sen

17:19

This is the Arnold way. Arnold > Abstraction

Sarah

@MikeMiller Who is mally?

Daniel Fischer

@Sarah I guess Gus Mally.

The Game

@Chris'ssis Is it useful to now the first values of $\zeta(n)$ ?

Like the first bell numbers

user112495

17:38

If I have a sequence {${\phi_k}$}$^{\infty}_{k=1}$ of step functions on $[a, b]$ and $\phi(x) :=$ $\sum_{k=1}^\infty \phi_k(x)$ converges $\forall x \in [a, b]$, is $\phi(x)$ a step function on [a, b]?

@DanielFischer Would you be able to help me with this?

Daniel Fischer

@user112495 Typically, the limit is not a step function. You can for example write any continuous function on $[a,b]$ in that form.

Chris's sis

@TheGame I posted above an inequality, that is

$$\sum_{k=1}^{\infty} \frac{k+1}{k^2(2k+1)}>\frac{\pi^2}{12}+\frac{1}{8}$$

@TheGame Now I ask myself a question: is it useful to recognize $\zeta(2)$?

The Game

Could be

Chris's sis

@TheGame Well, "Could be" is also my answer to your question. :-)

The Game

-_-

@Chris'ssis Anyway I was thinking about the inequality you posted some days ago

Chris's sis

17:45

@TheGame Which one?

The Game

$\sum_1^\infty\frac{2k-1}{k^{2s}}\le\zeta(s)^2$

Remember ?

Chris's sis

@TheGame Sure :-)

The Game

Now if we write $\sum_1^\infty\dfrac{2k^a-1}{k^{2s}}\le\zeta(s)^2$, what is the limit value of $a$ for that to be true ? $a\in\mathbb{R},s>1$

user112495

@DanielFischer Are there examples where this is not the case, though?

@DanielFischer And how could I justify that this statement is false?

Daniel Fischer

@user112495 Let $$\phi_k(x) = \begin{cases} 0 &, x < a + 2^{-k} \\ 2^{-k} &, x \geqslant a + 2^{-k}.\end{cases}$$ Check that $\sum \phi_k$ is not a step function.

The Game

17:51

@Chris'ssis $\sum_{k=1}^{\infty} \frac{k+1}{k^2(2k+1)}=\zeta(2)-2+\ln(4)$

and $\ln(4)-2>1/8$

Chris's sis

@TheGame Then you have a generalization ...

Prove that $$\sum_{k=1}^{\infty} \frac{(k+1)^s}{k^{2s} (k^s+(k+1)^s)}>\frac{1}{2}\zeta(2s)+\frac{1}{8}, \space s>1$$

I'd add and subtract $k^s$ in numerator ...

ccorn

That question ought to be a duplicate, but I cannot find a similar question.

The Game

@Chris'ssis I'm thinking about the generalisation I posted above atm :)

Chris's sis

OK

The Game

@Chris'ssis You can help me if you have the answer :)

Chris's sis

18:06

@TheGame After a jogging session I can barely think. ;)

robjohn

18:27

@Chris'ssis are you using these inequalities somewhere?

Chris's sis

No, not now.

robjohn

@Chris'ssis I don't know if the use of an integral in my answer disqualifies it...

Ooh... I got an upvote anyway :-)

Chris's sis

@robjohn Back. Nice approach. I tried to use integrals, but I had some trouble.

kiBytes

Hello there =)

Just wondering if any of you would be so kind to answer me an "easy" math quesiton

about "minimal" and "maximal" elements

of a set

The Game

@kiBytes Ask first :)

kiBytes

18:40

let me go ahead

c = { (x,y) € R^2 | 0 < y < 4 - x^2 }

robjohn

@Chris'ssis The actual minimum is much greater, but it seems to be hard to prove.

kiBytes

wolframalpha.com/input/?i=y+%3D+4+-+x%5E2%2C+y%3E0

Chris's sis

@robjohn Yeah, it is.

@robjohn I should get a badge for that one ... it's a very nice question.

kiBytes

from my point of view, I thought that it should have only one maximal element in x=0, y = 4

but it seems it have maximals in x € [0,2], y = 4 - x

(in my first presentation is 0 <= 4 <= 4-x^2 ) sorry about that

anybody?

*it has, sorry about that also xD

The Game

@kiBytes I may sound nooby, but how do you define the maximal element of a set of tuples ?

max{x|P(X)} is the greatest x in the set, but how do you define max{(x,y)|P(x,y)} ?

kiBytes

18:53

I believe that is the foundation of my doubt =D

The Game

@robjohn Is that usually defined ?

Chris's sis

@robjohn Did you see the added part of karvens?

@robjohn

@robjohn that part where he invoked the inequality can be done without pen and paper, it's obvious.

@robjohn $$\frac{k^s}{k^{2s}}> \frac{3\sqrt{3}}{2} \frac{1}{k^{2s}}\Rightarrow \frac{2 k^s}{k^{2s}}> 3\sqrt{3} \frac{1}{k^{2s}}, \space k>2$$

Universe

hello I have an question about Gauss Central Forward Interpolation..

Chris's sis

@robjohn but how about the first terms? I'm not sure I understood his point there.

Universe

anyone knows about Gauss Central Forward Interpolation..???

how I choose the central x0 in a even number of x's

Chris's sis

19:05

@robjohn that inequality is also true for $x=2$ hence the confusion. Yes, it works.

(he only wrote $x>2$)

Universe

I guess noone...

Chris's sis

@TheGame I never dedicated a song to you ;)

►

@robjohn Actually he used an inequality similar to what I posted here before having it solved.

Universe

atleast say that you dont know guys...

Chris's sis

That is $$\frac{1}{x(1-x^2)}\ge \frac{3\sqrt{3}}{2}, \space x\in(0,1)$$

Well, to say it directly, it's the same with $x\mapsto 1/x$

@robjohn I wonder if he was in chat in the last 24 hours ... just curious about that ...

Universe

anyone knows about Gauss Central Forward Interpolation..???
how I choose the central x0 in a even number of x's?

Chris's sis

19:19

@TheGame that one was a suggestion to use on my birthday that comes soon. Of course, no! :-)

The Game

@Chris'ssis I'm not sure how to take it :)

Chris's sis

@TheGame It's just funny, without any sense to me.

The Game

I find that music horrible :c metal isn't my type

Chris's sis

@TheGame That one was a proposal to add to my playlist, that's all.

robjohn

@Chris'ssis sorry... I just got out of the shower. I will look at his answer again.

Chris's sis

19:24

@robjohn Could you see if he visited the chat in the last 24 hours? He used the inequality I posted yesterday here.

robjohn

who?

Chris's sis

@robjohn karvens

robjohn

@Chris'ssis there is no way to tell if he has looked at a chat log, but if you look at his profile, he hasn't been on chat in 8 days

Chris's sis

@robjohn Yeah, you're right. Anyway, that guy is terribly good. It's the second time he manage to come up with a crazy awesome solution.

@TheGame I prefer this one ... (actually it's one of my favourite songs from all times)

►

(now that I see David's star in the video, I can say it is pretty similar to the logo of MSE if we use a bit our imagination )

The Game

Uh

Chris's sis

19:36

@robjohn that inequality put me in some troubles, so I need to work much more on inequalities.

The Game

@Chris'ssis Try the one I suggested :D

@Chris'ssis I'd be more like

►

Though i like yours too

Chris's sis

@TheGame It's fair you try to come up with a solution to that case and then try harder cases :D

The Game

@Chris'ssis I'm not sure I'm able to it though

The case $a=1$ was easy because of some identities

And we used a really simple bound $\frac{1}{k^{2s}}\le\frac{1}{k^{s+1}}$

Which means we could improve it a lot for higher values of $a$

Chris's sis

@TheGame I see. I didn't try it yet.

@TheGame Yeah, that's a nice song. Thanks.

Jasper Loy

Sometimes I feel that mathematics is meaningless.

Chris's sis

19:46

@robjohn all these inequalities I posted are really rare gems (I'd like to add some in my book, not too many).

The one with squared zeta is simpy too nice! It's amazing that you need to use no special tool, nothing, it can be done in a very easy way! That's fantastic!

At first sight it looks pretty ugly ... but in the end one realizes it's an amazing inequality.

Jasper Loy

When I write my books, I will include all the theorems that I know.

Chris's sis

@JasperLoy Would you like to write a book on integrals, series and limits? :D

Jasper Loy

@Chris'ssis Nope, they are not my cup of tea. =)

Chris's sis

@JasperLoy Oh, OK. :-)

Jasper Loy

@alizter You misspelled abbreviate.

Chris's sis

19:50

@TheGame did you see in your country some books very similar to Furdui's book?

The Game

@Chris'ssis That was the first non-school math book I ever bought

@Chris'ssis Why do you tell me that ??

I know that

You already told me

Jasper Loy

@Chris'ssis I am still waiting for you to tell me your name. You can email me one day when you are ready. =)

The Game

@Chris'ssis i'm just saying that it was the first non-school math book I ever bought, hence i cannot say if there are other similar ones in France

user112495

If $\phi_k(x) = \begin{cases} 0 &, x < a + 2^{-k} \\ 2^{-k} &, x \geqslant a + 2^{-k}.\end{cases}$, can anyone explain why $\sum \phi_k$ isn't a step function? Is it because there are infinitely many partitions where the function is constant over $[0, a+2^{-k}]$?

The Game

@Chris'ssis Uh ?? I never said that !

@Chris'ssis How did you come to understand that ?

Chris's sis

19:53

@TheGame You said " was the first non-school math book". There are also questions you may give in high school.

Jasper Loy

anon is very cold. He does not talk much to me. I think I won't talk to him anymore, lol.

Daniel Fischer

@user112495 A step function can attain only finitely many values.

The Game

by non-school I mean 'not a school math book' ie a manual etc

@Chris'ssis It's not a book your school will ask you to buy to follow classes

Chris's sis

@TheGame Ah, OK. That was a misunderstanding.

Jasper Loy

@chris So when are you going to quit your job and go back to school to study math?

The Game

19:55

I love the book, in fact ! @Chris'ssis

Chris's sis

@TheGame Really? Glad to hear that.

The Game

Yeah

Chris's sis

@TheGame How many pages did you read?

@JasperLoy First I need another job. Life here is not that easy, but with some tutoring and accounting I'm still pretty OK:-)

The Game

Not many yet, because 1) I don't have much time and 2) I don't like going to the next question without having understood the previous one first, and to me the book is quite difficult. @Chris'ssis

Jasper Loy

I have not studied math for so long that it's weird to do nothing but math every day next year.

The Game

19:56

About 5(ie exercices) + the annexes

I'd love to have more time

But I already have a lot of homework

Since the start of the schoolyear I've had more than 800 pages of chemistry

Jasper Loy

OK, I have decided to get Gratzer's More Math into LaTeX for my LaTeX book as it is the only one that focuses on mathematical typesetting.

Chris's sis

@TheGame I have met some problems with writing my book. As I said, I need every problem be special in some sense, not to meet it again and again on every page in a different form. I need to work more, to try to be more creative.

The Game

@Chris'ssis Such as ?

Jasper Loy

I have also decided to use pstricks for all my mathematical graphics as it has a large number of packages.

user112495

@DanielFischer Ah. Thanks. Just one more question. Would the following be correct as an example of a function $\psi : [0, 1] \rightarrow \mathbb{R}$ is not a step function, but $|/psi|$ is:

$\phi_k(x) = \begin{cases} 1 &, x \in \mathbb{R} \\ -1 &, x \notin \mathbb{R}.\end{cases}$

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$Mathematics$ Mathematics