Mathematics - 2013-06-07 (page 2 of 2)

Ethan

07:01

$$1+\sum_{n=0}^\infty\frac{2^nn!}{(2n+1)!}=\sum_{n=0}^\infty \frac{2^nn!}{(2n)!}$$

$$1=\sum_{n=0}^\infty\frac{x^ny^{\frac{n^2-n}{2}}}{(x^0+y^0)(x^1+y^1)(x^2+y^2)(x^3+y^3)(x^4+y^4)...(x^n+y^n)}$$

$|\frac{x}{y}|>1$

$$\frac{1}{\ln(q)}=\sum_{n=-\infty}^\infty\frac{2^n}{q^{2^n}+1}$$

$|q|>1$

$$\frac{1}{1-q}=\prod_{n=0}^\infty(1+q^{2^n})$$

$|q|>1$

$$80abc(a^2+b^2+c^2)=(a+b+c)^5+(a-b-c)^5-(a+b-c)^5-(a-b+c)^5$$

Dominic Michaelis

07:24

Does someone have a reference for me how to get used to pullbacks

Ethan

pullbacks?

Dominic Michaelis

in category theory

Ethan

sorry can't help lol

Lord_Farin

@DominicMichaelis Prove the pullback lemma yourself.

Loki Clock

@Dominic Are you comfortable with abstract algebra?

Dominic Michaelis

07:26

@Lord_Farin which one exacty? I mean more, that I have no feeling how to compute actually the pullback of two morphisms

Little Child

Good Morning, all ! :-)

Loki Clock

Morgunn

Dominic Michaelis

@LokiClock I don't know much about it but I like it :)

Lord_Farin

@DominicMichaelis Hmm... Well, "computing" is only relevant if you have a category at hand, to prove that it exists. Mostly, then, you can rely on the universal property.

@LittleChild A good morning to you.

Little Child

I was just doing limits by piece-wise and I hit a doubt

Dominic Michaelis

07:29

@Lord_Farin I was trying to prove that the only pullback stable epis in hausdorff are the surjections, and my first idea was totally right but i totally messed up what actually is the pullback so I failed ...

Lord_Farin

@DominicMichaelis In most $\mathsf{Set}$-like categories, the pullback is a subset (in this case, -space) of the product.

Loki Clock

@LokiClock You could try proving that the direct product of two groups is the categorical product first. Once you do that, after some though, finding the pullback should be easy.

@DominicMichaelis The last one is for you. I'm sleepy.

Dominic Michaelis

I am still a bit surprised about that writelatex.com/217665pqpzpn

The arrows in the first picture are the inclusions arrows

Identity restricted to the domain

Ethan

@LittleChild $$\int_{0}^1\frac{1}{x^x} \ dx =\sum_{n=1}^\infty\frac{1}{n^n}$$ sophomore's dream

Martin Sleziak

@Lord_Farin Pullback can be constructed using equalizer and product. (This is also mentioned on Wikipedia.) So this is true in any category where equalizers are subsets.

Little Child

07:35

@Ethan this is an identity that I must know ? :-)

Ethan

;[

Lord_Farin

@MartinSleziak I know. But I wasn't sure if I should throw this at @Dominic. :)

Dominic Michaelis

@LittleChild no you just need to know how to proof it ;)

Little Child

I havent started with integration, yet.

I am still doing limits

Ethan

@LittleChild $$\frac{1}{1-q}=\prod_{n=0}^\infty(1+q^{2^n})$$

|q|<1

Lord_Farin

07:36

@DominicMichaelis prove* (Sorry, grammar pet peeve.)

Dominic Michaelis

@Lord_Farin ah gee I make that mistake again

I thought julien healed me ^^

Ethan

28 mins ago, by Ethan

$$1=\sum_{n=0}^\infty\frac{x^ny^{\frac{n^2-n}{2}}}{(x^0+y^0)(x^1+y^1)(x^2+y^2)(x^3+y^3)(x^4+y^4)...(x^n+y^n)}$$

Little Child

wooof woof.. I am not a math major...

Dominic Michaelis

thatis not stuff for a math major

Ethan

$|\frac{x}{y}|>1$

Little Child

07:37

lol

Ethan

its baby stuff but still nice

Lord_Farin

@LittleChild With pride. I've had people call me a "grammar SS officer" :D.

Martin Sleziak

@Ethan Are you testing how long it takes until someone flags you?

Ethan

@MartinSleziak?

Little Child

@MartinSleziak Why would he be flagged ?

Martin Sleziak

07:38

Avoid constant and excessive links to Wikipedia and YouTube. These links would often open up and consume quite a lot of the conversation space.

Dominic Michaelis

spamming

Lord_Farin

@DominicMichaelis It has to sink in for a bit.

Loki Clock

I'm off to bed. Night, everybody!

Martin Sleziak

Don't you think that sending 10 math formulas which have nothing to do with the conversation going on with the chatroom does take a lot of conversation space?

Ethan

@LittleChild wana see a nice telescoping sum

Dominic Michaelis

07:39

But now I know that in Haus only the surjections are pullback stable epis even though there are much more epis in Haus

Lord_Farin

@LokiClock A good night to you. :)

Little Child

@Ethan Wanna see a deep programming loop ?

Dominic Michaelis

I go some minutes away need to buy something

Little Child

Ethan

@LittleChild $$\frac{1}{q+1}=\frac{1}{q-1}-\frac{2}{q^2-1}$$

Little Child

07:41

I need to go away for a few minutes as I need to buy something @DominicMichaelis

Ethan

$$\frac{1}{\ln(q)}=\sum_{n=-\infty}^\infty\frac{2^n}{q^{2^n}+1}$$

|q|>1

$$\frac{1}{q-1}=\sum_{n=0}^\infty\frac{2^n}{q^{2^n}+1}$$

Little Child

I have noooo idea what u are saying :-D

Ethan

it telescopes

isn't it nice

Little Child

I only know how to use telescopes to see stars

Ethan

what?

Little Child

07:42

I have no idea what those equations u are posting are

Ethan

do you have latex enabled in chat

Little Child

Yes yes I do but I have no clue what those equations are

how to solve

or prove

I concede defeat

Ethan

lol

Little Child

My turn !

Ethan

k

show me somthing

$$(a^4-2ab^3)^3+(a^3b+b^4)^3+(2a^3b-b^4)^3=(a^4+ab^3)^3 , \text{ Viete}$$

oh no wait, look at this guy

Little Child

07:44

Ethan

Jørgen Pedersen Gram

Jørgen Pedersen Gram (June 27, 1850 – April 29, 1916) was a Danish actuary and mathematician who was born in Nustrup, Duchy of Schleswig, Denmark and died in Copenhagen, Denmark. Important papers of his include On series expansions determined by the methods of least squares, and Investigations of the number of primes less than a given number. The mathematical method that bears his name, the Gram–Schmidt process, was first published in the former paper, in 1883. The Gramian matrix is also named after him. For number theorists his main fame is the series for the Riemann zeta function (t...

Little Child

Given the XML tree, find:

Ethan

"was killed by a bicycle" at the bottom of wiki page

Little Child

who lies on ancestor axis, preceding axis, following axis, descending axis

user1

@Ethan Are we sure it is the bicycle that killed him.

Ethan

07:46

it says so..

it says he was killed by a bicycle

;d

He died after being struck by a bicycle

user1

Well, it says at some point after being struck by a bicycle he died.

Little Child

It was a high speed bicycle

Ethan

he was attacked by a vicious bicycle

Little Child

He came to the wrong neighborhood, Mo*******r

Ethan

what

Little Child

07:47

Thats a meme.

knowyourmeme.com/photos/…

Ethan

i don't like it

Little Child

LoL !

@Ethan You good with limits ?

Ethan, I asked u something :-/

Ethan

@LittleChild define good

Little Child

Being able to clarify basic doubts = good

Ethan

limits is kinda vague

Little Child

07:51

limits involving absolute values and being solved with piecewise functions articulate ?

Maisam Hedyelloo

hi guys

Little Child

heya

would you stop posting random equations ?

Never Mind. Apologies.

@amWhy u there ?

Lord_Farin

@LittleChild For someone criticizing Dominic on his English, that's a liability to hypocrisy...

Maisam Hedyelloo

hi guys i need some referenceabout mathematical modeling of physical phenomenons such as wave, shocks and diffusion equation

Little Child

uh ??

I thought everyone was just having fun. No offense. :-/

user1

07:57

@Lord_Farin yo

Lord_Farin

@LittleChild I misunderstood your earlier comment, then. A drawback of text-only communication. Let us return to, cq. continue, having fun. :D

Little Child

This place is so much more hostile than StackOverflow, man !

Lord_Farin

@LittleChild Do you mean the main site?

Little Child

No, the programming site. S.O.

main site is S.E.

StackExchange

Lord_Farin

I was asking about "[t]his place".

Little Child

07:59

Any ways, I am out. Apologies if I hurt anybody's feelings --- unintentionally.

Especially @DominicMichaelis .

Lord_Farin

@LittleChild Oh, you surely didn't. Dominic doesn't mind some criticism on his English; he likes to learn.

No worries. :)

Oh, and goodbye!

Lord_Farin

08:13

I'm leaving, there's a lot of other stuff to do.

Bye all.

Dominic Michaelis

@LittleChild Oh you didn't hurt me :) As Lord_Farin says I want to improve me english :)

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