Mathematics - 2017-01-17 (page 5 of 5)

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user379685

23:00

Could someone help me with this integral wolframalpha.com/input/?i=1%2F(x*sqrt(x%5E2%2B4x-4)

in the most elementary way>

1/(x*sqrt(x^2+4x-4)) *

Akiva Weinberger

23:14

wolframalpha.com/input/…

Fixed the link

user379685

thanks but do you know how to get to this result?

PVAL-inactive

@AkivaWeinberger Oh my is that powerpoint?

Why not just read Milnor's paper?

Akiva Weinberger

I just took one of the first things that Google gave me

PVAL-inactive

Tautologically it is accessible to an undergraduate.

Akiva Weinberger

@user379685 I don't know. I would probably start by factoring the polynomial, or perhaps by completing the square

Socrates

23:17

can someone give me an excercise about set (in)equalities?

Akiva Weinberger

What's a set inequality

Socrates

something like $A\cap B=??$

or $\subset$

Akiva Weinberger

@Socrates Prove that $\{(x,y,z,n)\in{\Bbb N_{>0}}^4\mid x^{n+2}+y^{n+2}=z^{n+2}\}=\emptyset$

@Socrates Do you mean naïve set theory stuff? Like the type you'd solve by Venn diagrams

Socrates

well, it's ok, I try this

hopefully I learn something before I give up haha

isnt this fermat?

Akiva Weinberger

It is. I was joking

Socrates

23:23

I guessed so

but at first I thaught it's something else

well, some naive set theory stuff, but please no venns lol

i hate drawing venns

Akiva Weinberger

Prove that $(A\setminus B)\cup(B\setminus A)=(A\cup B)\setminus(A\cap B)$

That's the symmetric difference between $A$ and $B$ and is often denoted by $A\triangle B$

Socrates

meh, this one might be actually easier to prove with venn

lol

wow that was quick

But how do you make a venn proof that needs no further explanation?

Alessandro Codenotti

if you want something which is a real pain to show is that the symmetric difference is an associative operation

Akiva Weinberger

Is it really a pain? @AlessandroCodenotti

Alessandro Codenotti

it depends on your definition of pain I guess

it's more tedious than I'd like :P

Akiva Weinberger

23:38

Just point out that it's like addition in $\Bbb Z_2$?

$x$ will be in $A\,\triangle\,B$ iff $[x\in A]+[x\in B]$ is odd (using Iverson bracket)

which means that $[x\in A\,\triangle\,B]\equiv[x\in A]+[x\in B]\pmod2$

Alessandro Codenotti

yeah I see

Akiva Weinberger

So it essentially follows from the associativity of addition mod 2.

user379685

Is really no one able to help me with this integral? 1/(x*sqrt(x^2+4x-4))

Akiva Weinberger

Break up $[x\in(A\,\triangle\,B)\,\triangle\,C]$ and $[x\in A\,\triangle\,(B\,\triangle\,C)]$; you'll see they're both $[x\in A]+[x\in B]+[x\in C]$ mod $2$. QED.

@user379685 Have you tried completing the square?

user379685

yeah

Alessandro Codenotti

23:42

@AkivaWeinberger you can get examples of infinite rings where every element has finite order by using $\mathcal{P}(X)$ for an infinite $X$ with symmetric difference as addition and intersection as multiplication

Akiva Weinberger

@user379685 So you get 1/(x*sqrt((x+2)^2-8)), right?

user379685

yes

Akiva Weinberger

Substitute in u*sqrt(8)=x+2, factor out the sqrt(8) from the root and use some trig substitution probably

so you have 1/sqrt(8) * 1/((u*sqrt(8)-2)*sqrt(u^2-1)), I think. Probably some trig substitution works.

Socrates

@AkivaWeinberger I can't follow why we use mod 2

you mean because the structure is the same?

Akiva Weinberger

Prove that $[x\in A\,\triangle\,B]\equiv[x\in A]+[x\in B]\pmod2$, where $[P]=\begin{cases}1,&P\text{ is true}\\0,&P\text{ is false}\end{cases}$ (that's called the Iverson bracket)

But, yeah, essentially because the structure is the same

The set $\{\emptyset,A,B,A\,\triangle\,B\}$ forms a group congruent to $(\Bbb Z_2)^2$ under symmetric difference, for example.