Mathematics

3:00 AM

is that what your refering too?

or some other partial dt

you have putatively shown that at least one residue class has infinitely many primes, but in order to show all of the classes do you need to know that each and every limit is not zero (you are correct that existence is irrelevant), but your method (which is to show each limit is equal) doesn't work if convergence is an issue

Ethan

I know the limits wont be zero, but I don't know if they exist

anon

@Ethan how do you know each of them won't be zero?

Ethan

They could all oscilate between a finite value, but when summed together not oscilate

Because with the last line showing that the sum is equal to one, it implies one of them must converge to somthing positive ( as each term is itself always positive) and the total sum is positive

anon

(the limit of a ratio of two positive quantities can still be zero)

Ethan

3:02 AM

I know

but if they all were zero

then we wouldnt get one on the rhs

anon

right, that shows that they are not all simultaneously zero, which proves partial DT. to get full DT, you need each summand to be nonzero.

Ethan

I don't understand, but doesn't that imply atleast one summand is not zero

and the previous argument shows all the limits are equal

anon

um, what shows the limits are all equal?

Ethan

It shows there equal if the limits exist, as you mentioned

which was the problem

anon

they're

"It"?

Ethan

3:04 AM

$$\lim_{s\to 1}\frac{1}{\zeta(s)}\sum_{k=0}^\infty\frac{\Lambda(ak+b)}{(ak+b)^s}=\frac{1}{a}\sum_{\substack{\gcd(a,r)=1\\1\leq r\leq a}}\sum_{k=0}^\infty\frac{-\ln(ak+r)\mu(ak+r)}{(ak+r)}$$

Which shows that there is no dependence on b, provide the limit exists

which is what you pointed out, that I don't know it exists

anon

I identified an issue earlier: how do you know the limits, if they exist, are all equal, if we don't know that $\sum_{k=0}^\infty \frac{-\ln(ak+r)\mu(ak+r)}{(ak+r)}$ exists?

Ethan

Yes that is the problem

anon

I am not just talking about the limit on the left.

I just rewrote it, not two seconds ago.

Having to repeat myself to you over and over again is exhausting.

Ethan

sorry

anon

they're

Ethan

3:07 AM

Assuming it does exist am I okay? I will try to work on geting a proof of that

anon

Assuming each series $\sum_{k=0}^\infty$ on the RHS exists, then yes.

Ethan

I can show that they exist, when the modulus a is prime

anon

Well, I will be going home now.

Ethan

thanks for the help

I appreciate it alot

anon

cool

(it was mostly exhausting because there's so much latex involved, I should clarify)

BenjaLim

3:13 AM

@anon hey

Have you finished discussing with ethan?

Ethan

@BenjaLim

BenjaLim

yes?

Ethan

Why do polynomials of the form:

$(a+b+c)^n+(a-b-c)^n-(a-b+c)^n-(a+b-c)^n$

Take on, such a 'nice' expansion?

For example at $3,5$ we get,

$$(a+b+c)^3+(a-b-c)^3-(a-b+c)^3-(a+b-c)^3=24abc$$
$$(a+b+c)^5+(a-b-c)^5-(a-b+c)^5-(a+b-c)^5=80abc(a^2+b^2+c^2)$$

BenjaLim

No I don't think so.

Ethan

no what

BenjaLim

3:19 AM

no deep reason

Ethan

o

BenjaLim

you are asking why an identity like that is tautologically true

Ethan

what do you mean tautologically

BenjaLim

meaning to say that identity you wrote

is true even before you wrote it down :D

No I'm joking

Ethan

I don't understand lol

BenjaLim

3:20 AM

those identities are true directly from the axioms

Ethan

Is there a field of mathematics equiped with tools capable of determineing parametric equations for diophantine equations

BenjaLim

what do you mean?

Ethan

like finding polynomials satisfieing nice propertys

BenjaLim

No I don't think so.

Ethan

like lets say I wanted 3 polynomials whose cubes added up to a square

oh

@BenjaLim Do you know about p-adics?

BenjaLim

3:31 AM

no not really.

Ethan

I thought you did alot of commutative algebra

BenjaLim

yes but I haven't really studied completions.

@Ethan ok tell me what the p-adics are then.

Ethan

I don't know alot

BenjaLim

then why are you talking about them when you say << I don't know >> ?

Ethan

I was going to ask for help on proving an identity

related to p-adics

BenjaLim

3:36 AM

You want to prove an identity involving p-adics but don't know what they are?

Ethan

I didn't mean to say I don't know

I meant to say I don't know alot

BenjaLim

4:22 AM

@anon Hey, I seem to get myself in knots and can't figure out why in your post here the order of $k/(q+1)$ must be $q^2 - 1$. I am only able to prove that the order is greater than $q + 1$. It seems we should be able to deduce from $(k,q^2 - 1) = q+1$ that $(k/(q+1),q^2 - 1) = 1$ but I'm not seeing somehow.

@Sanchez hey

anon

you can't deduce that; (18,5^2-1)=5+1 but (18/(5+1),5^2-1)=3.

BenjaLim

yes.

@anon so why should the order of $k/(q+1)$ be $q^2 - 1$?

anon

perhaps it shouldn't

BenjaLim

@anon hmmm

@anon my answer is wrong then.

Sanchez

4:45 AM

@BenjaLim, hey

Gnintendo

5:01 AM

so now that I'm a couple weeks into this semester

I'm getting the impression that Differential Equations is just going to be a set of silly tricks to help solve DEs

BDub

@Gnintendo Is it upper or lower division DEs?

Gnintendo

6:21 AM

@BDub lower? I'm an undergraduate

this is my first DE class

(I'm a freshman)

Anthony Peter

6:41 AM

would anyone be willing to look over a simple undergrad paper for me?

Ethan

Anthony your 14 years old

Anthony Peter

Shhh*

Novice

Prodigy!

Anthony Peter

I just finished my paper for CA

It's my first math paper

Ethan

Ca?

Anthony Peter

6:44 AM

by CA I mean Complex Analysis not California aha

Ethan

lol I think you should study more

Anthony Peter

Study what more?

Why ahaha?

and N-nacci* numbers not just fibonacci

I just wanted to find a new way to go about things that hasn't already been exhausted

I dooo

It's tribonacci

tetranacci

pentanacci

so it suffices to say n-nacci

Ethan

what is your profile picture anthony?

Anthony Peter

A skull

Ethan

of what?

Anthony Peter

6:51 AM

It's a deformed skull

human

Ethan

why?

Anthony Peter

I dunno, it was the first thing that popped up in my downloaded pictures. I used it last year for something

Tai

8:00 AM

$f(x)$

can we use Latex here?

Orange Harvester

@Tai yes you can, see the starboard for more instructions. $\LaTeX$ support for chat

Chris's sister

9:32 AM

Hi all

Do you know some nice way to compute $\int_0^1 \frac{1}{x^4+1}$? I'm looking for something fast.

pourjour

9:48 AM

hi

how can use tex in word

2013

Mariano Suárez-Alvarez

10:26 AM

@Chris'ssister use a partial fraction decomposition of th eintegrand

Chris's sister

@MarianoSuárez-Alvarez: thanks. However, this way is a bit too long.

Mariano Suárez-Alvarez

oh you want a magic way

Chris's sister

@MarianoSuárez-Alvarez: hehe. Something like a magic way. :-)

Jonas Teuwen

Mathematica does it in a quiff.

Mariano Suárez-Alvarez

you need to adjust your definition of too long if that computation seems too long for you

Jonas Teuwen

10:38 AM

It is good to know how it is done, but otherwise I see no use in not just computing it with a CAS and verifying if that is right.

Unless it is for a course... (and it does not get much faster than that).

But a couple of trickies are to make it one to $0$ to $\infty$, then contours.

Old John

Hi @JonasTeuwen how are you today

Jonas Teuwen

I hardly could sleep because my legs were on firrrre!

Old John

@JonasTeuwen darn - sorry to hear that

Jonas Teuwen

I can handle the pain, not a biggie.

Plenty of spare vertebrae.

Old John

@JonasTeuwen Watching Murray - Djokovic match ... gripping stuff

Jonas Teuwen

10:46 AM

Some splendid spines they have eh?

Old John

@JonasTeuwen spines, legs, arms - not much about them that is not amazing

maybe their hairstyles :)

Jonas Teuwen

Sometimes it is.

Straight out of bed onto the playing court.

user19161

@JonasTeuwen I keep a cold pack in my fridge for all purposes, such as cooling my head.

Jonas Teuwen

Alright.

user19161

@OldJohn Maybe the hair is the only amazing thing. =)

Jonas Teuwen

10:53 AM

It is not actually on fire by the way, the nerve is compressed much higher, so cooling my legs will do... shit.

Old John

Time for more coffee - back later, folks

Jonas Teuwen

You had not enough? Good.

I have that one in my office.

I did not put it there.

user19161

12:07 PM

@JonasTeuwen Those who take whisky instead are called gods, like you!

Jonas Teuwen

Not while working eh.

user19161

1:02 PM

@JonasTeuwen I figured out why I got different image search results. It was not because of a different browser identity. It is because google is changing its search algorithm these days, and it seems the matter is not settled.

Orange Harvester

1:20 PM

@JasonBourne Google is changing its algorithms everyday, there is a reason why it is called machine learning. :-)

@OldJohn There was a time when my build was like that. wistful only 10-13 cm shorter. :P

Transcript for

$Mathematics$ Mathematics