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Joey Morani
Joey Morani
12:01 AM
Hey. Would -2x - x = -3x ?
or would it be -2 ?
 
user19161
@OldJohn Haha, if you ever come here it is really warm!
 
Old John
Old John
@JasonBourne I would love to visit Singapore or Indonesia - but I suspect it will not happen now
@JoeyMorani $-3x$
 
Joey Morani
Joey Morani
Thanks @OldJohn. That's what I thought :)
 
user19161
@JoeyMorani Why would it be -2?
 
user19161
@JoeyMorani If you ask this, that means you really need to revise all your algebraic manipulations and understand why they work. Knowing the answer to a single example won't help you!
 
user19161
12:07 AM
@OldJohn I would like to ask you if you have heard of the Hua Loo Keng Math School Textbook Series.
 
Old John
Old John
@JasonBourne no - I have not come across those - are they good?
 
user19161
@OldJohn Well, they seem great but are written in Chinese. I don't have copies with me now. They go all the way from years 1 to 12. Very Olympiad flavour. Great preparation for IMO and math degree.
 
user19161
I have seen some Japanese math textbooks too. I think the great Kosaku Yosida had a hand in writing/editing them.
 
Old John
Old John
@JasonBourne Pity - my Mandarin is fairly limited now :( - The best I managed was to read "he Lady in the Painting" in simplified characters
 
user19161
Is there a standard textbook for A Level Math in the UK? @old
 
Old John
Old John
12:12 AM
@JasonBourne Not sure - I have not taught A Level Maths for a few years, but there used to be several different "standards" depending on the exam board - confusing system
I used to use "Bostock and Chandler" for A Level Further Maths, and found it quite good, but I had to supplement it with lots of extra examples and explanations
@JasonBourne Do you know the folk-tale about the Lady in the painting?
 
user19161
@OldJohn No, your knowledge of Chinese history and culture might be better than mine. I am not really interested in anything to do with Chinese, other than their mathematics!
 
Old John
Old John
@JasonBourne I doubt that my knowledge would be greater!
 
user19161
@OldJohn I am going to bed. Good night!
 
Old John
Old John
@JasonBourne Goodnight! - sleep well
 
amWhy
amWhy
Hey all!
Any one around?
 
Will Jagy
Will Jagy
12:23 AM
There was a song about modelling, although that was not entirely clear at the time: youtube.com/watch?v=39YUXIKrOFk One of them said in an interview that they saw these guys in the gym who were in love with themselves.
 
Sanchez
Sanchez
Hi @amWhy
 
amWhy
amWhy
@Sanchez hello!
 
Sanchez
Sanchez
@amWhy, are you an undergrad at Wisconsin?
 
amWhy
amWhy
@Sanchez No, not an undergrad. I'm a grad student, but on a bit of a sabbatical...sort of ABD
 
Sanchez
Sanchez
@amWhy, oh wow.. in math?
What kind of math do you study?
 
amWhy
amWhy
12:28 AM
I'm sort of a generalist: I did earn a PhD in logic, but now working in algebra & math logic
 
Sanchez
Sanchez
Cool.
Do you know anything about model theory? @amWhy
 
amWhy
amWhy
@Sanchez Some
@Sanchez What's your area of interest?
 
Sanchez
Sanchez
@amWhy, nice. Number Theory. I heard that model theory is applied to algebraic geometry in recent years, do you know where I can find an exposition for that?
 
amWhy
amWhy
@Sanchez Number theory fascinates me!
 
Bel
Bel
Hi everybody
 
Sanchez
Sanchez
12:31 AM
Hi @Bel.
 
Old John
Old John
Time I slept - need to watch Murray trying to overcome Djokovic in the morning
 
Sanchez
Sanchez
Surely, number theory is awesome ;) @amWhy
 
Bel
Bel
I'm looking for help in algebra and specifically in quadratic form
 
amWhy
amWhy
@Sanchez Was it Gauss who called number theory the "Queen of Mathematics"?
 
Sanchez
Sanchez
Bye @oldJohn, that's definitely gonna be an intense match.
@amWhy, I think so.
 
Old John
Old John
12:33 AM
@Sanchez hope so! - bye for now
Bye @WillJagy
 
Sanchez
Sanchez
@amWhy, do you know if there's an exposition of model theory applied to algebraic geometry?
 
Bel
Bel
I have to implement a program that doesn't accept the form $B^T\times B$
 
Will Jagy
Will Jagy
Bye bye love
Bye bye happiness
Hello loneliness
Think I'm gonna cry.
@Bel, what quadratic form would that be?
 
Bel
Bel
and on the other hand I have to introduce trace((B^t\times B)^{-1})
in my program
so I would like to find the quadratic form for the trace_inv of B^tB
like for trace(B^T\times B)=\sum_{j=1}^mb_j^T\times b_j
where b_j is the j^th column of B
 
Will Jagy
Will Jagy
@Bel, you had better type some 2 by 2 and 3 by 3 examples into your question, where you are not using the words entirely correctly: math.stackexchange.com/questions/287750/…
 
Bel
Bel
12:39 AM
@WillJagy, you're right
But any trace of a matrix product could be written in the form of the sum of a quadratic form is that correct?
 
Bel
Bel
12:56 AM
@WillJagy, I have added an example in my post
 
user19161
So guys I happened to enter the OS guys room just now and read some of the stuff that has been said there. Please don't label people as "rude" or "intellectual snob" when they try to help you in math or tell you their honest opinion of something.
 
user19161
We don't need to help people only for them to call us "rude" or "intellectual snob". Go think about it.
 
user19161
Seriously, I am super upset by this now.
 
Ethan
Ethan
@amWhy can u help me
 
amWhy
amWhy
@Ethan yes? Hi!
 
Ethan
Ethan
1:00 AM
@amWhy you know about nt right''
number theory*
 
amWhy
amWhy
@Ethan not my specialty, but what's your question?
 
Ethan
Ethan
can u help me with my question, its the first one
on newest questions
 
amWhy
amWhy
I'll look...
 
Ethan
Ethan
I don't see anything wrong with my argument, except maybe the limits, but I think those are okay too
 
user19161
Anyway, if you call us rude or intellectual snob when we really are not so, we won't help you anymore. I am out of here.
 
amWhy
amWhy
1:04 AM
@JasonBourne Did that happen here?
 
Bel
Bel
Please can anybody take a look to my problem in this post math.stackexchange.com/questions/287750/…
 
amWhy
amWhy
@JasonBourne whatever - I guess you're ignoring me...
@JasonBourne bye bye
 
user19161
@amWhy Well, I am not sure what they are talking about exactly, but I happened to enter that room. So I am just making a general statement.
 
user19161
@amWhy No, please don't say people are ignoring you when they are not.
 
amWhy
amWhy
@JasonBourne Oh...in OS? What is OS?
 
user19161
1:06 AM
@amWhy Here enter this room chat.stackexchange.com/rooms/7018/os-guys and read for yourself.
 
amWhy
amWhy
@JasonBourne I agree with your general statement.
@JasonBourne it really bugs me when I try to help an asker, and they get belligerent. Really bugs me!
 
user19161
If you want to call people something, make sure you can back up that claim, and tell the person when the thing happens, not talk absolute rubbish behind their backs.
2
 
Johan Larsson
Johan Larsson
anyone doin' project Euler? I gave up on the new problem already
 
Will Jagy
Will Jagy
@Johan, I had some contact with Project Euler admins. They really, really, really want people to do the problems on their own, build up knowledge by experiment, and make something of themselves instead of asking elsewhere.
 
amWhy
amWhy
@JasonBourne starred!!
 
Johan Larsson
Johan Larsson
1:11 AM
yeah
 
Ethan
Ethan
Whats special about the zeta function in that it is used to give an explict formula for the prime counting function, I mean there are other generating functions that incode prime numbers? Why is it easier to work with the zeta function?
 
amWhy
amWhy
@JasonBourne What a drama!!
 
Johan Larsson
Johan Larsson
@WillJagy I'm not asking, I already gave up, also tired. But you are right of ocurse.
 
amWhy
amWhy
@JasonBourne backstabbing is one of the worst interpersonal "crimes" I can think of; it's so cowardly
 
user19161
@amWhy Well, maybe it is not as serious as I originally felt it to be. I don't want to exaggerate things either, considering that lines have been removed and we can't be sure what they said.
 
Will Jagy
Will Jagy
1:14 AM
@JasonBourne, @amWhy, any chance you can contact each other and fight it out by email? Official, standing offer: if you both email me, just check with my last name at ams.org/cml , I will forward email to the other, I have been doing this for years for Math Overflow and am trusted in this capacity. In particular, I will not post real names addresses here.
 
amWhy
amWhy
@JasonBourne I'm sorry about that. I take things too personally, and it's not fair to others when not valid.
 
Will Jagy
Will Jagy
PS I do not really care whether it is you too fighting or other people, i am not going to bother to check that other chatroom
 
anon
anon
but amwhy and jason are not fighting, they are lamenting other people's doings
 
Ethan
Ethan
@anon
 
amWhy
amWhy
@WillJagy We're not fighting...just observing.
 
Will Jagy
Will Jagy
1:16 AM
@anon, thank you. Offer still stands.
 
Ethan
Ethan
@anon Why is the zeta function used to give an explict formula for variants of the prime counting function? Why not use another function that encodes information about prime numbers?
 
user19161
@WillJagy (1) amwhy and I are not fighting (2) This place is the most appropriate place to talk about this as this happened in one of the SE chat rooms (3) I don't think we are going overboard here.
 
amWhy
amWhy
@anon We were lamenting the behavior we observe in others, not each other! If you knew me, Will, you'd know I'm not in the least a fighter!
 
Ethan
Ethan
Why is nobody talking about math?
3
 
user19161
@WillJagy Also (4) you don't need to worry when I am involved, I know my limits.
 
anon
anon
1:17 AM
@amWhy that's why I said "other people"
 
amWhy
amWhy
@anon I'm sorry, I meant to direct that to @Will - I have a habit of missing a moving target, when replying to comments :-)
 
Ethan
Ethan
Is anyone here familiar with analytic number theory?
 
anon
anon
@Ethan I take a small stab at that question, implicitly, in the beginning of this answer. Ultimately, it is because Fourier analysis is a natural thing to do (decomposing a function into a superposition of pure waves), and when we apply it to $\psi(x)$ we involve the zeroes of the Riemann zeta function into the mix.
 
Will Jagy
Will Jagy
@JasonBourne, all your points fine by me. Same with @amWhy. I make this offer as I'm about the only person in the world who still uses email. In a very few cases, that has helped straighten out some big problems. As there is actually no problem here, that is fine. Most of the time it has been mathematicians wanting to collaborate with some of the more paranoid, as opposed to arguing.
Taking a nap.
 
Ethan
Ethan
@anon why not use the zeros of some other function?
 
amWhy
amWhy
1:22 AM
@WillJagy Don't bother going there (the chat in question)...I won't go there again...not my cup of tea. I get easily distressed by distress or contention or whatever...so tend to avoid, when I can.
 
anon
anon
@Ethan You don't get to decide how math turns out. I can't make 1+1=3.
2
 
user19161
Anyway, I can't be sure what was said in that room, so sorry if I misinterpreted things. I have nothing to add.
 
amWhy
amWhy
@WillJagy thanks for the offer; that might help in other situations. Bases already covered, email-wise, with Jason...and like you, I still use email!!
@JasonBourne How does one remove a comment? I'm still trying to figure out this whole chat "thang."
 
Ethan
Ethan
@anon the vonmangoldt function satisfies $$\sum_{k=1}^n\ln(k)f(k)=\sum_{k=1}^n\Lambda(k)\sum_{j=1}^{[\frac{n}{k}]}f(jk)$$‌​, so one can find numerous functions that encode the vonmangoldt function, and thus the primes, into a sequence, why can't information be scavanged from using some other function like this, instead of the zeta function to get information about the primes
 
amWhy
amWhy
@Ethan Sorry Ethan, didn't mean to ignore you. I'm not terribly well-studied in analytic number theory...I'm good at reviewing proofs and with proof strategy, but when it's content I'm not super-familiar with, it takes me time to follow the logic. You've really grown in lots of ways, mathematically, and maturity, in a very short period of time. Kudos!
Anyone: How to remove a comment (not that I plan to undo what I say)...more useful yet, how does one edit a comment once it's posted. Oh...never mind, just saw that you can you can hover just to the left of a comment, click on the down arrow, and navigate to "delete" or "edit".
There, I've asked and answered my own question!
 
anon
anon
1:32 AM
@Ethan I don't really understand the connection between that identity and your statement that numerous functions encode the von mangoldt function.
 
user19161
@amWhy See the downwards arrow on the left? CLick on it.
 
amWhy
amWhy
@JasonBourne Thanks, Jas. I just asked and answered my question...right above anon's last post!
@JasonBourne Now I feel like a "newbie"...
 
BenjaLim
BenjaLim
@anon hey
 
anon
anon
It's not encoding the von mangoldt function that's of interest, it's capturing the growth of primes. And PNT is equivalent to $\psi(x)\sim x$ through basic ANT techniques.
hey benjo
 
Ethan
Ethan
ant techniques?
I understand that
but you could easily substitute a factor of 1/ln(k) which weights primes close to 1
 
amWhy
amWhy
1:35 AM
@Sanchez Sorry for the delay; no, I really don't know off-hand. But I can do a little digging.
 
BenjaLim
BenjaLim
@anon hmmm it seems to me with my comment here I can prove that the order of every element in the given finitie field is $q^2 - 1$.....
 
Ethan
Ethan
I mean one could consider the relation between the vonmangoldt function and zeta function as a special case of $$\sum_{k=1}^\infty \frac{\ln(k)x^k}{k^s}=\sum_{k=1}^\infty\frac{\Lambda(k)\text{Li}_s(x^k)}{k^s}$$
Where $x=1$,
Cant information about primes through other functions be drawn with complex analysis
I am asking why the zeta function is used
and not somthing else
 
anon
anon
You have it backwards I think. The zeta function's zeroes were discovered to be involved, not chosen to be used, in this context.
 
Sanchez
Sanchez
@amWhy, no problem, thanks!
 
BenjaLim
BenjaLim
@Sanchez Hey
 
anon
anon
1:38 AM
I would not be surprised though if some other way of encoding the growth of primes involved the zeros of some other special functions, so that zeta's zeros would not be the only mode of description available.
 
Sanchez
Sanchez
@BenjaLim, hey
 
Ethan
Ethan
If the derivative of the polylogarithm at s=0 is the lambert series for the vonmangoldt function
cant information be drawn from the polylogarithm to get info about primes?
 
BenjaLim
BenjaLim
@Sanchez I am unsure about something, it seems to me that with my comment [here]( math.stackexchange.com/questions/286429/… ) I can prove that the order of every element in the given finite field is $q^2 - 1$
 
anon
anon
@Ethan I don't know, can it?
 
Ethan
Ethan
@anon I don't know I am asking you
 
peoplepower
peoplepower
1:40 AM
@BenjaLim $r$ can equal 0, in general.
 
BenjaLim
BenjaLim
@peoplepower True I will have to come up with another reason then.
 
Ethan
Ethan
@anon I recall somthing called "tabularn theorems" or something like used by someone in an elementry proof of the pnt, where information is gained about the coeifients of specific power series
tabular?
I don't know
 
anon
anon
tauberian
 
Sanchez
Sanchez
Tauberian
 
anon
anon
I win.
 
user19161
1:43 AM
The great anon wins.
 
Ethan
Ethan
Why can't you use somthing along those lines, to obtain info from other functions that encode prime numbers
Is the zeta function easyier to work with or somthing?
 
anon
anon
Maybe you can. The zeta function is a very prototypical thing, and is probably what jumped out to people to work with more than anything.
 
peoplepower
peoplepower
@BenjaLim The only approach I am finding copies the other answer.
 
BenjaLim
BenjaLim
ok
 
peoplepower
peoplepower
@BenjaLim I might have something else, actually. I will check it.
 
BenjaLim
BenjaLim
1:49 AM
@peoplepower see the answer of anon.
 
Ethan
Ethan
@anon Also can't explict formulas for prime counting functions where the primes are in arithmetic progeressions be derived from identitys like
$$\sum_{k=0}^\infty \frac{\Lambda(3k+1)}{(3k+1)^s}=\frac{\zeta(s,\frac{2}{3})\zeta'(s,\frac{1}{3})-\zeta(s,\frac{1}{3})\zeta'(s,\frac{2}{3})}{\zeta(s,\frac{1}{3})^2-\zeta(s,\frac{2}{3})^2}?$$
Or is it more hard to work analytically with hurrwitz zeta functions then the Reimann zeta function?
I mean if you integrate both sides of that equation or a relation akin to it, and you invert the series you should get a formula for a generalized prime zeta function where the primes are congruent to some b modulo a
Like the one for the regular prime zeta function $$\sum_{p}\frac{1}{p^s}=\sum_{n=1}^\infty\frac{\mu(k)\ln(\zeta(ns))}{n^s}$$
 
peoplepower
peoplepower
@BenjaLim It's not visible; he must have deleted it.
 
BenjaLim
BenjaLim
@anon Why did you delete it?
 
anon
anon
@BenjaLim more or less the same as your answer (which I didn't read until after I posted)
 
Ethan
Ethan
@anon Can you get explict formula for generalized prime counting functions where the primes are congruent to some b modulo a, in the same matter you can do so for regular primes?
Where instead of working with the regular zeta function you use the hurrwitz zeta function?
 
anon
anon
1:58 AM
@Ethan draks asked this question at some point on MSE I think
 
Ethan
Ethan
I don't know much about Dirichlet L functions, but I know they can be given in terms of hurrwitz zeta functions
 
anon
anon
I don't sympathize with your preference of hurwitz zeta over Dirichlet L-functions, though. The latter is much more factory and fourier analytic.
 
Ethan
Ethan
I havn't ever used L functions or dirichlet characters in general
I don't really know anything about them
I will learn about them eventually
 
Sanchez
Sanchez
@Ethan, there are explicit formulas for Dirichlet L functions. Hurwitz zeta can be written in terms of Dirichlet L functions, so there is a version of that for Hurwitz zeta
 
Ethan
Ethan
Yes I know that
Their periodic so they can be decomposed into a series of weighted hurrwitz zeta functions
@Sanchez can you look at this for me please, math.stackexchange.com/questions/287782/…
 
Sanchez
Sanchez
2:03 AM
I will read it later.
 
Ethan
Ethan
I don't see how L functions are needed in proving dirichlets theorem, it seems like a couple convolution identities are all that is needed to prove dirichlets analytically
 
anon
anon
I had the opposite reaction to dirichlet's theorem: my impression was, "really, only some basic properties of these L-functions are needed?"
 
Ethan
Ethan
I think I can construct a 2 or 3 page proof of dirichlets, if my argument here:math.stackexchange.com/questions/287782/… is valid
 
BenjaLim
BenjaLim
@anon One can prove a special case on the infinitude of primes $p \equiv 1\mod{n}$ using ramification theory.
 
anon
anon
DT can also be phrased as, "the set of primes is dense in the profinite completion of Z"
 
Ethan
Ethan
2:07 AM
The only thing I want to be sure of is the limits I take, but I think those are ok too
Can you check it for me anon
its only like 5 lines long
3
Q: Does this imply Dirichlet's theorem on primes?

EthanDoes this identity imply dirichlets theorem? If $\gcd(a,b)=1$ , and $1\leq b\leq a$, then $$\sum_{k=0}^\infty\frac{\Lambda(ak+b)}{(ak+b)^s}=\frac{1}{a^s}\sum_{\substack{\gcd(a,r)=1\\1\leq r\leq a}}\zeta(s,\frac{I(r,b,a)}{a})\sum_{k=0}^\infty\frac{-\ln(ak+r)\mu(ak+r)}{(ak+r)^s}$$ Where $I(r,b,a)...

prime-numbers proof-strategy
 
BenjaLim
BenjaLim
@Sanchez hey
 
peoplepower
peoplepower
@BenjaLim I think that argument is erroneous, since $(ab,a^2)=(ab,a)=a$.
 
BenjaLim
BenjaLim
where did you learn your Lie theory from?
@peoplepower I don't get yout comment.
 
Sanchez
Sanchez
@Ethan, how do you prove the first identity?
@Benjalim, I read part of tom dieck and Brocker
 
BenjaLim
BenjaLim
@peoplepower what are $a$ and $b$?
 
Ethan
Ethan
2:10 AM
I found it a while back
can you tell me if the argument I made
is ok?
I am only worried about the limit I take
when the hurrwitz zeta function, is multiplied by that other sum
 
peoplepower
peoplepower
@BenjaLim It seems that you are using the assumption that $(m,n)=x$ implies $(m,n/x)=1$, which fails in general.
 
Sanchez
Sanchez
@Ethan it looks true if the identity is true
 
Ethan
Ethan
@Sanchez I can give like a $2\frac{1}{2}$page proof of the first identity, do you think that proof of dirichlets would be worth anything?
The proof doesn't use dirichlet characters or anything
 
BenjaLim
BenjaLim
@peoplepower Ah ok. I will see if I can correct it.
 
Sanchez
Sanchez
@Ethan, ah, wait
In the middle line
$$\lim_{s\to 1}\frac{1}{\zeta(s)}\sum_{k=0}^\infty\frac{\Lambda(ak+b)}{(ak+b)^s}=\frac{1}{a}\‌​sum_{\substack{\gcd(a,r)=1\\1\leq r\leq a}}\sum_{k=0}^\infty\frac{-\ln(ak+r)\mu(ak+r)}{(ak+r)^s}$$
 
Ethan
Ethan
2:16 AM
damnet that was what I was worried about
that limit?
 
Sanchez
Sanchez
Yes, RHS seems to diverge, no?
 
Ethan
Ethan
no the right hand side is finite
 
Sanchez
Sanchez
There is $s$ at the bottom
 
Ethan
Ethan
oops I forgot to write it as 1
I just fixed it, but that doesn't effect the proof
It doesn't matter if it didn't converge that would still prove
 
anon
anon
Sanchez is right: how do you know the resulting series converges?
 
Sanchez
Sanchez
2:20 AM
There are two issues here.
 
Ethan
Ethan
and it does converge
read the bottom argument
It doesn't matter if it didn't converge that would still prove the infinitude of primes
 
anon
anon
how?
 
Ethan
Ethan
If there were only finitely many primes
then the vonmangoldt sum would be finite
 
anon
anon
you mean primes in a certain residue class
 
Ethan
Ethan
which would mean that the ratio of zeta(s) over that sum
would be zero as s tends to 1
The prime powers are neglible
the s> or equal to 2
when its a prime power
 
Sanchez
Sanchez
2:22 AM
$\zeta(s, I(b,r,a)/a) \sim \zeta(s)$. Proof?
em wait, this is not the issue. This should be fine
 
Ethan
Ethan
$\zeta(s,q)$
q<1 in this case
 
Sanchez
Sanchez
Anyway, let me tell you why I don't believe it.
 
anon
anon
yeah, $\zeta(s)-\zeta(s,q)$ should be bounded as $s\to1^+$ when $q\in(0,1)$.
 
Ethan
Ethan
@Sanchez why?
 
anon
anon
in fact the difference has a limit as s->1, by comparison yadda yadda.
 
Ethan
Ethan
2:24 AM
Assuming the truth of the first identity
 
Sanchez
Sanchez
Oh wait.
 
Ethan
Ethan
I think sense the limit is the same I could go further to show that the dirichlet density of the primes congruent to b modulo a is is $\frac{1}{\phi(a)}$
 
anon
anon
@Ethan how does that show the infinitude of primes in a given residue class?
 
Ethan
Ethan
The vonmangoldt function
is zero if its argument is not a prime power
and ln(p) if its argument is a prime power =p^j
the prime powers are neglible as the denominator will be $\frac{1}{n^{2}}$ or smaller
which converges
while the non prime powers
will be $\frac{1}{n}$
 
Sanchez
Sanchez
@Ethan, I'm not sure. Let me say that I doubt your identity.
 
Ethan
Ethan
2:27 AM
@Sanchez I can post some numerical data for you
if you wana see
 
Sanchez
Sanchez
OK
 
anon
anon
please use full and complete English. a nonprime power cannot equal 1/n, for instance.
I am not going to play ANT mad libs with you :)
 
Ethan
Ethan
@anon I think I explained it to you weirdly
Let $ \Lambda(k) $ denote the von Mangoldt function:
$$
\Lambda(k) \stackrel{\text{def}}{=}
\begin{cases}
0 & \text{if $ k $ is not a prime power}, \\
\ln(p) & \text{if $ k = p^{j} $}.
\end{cases}
$$
ok
 
Sanchez
Sanchez
The usual dirichlet l function proof relies on $L(1,\chi) \neq 0$ for nontrivial $\chi$. This is a result with real content. Your method completely avoids dealing with this which makes me feel skeptical.
 
Ethan
Ethan
@sanchez is everything ok in the proof, Except the identity?
 
anon
anon
2:29 AM
@Ethan If the series on the RHS that we pointed don't converge, then you don't have a proof.
 
Ethan
Ethan
anon
 
anon
anon
Since your proof relies on talking about said series having a positive value.
 
Ethan
Ethan
read the bottom line
 
anon
anon
I did.
 
Ethan
Ethan
it shows that it converges to somthing positive
 
Sanchez
Sanchez
2:30 AM
In particular, Dirichlet L function proof involves writing your LHS as $\sum \chi(b)^{-1} \Lambda (s, \chi)$, each summand being different. Your sum did not distinguish this at all :/
 
Ethan
Ethan
The vonmangoldt function is always positive
and zeta(s) is always positive
the bottom shows it converges
 
Sanchez
Sanchez
@Ethan, almost, but let's say I believe what you say your post except the identity.
 
Ethan
Ethan
@Sanchez if everything is true but the identity, is my argument ok?
 
Sanchez
Sanchez
@Ethan, probably. Your are sketchy about $s \to 1$ part I mentioned. (You can't switch sums with limit that arbitrarily), but that's not an important issue here, i.e. it probably can be fixed as long as the identity is fine.
 
anon
anon
@Ethan it shows that the sum over all of the residue classes converges to something positive, but in order to go from that to the statement about the sums restricted to a given residue class coming out positive, you use the fact that the limits are the same independent of b, which you cannot conclude as given if the RHS part we are pointing to doesn't converge.
 
Ethan
Ethan
2:33 AM
@Sanchez do you understand what anon is saying? can you explain it to me?
The right hand sides does converge though
Useing the bottom argument
I show that it must converge
 
anon
anon
restricted to one residue class?
 
Ethan
Ethan
anon
the non coprime residue classes vanish
the ones that are coprime we don't know
all we know is that atleast one of them must converge
otherwise the right hand sum couldn't be equal to one
 
anon
anon
no
 
Ethan
Ethan
no what?
 
anon
anon
(x+-x) converges to 0 as x->oo, but neither x nor -x converges as x->inf. a sum of things converging does not imply any of those things converges.
 
Sanchez
Sanchez
2:35 AM
@anon, I don't think you need to worry about that convergence there.
His argument there may be wrong, but it's not the crux of his proof.
 
Ethan
Ethan
no sanchez if its wrong i want to know, it is the crux lol
I don't understand what hes saying
 
Sanchez
Sanchez
@Ethan, no, the crux is your identity, which shows that the effect of $b$ can be singled in some way.
 
Ethan
Ethan
$$\lim_{s\to 1}\frac{-\zeta'(s)}{\zeta(s)^2}=\lim_{s\to 1}\frac{1}{\zeta(s)}\sum_{k=1}^\infty\frac{\Lambda(k)}{k^s}=\lim_{s\to 1}\frac{1}{\zeta(s)}\sum_{k=0}^\infty\frac{\Lambda(ak+1)}{(ak+1)^s}+\frac{\Lambd‌​a(ak+2)}{(ak+2)^s}...\frac{\Lambda(ak+a)}{(ak+a)^s}=1$$
 
Sanchez
Sanchez
This is very different from the L function proof, where this cannot be done "uniformly" in certain sense.
 
Tim
Tim
What happened to MSE? math.stackexchange.com
 
Ethan
Ethan
2:39 AM
Now let me compare that to the other sum
for you
 
anon
anon
Yes, $$\lim_{s\to 1}\left[\frac{1}{\zeta(s)}\sum_{k=0}^\infty\frac{ \Lambda(ak+1)}{(ak+1)^s}+\frac{\Lambda(ak+2)}{(ak+2)^s}...\frac{\Lambda(ak+a)}{(‌​ak+a)^s}\right]=1.$$ How do you go from this to $$\exists b: \lim_{s\to1^+}\frac{1}{\zeta(s)}\sum_{k=0}^\infty\frac{\Lambda(ak+b)}{(ak+b)^s}~ \rm exists?$$
In addition, how do you conclude $$\lim_{s\to1^+}\frac{1}{\zeta(s)}\sum_{k=0}^\infty\frac{\Lambda(ak+b)}{(ak+b)^s‌​}=\lim_{s\to1^+}\frac{1}{\zeta(s)}\sum_{k=0}^\infty\frac{\Lambda(ak+c)}{(ak+c)^s}‌​$$ for any two $b,c$ coprime to $a$, if we don't know that $$\sum_{k=0}^\infty\frac{-\ln(ak+r)\mu(ak+r)}{(ak+r)}$$ ever even exists?
 
Ethan
Ethan
I know that one of them, must converge to somthing otherwise the total sum couldn't be 1
 
anon
anon
@Ethan why do you believe that?
@Ethan where did I say "for all b"?
 
Sanchez
Sanchez
@anon, all he needs is that $\lim_{s\to1^+}\frac{1}{\zeta(s)}\sum_{k=0}^\infty\frac{\Lambda(ak+b)}{(ak+b)^s}‌​$ is not equal to 0, to conclude.
 
Ethan
Ethan
Its not because of the first statement right?
 
anon
anon
2:43 AM
@Sanchez ah, true. so this would putatively show that at least one of the residue classes has infinitely many primes.
 
Sanchez
Sanchez
@anon, $$\lim_{s\to 1}\frac{1}{\zeta(s)}\sum_{k=0}^\infty\frac{\Lambda(ak+b)}{(ak+b)^s}=\lim_{s\to1}\‌​frac{1}{a^s}\sum_{\substack{\gcd(a,r)=1\\1\leq r\leq a}}\frac{\zeta(s,\frac{I(r,b,a)}{a})}{\zeta(s)}\sum_{k=0}^\infty\frac{-\ln(ak+r)\mu(ak+r)}{(ak+r)^s}$$
The quotient of $\zeta$ part is alright.
 
anon
anon
@Sanchez if you are replying to a specific message, please link to the specific message
 
Sanchez
Sanchez
So what you get is that if that limit exists for some $b$
Then the RHS limit would exist
and that for all other $b$, the limit on the left exists and are the same.
Also, if the limit does not exist for some $b$
The limit does not exist for all $b$ by the same reason.
 
anon
anon
@Sanchez how would the limit on the left existing show that it exists and is the same for each b?
 
Sanchez
Sanchez
where?
 
anon
anon
2:46 AM
do you know what the gray arrows are for in chat?
 
Sanchez
Sanchez
@anon, ah, I never noticed it. Thanks.
@anon, it's because the zeta quotient has limit 1.
 
anon
anon
@Sanchez sure, but what if the infinite series (i.e. the $\sum_{k=0}^\inf$ one) they are being weighted by don't converge individually at s=1 (or as s->1)?
 
Khromonkey
Khromonkey
what does this mean? Almost all natural numbers are very, very, very large.
 
Sanchez
Sanchez
Ah I see your point. Now I agree.
That's why I feel weird, I have been thinking that the series for each $r$ are the same.
Then yes, this is the main problem of the proof.
 
peoplepower
peoplepower
How should one not ping you @Ethan?
 
Ethan
Ethan
2:48 AM
what?
Im ethan
can you please stop
@Sanchez whats the main problem
 
anon
anon
If two users have the exact same username "Ethan," then pinging one will automatically ping the other. So either we are unable to ping the Ethan already in chat, or you must learn to live with being pinged by these messages, or one of you needs to add more to your username.
 
Sanchez
Sanchez
@Sanchez, convergence of those series.
 
peoplepower
peoplepower
Given that one has two years seniority over the other, I suggest a name change on the newer one.
 
Ethan
Ethan
whats wrong with the convergence?
@Sanchez whats wrong with the convergence?
 
anon
anon
@Sanchez I wouldn't necessarily say the main problem. It is still a detail that needs to be checked.
 
Ethan
Ethan
2:52 AM
anon what is wrong, I would very much appreciate it if you explained it to me
 
anon
anon
already did
 
Ethan
Ethan
If you mentioned it before can I get a thing linkining me to that message
which 1
 
anon
anon
this and the one immediately after
 
Sanchez
Sanchez
@anon, I think so, because I'm thinking the Dirichlet L function proof. If $L(1,\chi) \neq 0$ is a nontrivial thing, he can't just wave his hand off like this.
 
Ethan
Ethan
all I need to do is show the limit is not zero
 
anon
anon
2:55 AM
if you want to prove full DT, you need to show at least $\phi(a)$ different limits each either don't exist or are positive. this does not follow immediately from the limit of their combination existing and being positive.
 
Ethan
Ethan
Anon but all the sums are strictly positive
 
anon
anon
do the limits each exist, and are they all positive? the sums being positive doesn't gaurantee this.
 
Ethan
Ethan
They must converge to somthing
otherwise
we couldn't have 1
why does it matter if they exist? if they oscilate between a finite value I still have Dt
 
anon
anon
@Ethan False. We can have $\lim[ f(x)+g(x)]$ exist while neither $\lim f(x)$ nor $\lim g(x)$ exists.
 
Ethan
Ethan
why does it matter if they exist?
oh I see, if they don't exist
then they cant be all equal?
 
anon
anon
2:58 AM
that's right.
 
Ethan
Ethan
damnet
 
anon
anon
you can get partial DT though
 
Ethan
Ethan
If its a prime I know
I wrote that proof up already
because then the residue classes are 1,2,3,..p-1
and I can p explictly
so I can get the other sums
 
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