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!

@JasonBourne I would love to visit Singapore or Indonesia - but I suspect it will not happen now
@JoeyMorani $-3x$

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.

@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.

@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

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!

@JasonBourne I doubt that my knowledge would be greater!

user19161
@OldJohn I am going to bed. Good night!

@JasonBourne Goodnight! - sleep well

Hey all!
Any one around?

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.

Hi @amWhy

@Sanchez hello!

@amWhy, are you an undergrad at Wisconsin?

@Sanchez No, not an undergrad. I'm a grad student, but on a bit of a sabbatical...sort of ABD

@amWhy, oh wow.. in math?
What kind of math do you study?

12:28 AM
I'm sort of a generalist: I did earn a PhD in logic, but now working in algebra & math logic

Cool.
Do you know anything about model theory? @amWhy

@Sanchez Some
@Sanchez What's your area of interest?

@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?

@Sanchez Number theory fascinates me!

Hi everybody

12:31 AM
Hi @Bel.

Time I slept - need to watch Murray trying to overcome Djokovic in the morning

Surely, number theory is awesome ;) @amWhy

I'm looking for help in algebra and specifically in quadratic form

@Sanchez Was it Gauss who called number theory the "Queen of Mathematics"?

Bye @oldJohn, that's definitely gonna be an intense match.
@amWhy, I think so.

12:33 AM
@Sanchez hope so! - bye for now
Bye @WillJagy

@amWhy, do you know if there's an exposition of model theory applied to algebraic geometry?

I have to implement a program that doesn't accept the form $B^T\times B$

Bye bye love
Bye bye happiness
Hello loneliness
Think I'm gonna cry.
@Bel, what quadratic form would that be?

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

@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/…

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?

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.

@amWhy can u help me

@Ethan yes? Hi!

1:00 AM
@amWhy you know about nt right''
number theory*

@Ethan not my specialty, but what's your question?

can u help me with my question, its the first one

I'll look...

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.

1:04 AM
@JasonBourne Did that happen here?

Please can anybody take a look to my problem in this post math.stackexchange.com/questions/287750/…

@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.

@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.

@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

anyone doin' project Euler? I gave up on the new problem already

@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.

@JasonBourne starred!!

1:11 AM
yeah

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?

@JasonBourne What a drama!!

@WillJagy I'm not asking, I already gave up, also tired. But you are right of ocurse.

@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.

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.

@JasonBourne I'm sorry about that. I take things too personally, and it's not fair to others when not valid.

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

but amwhy and jason are not fighting, they are lamenting other people's doings

@anon

@WillJagy We're not fighting...just observing.

1:16 AM
@anon, thank you. Offer still stands.

@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.

@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!

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.

1:17 AM
@amWhy that's why I said "other people"

@anon I'm sorry, I meant to direct that to @Will - I have a habit of missing a moving target, when replying to comments :-)

Is anyone here familiar with analytic number theory?

@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.

@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.

@anon why not use the zeros of some other function?

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.

@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.

@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."

@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

@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".

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.

@JasonBourne Thanks, Jas. I just asked and answered my question...right above anon's last post!
@JasonBourne Now I feel like a "newbie"...

@anon hey

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

ant techniques?
I understand that
but you could easily substitute a factor of 1/ln(k) which weights primes close to 1

1:35 AM
@Sanchez Sorry for the delay; no, I really don't know off-hand. But I can do a little digging.

@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 $q^2 - 1$.....

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

You have it backwards I think. The zeta function's zeroes were discovered to be involved, not chosen to be used, in this context.

@amWhy, no problem, thanks!

@Sanchez Hey

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.

@BenjaLim, hey

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?

@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 $q^2 - 1$

@Ethan I don't know, can it?

@anon I don't know I am asking you

1:40 AM
@BenjaLim $r$ can equal 0, in general.

@peoplepower True I will have to come up with another reason then.

@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

tauberian

Tauberian

I win.

user19161
1:43 AM
The great anon wins.

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?

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.

@BenjaLim The only approach I am finding copies the other answer.

ok

@BenjaLim I might have something else, actually. I will check it.

1:49 AM
@peoplepower see the answer of anon.

@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 $$\sum_{p}\frac{1}{p^s}=\sum_{n=1}^\infty\frac{\mu(k)\ln(\zeta(ns))}{n^s}$$

@BenjaLim It's not visible; he must have deleted it.

@anon Why did you delete it?

@BenjaLim more or less the same as your answer (which I didn't read until after I posted)

@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?

1:58 AM
@Ethan draks asked this question at some point on MSE I think

I don't know much about Dirichlet L functions, but I know they can be given in terms of hurrwitz zeta functions

I don't sympathize with your preference of hurwitz zeta over Dirichlet L-functions, though. The latter is much more factory and fourier analytic.

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

@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

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/…

2:03 AM

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

I had the opposite reaction to dirichlet's theorem: my impression was, "really, only some basic properties of these L-functions are needed?"

I think I can construct a 2 or 3 page proof of dirichlets, if my argument here:math.stackexchange.com/questions/287782/… is valid

@anon One can prove a special case on the infinitude of primes $p \equiv 1\mod{n}$ using ramification theory.

DT can also be phrased as, "the set of primes is dense in the profinite completion of Z"

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

Does 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 $I(r,b,a)... @Sanchez hey @BenjaLim I think that argument is erroneous, since$(ab,a^2)=(ab,a)=a$. where did you learn your Lie theory from? @peoplepower I don't get yout comment. @Ethan, how do you prove the first identity? @Benjalim, I read part of tom dieck and Brocker @peoplepower what are$a$and$b$? 2:10 AM I found it a while back can you tell me if the argument I made is ok? I am only worried about the limit I take when the hurrwitz zeta function, is multiplied by that other sum @BenjaLim It seems that you are using the assumption that$(m,n)=x$implies$(m,n/x)=1$, which fails in general. @Ethan it looks true if the identity is true @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 @peoplepower Ah ok. I will see if I can correct it. @Ethan, ah, wait In the middle line $$\lim_{s\to 1}\frac{1}{\zeta(s)}\sum_{k=0}^\infty\frac{\Lambda(ak+b)}{(ak+b)^s}=\frac{1}{a}\‌​sum_{\substack{\gcd(a,r)=1\\1\leq r\leq a}}\sum_{k=0}^\infty\frac{-\ln(ak+r)\mu(ak+r)}{(ak+r)^s}$$ 2:16 AM damnet that was what I was worried about that limit? Yes, RHS seems to diverge, no? no the right hand side is finite There is$s$at the bottom 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 Sanchez is right: how do you know the resulting series converges? 2:20 AM There are two issues here. 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 how? If there were only finitely many primes then the vonmangoldt sum would be finite you mean primes in a certain residue class 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 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$\zeta(s,q)$q<1 in this case Anyway, let me tell you why I don't believe it. yeah,$\zeta(s)-\zeta(s,q)$should be bounded as$s\to1^+$when$q\in(0,1)$. @Sanchez why? in fact the difference has a limit as s->1, by comparison yadda yadda. 2:24 AM Assuming the truth of the first identity Oh wait. 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$\frac{1}{\phi(a)}$@Ethan how does that show the infinitude of primes in a given residue class? 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$\frac{1}{n}$@Ethan, I'm not sure. Let me say that I doubt your identity. 2:27 AM @Sanchez I can post some numerical data for you if you wana see OK 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 :) @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 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. @sanchez is everything ok in the proof, Except the identity? 2:29 AM @Ethan If the series on the RHS that we pointed don't converge, then you don't have a proof. anon Since your proof relies on talking about said series having a positive value. read the bottom line I did. it shows that it converges to somthing positive 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 :/ The vonmangoldt function is always positive and zeta(s) is always positive the bottom shows it converges @Ethan, almost, but let's say I believe what you say your post except the identity. @Sanchez if everything is true but the identity, is my argument ok? @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. @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. 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 restricted to one residue class? 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 no no what? (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. 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. no sanchez if its wrong i want to know, it is the crux lol I don't understand what hes saying @Ethan, no, the crux is your identity, which shows that the effect of$b$can be singled in some way. $$\lim_{s\to 1}\frac{-\zeta'(s)}{\zeta(s)^2}=\lim_{s\to 1}\frac{1}{\zeta(s)}\sum_{k=1}^\infty\frac{\Lambda(k)}{k^s}=\lim_{s\to 1}\frac{1}{\zeta(s)}\sum_{k=0}^\infty\frac{\Lambda(ak+1)}{(ak+1)^s}+\frac{\Lambd‌​a(ak+2)}{(ak+2)^s}...\frac{\Lambda(ak+a)}{(ak+a)^s}=1$$ This is very different from the L function proof, where this cannot be done "uniformly" in certain sense. What happened to MSE? math.stackexchange.com 2:39 AM Now let me compare that to the other sum for you 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? I know that one of them, must converge to somthing otherwise the total sum couldn't be 1 @Ethan why do you believe that? @Ethan where did I say "for all b"? @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. Its not because of the first statement right? 2:43 AM @Sanchez ah, true. so this would putatively show that at least one of the residue classes has infinitely many primes. @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. @Sanchez if you are replying to a specific message, please link to the specific message 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. @Sanchez how would the limit on the left existing show that it exists and is the same for each b? where? 2:46 AM do you know what the gray arrows are for in chat? @anon, ah, I never noticed it. Thanks. @anon, it's because the zeta quotient has limit 1. @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)? what does this mean? Almost all natural numbers are very, very, very large. 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. How should one not ping you @Ethan? 2:48 AM what? Im ethan can you please stop @Sanchez whats the main problem 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, convergence of those series. Given that one has two years seniority over the other, I suggest a name change on the newer one. whats wrong with the convergence? @Sanchez whats wrong with the convergence? @Sanchez I wouldn't necessarily say the main problem. It is still a detail that needs to be checked. 2:52 AM anon what is wrong, I would very much appreciate it if you explained it to me already did If you mentioned it before can I get a thing linkining me to that message which 1 this and the one immediately after @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. all I need to do is show the limit is not zero 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. Anon but all the sums are strictly positive do the limits each exist, and are they all positive? the sums being positive doesn't gaurantee this. 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 @Ethan False. We can have$\lim[ f(x)+g(x)]$exist while neither$\lim f(x)$nor$\lim g(x)\$ exists.

why does it matter if they exist?
oh I see, if they don't exist
then they cant be all equal?

2:58 AM
that's right.

damnet

you can get partial DT though

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