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12:00 AM
@PeterTamaroff dafuq! - you won?
 
@OldJohn Yes.
But I am like 2 - 14
And I lost most games because I ran out of moves.
 
That's nothing - I have been teaching my dog to play chess ...
... but I'm beating him 3-2 :))))
 
@OldJohn Your dog should learn how to rage quit.
He would win every time
Everyone knows pets are always allowed to ragequit.
 
@PeterTamaroff :)
Time I slept too - g'night all
 
@OldJohn I'm still lurking around BCT
@OldJohn Is it like 12 pm there?
 
12:04 AM
@PeterTamaroff yeah
BCT is good to think about
I have forgotten almost all I knew, though :(
 
Humph! I tried to cap yesterday, but missed by 40. It was a good day for answering, nonetheless.
 
@robjohn I can't remember the last time I capped. It was probably sometime last year.
 
@MikeSpivey Since becoming a mod, it has been more of a challenge. More distractions.
 
@robjohn What exactly does a mod spend his time doing? Responding to flags?
 
@MikeSpivey responding to flags, considering infractions, emailing users, migrating questions, replying on meta, etc.
 
12:19 AM
@OldJohn, all lipids normal. There is another item, Hemoglobin A1c which is considered the more important of the two ordinary tests for diabetes. They want 4.8-5.6 thingies, which I got eight out of ten times, but in 2010 Dec it was 5.9 and this time 6.0. I think we should just get to throw out our one worst test, like dropping the lowest quiz score.
 
@MikeSpivey we have meetings and talk on mod chat regarding things.
 
@robjohn, do you get up and talk about the causes of your addictions?
 
@WillJagy Hi. My name is robjohn and I am an MSE mod...
 
The first step is admitting you have an addiction :)
 
@robjohn, that's it, own it. It's an addiction, but it's your addiction.
 
12:22 AM
@robjohn Do you know what the latest is on MO moving to the SE network?
 
@WillJagy thank you for sharing.
@MikeSpivey I've not been keeping abreast of that. Sorry.
 
@MikeSpivey, a month ago Anton said he was hung up with the legal stuff. There will be a number of guarantees enforceable by law. At the same time, nobody ever made any serious headway on programming an alternative (a huge task to do it right) and things would eventually fall apart on the 1.0 platform, so there is no real alternative. My guess is they can migrate as soon as everything is signed. Maybe Anton will have extra time over the winter break.
 
@WillJagy Thanks, Will. What sort of guarantees?
 
@MikeSpivey reading this, it doesn't look as if a merger will occur.
 
user19161
Addiction is not well-defined. What is addiction? That is a very deep question.
 
12:31 AM
hi! I'm new in the chat... but does anyone who whether a constant speed geodesic obtained in a coordinate frame remains a constant speed geodesic in another? is it still a geodesic at all in another frame ? Not sure if the chat is a place to ask
 
@MikeSpivey MO seems to think that MSE is a pariah
 
user19161
I have decided not to try to cap anymore. So I won't be aiming for the epic badge which 17 people already have.
 
@JasperLoy capped yet?
 
user19161
@robjohn Nope, I have not answered any question today, and only answered one yesterday.
 
@JasperLoy yet you got 165 yesterday :-p
 
12:33 AM
@robjohn That link seems to be a rant by one disgruntled person, rather than an MO consensus.
 
user19161
@robjohn Yes, due to the synchronicity of various factors.
 
@MikeSpivey I hope so. I don't like to think that we are seen that way over there.
 
user19161
Ooh, I got 5 stars thanks to Riemann and Darboux.
 
@JasperLoy favorites?
 
user19161
@robjohn I mean chat stars, look to the right. -------------------------->
 
12:35 AM
@robjohn Really? I've not gotten that impression. There are lots of people active on both sites. Some of the top users over there (e.g., Qiaochu, Mariano, JDH, MattE, David Speyer, Pete Clark), have participated quite heavily here at times.
 
@JasperLoy Ah, I see.
 
user19161
@robjohn Well, perhaps MSE also thinks that MO is pariah.
 
user19161
See, MSE can accept all questions and MO only research level questions, and some questions on MO are pretty trivial, so MO is the real pariah, not MSE. QED.
 
@MikeSpivey I didn't say we were thought of that way, I said that I'd hate to think we were.
 
@MikeSpivey, the best info on what is actually happening is meta.mathoverflow.net/discussion/1416/5/mathoverflow-20 There are plenty of other threads warning against doing it, but it is happening. The idea makes me nervous too, but it is not realistic to thing that Anton and Scott Morrison can maintain the software side indefinitely. Part of the panic over this was bad dealings with someone named Jeff Atwood, who is now out of the picture.
 
12:36 AM
@robjohn Fair enough. :)
 
user19161
@WillJagy Oh well, without Jeff there would be no MO or MSE.
 
@DumbCow Books are there.
 
@JasperLoy, my sense was that Atwood was something of a Bill Dubuque figure, influential, important, but dictatorial.
 
Bill Dubuque was dictatorial?
 
@skullpatrol his statement says that Atwood was
@skullpatrol I guess it could be read either way regarding Bill
 
12:43 AM
@robjohn Has Atwood ever come into chat?
 
@WillJagy I believe Jeff was one of the original founders of the SE network. Regardless, he was the primary liaison between the SE folks and math.SE in the early days.
 
I think it fits both parties. I liked Jyrki Lahtonen's short summary, that Bill had a very precise notion of how MSE should work, and got upset when people did not go along with that.
 
@skullpatrol He was a Stack Overflow guy. He never asked or answered questions on the main site - only meta.
 
@MikeSpivey Thanks.
 
@skullpatrol Here
Jeff Atwood is not just some guy on StackExchange. Rather he can be said to be "the guy".
 
12:47 AM
@JayeshBadwaik Yes, Jeff Atwood and Joel Spolsky
 
To all, my only interaction with Atwood was an attempt to get him and Pete L. Clark communicating by email. The results were less than stellar. However,i continue to think that really explosive stuff should be discussed between two opponents (or two allies) and not exposed to 50 random pairs of eyes until people can calm down a bit and use clean language. I have no reason to doubt Atwood's importance in this.
 
Why is he holding his baseball cap like that?
 
Does anyone know of a program I can use to solve large systems of linear equations?
 
@skullpatrol almost 2 years ago was the last I can see. Before I was here.
 
@robjohn Thanks for looking :)
 
12:56 AM
Does anyone know of a program comatible with common functions I can use to solve a large system of linear equations, I cant use wolframalpha, there data input is too small
any 1
 
@Ethan scilab, matlab.
 
Are those free?
 
scilab is free.
 
sec
 
12:58 AM
Im not very tech savvy, but I need somthing alot stronger then a calculator
 
scilab is powerful
trust me
 
oh
 
you will have to learn it somewhat
 
Do i have to learn alot
is it a bunch to learn
 
I am not sure. I already knew a lot of programming when I learned scilab.
so I cannot comment.
 
12:59 AM
If I give you a five by five system could you paste in for me?
ln(5)*g(1)-zeta'(s,1/5)=f(1)*g(1)+f(3)*g(2)+f(2)*g(3)+f(4)*g(4),
ln(5)*g(2)-zeta'(s,2/5)=f(2)*g(1)+f(1)*g(2)+f(4)*g(3)+f(3)*g(4),
ln(5)g(3)-zeta'(s,3/5)=f(3)*g(1)+f(4)*g(2)+f(1)*g(3)+f(2)*g(4),
ln(5)g(4)-zeta'(s,4/5)=f(4)*g(1)+f(2)*g(2)+f(3)*g(3)+f(1)*g(4),

solve for f(1),f(2),f(3),f(4)
 
ahh, that's not a completely straighforward job
what is g(1)? a function? is f(1) also a function?
you might need more analytical tools than numerical then..
 
yes
there constants
just assume
they are
I just need f(a) in terms of the other functions
f(a) in terms of anything but other f(a)'s
 
you should be able to do that with simple manual elimination I guess....
for four equations, it would not be too big of a problem
 
It wont even fit in the input bar, on wolfram alpha
I have an account
but when I try to upload it
it just treats it like a table of data
and doesn't solve as it would normally
 
I am not sure I get your problem.
 
1:06 AM
@GarbageCollector: would you like a hat?
3
 
@Ethan I guess you have a multiple zeta function. And you want to find solutions, but you have not specified $s$.
 
s doesn't matter
I Just want
f(a)
in terms of the other terms
not other f(a)'s
you can assume everythings constant
just let zeta'(s,q) be a constant not dependent on s, only q, which is already defined
 
okay
then, you can do this manually
 
thanks
 
using pen and paper
 
1:07 AM
lmao
a system of 4 equations in 4 unknownsa
 
i dont think so, that would take forever
I dont have that kind of time
 
it would have been done in the time you spent chatting here
 
and im likely to make an error
doubt it
lol
 
and that skill acquired in proper book-keeping would have been handy in future
 
1:09 AM
I already have scrap paper all over my room
 
@Ethan what a lovely decor!
 
lol
 
hi all
 
@DanielR hey there
 
ln(5)*g(1)-z(1)=f(1)*g(1)+f(3)*g(2)+f(2)*g(3)+f(4)*g(4),
ln(5)*g(2)-z(2)=f(2)*g(1)+f(1)*g(2)+f(4)*g(3)+f(3)*g(4),
ln(5)g(3)-z(3)=f(3)*g(1)+f(4)*g(2)+f(1)*g(3)+f(2)*g(4),
ln(5)g(4)-z(4)=f(4)*g(1)+f(2)*g(2)+f(3)*g(3)+f(1)*g(4),

solve for f(1),f(2),f(3),f(4)
can someone solve this for me pleasee
 
1:11 AM
@Ethan you are not looking for numeric calculation here btw, so scilab is of no use. Probably sage math will be better. If you are looking for some software to enable you to solve bigger such problems in the future, symbolic manipulation software like sagemath, mathematica etc will be much more useful. You can also learn sympy and numpy, which have a higher learning curve, but are much more powerful.
 
solve for f(1),f(2),,... in terms of anything but other f(x)'s
 
I will write it as a matrix system and use Mathematica to solve it.
 
I don't have any real technical expierence with computers
 
If you give me a cookie.
 
okkk
il give you an identity
:d
 
1:13 AM
Oops, I scared off the GarbageCollector.
2
 
Are you familiar with matrices or linear algebra perchance?
 
Fairly
Not really tho
I forget most of it
 
...
 
Il give you uhm
Il give you a cool identity
if u do it 4 me
plzzz :o
ln(5)*g(1)-zeta'(s,1/5)=f(1)*g(1)+f(3)*g(2)+f(2)*g(3)+f(4)*g(4),
ln(5)*g(2)-zeta'(s,2/5)=f(2)*g(1)+f(1)*g(2)+f(4)*g(3)+f(3)*g(4),
ln(5)g(3)-zeta'(s,3/5)=f(3)*g(1)+f(4)*g(2)+f(1)*g(3)+f(2)*g(4),
ln(5)g(4)-zeta'(s,4/5)=f(4)*g(1)+f(2)*g(2)+f(3)*g(3)+f(1)*g(4),

solve for f(1),f(2),f(3),f(4)
in terms of anything but other f(x)'s
 
@Ethan, you might like to recall Gaussian elimination from linear algebra
 
1:17 AM
I know how to do ity
but it will take a long time to do by hand
And I dont have a good enuf calculator to solve it
 
It's only reducing a 4x4 matrix
 
Could you do it for me
please]
 
Gaussian elimination - wouldn't we need actual numerical values for the g's to do that?
 
just assume everythings a constant the functions etc
treat em like constants
ln(5)*g(1)-zeta'(s,1/5)=f(1)*g(1)+f(3)*g(2)+f(2)*g(3)+f(4)*g(4),
ln(5)*g(2)-zeta'(s,2/5)=f(2)*g(1)+f(1)*g(2)+f(4)*g(3)+f(3)*g(4),
ln(5)g(3)-zeta'(s,3/5)=f(3)*g(1)+f(4)*g(2)+f(1)*g(3)+f(2)*g(4),
ln(5)g(4)-zeta'(s,4/5)=f(4)*g(1)+f(2)*g(2)+f(3)*g(3)+f(1)*g(4),

solve for f(1),f(2),f(3),f(4)
 
Okay, here is the system: $$(\log5)\begin{pmatrix}g_1\\g_2\\g_3\\g_4\end{pmatrix}-\begin{pmatrix}z_1\\z_2\\z_3\\z_4\end{pmatrix}=\begin{pmatrix}g_1&g_3&g_2&g_4\\g_2& g_1&g_4&g_3\\g_3&g_4&g_1&g_2\\g_4&g_2&g_3&g_1\end{pmatrix}\begin{pmatrix}f_1\\f_2\\ f_3\\f_4\end{pmatrix} $$
 
1:19 AM
ok
how do i solve that thing
well i mean
can you solve it :d
 
You multiply by the inverse matrix.
 
ye but I dont have the inverse lol
anon
 
@robjohn Oh. Thank you very much. It is very nice modification. :-)
 
That's where Mathematica comes in.
 
can you plzz do it for me
il give you somthingg
plz
 
1:21 AM
Settle down bro.
 
@BrianMScott
 
 
now just multiply that matrix by the LHS of the equation anon posted above
 
Anon, do you know alot about linear algebra? I think I can give a very elementry proof of dirichlets if I can show a certain system of equations is always solveable
Also heres another identity
lambda is the vonmangoldt function
let s->1, and you prove the infintude of primes congruent to 2 modulo 3
lol
I can get alot of those with out using dirichlet characters and other non sense, just algebra and some periodic series manipulations
anon?
can you gime the solutions to that matrix
plz
 
I advise a different tact.
 
1:31 AM
those are the solutionS?
fuck
 
Yes, at the bottom is the column vector with components f1,f2,f3,f4.
 
nice of you to do that for him btw @anon
 
I don't see a 1 at the end of the numerator on those, is that implied?
 
oh crap, it stretches to the right
that's like half of it...
 
omg
 
1:33 AM
@ethan, i think he clipped some off by mistake :P
 
lolll
anon what do you do for a living?
are you a student somewhere?
 
 
each row is f(1),f(2),.. etc?
right
 
I wouldn't say I do anything "for a living" (that would be, like, a career), but I just finished a job as a security guard and will be tutoring next semester. Yes I am a student.
 
oh
 
1:37 AM
the first row is f(1), the second f(2) etc
 
thanks alot lol
Have you ever published anything? do you do your own work?
or are you just studdying?
 
No I have not published anything, I often work on my own, yes I study.
Unless you count MSE publishing.
 
What kind of work do you do?
I mean research
lol
 
Right now I just got back to looking at schur-weyl duality and tensor symmetry (chapter 9 of Procesi's Lie Groups). I was looking at Monsky's theorem a bit earlier. I will (hopefully) be presenting both of these topics at an informal conference on the 29th.
 
lol lost me at looking
Do you know anything about dirichlet L series?
 
1:40 AM
My interests at the moment lie in combinatorics, group theory, representation theory, number theory, and some categorical flavor.
 
"Nothing general appears to be known about or , although it is possible to express all in terms of known transcendentals (Zucker and Robertson 1976)" on this page mathworld.wolfram.com/DirichletL-Series.html
when they say that
do they mean the functions cant be expressed in terms of other functions? like the hurrwitz zeta etc
 
@Ethan You could say that.
 
or just that the constants cant be evaluated?
 
They probably mean expressed in terms of other functions.
 
I think I can express many of those
in terms of other functions
are you sure they dont mean constants?
 
1:43 AM
constants are particular values of functions typically...
 
Yes well they say for integers
2n, 2n-1
 
even, odd
 
I think I can give many of those in terms of large expressions involving hurrwitz zeta functions
for example in this previous sum
re-write the vonmangoldt sum
in terms of primes
integrate both sides wrt to s
you should get an euler product
composed of primes congruent to 2 modulo 3
allign the congruence classes of the L functions up with weighted values of these vonmangoldt sum
*s
I can give many of these for ak+b
the case for modulo 5 , b=1,2,3,4, looks like they should each take up a quarter of a page though
the expressions were encoded in that large system I had you solve
These vonmangoldt dirichlet series' were generalized cases
of the previous paper I showed u
that 1
I just subtracted the zeta function weighted with a constant and let s->1, then re-wrote the harmonic series with a logarithm and eulers gamma constant (sense they both approach the same thing as n->infinity)
which can be re-written in terms of prime sums like I said before
Because the L function has an euler product, I can break up each of its values into congruences because the dirichlet characters are periodic
and then use previous identitys to obtain each euler product
weight them appropreitly
and then I should have the L function
but written in terms of hurrwitz zetas
functions
and logarithms etc
actually prolly not logarithms, those should disapeer when integrated with respect to s
 
prolly?
 
probabley
are you questiong the spelling or weather or not they disapeer
lol
 
1:56 AM
spelling, sorry
:)
 
brb
also the second coeiffient
i mean the coeiffient on the logarithm of the second identity on the second page, is wrong
it shouldnt be 3/2
i think it should be 5/2
i made an error
Oh anon I also have a nice identity for calculateing chebyshevs second function
The main terms already starts out at n/2
it instantly shows that phi(n)=O(n)
hellooo
am i alonee
 
ping anon
 
lol what does that mean
why no 1 talk
 
That means you can send him a message by putting "@" in front of his user name
 
wtf
its ok
 
leo
2:04 AM
@Ethan means write @anon in order to invoke anon
 
@Ethan like this :)
 
i get it
hey u guys wana see a cool series
I dont know If i showed you these two before
 
sure
 
sec
 
phi(n)<=n is already automatic
 
2:05 AM
its assymptotic to n
thats the prime number theorem
im just saying i have an identity
were the main terms isn't anything that has to do with primes and such
its an expression involving the logarithm of the barnes g function and gamma function
I should re-write in terms of factorials, I guess sense its only integer valued
 
oh, you mean pi. then it's asymptotic to $n/\log n$, not $n$, and the asymptotic relation $\sim$ is not the same as $f\in O(g)$ (big-O notation)
 
chebyshevs second function
is assymptotic to n
anyway heres the two series skull
I showed anon b4 i think
 
oh, then you want psi
 
ye whatever
i forget the letters and sht
llol the first 1 is actually 1/2
I think
 
Your grammar and spelling ethos is decidedly representative of your generation.
This is why I was wondering if you were in chat on your phone ...
 
2:08 AM
anyway they converge double exponentially, the first 10 terms should give you hundreds of digits of accuracy
no
im 17
lol
I have more general formula, i thought i could get a fast converging series for somthing not rational as thats kinda useless, but I end up geting double series that cant be simplifyed
but there cool to look at lol
 
*they're *simplified
 
stop correcting my spelling
 
Why?
 
its an online chat room
who cares
lol
 
leo
@Ethan Chat Guidelines many people here cares about
 
2:13 AM
i dont see anything about abreivations
who cares
im not useing l33t speak
if i forget an apostrophee or spell apostrophe wrong, who cares
 
Well, ignoring you is always an option. :-)
 
@leo Duuuuuude.
 
leo
@PeterTamaroff have you noticed that we have been violating point 6 of Chat Guidelines
 
"Please try..."
We tried. Didn't work, amigo.
 
Its have leo, not hav
 
leo
2:17 AM
@PeterTamaroff Oh, well... :D
 
no l33t speak
 
leo
@Ethan thanks
 
no
u say no thanks
 
@Ethan Don't play the smart guy, now.
 
?
 
leo
2:18 AM
@Ethan what did I say?
 
what did i say?
 
@leo So how is it going?
 
leo
@PeterTamaroff about to eat :-)
 
@leo What?
 
leo
@PeterTamaroff Qué que voy a comer?
 
2:24 AM
@leo Claro.
???
 
leo
@PeterTamaroff :-)
 
@leo Plátano frito? Pollo? Arroz?
 
@DumbCow The books are there. =)
 
leo
@PeterTamaroff si, frijoles, ensalada con lechuga, cebolla tortillas tostadas y picadillo
 
@leo Cual es la comida típica de ahi?
 
leo
2:33 AM
@PeterTamaroff y eso? :-P
 
leo
@PeterTamaroff esa. Se llama casado. Básicamente es arroz, frijoles ensalada y una carne.
es bueno :-)
 
@leo Ya me imagino!
 
leo
Una vez escuché que CR es el tercer país del mundo donde la gente es más longeva y se supone que es por la alimentación. El primero es Japón. El segundo no me acuerdo
 
@leo Ha! Pero que buen dato.
 
leo
2:37 AM
Ya regreso =)
@PeterTamaroff cool page!
@PeterTamaroff as visto este
ahora sí ya casi regreso...
 
@leo Hmm, no. No soy un gran lector.
 
Merry Christmas.
 
leo
2:54 AM
@skullpatrol same to you my friend :-)
 
:-D
 
leo
@PeterTamaroff yo no lo he leído. Pero suena interesante
 
@leo Si. Igual suelo ser indiferente a ciertas cosas.
No digo solo por los temas de ese libro. En general, soy muy indiferente =P
 
leo
@PeterTamaroff Yo también. Casi toda la gente que tiene algo que ver con mate o con alguna ciencia dura es indiferente a ese tipo de cosas.
 
@leo Hmmm. Puede ser.
 

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