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9:03 AM
what kind of number theory do you do, @teadawg1337?
 
@Balarka Nothing too complicated, I only know the basics
 
ok, elementary number theory. that stuff can be fun too. you know about diophantine equations?
 
Mhm, I came up with a variant of Brocard's problem; taking away the restriction to pairs, there's three ways to sum distinct factorials to get a perfect cube, which is the same amount of solutions conjectured for Brocard's problem
 
ah, yeah, you posted that stuff here
 
I still don't have a definitive answer about factorials being Diophantine equations, though. I'm aware of Florian Luca's article, which (assuming the abc conjecture is true) proves that any Diophantine equation of the form $n!=P(x)$ with a degree of at least two has a finite number of solutions
I just don't know if my particular equation is Diophantine...
 
9:13 AM
@skullpatrol Thanks. Can we call such person as a tantalizer?
 
@Venus I'm not skullpatrol, but the answer to your question is yes
 
@teadawg1337 sure it's diophantine
 
anyone got any ideas about math.stackexchange.com/questions/1059379/… ?
 
@Balarka But the factorial function isn't an algebraic function, is it?? If $P(x)=n^3$, how is the whole equation Diophantine if one sums at least two factorials?
 
diophantine equation is just a name for finding integer solutions to many-variable equations
your curve needn't be algebraic
 
9:22 AM
@teadawg1337 Thank you Tea
 
@Venus My pleasure :)
(you're welcome)
@Balarka Hmm... That's what I thought at first, but I've been getting conflicting answers when I ask...
 
it's just a name
who cares if it's not diophantine?
;)
 
@Balarka This article is what made me confused about my variant being a Diophantine equation
 
i don't care if it's not diophantine
 
I know you don't, but I'm dying to figure out if my variant is just an example of what Luca showed in the article I linked to
 
9:32 AM
good evening everyone
 
what is the problem, @teadawg1337?
i mean, what's the equation explicitly.
it was $x! + y! = n^3$ wasn't it?
 
How many perfect cubes can be expressed as the sum of two or more unique integer factorials without subtractions, divisions, or multiplications of factorials?
There's no limit on the amount of factorials being summated
 
"without subtractions, divisions, or multiplications of factorials?" i don't understand this.
 
I threw that in to avoid ambiguity. I intended for it to mean that only addition can be used
 
oh sure
 
9:36 AM
?
 
i am pretty sure finiteness of $x! + y! = n^3$ can be concluded from abc-conjecture.
 
Yes @Venus "tantalizer" is one possible choice. You see temptation is a big part of the false hope giver's "game."
 
r9m
@Chris'ssis see your integral here :) I remember you asking it in chat !! :D
 
But the abc-conjecture hasn't been proven yet (Mochizuki's proof uses terminology he probably came up with himself)
 
r9m
@Venus Nice solution there ! :D (+1)
 
9:40 AM
yes, i believe finiteness of x! + y! = n^3 is far out of the reach of diophantine number theory techniques
so i don't think you can prove it in a whim, @teadawg1337 :)
 
For now, I'm stuck in the unknown. And I hate being stuck in the unknown.
 
unknown?
 
Relative to me, at least. The solution is beyond my current knowledge
 
well i told you i think there's no possible approach that'd derive finiteness
you can only conclude some finitude results by invoking strong abc.
 
@skullpatrol What else the other words for "tantalizer"?
@r9m Thanks ^^
 
9:45 AM
Use a thesaurus
This^ is learning :-)
@Venus
 
@Venus temptress (unsure), enticer, that's all I can think of off the top of my head
 
@teadawg1337 you know some algebra?
 
Huy
@teadawg1337: What the hell does a temptress or enticer have to do with giving false hopes? They're synonyms for a seducer...
 
@Balarka Linear algebra? Hardly
 
nah. abstract algebra.
you should study some of that
 
9:52 AM
The most efficient approach is to build-up your vocabulary and learn many words @Venus
Like I said @Huy in depends on the context that the false hope was given.
 
@skullpatrol Fine
@teadawg1337 No!
I have a problem with my internet
 
Huy
@skullpatrol: I really doubt @Venus was looking for someone giving "false hopes" in that context.
 
@Huy Indeed, I'm too tired to be a thesaurus right now
 
@Huy I'm looking for a word for someone who always gives promises but never keeps it
 
Promises about what?
What you call that person depends on the context in which that person is giving false hope @Venus
 
9:59 AM
@Venus I've got two legitimate synonyms in that case: reneger and flake
 
Huy
@skullpatrol: Not really. There is surely a word for that kind of person but I can't recall it just now.
Reneger is actually pretty good.
 
Con man
Heart breaker
Dream maker
 
Huy
Con is way too strong imo, not keeping a promise doesn't mean they did it intentionally, and to con (imo) is intentional.
 
r9m
LOL ;)
2
 
@r9m haha, yeah! That one is not really mine, but it's taken from Inside Interesting integrals ...
@r9m I think @N3buchadnezzar also posted some kind of generalization.
 
r9m
10:04 AM
@Chris'ssis ah ! :) you can post your solution there if you want ! (the accepted answer already did it with trig substitution .. that's how I did it ..)
 
@skullpatrol Is that a Pat Benatar song?
 
Yep
 
r9m
@Chris'ssis OO ! link please ! :D
 
@tea love taker Don't you mess around with me .... Is the next line :)
 
@r9m I knew how to do it in the first seconds I saw it. :-)
@r9m let me see ...
 
r9m
10:06 AM
@Chris'ssis :-) ... I took 40 mins :P
 
@r9m I don't believe you. ;)
 
r9m
@Chris'ssis ?! why ?! I posted it in chat (the substitution within 40 mins of your posting it didn't I ?! :O)
 
@skullpatrol I've heard the song before, lol
 
@r9m I meant you did it much much faster. :-)
 
r9m
@Chris'ssis nah ! It too me straight 40 mins ! no more no less .. it was a baffling one ;)
 
10:09 AM
I'm uncomfortable seeing all of these emoticons...
 
r9m
@teadawg1337 write a script that eliminates all the emoticons ;)
 
@r9m Meh, too lazy.
 
@r9m Have you ever estimated the number of hours spent on doing integrals, series and limits? Looking back I'm inclined to say: "wow, I worked crazy much & crazy hard". :-)
 
r9m
@Chris'ssis :D
 
Why would you @teadawg1337 be uncomfortable with seeing many emoticons?
 
10:12 AM
@r9m Apart from the results I got, I'm very glad I could keep my way and steadily worked very hard, every day (even in those not-so-good days).
 
@skullpatrol I was trying to be sarcastic, I failed miserably
 
r9m
@Chris'ssis That's a great reward in itself :-)
 
@r9m Indeed. :-)
 
@teadawg1337 icic :D
good one
 
r9m
@Chris'ssis did you know what was special about yesterday ? :)
 
10:15 AM
@r9m Definitely!!! Ramanujan's birthday ...
:-)
 
r9m
@Chris'ssis :D I just found it out ! poor me googled it late
2
 
hehe, I couldn't miss that :D
 
Star for Rama ^ let the world know :D
 
I don't quite find Ramanujan's works interesting.
 
Isn't Ramanujan's birthday celebrated as National Math Day in India @r9m?
 
10:20 AM
@Balarka gasps
 
r9m
@Venus YES !!
 
@teadawg1337 I know he discovered modular forms, but it's, eh, in a completely weird language.
 
@Balarka Yes, the language of genius
 
the language of integrals and series
there are lots of geniuses out there who never worked with integrals his whole life
 
@teadawg1337 Flake: a person who you cannot trust to remember things or to do what they say they will do, or someone who behaves in a strange way
@teadawg1337 We have Flake & tantalizer, but which one is more common in English?
 
Huy
10:24 AM
clearly flake, I've never heard tantalizer before. @Venus
 
@Venus Probably flake, tantalizer is somewhat archaic (to
the masses
 
r9m
@Venus see this and here :)
 
Grothendieck is a cool guy.
So is Gromov.
 
Huy
Not as cool as Galois.
 
I'm against comparing mathematicians.
 
Huy
10:26 AM
I'm not.
Therefore I'm cooler than you are.
 
@Huy @teadawg1337 I guess so, flake! :D
 
r9m
@Huy that guy died in a duel fighting for his honor in front of his girl ! He might be a genius in mathematics and language .. I don't consider him cool ! :P
 
@r9m Thanks... :-)
 
Huy
@r9m: Source that it was IN FRONT of his girl?
 
10:27 AM
@Venus Reneger is another fantastic word, means the same
thing and sounds awesome
 
@r9m He was a rebel, which is not cool.
 
@teadawg1337 OK
 
Huy
@BalarkaSen: He fought for what he believed in.
 
Not mathematics, thus not cool.
:P
 
Huy
There are different things in live than mathematics. One day you will grow up and realise it too.
 
10:28 AM
@Balarka \o
 
@Huy I wish not to grow up then.
 
Huy
Everyone wishes that.
 
I've come to the opposite conclusion, there's more to mathematics than there is to life
 
@Studentmath \o
coincidentally, "\o" looks like "yo".
:P
 
r9m
@Huy nah ! I was wrong ! The wiki page doesn't mention it !
 
Huy
10:31 AM
@r9m: Flake.
 
What would o/ look like then?
 
I'm reading Gromov hyperbolicity for time pass.
Cool stuff.
 
r9m
@Huy that's a local cigarette brand ! want one ?
 
LEL @r9m
 
Huy
@r9m: I don't smoke. It stinks.
 
10:32 AM
Mathematicians should stink @Huy
 
Huy
Maybe that's just what you need to tell yourself to keep going.
 
r9m
@BalarkaSen :P
 
@Balarka Excuse you, I always smell fantastic
 
@Huy I stink, I assure you that.
 
I've got a question. If I look at the ordered set $N\times N$ and place on it the cofinite topology, and then I look at $N_{cf} \times N_{cf}$ (with the box/product topology, doesn't matter - it's a finite product so they are the same). Won't these two topologies be the same?
 
Huy
10:34 AM
@BalarkaSen: Don't worry, I have no doubts at all.
 
r9m
@BalarkaSen oi ! that's not a thing to be proud of ! :P
 
sure it is
 
r9m
:P lol ... I have to find a word for that :P
 
Huy
Proud skunk?
2
 
The open sets in the first one are those that have finite complement, and thus each has finite complement in every $N$, and thus they are subsets of $N_{cf} \times N_{cf}$, And the latter has open sets as the sets of which complement is finite in $N$, but thus the product has the complement finite as well so it's i n $N\times N$ too. What am I doing wrong?
 
r9m
10:35 AM
@Huy :P LOL
 
@Huy I like that.
 
r9m
@Huy skroud does not sound right ! perhaps a punk ? (= proud + skunk)
 
I like very much the definition of the mainfold given here (even for me that knows almost nothing about this area)
3
Q: Simpler definition of manifold

saadtaameI'm new to topology but I must understand how it works to progress in my research. First, can anybody point me to a document that introduces topology in a "gentle" way. What pre-requisites do I need to understand the topic? I need an intuitive understanding of manifolds for now.

 
Huy
@Chris'ssis: Which one?
 
@Huy The chosen one.
 
r9m
10:38 AM
FAINTS !!
 
It's just an n-surface with every point having an nbhd homeo to R^n
@Chris'ssis what it says is essentially what i said above
 
when you "zoom" onto it, you're essentially trying to find a correct neighborhood
 
Huy
@Chris'ssis: I read all the answers and they don't really differ significantly from ones I've learnt?
 
@Huy I only read that one.
 
10:42 AM
and when you feel it "looks like" R^n @Chris'ssis, it means that there is a continuous bijection from the nbhd onto R^n
it's good to know that you're even looking at this stuff @Chris'ssis
 
@BalarkaSen Right.
@BalarkaSen Yeah, it seems interesting, that kind of abstract thing I love.
 
if you love it, learn it
 
Huy
@Chris'ssis: You should study GR or DG with me. :D
 
no one wants to study GR with you @Huy
i've never studied manifolds @Chris'ssis but if you do, i'd be glad to know you're doing something other than integrals.
well, obviously, @Huy. i stink.
 
Huy
I second that.
 
10:50 AM
@Studentmath $A = \mathbb{N}\setminus \{0,1\}$ has finite complement, hence is open in the cofinite topology of $\mathbb{N}$. Hence $A\times A$ is open in the product topology.
 
And yet $A\times A$ has infinite complement, so it's not in $N\times N$ with cofinite topology. I get it now, thanks @Daniel!
 
11:14 AM
@Huy ;)
@BalarkaSen It's never too late. :-)
 
Huy
@Chris'ssis: ;)
 
11:27 AM
@BalarkaSen By the way, I don't do only integrals. It's true that I'm mainly focused in this period of time on integrals, series and limits because I wanna publish a book.
 
r9m
Asked on main ! :-) interesting problem !
 
She also does sums :3
 
@r9m Erdos studied & solved such problems.
 
r9m
@Chris'ssis oh ! links and reference on main please then ! :D
 
@r9m I have such problems in one of my books (by Erdos). I have it somewhere ...
 
r9m
11:31 AM
@Chris'ssis interesting ! :D Book title please :) I'll get it from Santa if its a problem collection ;)
2
 
@r9m It's a Romanian book ..
 
r9m
@Chris'ssis well .. Santa will get that for me form Romania then ! ;) what's the name of the book ? :-)
 
@r9m Problems in arithmetic and number theory (Laurentiu Panaitopol and Alexandru Gica)
 
r9m
@Chris'ssis Thanks ! :D
 
r9m
11:37 AM
@Chris'ssis tHANKS ! :D
 
This problem has just been active:
21
Q: Evaluating $\int_0^1 \log \log \left(\frac{1}{x}\right) \frac{dx}{1+x^2}$

Kirthi RamanShow that $\displaystyle{\int_0^1 \log \log \left(\frac{1}{x}\right) \frac{dx}{1+x^2} = \frac{\pi}{2}\log \left(\sqrt{2\pi} \Gamma\left(\frac{3}{4}\right) / \Gamma\left(\frac{1}{4}\right)\right)}$ This question was posted as part of this question: Solve the integral $\displaystyle{S_k = (-1)^k...

Does anyone here have a better answer for that?
 
@r9m Around $7$ euro in my country. :-)
 
Hello guys! I have recently found solutions to Mordell Equation for the case k=(-1) using elementary methods. This has only been done using UFD or gaussian integers and no elementary solution has been published yet. Do you think that I should consider publishing it anywhere? Will that be accepted or is it too trivial?
 
r9m
@Venus A better answer ?! :D You must be joking ! (I'm familiar with the new proof my M.N.C.E)
 
@r9m I mean an easier one. I think it'd be the better one
 
r9m
11:40 AM
@Chris'ssis hmm .. ! that's not my problem ! I'll let Santa deal with them rough edges ;)
@Venus they are called Malmsten Integrals ! google 'em :D
 
@r9m :-)))
 
I found that I think the better one but the proof doesn't belong to me
 
r9m
@Venus post it !! I'll upvote ! :D (I love new answers to nice questions !!)
 
Ughhhhhhh
Long solutions here ...
22
Q: Evaluating $\int_0^1 \log \log \left(\frac{1}{x}\right) \frac{dx}{1+x^2}$

Kirthi RamanShow that $\displaystyle{\int_0^1 \log \log \left(\frac{1}{x}\right) \frac{dx}{1+x^2} = \frac{\pi}{2}\log \left(\sqrt{2\pi} \Gamma\left(\frac{3}{4}\right) / \Gamma\left(\frac{1}{4}\right)\right)}$ This question was posted as part of this question: Solve the integral $\displaystyle{S_k = (-1)^k...

 
@r9m But it doesn't belong to me. The one who has the answer is also a user here
 
11:43 AM
@Chris'ssis Well, you rarely do anything other than integrals here
 
r9m
@Chris'ssis they are pretty much how Adamchik did them :)
 
@BalarkaSen Integrals, series and limits (rarely some inequalities). That's because I wanna publish a book on this topic.
3
 
@BalarkaSen Maybe you can answer my previous question....
 
@MathGod what question?
 
@BalarkaSen scroll up
 
11:45 AM
Chris you are a genius. How long did it take You to learn group theory]?
 
@MathGod Try to post the main idea here first & see how community here responds to your method
 
@beginner She doesn't do group theory.
@MathGod what is "Mordell equation for k = -1"?
 
@Chris'ssis That's incredibly cheap
 
r9m
@Venus The already posted solutions are similar to what V. Adamchik did ! what you can do is ask him if you can post the solution here and then post it :D
 
Hello people :)
 
11:47 AM
I have learnt heaps of math since I was grounded ithink balarka . how have you been balarak
 
you were grounded?
 
@r9m The proof belongs to Anna, I feel bad if I post it here without her permit
 
Yeah that is why I was gone. I guess you didn't mis me hehe
 
maybe i should ground myself sometimes
 
r9m
@Venus is that so ?! hmm :) ask her if she's willing to post it .. if not whether you are allowed to do the same on her behalf ;) .. else send me the link of her solution to me (I'll half that out for you .. volume wise ;) .. )
^ that's how I roll :P
 
11:50 AM
@r9m How to reach her? I don't know
 
I am trying to prove the general case that Z/a maps isomorphically to z/a_1 X z/a_2 X ... X z/a_i, where a = a_1a_2...a_i
 
oh boy
you've already started group theory, @beginner?
 
Ted gave me a starter but I haven't worked with ker before
I guess, I don't know if that above is group theory
 
@r9m See here
 
are you studying from a book or something?
 
r9m
11:51 AM
@Venus I'm in touch with her through Quora ! I can contact her on you behalf .. else you can always start a chat room :)
 
Pdfs but I don't know which this came from
 
I don't understand the commenter to my answer, am I wrong? : math.stackexchange.com/questions/1078370/…
 
I think it was just under Chinese remained theorem for modular arithmati
 
r9m
@Venus that's not her solution ! -_-
 
@beginner right it is chinese reminder theorem
but i am not convinced that you are familiar with that much group theory
 
11:52 AM
@r9m What did you mean? Scroll down
 
What should I be familiar with for it?
 
@TheArtist $\sqrt{4}\neq-2$
 
@beginner Since you are talking about isomorphisms, i presume you know what an isomorphism is?
 
It is a bijective homomorphism
Defined on a singe group operation
single
 
@TheArtist On real numbers, the square root is only defined for positive numbers
 
11:53 AM
@Hippalectryon oh lol thanks :D
 
r9m
@Venus what do you want me to look at in that page ?
 
OK, just an yes would do. Are you interested in doing an exercise, @beginner?
 
Definitely :)
 
@r9m I don't get it, what do you mean? Is the link broken?
 
@BalarkaSen The equation y^2=x^3+k is called Mordell equation. In particular, for k=-1, we have y^2+1=x^3. The only known methods to solve this (and most of the Mordell Equations in general) is using U.F.D. (Unique Factorization Domain) and sometimes Gaussian Integers (complex numbers). There hasn't been any published elementary proof of the case k=-1 and I have apparently solved it using elementary methods. Shall I consider publishing it?
 
r9m
11:55 AM
@Venus no no .. I mean I am familiar with that solution as well ! its nothing new !
 
I know what a mordell equation is, just weren't familiar with what your k is.
What is your elementary solution, @MathGod?
 
@r9m Really? I thought it a new one since no one answers that question using this way
 
r9m
@Venus look at Felix Marin san's answer carefully :)
 
@BalarkaSen I have used the Pell - Fermat equation.
 
Show me the explicit proof, not names :P
 
r9m
11:57 AM
also the regulator can be really a lot of things :-) that's whats nice about this method !
anyway ,, BBL (out for shopping)
 
What was the exercise @balark
 
Right, you know about group presentations, @beginner?
 
Maybe not by name
 
@r9m Slightly different, but I think it's essentially the same
 
Not by that name I men
Mean*
 
11:59 AM
@beginner Just a retarded. If I were a genius I would have graduated form a celebre uni ...
 
OK... consider the set {1, x, x^2, x^3, ..., x^(n-1)}
 
@r9m It should be Felix Marin sama :D
 
Ok yep
 
Equip this set with group operation $\cdot$ as $x^a \cdot x^b = x^{ab}$
 
@r9m I'm stalking her btw :D
 
12:01 PM
This is hard to describe ugh. You know what generators are, @beginner?
The group I am talking about is <x | x^n = 1>
 
@Hippalectryon where have I used the square root of a negative number?
 
Generators reach all elements of a cyclic set?
 
i.e., the group {1, x, ..., x^(n-1)} with operation x^a \cdot x^b = x^(ab) and x^n = 1
OK ^ that group.
 
Ok yep
 
@TheArtist Oops, I meant it's defined from positive reals to positive reals
 
12:02 PM
@beginner prove that this group is isomorphic to Z/n
 
And it is though?
 
It is isomorphic to Z/n
 
@BalarkaSen Well, it's a long proof to fit in this chat (not joking), and you know, for obvious reasons I wouldn't consider it safe to make it public until.... Maybe you can give me your mail or something (if you wish) so that I can send it to you.
 
?
 
12:04 PM
@beginner Prove it.
 
ok
 
@MathGod Nah. I don't think it's serious enough to be published. There's a question by Alizter that asks for an elementary proof of it maybe you can just post it as an answer there.
 
@Hippalectryon ohhh I got it. I was being very stupid there
 
Here. There's an elementary proof there using quadratic residues.
 
@BalarkaSen Anyways, thanks for your opinion.
 
12:09 PM
i need to leave. when you're finished with that exercise, @beginner, prove me that Z/(mn) is isomorphic to Z/n \times Z/m where gcd(m, n) = 1 (hint : use the previous exercise).
 
@BalarkaSen He can publish anything he wants
A new bounty, is anyone interested in answering it?
5
Q: Closed form of $\int_{0}^{\eta}\cos nt\log\left(\frac{\cos(t/2)+\sqrt{\cos^2(t/2) -\cos^2(\eta/2)}}{\cos(\eta/2)}\right) dt$

user_of_mathI am reading a paper (sorry, no e-copy) with a number of infinite series, in which each term of the series is an integral of a complicated transcendental function like the one in the title. There are no fewer than a dozen of these infinite sums of integrals in the paper. For instance, the auth...

 
user134936
@MathGod Can you help me with this question?
 
user134936
@BalarkaSen I don't think quadratic residue can be considered elementary. It's put in the same category as U.F.D. and Gaussian Integers.
 
12:33 PM
@r9m It's curious that my last integrals wasn't evaluated by any expert integrals I know ... I also have some version containing series too. They really look like the ones of Ramanujan.
 
Huy
10°C for Christmas, but global warming was a lie!
 
this year
 
Huy
It gets worse every year. ^_^
 
So does this guy -_-
 
Huy
@skullpatrol: I checked out the link in the $\hbar$, but closed it immediately. What is it about?
 
12:41 PM
He is an ultra finite mathematician.
 
Huy
I don't know what an ultra finite mathematician is.
 
He does not use the irrational numbers ...
 
One who believes no numbers $> 10$ exist. The bound could in some cases be larger than $10$.
 
that^ too
 
1:04 PM
hahahaha, what an amazing new series I've developed!!!
Yumi, Yumi, Yumi
 
@Chris'ssis Did you mean "Yummy"?
 
1:31 PM
@DanielFischer lol, yeah :-)
 
tasty?
 
@skullpatrol Right
I'm working in the garden now. Back later on.
(with my new fence)
 
later
 
So I apparently passed out a few hours ago, because I don't remember falling asleep at all...
 

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