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14:00
Mmmmm, I believe it is more or less the standard proof. Written out colloquially.
$X$ be a metric space, $x, y \in X$ be two arbitrary points. $S_r(x)$ and $S_{r'}(y)$ be two open balls centered at $x$ and $y$. Pick $r$ and $r'$ such that $r + r' < d(x, y)$. Then if $z \in S_r(x) \cap S_{r'}(y)$, $d(x, y) \leq d(x, z) + d(z, y) < r + r'$, contradicting the assumption. Hence $S_r(x) \cap S_{r'}(y)$ is null.
@DanielFischer ^
What other main areas of math are there than Algebra, Analysis, Topology and Stat?
@Committingtoachallenge Number theory!
Whaaaaa!
You can't possibly forget that!
That is a sub category of Algebra?
14:04
@BalarkaSen Pretty much the standard proof. Though it's most common to use $r = r' = \frac{1}{2} d(x,y)$.
They intersect, but a lot of number theory isn't a concern of algebra @Committing
Of Analysis, @BalarkaSen ;)
@DanielFischer of analysis? you mean NT is a subset of analysis? that's so not true.
@BalarkaSen ";)"
I'm offended. No ";)" taken.
14:06
@DanielFischer. In regards to metric tensors, $g_{ab}$, do you know what the subscripts represent?
saying analysis is a superset of nt. grumph humph.
@DanielFischer Ah.
@GustavoMontano "Metric tensors" as in "Riemannian metric"?
Yes. That is what I mean.
@r9m Hello!
Brian M. Scott is back :D !
14:08
hip hip hooray
@GustavoMontano You express it in local coordinates, and $g_{ab}$ is the inner product of $\frac{\partial}{\partial x_a}$ and $\frac{\partial}{\partial x_b}$.
What is $x_a$ and $x_b$?
r9m
r9m
@BalarkaSen Hello :) Hasse identity for Reimann Zeta function is $\displaystyle \zeta(s) = \frac{1}{s-1}\sum\limits_{n=1}^{\infty}\frac{1}{n+1}$
I'm a little paranoid now that I've seen these sub/super scripts and Einstein summation.
@r9m That diverges.
14:10
@GustavoMontano The $a$-th and $b$-th components of the local coordinates.
r9m
r9m
@BalarkaSen Hello :) Hasse identity for Reimann Zeta function is $\displaystyle \zeta(s) = \frac{1}{s-1}\sum\limits_{n=0}^{\infty}\frac{1}{n+1}\sum\limits_{k=0}^{n}\frac{(‌​-1)^{k}\binom{n}{k}}{(k+1)^{s-1}}$ (sorry)
Isn't that a fancy application of Euler acceleration?
See, I have just derived the length of a curve by $\mathcal{L}(\gamma)=\int_{a}^{b}\sqrt{\left(\frac{du}{dt}\right)^{2}E+2\frac{du‌​}{dt}\frac{dv}{dt}F+\left(\frac{dv}{dt}\right)^{2}G}dt$. How would one introduce the metric? I am assuming that you can. @DanielFischer.
r9m
r9m
@BalarkaSen yes .. indeed
14:14
@r9m I knew it YES
By this link en.wikipedia.org/wiki/Geodesic#Affine_geodesics . I am led to believe that it is the expression under the square root.
r9m
r9m
@BalarkaSen :D
@GustavoMontano Then $u = x_1,\, v = x_2$, $E = g_{11},\, F = g_{12} = g_{21},\, G = g_{22}$.
Ah! Ok, I understand the $x_i$! Oh, so the metric is strictly an inner product.
I guess it is, a way of giving 2 vectors a scalar...
@GustavoMontano Not quite, it's a family of inner products. One on each tangent space, and depending smoothly on the base point.
14:17
So every tangent space has its own "fill this in" and the metric is the collection of these "fill this in"?
Fill this in = Inner product?
Brrrr, I'm a little confused.
Oh ok! And I guess this works to facilitate the whole idea of invariant distances etc.
writing this down
@r9m How's number theory going?
r9m
r9m
@BalarkaSen good good not bad :)
What are you learning at the moment?
14:21
@DanielFischer. I must thank you! You have been a great deal of help.
You're welcome.
r9m
r9m
@BalarkaSen haven't learnt anything new .. (there are no classes 'coz of the upcoming exam) .. I'm trying to learn stuff from here and there (like Apostol, and other books)
@MikeMiller!
mystical greetings upto homeomorphism
14:23
that could wildly change the greeting, and is not at all in the spirit of the mystical greeting itself
I am closing in at Baire category theorem. What is it about?
Dense sets.
I am completely unsatisfied.
=P
What about dense sets?
Huy
Huy
@BalarkaSen: You're doing functional analysis at last?
@Huy Baire category is often introduced in topology courses.
Huy
Huy
14:29
Oh. I never did topology except for the compulsory course.
I only learnt about Baire category in functional analysis.
For shame, it gets fun after the compulsory course.
Huy
Huy
@MikeMiller: But I already disliked the compulsory course.
I didn't dislike mine because we took a nonstandard approach, but I find the basic material impossibly dull, @Huy
Huy
Huy
@MikeMiller: Our prof wasn't allowed to keep doing research at our university after the semester so I think he just stopped caring.
@Huy No, topology.
14:32
:(
Huy
Huy
@BalarkaSen: Boring. :P
throws table at @Huy
general topology is extremely fun
Huy
Huy
How would you know
you don't even know Baire category
:P
Well, I am just studying it
And it looks fun enough.
Huy
Huy
"it looks fun" -> "it is extremely fun"
14:33
Unlike analysis, which looked like a bunch of rigorous BS when I started doing them.
@Huy I am having a good deal of fun doing the exercises.
Huy
Huy
Good for you.
@Huy I'll be doing a bit of functional analysis after topology, I think.
Is it fun?
Huy
Huy
@BalarkaSen: Let $(M,d)$ be complete and $(f_n)_{n \in \mathbb{N}}$ be a sequence of continuous functions $f_n: M \to \mathbb{R}$ and assume that the limit $$\lim_{n \to \infty} f_n(x) =: f(x)$$ exists for all $x \in M$. Is the set of points where $f$ is continuous dense in $M$?
@BalarkaSen: That's the first question I remember from functional analysis and I really liked it.
And Riesz' representation theorem still blows my mind, but I don't know if it would do the same to you.
@Huy Which one? The anti-isomorphism of $H$ and its dual?
Huy
Huy
@MikeMiller: The "simple" one, that for all $\ell \in \mathcal{H}^*$ there is exactly a $y \in \mathcal{H}$ such that for all $x \in \mathcal{H}$ $$\ell(x) = (y,x)_{\mathcal{H}}.$$
14:40
Sure.
I was just checking between that one and the one about measures.
Huy
Huy
@MikeMiller: It was introduced in our functional analysis 1 course and I found it really impressive but it had no further application so I was a bit disappointed. Little did I know it would return in functional analysis 2 and with such significance. :o
But noted.
@Huy Haven't studied complete metric spaces yet.
I don't know that much functional analysis. The basics, some stuff with distributions, some stuff with operator algebras, but no more.
Huy
Huy
@BalarkaSen: The problem was solved by Mr. Baire, using categories named after him. :P
Hmmm, is it evaluated so far? $$\int_0^1 \left(\frac{\log(x) \log(1+x)}{x}\right)^3 \ dx$$ May I say I'm the first one that evaluated it? No, not really, you never know from where a hidden paper comes to surface.
14:42
@Huy :P
OK, noted then.
"Category" here just means "a certain class of sets"
It's not the algebraic sense.
I figured that.
Huy
Huy
@MikeMiller: Did you not learn about Sobolev spaces then? :o
Huy
Huy
14:44
@MikeMiller: How so? Did you not like the basic stuff?
No, I liked it, I just pursued different interests.
Huy
Huy
I see.
OK, let's begin with something easier though ...
Do you like algebra at all, @Huy?
14:46
$$\int_0^1 \frac{\log^2(x) \log(1+x)}{x} \ dx$$
and then
$$\int_0^1 \frac{\log^2(x) \log^2(1+x)}{x} \ dx$$
and then
Huy
Huy
@MikeMiller: Some things in algebra, I do like, but I didn't pursue it further after the basic abstract algebra course. I am really interested in mathematical physics and unfortunately, real algebra isn't very common in physics.
$$\int_0^1 \left(\frac{\log(x) \log(1+x)}{x}\right)^2 \ dx$$
@MikeMiller How much topology is actually needed to study algebraic topology?
Wait, you have seen nothing yet!!!
@BalarkaSen Not much. I actually find point-set topology terribly dull.
14:48
Do you remember this one here?
9
Q: Closed form of $\int_0^1(\ln(1-x)\ln(1+x)\ln(x))^2\,dx$

Chris's sisI remember that some time ago I was asking this question Evaluate $\int_0^1\ln(1-x)\ln x\ln(1+x) \mathrm{dx}$ , and now, while I was making a review, I asked myself if we can get the closed form of $$\int_0^1(\ln(1-x)\ln(1+x)\ln(x))^2\,dx$$ by using the similar tools as in that proof. The probl...

that no one computed it?
Now, see this one ...
@Huy You might like C*-algebras. I took a course in it and enjoyed a lot. It's serious functional analysis, with a seriously algebraic taste.
Find the closed form of $$\int_0^1\left(\frac{\ln(1-x)\ln(1+x)\ln(x)}{x}\right)^2\,dx$$
Can someone stomp on this question really hard?
2
It's getting upvotes from the network hot list. Users farming rep by posting funny pictures from the internet is not what the site is for.
I also wonder how we connect $$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{H_n^{(6)}}{n}$$ to $$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{H_n^{6}}{n}$$ and then compute them. I really doubt any paper on earth can tell that.
@MikeMiller Shocked
14:53
Not to mention that the OP has quite a history on the site, and this just might given them close/reopen privilege -- on the basis of posting pictures.
@Chris'ssis :O
@CareBear I guess there's a generalization ?
@CareBear What should I do ? I can't really flag it was low qual
Well, i can after all I guess
@TheGame You can DV to decrease the hotness; I did the same with answers.
I have already downvoted
Mostly, my appeal is to those with 3K rep.
I should edit the question also I guess
14:57
I edited it, making the title more specific
"incidentally" introducing mathjax in title, which excludes the question from the network list.
@CareBear editing as in removing all the pics
But that would be going too far.
@TheGame This was my first version solution to Au-Yeung series (my work, my ideas, my creative mind)
Removing the content of a question. It's a bad question, and (I think) asked in bad faith, but I would not blank it out.
@Balarka You're shocked that different field of math have overlap?
15:00
@CareBear
@TheGame then a mathematician proved that logarithmic integral result in a very elementary way and put all in an article that is going to be published soon.
@CareBear Is that ok ?
@Chris'ssis mk
@TheGame I don't like this; looks like a complete rewrite of the question.
@CareBear Isn't it the code of the question ?
@CareBear Good to know MathJax precludes a question from being a hot network question
15:01
@CareBear Basically all the images other than the first one are totally useless
@TheGame I guess you are right.
The other images are indeed redundant. But then the title should again be edited since it's no longer about a captcha.
What leads you to believe the question is asked in bad faith?
6
Q: Why does this captcha contain $\pi(x) - \int_0^x \frac{dt}{\ln t}=O(x^{1/2}+\epsilon)$?

VividDI found the following problem in a captcha: What does it mean, and what would the solution be ?

@MikeMiller No, I am shocked at the fact that algebraic topology needs little topology, regardless of the name.
I can't help but LOL @CareBear
Still trying hard not to upvote.
@MikeMiller OP is active in number theory tag, yet pretend to be unaware of the Riemann Hypothesis.
15:05
Good point, @CareBear
@CareBear Thanks for noticing that, anyway
@Balarka Certainly you should keep studying it, as you do need quite a bit of basic knowledge.
This question was definitely not asked the way it had to, even if we suppose there wasn't any bad intent
@MikeMiller I am doing Simmons. What d'you think?
I haven't read any general topology books.
15:09
Oh, OK.
Does everyone know their optimal study method?
There is no optimal study method.
What do you mean by "optimal study" in any case?
Time optimal information assimilation
15:13
Heh?
I use the method of prolonged staring.
Someone famous suggested that method didn't they? Look at it for 15 minutes
Sleep on what you have studied. Poincare method.
Rewrite every sentence in the text as relations between nouns. It is time consuming, but in the qualitative sciences where there is a lot of redundancy and faulty reasoning it works well.
@robjohn If we put together our old work, we prove elementarily that $$\displaystyle \sum_{n,k\ge 1}\frac{2}{k^4(k+n)^2}=-2\zeta^2(3)+\frac{25}{6}\zeta(6)$$
Actually, it's enough to prove this one $$\zeta(4,2)=\zeta^2(3)-\frac43\zeta(6)$$ that is done by our old work.
@robjohnit looks like no one there knows how to do that elementarily.
15:30
@TheGame
@BalarkaSen
-_____________-
15:42
@robjohn in this proof, we consider all until we get the relation $(2)$
@robjohn then we consider your approach to the main series that was done elementarily, here (where you used $\displaystyle \sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^an^b}-\frac1{k^a(k+n)^b}-\frac1{k^b‌​‌​(k+n)^a}=\zeta(a+b)$)
Hence, we conclude that $$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}=\zeta^2(3)-\frac{\zeta(6)}{3}$$
Q.E.D.
Huy
Huy
15:56
I just sneezed and my PC turned back on.
@Chris'ssis They should make a special star system for your posts >:c
@Chris'ssis I just can't star them all xD
@TheGame Thanks. What I posted above is more than mind-blowing since many mathematicians don't even know, imagine, dream that it is possible to compute that without using complex analysis, but only elementary tools.
@Chris'ssis Looks good...
@robjohn Your work there is GORGEOUS.
16:17
Do you have idea how to find mapping between $H \oplus H$ and $H \otimes \mathbb{C}^2$ that establishes isomorphism? $H$ is some Hilbert spaces over $\mathbb{C}$.
Huy
Huy
@Cortizol: Does one always exist?
@Huy I didn't understand you
@Committingtoachallenge I think the famous person you mean is Richard Feynman? and I the the "method" was the Feynman Algorithm
As for my study methods, I've found that actually keeping track of when you're studying can be helpful. The first time I did that I realized how much time I waste and realized I wasn't studying as much as I thought I was.
I'm now doing 25 minutes with a 5 minute break 4 times and then a longer break, about 2-4 times a day (depending if I'm working or not)
I've also stopped writing everything I read. I used to do this and realized it was just rote which wasn't really learning. I now sit down with a book and work through everything in my head. I'll work through a tough proof, understand each step, then take a break and then work through it all in my head to make sure I really understood it.
As for whether or not these methods are optimal for me, I'm not sure. But they seem to be improvements.
 
1 hour later…
17:35
I hope I'll be able to finish soon my master theorem on generalized harmonic numbers ...
17:45
I have a copy of Hardy's introduction to number theory :D
6th edition so there is a section on elliptic curves added by andrew wiles :D
I just realized I can do this one in more ways
$$\int_0^1 \frac{\log^2(x)\log(1+x)}{1+x} \ dx$$
@Alizter Good book, that one.
Topology! Topology! Topology!
Yes, I am the pelicans again.
@BalarkaSen hehe
Not again @Sawarnik
@BalarkaSen Who is he? :O
17:53
@BakralaSen Who else can you be except Sawarnik?
@Sawarnik why are you balarka?
@BalarkaSen Why, I am Bakrala! I don't know about Saw?
I am so confused
@BalarkaSen Though if you notice carefully, neither our name nor our avatars match .. si we are unique :)
There is a curious phenomenon that happens with that integral though ...
17:55
I am not going to tolerate this nonsense, @BakralaSen
@BalarkaSen Oww, what are you going to do? :O
I will ping a mod if you don't change your avatar and username pronto.
Trying to show tha this is true:

$$\sum _{i=0}^\infty (1-p_{i})\prod _{j=0}^{i-1}p_{j}=(1-\prod _{j=0}^{\infty }p_{j})$$
@BalarkaSen I told you they were different :O
My avatar is strangely pixelated :P
@Kasper what are $p_j$?
17:57
@robjohn Can you have a look at this guy : math.stackexchange.com/users/184352/bakrala-sen? I don't wish to be mimicked.
@Alizter oh sorry, forgot to mention that $p_j$ are real numbers $0<p_j<1$ and also $0<\prod _{j=0}^{\infty }p_{j}<1$ (by assumption)
The damned username, @BakralaSen
Change it.
@Kasper if $0<p_j<1$ how can $0<\prod p_j$?
@BalarkaSen After 30 days :P
17:59
Oye.
But I am Bakrala and you Balarka, so no problem actually? :D
No it is a problem.
@Alizter the $p_j$ are chosen so that $\prod p_j$ converges to some $p>0$.
@Kasper if $p_j<1$ this cannot be true surely?
well $p_j$ must converge sufficiently rapidly to $1$.
18:01
@Kasper I am trying to imagine this
Hi @Balarka
Hi @Alitzer @Kasper
kicks @Sawarnik
@Kasper ok I see it
@BalarkaSen Or threw a table?
ignores
18:03
:O
@Alizter For fun :D
great, do you have any ideas for this one?
$$\sum _{i=0}^\infty (1-p_{i})\prod _{j=0}^{i-1}p_{j}=(1-\prod _{j=0}^{\infty }p_{j})$$
The only thing I got is that this is equivalent with:
$$\sum _{i=0}^\infty \prod _{j=0}^{i-1}p_{j}-\prod _{j=0}^{i}p_{j}
=(1-\prod _{j=0}^{\infty }p_{j})$$
@Kasper see if you can take a partial sum up to n on the left and a partial product up to m on the right and play around with trying to equate them. Then magically limits might work?
@Kasper That looks very telescopy, doesn't it?
@Kasper It's a telescoping sum
@DanielFischer I hope you are right, let me think
oh shit you guys are right, feel so stupid now, but thanks !
18:16
$$\int_0^1 \frac{\log^2(x)\log(1+x)}{1+x} \ dx=2\sum_{n=1}^{\infty} (-1)^{n+1}\frac{H_n}{(n+1)^3}$$ Q.E.D. (since I have at hand the proper generating function)
My point is somewhat different though, and it is related to another proof of this one that shows that one can avoid tough calculations when computing some ugly series.
Fun fact: First time I am hearing @Chris'ssis say "series" and "ugly" in the same sentence.
18:31
@Alizter It's just a way of spoiling series ... :-)
18:52
Ah, this guy is Tunk-Fey ...
5
A: Find the closed form of $\sum_{n=1}^{\infty} \frac{H_{ n}}{2^nn^4}$

M.N.C.E.Here is a solution that does not rely (too much) on softwares. I will be using the known values of the sums $\small{\displaystyle \sum^\infty_{n=1}\frac{H_n}{n2^n},\ \sum^\infty_{n=1}\frac{H_n}{n^22^n},\ \sum^\infty_{n=1}\frac{H_n}{n^32^n}}$. Let $$\mathcal{S}=\sum^\infty_{n=1}\frac{H_n}{n^42^n}...

It seems he gave up the previous account (there were some answer to correct, that means a lot of work to do).
Now, here is an interesting thing to note ...
In this answer he says "I will gladly provide a detailed solution for $\sum^\infty_{n=1}\frac{H_n}{n^32^n}$ too if there is a need."
On the other hand, in this answer (his last one)
6
A: What is a closed form for ${\large\int}_0^1\frac{\ln^3(1+x)\,\ln^2x}xdx$?

M.N.C.E.I will be using the following results: $$2\sum^\infty_{n=1}\frac{H_n}{n^q}=(q+2)\zeta(q+1)-\sum^{q-2}_{j=1}\zeta(j+1)\zeta(q-j)\tag1$$ $$\sum^\infty_{n=1}\frac{H_n}{n^22^n}=\zeta(3)-\frac{\pi^2}{12}\ln{2}\tag2$$ $$\sum^\infty_{n=1}\frac{H_n}{n^32^n}={\rm Li}_4\left(\tfrac{1}{2}\right)+\frac{\pi^4...

84
A: Completion of rational numbers via Cauchy sequences

kahenUpdate 1 This should be the final edit. Lots of typos have been corrected and the presentation in the section on existence has been greatly improved. It should be very close to a form that can be used as a basis for a project for students. Update 2 Even more typos corrected. Added example of n...

he says "roofs of $(1)$, $(2)$ and $(4)$ can be found [here][1], [here][2] and [here][3] respectively. Unfortunately, there has not been a mathematically sound proof of $(3)$ on MSE as of now."
Also in some proofs in the previous account he had some problems with that series.
@Chris'ssis I have a feeling that it is the same person. Usually for upvotes. If a diamond mod cross references the ips of the account it will be clear that they are the same person. But moderators will only do this if there is serious abuse. Nothing major or provable here.
@Chris'ssis tunk-fey is like 28
@Alizter Believe me, I don't need a proof of that, I simply know, I'm good at doing that (recognizing some patterns).
In the above article, he defines transitivity as $\forall x,y,z \in K : x \leq y \implies x+z \leq y+z$. Is this equivalent to the usual definition?
18:57
@Chris'ssis it seems Anastasiya-Romanova caught on.
@Alizter yeap
@Chris'ssis those kinds of integrals are also related to limit points of the derivatives of the beta function.
@AbstractionOfMe There are so many books treating this that it is not necessary for him to reproduce it here.
The strange thing is that people ask about proofs of standard theorems in this site when they can be looked up in over 9000 books.
Asking about solutions to exercise problems is fine as these are not found in the mathematical literature.
@JasperLoy I don't have any books
@TheGame It is time to get some then, lol. Or visit the library. 16 year olds should not be studying the Riemann Hypothesis anyway. They should be enjoying ice cream with their boy friends and girl friends, lol.
19:01
Nooo :c
throws girl friends to the bin
gives ice cream to a random guy
That is the problem with internet learning. One ends up with a bunch of random facts rather than a coherent set of theorems from a book.
@JasperLoy Internet books?
@JasperLoy
1 hour ago, by Alizter
I have a copy of Hardy's introduction to number theory :D
@JasperLoy wat
One should not think that one can learn from using the internet alone. It is just a supplement.
19:05
@Alizter that stuff requires very high skills ...
@Chris'ssis hence why I could not proceed
It's like reading einstein's introduction to relativity :c
It's awful
@TheGame Relativity $\ne$ Math
@Alizter But involves advanced maths
Just like quantum physics
@TheGame yes, that book loses something in the translation >:c
It is written in advanced Britsh English
Originally it must have been German.
19:11
@Chris'ssis Consider $$\int_0^1 \frac{\log^a (x)\log^b (1-x)}{x^c}dx=\lim_{(x, y)\to(0, 0)}\operatorname{B}(x+a-c+1, y+1)$$
or something similar
But according to the author it is accessible to any bright teenager with a high school education.
@JayeshBadwaik The good ones have usually been published as books. The rest are usually not as polished.
@JasperLoy That's true too, but I hope it goes the other way soon, with books being published on internet, rather than by publishers.
Obviously, publishers will fight against that.
19:18
I have a medical condition that makes it very difficult for me to post long answer
?????
I can't sit still for long.
And so ?
You can go for a walk, come back, resume your answer
How convenient.
I don't see what medical condition would prevent one from writing long answers
Attention deficit hyperactivity disorder predominantly inattentive (ADHD-PI), also called attention deficit disorder (ADD), is one of the two types of attention deficit hyperactivity disorder (ADHD). The term was formally changed in 1994 in the new Diagnostic and Statistical Manual of Mental Disorders, fourth edition (DSM-IV), to "ADHD predominantly inattentive" (ADHD-PI or ADHD-I) - though the term attention deficit disorder is still widely used. 'Predominantly Inattentive' is similar to the other subtypes of ADHD except that it is characterized primarily by inattentive concentration or a deficit...
19:21
@IceBoy If she can write them down on paper, she can write them down on the website
@IceBoy There seemed to be enough attention for an excuse.
I was very annoyed today, and I'm specially very annoyed when someone, no matter who, try to explain somehow the provenience of my work, even if that thing is done in a very elegant way. I understand that for some is hard to accept that someone with no background works on articles and try to publish things like a book, then come up with amazing proofs for some celebre problems, but this is the reality, and nothing can change that, I have no background in mathematics, and all my work is original.
Of course, I don't plan to convince anyone to believe what I say, but it's good to know that reality may be the way I present it.
Tell about me anything you want but not about my work. My work is saint to me.
@Chris'ssis We all know you received those answers for Christmas -__- Santa is really good at maths
:P
@TheGame :-)))
Saint Nick :D
19:29
If Christmas is only once a year, it's because Santa needs some time to solve all the integrals :D
19:40
Hi, I have one question please about the order of $\bar{p}$ in $Z/nZ$ is $n/gcd(p,n)$: I know that the order is the smallest integer such that $kp$ is divisible by $n$, so is $n/gcd(p,n)$ because is the smallest so we have to choose $gcd(p,n)$? Am-I right? Thanks
@BalarkaSen For you I guess ^

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