Mathematics - 2014-08-20 (page 3 of 4)

@BalarkaSen That depends on the definition of countable, about which I have no idea. Is it to do with the cardinality of the set, as in counting how many elements it has?

Balarka Sen

@Khallil OK, then don't try it. But I can tell you the definition of countability : a set finite of infinite is called countable if you can enlist the elements of the set in a particular order.

it has everything to do with cardinality of infinite sets. whether or not R is countable is equivalent to asking whether or not you can "count" the elements of R

Khallil

@BalarkaSen By that definition, $\mathbb{N}$ is countable since we're just increasing the value of each element by 1 as we go along (from left to right in the set and the real line).

@BalarkaSen That sounds long!

Balarka Sen

yes, it is.

N is countable and we say that cardinality of N is $\aleph_0$

for what it's worth, you cannot count R, as we cannot do P(N) either. Both has cardibnality bigger than $\aleph_0$.

Khallil

@BalarkaSen Why can't we count $\mathcal{P}(\mathbb{N})$?

Balarka Sen

3:21 PM

Try counting.

The formal reasoning would be overkill at your stage.

Khallil

I'll avoid it until I'm ready. Give me a day (or $\aleph_0$). ^_^

Balarka Sen

rationals are countable however.

i wonder if you can produce a proof of that.

Khallil

It's not a matter of whether I can or not, I can. It's solely dependent on whether I can pull myself into doing it. That's true for everybody. Everything takes dedication, and only the most persevering of us make the breakthroughs. Of course, the amount of dedication required depends on how naturally affine we are to the matter at hand.

Or at least, that's my optimistic view!

Balarka Sen

i am a realist, thank you.

Khallil

So what's your opinion of it all?

Balarka Sen

3:25 PM

of what?

Sawarnik

$\mathbb{N}\times\mathbb{N}$ is countable.

Balarka Sen

@Sawarnik yes, it is.

finite direct product of countably infinite sets are always countable.

Khallil

Of whether or not we all have the potential to make breakthroughs, and if we do, then is it just a matter of persevering enough to do it?

Sawarnik

@Khallil And luck.

Khallil

@Sawarnik If you believe in it (which I myself, am on the fence about).

Sawarnik

3:27 PM

@BalarkaSen I liked the informal explanation for that, but the formal one was comparatively quite tough.

Balarka Sen

@Khallil to be frank, i am not sure.

Khallil

I've got a question. A philosophical one.

Balarka Sen

oh @Sawarnik, i read some stories. The Sussex Vampire was good, and the story of the hydrolic engineer in the Engineer's thumb was action-packed.

Sawarnik

:D

Sussex Vampire ... hmm ... I forgot the contents.

Balarka Sen

What is your opinion of The Man With Twisted Lip?

Sawarnik

3:30 PM

@BalarkaSen Good enough.

Khallil

If you like math, is it better to like it because it's fun, or to like it because you're good at it (i.e. every testable obstacle, you've passed with flying colours)?

Sawarnik

You read Charles Augustus something?

Balarka Sen

I couldn't find it.

Sawarnik

[And where did you get hold of that mega book?]

Balarka Sen

@Sawarnik heh heh

from the opium den.

Sawarnik

3:31 PM

:/ Opium den probably wouldn't have it.

Balarka Sen

=P

Sawarnik

So from where! Your library?

@BalarkaSen en.wikipedia.org/wiki/… . Dont read the plot.

Balarka Sen

will have to look for it.

i won't read it, don't worry.

Khallil

Have you watched Death Note, @BalarkaSen and @Sawarnik?

Balarka Sen

nahp

Sawarnik

3:33 PM

nahp

Khallil

Do you watch anime at all?

Balarka Sen

a little

i watched naruto once upon a long time ago, as well as Dragon Balls.

Sawarnik

nahp

@balarka And the Dancing Men is nice as well.

Balarka Sen

noted

Khallil

Time to start some set theory.

Balarka Sen

3:39 PM

i am looking for a comment by Noam D. Elkies in mathoverflow which I am going to use as a reference for a another comment which I am going to write down in MO

Sawarnik

All I have been studying in math these days is olympiad level geometry :|

Balarka Sen

bleh

Khallil

Given that $\left| A \right| = m$ and $\left| B \right| = n$, then is it true that $\left| \mathcal{P}(A \times \mathcal{P}(B)) \right| = 2^{m \cdot 2^{n}}$?

Sawarnik

@BalarkaSen I understand.

@Khallil I think yes.

Khallil

@Sawarnik Me too! (Smooth edit there!)

Sawarnik

3:43 PM

:D

@BalarkaSen Naruto is nonsense.

@Chris'ssis Obviously 2.

Balarka Sen

@Khallil is that not patently obvious?

@Sawarnik i don't recall it. i watched it when i was 7 or 8 or something like that.

Sawarnik

@BalarkaSen Wow, Balarka remembers similar triangles. :O :O :O Yay.

Balarka Sen

no, i haven't studied similar triangles.

as you'd see when you click the starred message, i solved it by pitagoras.

Sawarnik

Oh :(

Khallil

@Sawarnik It's too late to take that statement back.

Lord Magento Uzumaki will take you down!

Sawarnik

3:48 PM

@Khallil That's why I called it nonsense.

Any sensible guy would agree with me.

Balarka Sen

mangas are somewhat nonsense.

Sawarnik

true.

Khallil

I didn't think it was that obvious, @BalarkaSen!

Balarka Sen

@Khallil what is the cardinality of P(A) \times B?

oops, I meant A \times P(B)

Sawarnik

@Khallil It was a bit obvious.

Khallil

3:51 PM

@BalarkaSen Their respective cardinalities multiplied together.

Balarka Sen

and what is that?

Khallil

@Sawarnik Hey, I'm still getting the hang of set theory!

@BalarkaSen It is $m \cdot 2^{n}$. Oh, I see.

Balarka Sen

so, by our previous experience, what does that tell you about the power set?

Khallil

It's cardinality is $2^{m \cdot 2^{n}}$.

Balarka Sen

right.

there you go.

Khallil

3:53 PM

Be back in a bit. I'm going to have some lunch. How could I have delayed it until this time!?

Sawarnik

@Khallil What time there?

Khallil

4:15 PM

@Sawarnik Right now it's 17:15. I'm guessing it's near 21:30 where you are, right?

Sawarnik

@Khallil Yes.

IST is +5:30 hrs, I think so you are off by 15 mins.

Khallil

@Sawarnik Oh, nice! I was close!

MickLH

Close only counts in horseshoes and hand grenades, and numerical methods

Sawarnik

:D

@MickLH ?

Ok bye.

MickLH

lol peace

Khallil

4:51 PM

What's the power set of the empty set, @BalarkaSen?

The power set is the set of all subsets of a set.

So I'm thinking that $\mathcal{P}(\varnothing) = \{ \varnothing \}$.

robjohn

@Chris'ssis Okay... I posted my full proof here

Chris's sis

@robjohn Very interesting the idea in $(3)$. Actually that helped you to employ the methods of complex analysis.

robjohn

@Chris'ssis Then I can convert it into a nice contour integral that can be collapsed.

Chris's sis

@robjohn Yeah, I got that. (+1)

robjohn

@Chris'ssis It also uses a bit of Fourier analysis, too

@Chris'ssis It was different enough from the other answers, that I thought it warranted such a late answer.

Chris's sis

5:02 PM

@robjohn It's never too late for another answer.

robjohn

@Chris'ssis Yeah, but sometimes all the good approaches have been taken.

@Chris'ssis Plus, I didn't use the digamma or polylogarithm functions :-)

Chris's sis

@robjohn Or sometime no, especially when it's about research.

robjohn

@Chris'ssis no complex analysis at all?

Chris's sis

@robjohn No. It's an amazing proof (something unseen before).

robjohn

@Chris'ssis I should work on one without complex analysis.

Chris's sis

5:06 PM

@robjohn I like your attitude!!! :-) That proof can be given to high school.

Khallil

I've got some problems with $\varnothing$ as an element and a subset.

Is $\varnothing$ an element of the power set of any set?
If so, is it also a subset of the power set of a set, or is it instead, $\{ \varnothing \}$ that's a subset of the power set?

robjohn

@Chris'ssis your proof that we haven't seen yet?

rehband

@Khallil The empty set is an element of every power set :)

Khallil

@rehband Is it also a subset of every power set, or is $\{ \varnothing \}$ a subset of every power set?

user1534664

Why can ratio's (3 : 4) be represented as fractions (3/4)? I just can't seem to grasp it

rehband

5:11 PM

@Khallil Saying that $\varnothing$ is an element of the power set of some set $X$ is the same as saying that $\varnothing$ is a subset of $X$

Khallil

@user1534664 I didn't think they could. If you've got the ratio $3:4$, in total, there are 7 parts, so you'd have $\frac{3}{7}$ representing the 3 parts, and $\frac{4}{7}$ representing the 4 parts.

@rehband Sorry, I wasn't clear enough. Is $\varnothing$ also a subset of the power set of some set? It's not, is it?

user1534664

yeah that's why I'm confused, because when solving proportions people tend to turn the ratio into a fraction but they don't seem to be the same thing

Khallil

@user1534664 Is there an example of this that you can get a hold of? I really don't get it either. It doesn't make sense.

Chris's sis

@robjohn Take this one (it's my initial proof that comes from personal research). Then, there is another version where a great mathematician contributed and came with a very brilliant proof to that logarithmic integral). You'll see that version when our article is released (it's amazing).

user1534664

one moment

rehband

5:14 PM

@Khallil $\varnothing$ is an element of every power set. It can also be a subset of a power set

@robjohn take it

@Chris'ssis got it

@user1534664 I don't really want to watch the whole thing.

user1534664

lol

Chris's sis

5:16 PM

@robjohn I love so much that logarithmic integral, it's so powerful!

Khallil

@user1534664 Oh, I got it. Yea, apparently they are just two ways of writing the ratio, but I'd really go with the $a:b$ notation so as to avoid confusion with actual fractions. As he said, he got it from Wikipedia which isn't the most trustworthy of sources.

rehband

@Khallil Consider the set $X=\{1,2\}$. Its power set is $P(X)=\{\varnothing, \{1\},\{2\},\{1,2\}\}$

Now $\{\varnothing\} \subset P(X) \iff \varnothing \in P(X)$ which is true

Khallil

@rehband Got it. Thanks!

Chris's sis

@robjohn Also the proof by Ovidiu Furdui is very nice although it uses dilogarithm. I don't know if you have seen his book (too nice to be true). I mean if you study it well you are about to become a little god in analysis. :-)

user1534664

@khalil ahh thanks but what's the use of using it in the fraction notation?

it just seems to complicate things

Chris's sis

5:22 PM

@robjohn I said that in general, you're already a god in analysis. :-)

Khallil

That convolution of power sets was tricky because I didn't recall the actual definition of a subset.

robjohn

@Chris'ssis I will continue the work I was doing on $$\sum_{j,k,n=1}^\infty\frac1{jk(j+n)(k+n)}$$ I think that can lead to something, too.

That is simply the same sum in different clothes

Chris's sis

@robjohn This looks like Sandham's proof that @r9m mentioned.

robjohn

@Chris'ssis That was the first way I was working on it, then I saw the Fourier analysis approach and went with that.

rehband

Chris or robjohn: I'm investigating $$\lim_{n\to\infty} \left( \left(\sum_{k=n+1}^{2n}2\sqrt[2k]{2k}-\sqrt[k]{k}\right)-n\right)$$. Is there a clever trick for this type of sum? Could it ever be useful to look at $$\sum_{k=n+1}^{2n}\log(2\sqrt[2k]{2k})-\log(\sqrt[k]{k})$$?

Chris's sis

5:27 PM

@robjohn

@robjohn @r9m said this one was published by Sandham.

@rehband wow, where do you have this one from? I mean both limits.

rehband

@Chris'ssis The book of course. :) Unfortunatly some of these don't have solutions/hints. I've tried using some strategy that was used in the previous exercises but nothing seemed applicable

I feel like there might be something good to learn from this problem but I can't find a way of solving it.

Chris's sis

@rehband I solved that first one a long time ago. I talked about it with someone these days.

rehband

@Chris'ssis Nice! How did you do it?

Khallil

Can I ask you a question on sets, @rehband?

rehband

@Khallil Of course!

Khallil

5:34 PM

The question is to find the cardinality of the set $\{ X \subseteq \mathcal{P}(A) : |X| \leqslant 1 \}$, given that the cardinality of the set $A$ is equal to $m$.

robjohn

@rehband That should be $\log(2)^2$, no?

rehband

@robjohn Yes, how did you find that??

Khallil

@rehband I think it's the same as the cardinality of $\mathcal{P}(\mathcal{P}(A))$. Is that right?

Chris's sis

@rehband

robjohn

@rehband when $n$ is very big, $n^{1/n}\sim1+\frac{\log(n)}{n}$

rehband

5:37 PM

@Chris'ssis Fantastic, thank you! Can't wait to look over that

@robjohn That's good to know, I didn't know about that last term

Chris's sis

@rehband Welcome! :-)

robjohn

:17250563 $n^{1/n}=1+\frac{\log(n)}{n}+O\left(\frac{\log(n)^2}{n^2}\right)$

rehband

@robjohn Nice, thx!

Chris's sis

@robjohn Yeap.

robjohn

The error term in that proof is a bit optimistic, but the difference doesn't really matter.

Chris's sis

5:39 PM

@robjohn Perhaps I did all in a hurry. Initially I did it without pen and paper.

@robjohn I remember that last year I posted it here 2 or 3 days and no one answered it. Maybe no one took interest since it has a certain appearance. :-)

robjohn

@Chris'ssis The thing I just answered with $\log(2)^2$? I must not have seen it

Chris's sis

@robjohn Yeap.

rehband

@Khallil If $A$ has the cardinality $m$, then the power set of $A$, $P(A)$, has the cardinality $2^m$. The set $\{ X \subseteq \mathcal{P}(A) : |X| \leqslant 1 \}$ is a certain subset of $P(A)$, so its cardinality is at most $2^m$. It cannot be $|P(P(A))| = 2^{2^m}$ :)

Khallil

@rehband Yep, I realised that a few minutes ago!

robjohn

@Chris'ssis Had I seen it, I would have answered using the same idea.

Chris's sis

5:44 PM

@robjohn Yeah, Taylor series rulez ... :-)

rehband

@Khallil The cardinality of $\{ X \subseteq \mathcal{P}(A) : |X| \leqslant 1 \}$ is the number of sets contained in the power set of $A$ with one or less elements.

Khallil

@rehband Hold on.

@rehband So would it be correct in claiming that $| \{ X \subseteq \mathcal{P}(A) : |X| \leqslant 1 \} | = 2^{m}$ as a result of $|A| = m$?

rehband

@Khallil We know that $|P(A)| = 2^m$. The set $\{ X \subseteq \mathcal{P}(A) : |X| \leqslant 1 \}$, call it $Q$ for short, is a subset of $P(A)$, that is, every element in $Q$ is an element of $P(A)$. Not every element of $P(A)$ needs to be in $Q$ though.

@Khallil So then we have that $|Q| \leq |P(A)|$

Chris's sis

@robjohn I think that question might be given to some contest, maybe Putnam?

Khallil

We're looking for the cardinality of the set containing the subsets of the power set of $A=\{ x_1, x_2, \dots, x_m \}$ with one or less elements, so we're essentially just looking for the cardinality of the set $\{ \{ \{ x_1 \} \}, \{ \{ x_2 \} \}, \{ \{ x_3 \} \}, \dots, \{ \{ x_m \} \} \}$, right?
Your explanation does make sense, so thank you! I'm just trying to make sense of my own more explicit argument. ^_^

robjohn

5:57 PM

@Chris'ssis seems too easy, but maybe I have become jaded...

rehband

@Khallil We are looking for the cardinality of the set $\{\varnothing ,\{x_1\},...,\{x_m\} \}$. These are all elements of $P(A)$ which have one or zero elements.

Chris's sis

@robjohn I initially gave it to some students and no one did it.

Khallil

@rehband Oh, sorry! I forgot $\{ \varnothing \}$.

robjohn

@Chris'ssis what level of student? Putnam age students?

Chris's sis

@robjohn Well, it's true that for those guys something harder is needed. However I bet this problem would produce many surprises.

Khallil

6:02 PM

@rehband You've forgotten that $\{ \varnothing \} \subseteq \mathcal{P}(A) \implies \varnothing \in \mathcal{P}(A)$.

rehband

@Khallil I don't understand. What do you mean? :)

Khallil

6 mins ago, by rehband

@Khallil We are looking for the cardinality of the set $\{\varnothing ,\{x_1\},...,\{x_m\} \}$. These are all elements of $P(A)$ which have one or zero elements.

The set you wrote down should've had $\{ \varnothing \}$ as an element as opposed to $\varnothing$, should it have not?

rehband

@Khallil No, $\varnothing = \{ \}$. $\varnothing$ itself is a set.

Khallil

@rehband So it's both an element and a subset of every power set?

rehband

@Khallil Yes

Chris's sis

6:12 PM

@r9m How are you doing? Where are ya? :-) I hope you're fine.

rehband

@Khallil $\varnothing \subset P(X) \iff$ every element of $\varnothing$ in also an element of $P(X)$, which is true since there is no element in $\varnothing$.

@Khallil $\varnothing \in P(X) \iff \varnothing \subset X \iff$ every element of $\varnothing$ is an element of $X$. Also true.

Chris's sis

6:32 PM

What the heck ... math.stackexchange.com/questions/876845/… Someone posted here a question I thought I created these days ...

Maybe he/she took it from here ...

Khallil

@Chris'ssis When did you post it on here? The question was posted 27 days ago.

Chris's sis

Oh, that guy has some hard stuff.

Khallil

@Chris'ssis Yea, I noticed. That's why I was skeptical of him copying your question.

Chris's sis

6:51 PM

For example, the proof to the Theorem 3 here is standard - math.stackexchange.com/questions/875076/…

This integral can be computed in a simpler way though $$\int_{0}^{1}\!\!\int_{0}^{1}
\left\{\frac{1}{x y}\right\}^{2} \:\mathrm{d}x \mathrm{d}y = 1-\gamma+\dfrac{\gamma^2}{2}-\dfrac{\pi^2}{24}+\ln(2\pi)-\dfrac{\ln^2(2\pi)}{2}$$

Anyway, I like that guy.

People like this one must be encouraged permanently, this is the way the art questions should look like.

This is another nice result (I met it in the past?) $$\int_{0}^{1} x^{s}\left\{1/x\right\}^{2}\:\mathrm{d}x = -\frac{2\zeta(s)}{s(1+s)}-\frac{\zeta(1+s)}{1+s}-\frac{1}{1-s} $$

but it can be hugely enriched (it's very easy to prove).

rehband

@Chris'ssis Earlier today I posted a question about that limit I asked you about an hour ago. math.stackexchange.com/questions/904252/… Wanna post your nice solution there?

Chris's sis

@rehband No.

rehband

@Chris'ssis Hehe, ok no worries. I'll just leave it then

Chris's sis

@rehband Maybe some come up with new ideas. I'm just a tiny bit active on main. Most of the time I didn't even read the things that are posted there. I already have my own questions to answer, they are a lot.

rehband

@Chris'ssis True

@Chris'ssis I understand, respect for the hard work

Chris's sis

7:00 PM

Thank you. It really requires a lot of very hard work.

I don't wanna discourage anyone, but very hard work means very hard work.

(study, research, study, research ... and so on)

rehband

@Chris'ssis I have no doubt, I love the commitment! I wanna be like that!

Chris's sis

@rehband Yeah. No need for pride in mathematics. When I meet someone that knows more than me, I'm happy to listen, to learn, to recognize his /her efforts. There is no way to become better without putting the pride away.

rehband

@Chris'ssis That's very good advice

@Chris'ssis You remind me of Paul Erdos lol

Chris's sis

@rehband hehe, you know, I was already told that! :-)

Balarka Sen

@rehband The only teensy weensy difference being Paul Erdos was a combinatorist =P

rehband

7:09 PM

Haha

Chris's sis

@BalarkaSen :-)))))

Balarka Sen

@Chris'ssis =)

Erdos was my inspiration for a brief period of time, but not until coming to know that his works on combinatorial graph theory beats his works on number theory.

le sigh

Khallil

Yo, @BalarkaSen!

Balarka Sen

@Chris'ssis if i recall correctly, i heard either you or r9m speaking abou Martin Kneser a few days ago. am i right?

@Khallil Ollo

Chris's sis

@BalarkaSen Not me.

Balarka Sen

7:16 PM

ok, then probably r9m

Chris's sis

@BalarkaSen btw, I didn't see @r9m in the last period of time ...

Balarka Sen

yeah, me neither

probably his tests are on the way

Khallil

@BalarkaSen I'm still stuck on some question I was doing earlier. I can't determine whether $|\{ X \subseteq \mathcal{P}(A) : |X| \leq 1 \}| \leq |\mathcal{P}(A)|$ or whether the two are equal.

Balarka Sen

Replace P(A) by some set B

Khallil

Oh wait, I've got it!

Balarka Sen

7:19 PM

neither is true.

Chris's sis

This question is pretty enjoyable too math.stackexchange.com/questions/902905/…

Khallil

@BalarkaSen Neither?

Balarka Sen

Answer is |P(A)|+1

You forgot the null set

Khallil

@BalarkaSen Heh? How'd you get that?

Balarka Sen

Wait a sec

right, right, P(A) can't be treated the same as any other set.

but of course the answer is > |P(A)|. let me eat first

Chris's sis

7:21 PM

And this one can be done pretty easy (it's cute too) math.stackexchange.com/questions/866382/…

Khallil

@BalarkaSen How so? I thought that $\{ X \subseteq \mathcal{P}(A) : |X| \leq 1 \}$ is a subset of $\mathcal{P}(A)$, so it's cardinality is at most the cardinality of $\mathcal{P}(A)$ ...

Oh, enjoy!

Oh, no I'm wrong. All of the sets $X$ are subsets of $\mathcal{P}(A)$, but we're looking at the set of all $X$ of that form.

Balarka Sen

Back.

Right, you're wrong.

Khallil

Right.

Balarka Sen

No, wrong.

=P

Khallil

Gosh, I'm so confused.

Balarka Sen

7:29 PM

Isn't that some transformed pikachu?

raichu, i think, it's called.

Khallil

Jolteon! It's the evolved form of Eevee after giving it a Thunderstone.

Balarka Sen

Ah, right. what does eevee look like again?

Khallil

That qt$\pi$. ^_^

Balarka Sen

the pokemon i liked the best is probably called dark something. it puts stuffs in deep sleep.

Khallil

Darkrai! (bulbapedia.bulbagarden.net/wiki/Darkrai_(Pokémon))

Balarka Sen

7:33 PM

deep sleep with horrible dreams.

yeah, that one.

Khallil

It's got the move Dream Eater which depletes your HP (Health points) whilst your asleep.

Balarka Sen

some guy somehow inserted it in a pokeball, no?

i wonder how he did that.

Khallil

I'm not sure who it was, but I'm pretty sure it was a professor. The Pokémon world is way more advanced than ours. They store physical objects, digitally. Also, I think they have portals to other dimensions in their bags. You can seemingly store an infinite number of rare candies in there for some reason.

Balarka Sen

he participated in some pokemon competition though.

you are reminding me of digimons.

=P

Khallil

Digimon was awesome! It was part of my halcyon days.

Balarka Sen

7:36 PM

yeah

Khallil

I feel all warm inside.

^_^

Balarka Sen

digimon was way better than pokemons. i forgot most of it though.

Khallil

It had a great storyline. The interactions between Tai and Matt were great. The main themes were full of hope as well, with the crests, tags and digi-eggs.

Balarka Sen

yes. there were many versions of digimons though.

Khallil

Not as many Digimon as Pokémon, however.

I think there are almost 700 Pokémon now.

Balarka Sen

7:39 PM

whoa

@Khallil do you recall that manga about football? some leagues or whatever, i don't recall the name.

Khallil

@BalarkaSen Inazuma eleven? Area no Kishi (Knight in the Area)? I've seen loads.

Balarka Sen

Right, that's the one!

Inazuma eleven.

Khallil

I haven't actually seen that one. It just sprung to mind for some reason.

Was it any good? I thought Area no Kishi was good, but less centred on the sport than the tragedy.

Balarka Sen

it was awesome. i used to watch that everyday.

Khallil

I might give it a chance. I might also give the IT crowd a chance.

Balarka Sen

7:43 PM

but more than anything, i liked dragon balls the best among all the other mangas.

cirpis

so messing with integration and taylor series i found this identity: $$\frac{1}{1*3}-\frac{1}{3*3}+\frac{1}{5*9}-\frac{1}{7*27}+ ... + \frac{(-1)^n}{(2n+1)*3^{n+1}}+...=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)*3^{n+1}}‌=\frac{\pi}{6}*\tan(\frac{\pi}{6})$$. How would one go about proving the sum without prior knowledge of the integral that leads up to this?

Khallil

@cirpis Manipulation?

@BalarkaSen Dragonball was great. I never read the manga, but I watched the anime. It was a nice adventure to say the least.

Balarka Sen

i did watch the anime, but i recently read the manga too.

k, i gotta run. writing some kind of a survey on riemann hypothesis upon request of some guy in a forum.

Khallil

Cool! Have fun!

^_^

cirpis

@Khallil what manipulations tho?

Khallil

7:58 PM

Chris's sis

8:21 PM

By the way, in those answers one question proposed by Ovidiu appears.

gbox

Anyone can help me prove Rank(AB)=Rank(B) iff Rank(A)=n?

Chris's sis

@Alizter I have a very nice question for you.

Alizter

@Chris'ssis ok :)

Chris's sis

@Alizter It's for beginners. This one $$\int_0^1 \left(\left\{\frac 1x\right\}-\frac12\right)\frac{\log(x)}{x}dx$$ If you like it, give it a try. It's pretty easy, all flows naturally.

Alizter

@Chris'ssis i will have a go later. :)

Chris's sis

8:31 PM

@Alizter No hurry. Whenever you want.

James Fielder

9:17 PM

Hi there, I'm trying to comment on the answer to a question I posted earlier but don't seem to be able to. Is there a particular amount of reputation I need to be able to comment? Or am I missing something obvious.

Ah, there was some oddness with cookies in my browser, can now comment! Sorry to bother anyone.

skullpatrol

np

thanks for asking :)

Khallil

9:48 PM

This shall serve as the chat reviver, forevermore.

Courtesy of our favourite voice, Morgan Freeman.

MrWho

10:11 PM

@robjohn I don't really like Putnam or Competitions as a motivation to learn math, but the reward given to the one who stands first is intriguing

Chris's sis

10:35 PM

lollll, I had so fun reading some comments, like the one here math.stackexchange.com/questions/875076/…

"Very good job! achille hui : congratulations! Your method is interesting! I accept your answer. Thank you! Let me give a different approach." – Olivier Oloa

:-)))))

Let me give a different approach.

Olivier Oloa asked the question and then answered it.

skullpatrol

:D)))

Jasper Loy

@Khallil Because I seek the truth, though I may never find it.

skullpatrol

the enjoyment is in the search...

Khallil

@JasperLoy That's the right way of going about things! I've considered a bunch of other religions too, despite the pressures from my family to 'return' to Islam. None of them appeal to my method of reasoning.

skullpatrol

we all have different ways of thinking

Khallil

10:46 PM

You are a wise person, @skullpatrol. ^_^

The universe is damn crazy.

skullpatrol

thnx ^_^

hello professor @TedShifrin

now there^ is a title of wisdom

Khallil

Indeed!

Transcript for

$Mathematics$ Mathematics