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01:00 - 19:0019:00 - 00:00

1:48 AM
@KarlKronenfeld Thanks for the hint.
 
@PedroTamaroff Were you able to take it from there?
 
@KarlKronenfeld I didn't think about it much more, but I was kinda convinced.
I can think about it tomorrow, though.
 
There are some details to resolve once you've derived my contradiction, so it is worth taking a close look eventually.
 
Dedekind domains are ugh.
@KarlKronenfeld Mike posted a cool problem here a few day ago. Field theory/
If $F$ is a field, it is finite if and only if $F^\times$ is finitely generated.
Turns out that there is a commutative ring generalization. Any idea if this holds for noncommutative rings?
 
no idea, I would guess no.
 
2:02 AM
me too
Oh and $F^\times$ means multiplicative structure of $F$ in case you think it's the group of units.
=P
 
 
2 hours later…
3:34 AM
@DanielFischer I made two questions on main.
 
4:07 AM
good night folks
is there a simple way of proving the only powers of 2 and 3 with diffenrence 1 are 8 and 9 without proving catalan?
 
4:38 AM
nevermind, I got it
 
 
1 hour later…
6:03 AM
We have 1000th question on the site.
 
6:39 AM
Hip, Hip, Hooray!
 
7:05 AM
Greetings
@robjohn that integral I showed you yesterday is finally a simple application of beta function integral. I'll show you my proof as soon as I put things on latex.
 
@Chris'ssis which integral? I remember sums, but I am not remembering an integral.
 
@robjohn the one with double log
 
@Chris'ssis Oh, I should look back at that...
 
@robjohn Initially I thought it is harder but no ...
 
7:24 AM
@robjohn All gets reduced to the simple fact that our integral equals $$\int_0^{\pi/2} \log(2\sin^2(x)) \log(2\cos^2(x)) \ dx$$ Then, expanding this, we're almost done.
 
@Chris'ssis I know... I am working on it...
 
@robjohn much more interesting is the version with 3 logs.
 
using
$$
\begin{align}
\log(1-\cos(x))
&=\log(2\sin^2(x/2))\\
&=2\log(2\sin(x/2))-\log(2)\\
&=2\log(e^{ix/2}-e^{-ix/2})-\log(2)-i\pi\\
&=2\log(1-e^{-ix})-\log(2)-i\pi+ix\\
&=2\log(e^{-ix/2}-e^{ix/2})-\log(2)+i\pi\\
&=2\log(1-e^{ix})-\log(2)+i\pi-ix\\
&=-2\left(\frac{\cos(x)}1+\frac{\cos(2x)}2+\frac{\cos(3x)}3+\dots\right)-\log(2)
\end{align}
$$
and
$$
\begin{align}
\log(1+\cos(x))
&=\log(2\cos^2(x/2))\\
&=2\log(2\cos(x/2))-\log(2)\\
&=2\log(e^{ix/2}+e^{-ix/2})-\log(2)\\
&=2\log(1+e^{-ix})-\log(2)+ix\\
&=2\log(e^{-ix/2}+e^{ix/2})-\log(2)\\
&=2\log(1+e^{ix})-\log(2)-ix\\
&=2\left(\frac{\cos(x)}1-\frac{\cos(2x)}2+\frac{\cos(3x)}3-\dots\right)-\log(2)
\end{align}
$$
it should be easier
 
@robjohn I think the only problem would be that product of series.
 
@Chris'ssis well, most of the terms vanish by orhogonality
 
7:40 AM
Indeed.
 
is it $-\frac\pi2\zeta(2)+\frac\pi2\log(2)^2$?
 
@robjohn do you refer to the value of the integral?
 
@Chris'ssis just doing the integral in my head, so I may be missing some things.
2
 
@robjohn $$-\frac\pi2\zeta(2)+\frac\pi2\log(2)^2$$
 
@Chris'ssis Ugh... I missed the factor of $\pi$ on the first term... yeah since the integral of $\cos^2(x)$ over the whole circle is $\pi$ and not $1$
@Chris'ssis Yeah... it is pretty easy.
 
7:48 AM
@IceBoy You're funny. I like your style but it looks like lots of users here are too serious. For @WillHunting, congrats for 1000+ rep!!
 
I tried it in Mathematica, and it turns out to be a good way to warm my lap :-)
 
@robjohn lol, I know ... :-)
 
@Chris'ssis my laptop is heating up, but no answer yet.
 
Huy
I want to find a maximal set $A \subseteq \mathbb{N}$ such that $$\forall x,y \in A: x+y \notin A.$$ I'm currently thinking about $$\mathbb{P} \cup \{1\} \setminus \{2\}.$$ Is this set maximal? If so, is it the unique solution?
 
@robjohn Mathematica is helpless there.
 
7:50 AM
@Chris'ssis it seems so... I killed the process so that I wouldn't burn my legs.
 
:-)))))
 
Thanks @Anastasiya :D
 
let my laptop cool off
@IceBoy There are a lot of people on the site who are not on chat and they don't understand about Jasper.
@Huy $A$ is odd numbers?
 
Huy
@robjohn: Oh. That's bigger, yeah.
@robjohn: Do you have any idea how to show whether the solution is unique or not?
 
@IceBoy This site needs more users like you. Lots of users here are like as straight as a ramrod. I don't know that is the correct way to describe it in English
 
7:55 AM
@Huy any subset of a solution is a solution...
 
Huy
@robjohn: No, it has to be a maximal set.
 
@Huy the integers that are $2$ mod $4$ is another.
 
I found it very insensitive to remove the "celebration" tag.
:(
 
Huy
@robjohn: Nice, thanks.
 
@Huy I don't think it is maximal, though... if you include $1$ it might be
 
Huy
8:00 AM
@robjohn: Couldn't I also skip every second odd number and then take the union?
 
@Huy integers that are $1$ or $2$ mod $4$ no, that doesn't work
Now that I think of it, integers that are $2$ mod $4$ is maximal...
if we include $1$ then $1+1$ is in the set
 
Huy
@robjohn: Yeah, because $1+1=2$.
Yeah, that's what I just thought.
 
So the answer is not unique
 
Huy
Aye.
 
8:16 AM
$$\int_0^{\pi/2} \log(1-\cos(x)) \log(\cos(x))\log(1+\cos(x)) \ dx=\space:-)$$
 
8:31 AM
@Chris'ssis at least $\log(\cos(x))$ has a nice form involving $\cos(2nx)$
 
Yeap.
 
 
1 hour later…
9:57 AM
words = left
numbers = right
 
10:22 AM
?
??
???
 
Huy
1331
 
???
??
?
 
Huy
numbers = right
words = left
 
 
1 hour later…
11:45 AM
Yeap.
 
Hello. With the help of math.se user Leo for the first draft I concocted a question that's stored here: http://meta.math.stackexchange.com/a/4668/140933
I'd like to know if there's redundant info and if it's understandable to people with no roleplaying games knowledge. Thank you.
 
12:09 PM
“We start off with high hopes, then we bottle it. We realise that we’re all going to die, without really finding out the big answers. We develop all those long-winded ideas which just interpret the reality of our lives in different ways, without really extending our body of worthwhile knowledge, about the big things, the real things. Basically, we live a short disappointing life; and then we die. We fill up our lives with shite, things like careers and relationships to delude ourselves that it isn’t all totally pointless.”
I've just remembered this great quote while reading some about Trainspotting 2.
 
Let me argue that, without a career, life would be significantly shorter.
(I bottled the idea of a relationship, on the other hand)
 
12:24 PM
@Chris'ssis Well, the meaning of life is this. Just enjoy yourself for a few decades on earth and then die.
4
 
@WillHunting True.
 
Of all chats, you! How dare you reject the fact that the meaning of life is a number and it's 42?
(Having a mathematician friend, I know most math isn't about numbers. The thing still looks sad to me.)
 
I'm preparing to write up some proofs, I feel I need to be more productive in terms of proofs.
 
@Chris'ssis How many a day?
 
@Alizter I like more to create things than to solve them. It depends, some proofs require a lot of time.
 
12:33 PM
@Chris'ssis how was that general bessel sum?
 
@Alizter I didn't continue the work on it.
@Alizter I recommend you to work on my last integral above. It's very nice.
 
1:23 PM
@WillHunting This answer seems Jasper-esque. Did you write it?
 
@ParthKohli No. I don't write "Hint" in my answer. That says nothing.
If I give a hint, I just give it. No need to say "Hint" anymore.
 
@ParthKohli That was T.Bongers.
 
1:56 PM
$J_2(\pi)$ is annoyingly close to $1/2$
@Chris'ssis $$\sum_{n=1}^\infty \frac{J_k(2n)}{n^k}$$
$k=0$ is either really small or zero.
$k=1$ is $1/2$
 
@Will: Hint: Will
 
2:44 PM
@Nick!
 
@BalarkaSen: Balsensai!
Now, we're both factorials :D
 
(@Nick)(@Nick - 1)(@Nick - 2)(@Nick - 3) ... (3)(2)(1)
 
What if @Nick isn't an integer, @BalarkaSen?
 
Gamma(@Nick-1)
 
^_^
 
2:49 PM
@robjohn see above my way
 
@Nick! $ = \int_{0}^{+\infty} x^{\text{@Nick}} e^{-x} \text{ d}x$
 
@Khallil: I feel very violated. Don't you dare to integrate me again!
 
I Laplace'd you, @Nick. =P
At $s=1$.
 
lol
@BalarkaSen: The following is more closer to the truth:$$\prod^{\text{@Balarka}}_{\text{@Nick} = 1} \text{@Nick}$$
Balsensai is the product of everything I have and could learn in my lifetime.
 
Is $|\text{@Nick}| < 1$, @Nick?
 
2:59 PM
@Khallil: It would be an insult to @Balarka if I was.
 
Stop underestimating yourself, @Nick.
 
@Khallil: I'm not underestimating myself. I'm just putting @Balarka closer to $\infty$
@Khallil: You've probably heard this question before but if you haven't: How can you distribute 9 pigs into 4 pens so that there are an odd number of pigs in each pen?
 
@Nick I'm countably finite thank you.
 
@BalarkaSen: Then, who in this room is the sleeping eight?
 
And these puns are getting old.
 
3:05 PM
I elect @robjohn to be $\infty$
 
Me too.
 
It's not possible, @Nick.
 
@Khallil: Why not?
 
How can 4 odd numbers add up to an odd number?
 
@Khallil: Why do you assume the pens to be disjoint?
 
3:08 PM
Oh.
So you can have pens within one another?
 
Penception
 
Aren't there other combinations?
 
@Khallil: There are 88 possible answers considering joint pens.
 
1 in the middle one. 2 in the one just outside. 2 in the one just outside that. Then 4 in the outermost pen.
Oh, nice.
How'd you work that one out?
 
@Khallil: I didn't. This is a very very very famous question here on M.SE
 
3:10 PM
Sorry, @Nick. I hardly spend any time on here.
 
@Khallil: The best answer with 101 upvotes is "This question is silly. You can't fit nine pigs into a pen. Only ink goes in pens. You have to liquidize the pigs into pig ink, and then you can divide the pig ink equally into four portions."
 
Slow clap
Ha ... ha … ha
 
It has so many jokes. I love the answer with $\zeta(s)=\displaystyle \sum_{n=1}^\infty n^{-s}$
 
I don't find it funny.
However, I can see where the humour would lie in those comments.
 
@Khallil: You don't find having 2/3 rd of a pig funny?
 
3:17 PM
Kinda sad. Much like the murders of chicken, cows etc.
It's a necessary evil.
 
You're not vegetarian, are you?
 
Evil doesn't amuse me (that often).
Nope. I'm not.
Are you?
 
I used to be but being vegan wasn't healthy.
 
Vegan is too much
 
Now, I don't like to consider myself a non-vegetarian either.
Right now, I am technically a pollo-pescateric eggetarian.
And no, that does not mean I eat the eggs of sea chickens.
 
3:20 PM
Sorry, I just find all of these labels to be so pretentious.
Identifying with one just seems so worthless.
 
It's just meta tag. Keywords are faster than saying an entire paragraph about what you are.
 
Huy
What's up, @Khallil. Just got back from an awesome football session.
 
Now that I think about it, all nouns are meta-tags.
I should stop thinking now.
@Khallil: Are you familiar with the product rule in calculus? [$(ab)' = ab' + a'b$]
 
Yea, @Nick.
 
@Khallil: how do you extend that for n functions?
 
3:31 PM
Hey, @Huy. I think I'm going to play football with some friends of some friends tomorrow night.
Liebniz's rule.
 
Like $\left(\prod_{i=1}^{n} u_i\right)'$
 
Oh.
 
Leibniz
 
How do you express it in prod
 
Not Liebniz, @Khallil
 
3:32 PM
I don't care, @BalarkaSen. Hardly anybody I know irl can even say my name properly.
 
Do you just mean the first derivative, @Nick?
 
@Khallil: Yes
 
It's pretty simple.
 
It's easy, @Nick
 
3:33 PM
I don't think there's a nice form to it.
 
Like $$(abc)' = a'bc + ab'c + abc'\\
(abcd)' = a'bcd + ab'cd + abc'd + abcd'$$
 
It's in it's simplest form like that, @Nick.
 
You can produce a recursion.
 
For however many functions a,b,c,... you choose.
What do you mean by that, @BalarkaSen?
Be back in a short while.
(Also, I was on MHB yesterday, looking to see what kinds of threads are posted in the 'more accessible' sections like precal etc.)
 
Huy
What's MHB?
 
3:36 PM
Another forum.
So what do you think of them, @Khallil? The forum?
I personally like it. It's well-maintained, and the admins are almost always active. The traffic is a bit low though.
 
@DanielFischer What an unfortunate name.
 
@ParthKohli: No, Lemony Snicket is an unfortunate name.
@BalarkaSen: $$\left(\prod_{i=1}^{n} u_i\right)' = \sum_{j =1}^{n} \text{something}$$ What is that "something"?
 
@ParthKohli He inherited the surname from his parents.
 
$$\text{something} = \left ( \prod_{i=1}^n u_i \right) \frac{u_j'}{u_j}$$
 
Ooh. Thanks! :D
 
3:47 PM
Oh, that makes sense.
What do I think of them? As you said, the traffic is quite low, but it seems like a good place to talk.
I know of another forum which I was a member of, where the traffic was quite high, but the majority of users just wanted help with exam questions and homework questions, @BalarkaSen. So I guess that's the trade-off.
 
Yep. And in MHB it's rule-violation to help with homeowork assignments with no context or attempt.
 
@BalarkaSen: Isn't M.SE like that as well?
 
MSE is not a forum.
 
@OlivierOloa: I am amused by how almost identical our reps/badges are right now (though I think you'll sweep past me very soon).
 
hai @TedShifrin
 
3:56 PM
@TedShifrin!
 
Hi ... Just dropping by for a sec. ....
 
@BalarkaSen!
 
@Parth!
 
How're you?
Are you done with exams?
 
Fine.
 
4:04 PM
@BalarkaSen For a solid book on general topology I wanna recommend you Willard's General Topology.
 
No thanks ;)
 
@BalarkaSen At least take a look at it.
 
How to show that the set $\{(x,y)|x,y\in \mathbb R \text{ and } (x^2-y^2)^2=x^2+y^2\}$ is bounded?
 
@Sush What have you tried?
 
@PedroTamaroff, I tried $(0,0),(1,0),(-1,0)$.
 
4:15 PM
But that is not bounded.
 
@PedroTamaroff, OMG! What are you saying!
 
That is not bounded.
In polar coordinates, write it has $r^2 \cos^2 2\varphi=1$.
It is clear it is unbounded now, yes? @Sush
 
4:34 PM
write it as a quadratic in y^2, compute the discriminant as 8x^2+1, which is always positive
 
Huy
I want to provide an optional course for pupils at my high school in mathematics. I only get 2 lessons (45 minutes each) per week and it should be something I can cover within half a year. The pupils just started learning some vector geometry and differentiation, no integration yet. Does anyone have some nice ideas?
 
@Huy Geometric series, complex numbers
 
Huy
@Alizter: Those topics are usually covered in the normal mathematics lessons already at some point.
(just a bit later)
 
number theory that usually isn't taught in schools?
such as modular congruence
 
Huy
I don't know anything about number theory. xD
 
4:43 PM
then teach them integation
solving cubics
limits
logic
 
Huy
Integration will be taught in standard curriculum as well.
There is already a course about logic/axiomatic method.
 
@Huy So because they are going to learn it you are not going to teach it?
 
Huy
I don't think solving cubics or doing limits is at all appealing to high school pupils.
 
Yeah so teach them about integration
 
Huy
@Alizter: It is an optional course, the appeal being to learn something you are not taught in the standard curriculum.
 
4:45 PM
talk about finding volumes
@Huy There are many areas that will not be taught in standard curriculum
but if the basics arn't down how are you going to show them anything interesting?
 
Huy
I know that but it should be a subject where they don't need to learn integration first if they don't know any yet.
 
number theory
geometry
ones that tease their knowledge
not a boring do this do that
Make them think outside the box
 
@anon why does positive discriminant impliy unbounded?
 
Huy
@Alizter: I need to find something nice in those areas. Geometry is a very broad subject. It needs to blow their minds and still have some kind of application, in my opinion.
 
Seems that people downvote 100k congratulaory post too, lol.
 
5:01 PM
@anon, $(x^2+y^2)^2=x^2-y^2$ also has the same discriminant $8x^2+1$, but is bounded !!!
Please help!
 
5:13 PM
@Alizter "modular congruence"?
i've heard of modular arithmetic and congruence both but admittedly never saw them stringed together. =P
@Sush why not try polar?
 
@BalarkaSen, what is polar?
 
$(x, y)=(r\cos(\theta), r\sin(\theta))$, $x^2 + y^2 = r^2$ and $x^2 - y^2 = r^2(\cos^2(\theta) - \sin^2(\theta)) = r^2 \cos(2\theta)$
@Sush Polar parameterization of coordinates.
Your curve is $r^2 \cos^2(2\theta) = 1$
And that is surely unbounded.
 
@BalarkaSen, frankly speaking, I know almost no trig.
 
That is bad. Learn some.
Polar coordinates is useful while dealing with these stuff.
 
@BalarkaSen, Which book is good?
 
5:18 PM
@WillHunting Yeah, that's the first time I've seen something like that.
Two close votes too!
 
@Sush I dunno, I've learned it from precalc textbooks.
 
OK!
@BalarkaSen, will you please help me with $8x^2+1$ argument?
 
nah, i don't get what anon was trying to say
 
@Huy: Projective geometry?
 
hmm!
 
5:41 PM
@ParthKohli Strictly speaking, it is off topic, lol. But no need to be so uptight, lol.
Guys, I am aiming for the Marshal Badge. I need 500 flags. I will flag the hell out of X, Y and Z!
Hey @TedShifrin do you have a favourite math book, just one?
 
Why are you so concerned about internet badges and points, @Will?
 
@Khallil It's just for fun.
@Khallil Why do people play games? Just for fun.
 
The two aren't the same.
 
It's the same. In a way, this site is also a game to me.
 
However, they are similar so I won't pursue this any further.
 
5:53 PM
Yeah, you sound overly concerned, lol.
@Khallil How are classes so far?
 
Yep, it's just been something that I've been thinking about for a while. I was in a similar position a short while ago. Classes begin in just over a week and I'll be getting there in just under a week. ^_^
 
I will spend Oct improving my English, Nov learning French and Dec learning German.
Then I will spend the whole of next year studying math.
 
Sounds like a plan. I'll try to improve all of them, concurrently, throughout the remainder of this year and the next.
 
$$\int_0^1 \left(\frac{\log(1-x) \log(1+x)}{1+x}\right)^2 \ dx =\space :-)$$
 
6:08 PM
0.221138
 
sage segfaulted when i typed it in
 
6:33 PM
2
Q: Derivation of power method

user96402POWER METHOD Let $x_0$ be an initial approximation to the eigenvector. For $k=1,2,3,\ldots$ do Compute $x_k=Ax_{k-1}$, Normalize $x_k=x_k/\|x_k\|_\infty$. Then $\|x_k\|_\infty$ approaches the dominant eigenvalue and $x_k$ approaches the corresponding eigenvector of the matrix $A$. My que...

 
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