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3:00 PM
you have pointwise convergence and the limit of the function sequence is continuous
he doesn't say that $f_n$ is continuous
 
this ssucks I don't undersatnd my group thoery notes
I think its because i have a cold
 
user19161
@user58512 Yeah you should take some meds and take a nap.
 
3:28 PM
hey
 
hi
 
I posted a question and got an answer but I still dont get it
0
Q: Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet

DaveShow that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet First, I showed that $(m^2 - n^2, 2mn, m^2 + n^2)$ is in fact a Pythagorean triplet. $$\begin{align*} (m^2 - n^2)^2 + (2mn)^2 &= (m^2 + n^2)^2 \\ &= m^4 -2m^2n^2 + n^4 + 4m^2n^2 \\ &= m^4 + 2m^2n^2 + n^4 \\ &...

 
I bought spinach but it tastes bad
 
@Dave that's what the comment section underneath answers is for
 
3:31 PM
I can't tell if its off or I just forgot what its supposed to be liek
 
although you're welcome to engage with chat as well
heh
 
yeah anon but if i post a comment it wont bump the thread so it wont go any further ?
 
@Dave, your calculation doesn't make sense
 
yeah i figured that user58512, I dont understand what im doing
 
$$\begin{align*} (m^2 - n^2)^2 + (2mn)^2 &= (m^2 + n^2)^2 \\
&= m^4 -2m^2n^2 + n^4 + 4m^2n^2 \\
&= m^4 + 2m^2n^2 + n^4 \\
&= 1\end{align*}$$
that's what you wrote
 
3:33 PM
yeah
I tryed to proove that
 
Greetings
 
Can anybody show me how to work out $$\int_{\gamma} \frac{ze^{i \pi z}}{z^2+4}dz$$ where $\gamma$ is the path $\gamma(t)=Re^{it}$?
 
if I apply a^2 + b^2 it equals c^2
and it works
 
okay, @Dave. Try expanding $(m^2-n^2)^2$ out first of all. What do you get?
 
$$\begin{align*} && (m^2 - n^2)^2 + (2mn)^2 - (m^2 + n^2)^2 \\
&=& m^4 - 2m^2n^2 + n^4 + 4m^2n^2 - m^4 - 2m^2n^2 - n^4 \\
&=& m^4 + n^4 - m^4 - n^4 \\
&=& 0\end{align*}$$
that's how to actually do it
to prove P = Q prove P-Q = 0
 
3:34 PM
@DominicMichaelis hi
 
or how to bound the integral $\int_\gamma |e^{i\pi z}||dz|$?
 
@Charlie hi sweetie
 
@SamuelHandwich what does the residue formula say for that integral?
@SamuelHandwich in that integral, what can the integrand be simplified to?
 
@anon sorry i mean for $0<t<\pi$.
 
@DominicMichaelis Excuse me... "sweetie" ???
 
3:35 PM
@DominicMichaelis how are you Dominichen?
 
erm, that second comment of mine may not be so helpful
 
@Charlie i am fine and you ?
@skullpatrol i talked to charlie
 
@DominicMichaelis ganz gut
 
That is nice to hear :)
 
@skullpatrol relax skull
At least he is being nice
 
3:38 PM
and I'm not?
 
@skullpatrol you are always nice :D
Don't worry
 
Thanx
 
@skullpatrol ;D
 
@DominicMichaelis Pardon my comment: Excuse me... "sweetie" ???
 
yeah i nevermind :)
 
3:46 PM
Hi @sabertooth did you loose your tiger?
 
@skullpatrol not quite
 
@sabertooth Welcome to the room Rarely if ever expressible as a ratio of integers.
 
@sabertooth did you find wolverine?
 
@Charlie wolverine? what is that
 
@sabertooth X-men, of course
 
3:55 PM
:p
 
@anon laugh all you want kid.
 
@Charlie my bad
 
Wolverine is the best, shut up
@sabertooth don't worry sabertooth doesn't like him, anyway
 
Hi! Is it true that any simply connected open set in $\mathbb{R}^n$ is contractible?
 
4:10 PM
without any topology lecture i would say yes
 
oh that's what "graded on a curve" means freakonomics.com/2013/02/20/how-to-game-a-grading-curve
 
@DominicMichaelis do you know how to prove it?
 
it has nothing to do with curves at all
 
@nimza the fundamental group is trivial
 
@DominicMichaelis how to prove this then?
 
4:13 PM
do you have any advice, what should I do wwhile I am too ill to study?
 
@DominicMichaelis for convex domains it is ok, but for general simply connected I have no idea
 
@user58512 Most Profs would just fail the entire class.
 
you only need star domains for the proof not convex ones
 
@DominicMichaelis yup, star ones are ok too
 
open and connected in mathbb R^n implies path connected
 
4:15 PM
aha
 
are we talking about a special topology ?
or any ?
 
@DominicMichaelis standard
 
ok than open implies that it's dimension is n too
open + connected
dimension of non connected sets are ugly
 
open+connected, what's next?
 
@TobiasKildetoft It is a nice question!
Very interesting!!
2
Q: Sudokus as composition tables of finite groups

Tobias KildetoftIf $G$ is a finite group then the composition table of $G$ is a latin square (ie, each row and column contains each group element exactly once). Assume now that $|G| = n^2$ for some natural number $n$. We can then split the composition table for $G$ into $n^2$ $n\times n$ squares, and we can ask...

 
4:21 PM
@awllower I thought so too. I am not sure how I came to think of it actually, as I almost never think about composition tables for groups
 
Haha
I think it could be formulated as certain conditions on the operation of the group. Per chance it could be related to representations?
 
probably not representations
 
Then we shall obtain a class of groups called "sudoku groups".
 
but certainly to the action of the group on itself
 
Then this is a representation.
right?
 
4:26 PM
sure, but it is not obvious why it helps to extend this action to some vector space
 
I see what you mean now.
But I think that representations might help in yet another way.
 
@DominicMichaelis I'm not sure it is true at all math.stackexchange.com/a/165529/16273
 
I guess it might come into play for the non-cyclic groups of order $p^2$ for a prime $p$
since those are vector spaces in a natural way
 
yes
And maybe one can relate somehow the composition table and character table?
 
if it is simply connected there aren'T any holes in or ?
 
4:30 PM
At least the "sudokability" of the two tables?
 
@awllower the character table will never be sudokable
 
because of the trivial character?
 
@DominicMichaelis no, why? what about solid torus?
 
@awllower yeah
also that a lot of irreducible characters tend to vanish a lot of places
or have non-trivial kernels
 
@DominicMichaelis ah yep, not simply connected
 
4:32 PM
Yes
So we shall think about a way of transforming "sudokability" to some other properties.
 
@awllower we get a lot of pairs of subsets $A,B$ each of size $n$ such that $AB = G$
 
mh i don't know how to proof sry
 
So?
@TobiasKildetoft I mean how is this related to the discussions?
 
@awllower to what discussion?
 
to the sudokability of groups
 
4:38 PM
the existence of those subsets is a consequence of sudokability
 
Very well!
 
I guess one can actually get the sudokability from the existence of such subsets, if we just get the conditions right
 
We need disjoint pairs A_i, and B_i, with A_iB_i=G
BTW, is this answer too long?
http://math.stackexchange.com/questions/306675/legendre-symbol-what-is-the-proof-that-it-is-a-homomorphism/306751#306751
 
@awllower, yeah the connection with permutation is amazing, but I don't quite understand it
 
Which part?
 
4:43 PM
Zolotarev
 
Oh
Actually that construction is easy, if one knows that there is only one non-trivial homomorphism from Z_p^* to {-1,1}
Hm, wait: I am not so sure about this...
In any case, one could proove the construction by use of a primitive root modulo $p$.
 
@awllower yes, there is a unique such homomorphism, since the domain is cyclic, so it has a unique subgroup of index $2$ (and a non-trivial homomorphism must be surjective)
(but of course, that uses the existence of a primitive root)
 
Very well: I had it right.^^
BTW, a saying in the book on finite groups by IMIssacs is quite interesting:
representations: homomorphisms to GLV
old group theory: homo to S_n
transfer: homo to subgroups.
I wonder if it is true...
http://math.stackexchange.com/questions/309251/is-it-true-that-the-book-calculate-primes-has-found-the-pattern
 
Why is there an upvote on that question?!
 
I thought this question was interesting?
 
5:00 PM
Hi all
 
Hi
 
hi
 
Excellent work!!the answer by robjohn:
http://math.stackexchange.com/questions/309023/prove-that-if-an-mid-bn-then-a-mid-b/309025#309025
 
I'm trying to elementarily compute $\lim_{n\to\infty}\left(\int_0^1 \frac{1}{1+x^n}\,\mathrm{d}x\right)^n$. (high school question)
 
I am sleepy now.
Later everyone!
 
5:08 PM
You keep using that phrase, "high school question," 1.) as if it's true (it's not), and 2.) as if it matters.
 
@Arkamis: do you suggest I'm a lier?
 
the limit is clearly 1
 
I suggest that you need to re-calibrate your estimation of difficulty and/or the talents of high school students.
While a question like that is most certainly accessible to some high school students, those students are exceptional.
And a question of that difficulty is well above the average difficultly level of a true high-school level problem. Even though it might not require techniques which are, individually, more complicated than what a typical HS calculus student knows, it is the comprehensive understanding that becomes more difficult
Fundamentally, driving a Formula 1 car is the same combination of actions as driving my SUV; but Monza is not an every day commute.
 
seems like it's actually 1/2
 
A high school student that can solve that problem is definitely advanced in her knowledge, and upon attending university will almost certainly start at least 1 or 2 classes ahead of her peers.
 
5:14 PM
@Chris'ssisterandpals, do you have a non-elementary way to do it?
how do you turn it into a n-variable integral?
 
@Arkamis: actually this problem is created by my brother and guess what? The requirement to me is to evaluate it by only using high school knowledge and NO PEN AND PAPER. Now, if you say again that I'm a lier then I understand you.
brb
 
@Chris'ssisterandpals That means nothing. As I said, driving a formula 1 car requires only pressing a gas pedal, shifting, and steering. That doesn't mean it's a car that a high-schooler can drive.
 
@Chris'ssisterandpals, would $$\left(\int_0^1 \frac{1}{1+x^2}\,\mathrm{d}x\right)^2 = \int_{[0,1]^2} \frac{\mathrm{d}x\mathrm{d}y}{1+x^2+y^2}$$?
 
@user58512: I don't know what you did there.
 
@Chris'ssisterandpals, it mighr not be right
I just want to know how to turn it into a multidimensional integral
 
5:27 PM
@user58512: yes, the limit seems to be $1/2$.
 
I have no idea how to show it
this is a very interesting problem
 
@user58512: hold on! Chris has some thousands of such problems! I'm thinking to make a mega compilation, but this will take a lot of time (and for many problems I don't have solutions yet). I love so much these little wonders.
 
that would be nice, especially when I have some more skill at solvinng them
 
What do we need here?
@Chris'ssisterandpals Oh, OK.
@Chris'ssisterandpals Would L'Hôpital count as High School?
 
@PeterTamaroff: we learned L'Hôpital in the penultimate year in high school.
 
5:40 PM
@Chris'ssisterandpals Good.
And Leibnit'z Rule?
 
Also Leibniz's rule. (the last year)
 
I think you can tweak the integral into some trigonometric form
Using $\tan$
 
Yes, I think we can do it. Do you think this helps? (let me check something)
 
I'm still working on it, FYI @CHris
 
Nice! :-) Sister here.
 
5:51 PM
We have that with $x\mapsto x^{-1}$
$$\int_0^1 \frac{dx}{1+x^n}=\int_1^\infty \frac{x^{n-1}}{1+x^n}\frac {dx}x$$
@Chris'ssisterandpals Ah?
 
OK, and then?
 
I'm thinking =)
 
@Chris'ssisterandpals Ok, we all agree that $\to \frac 1 2$ is the correct answer, yes?
 
user19161
5:57 PM
"Ad hominem" is too hard for me. I avoid Latin at all costs.
 
@PeterTamaroff: we don't know this at the beginning. The problems doesn't tell us the value of the limit. This values is what W|A says.
 
@Chris'ssisterandpals Well, we are good guessers.
What does W|A say?
 
It says that the value of the limit is $1/2$.
 
how do you know the answer is 1/2
 
It's very simple.
Take a finite value in W|A for $n$ and see what values W|A approximates. Well, a larger value for $n$.
 
6:00 PM
OK, another "solution" is to use an infinite series.
 
this is such a difficult problem
 
hi folks, I have a question.
 
Integrate[1/(1 + x^20), {x, 0, 1}]^20=0.514220...
 
Is it ok to show the copyrighted material in SE as in tex.stackexchange.com/questions/99124/… ?
 
$$\sum_{k=0}^\infty (-1)^k \frac 1 {kn+1}$$ is your integral
 
6:01 PM
@GarbageCollector, that is very likely acceptable under fair use
 
@GarbageCollector Yes, that's fine.
 
@MarianoSuárez-Alvarez Hola!
 
@PeterTamaroff: how did you get that form?
 
@user58512 Confirmed. thanks.
@Arkamis Ok. Thanks
 
$$\frac{1}{1+x^n}=\sum (-1)^k x^{nk}$$
@Chris'ssisterandpals
 
6:03 PM
@PeterTamaroff: Ah, got it.
 
how does that help?
this doesn't converge for x=1
 
@user58512 Deal with it, bro.
@Chris'ssisterandpals I might have another idea.
Let's find antoher method that doesn't work! =)
 
hehe, let's find it :D
But hold on!
This way you told isn't bad. It may work. I need to work on it a bit more.
 
I think findiing tight bounds and use squeeze would be great.
But it is kinda tough.
 
Agree. To find tight bounds enough seems a difficult task here. I'm trying to see something different. (back a bit later)
 
6:20 PM
@Chris'ssisterandpals Hey.
Using the series expansion looks kinda promising.
 
@PeterTamaroff: yeah, it works!!! Thanks!!!
 
It did?
 
Back immediately.
@PeterTamaroff: the point is to make the guess that $\lim_{n\to\infty}\int_0^1 \frac{dx}{1+x^n}\, \mathrm{d}x=1$. That is the case $1^{\infty}$. Then we may use the limit $\lim_{x\to0} (1+x)^{(1/x)}=e$
 
6:24 PM
@Chris'ssisterandpals Yes.
@Chris'ssisterandpals OK.
Oh, wait.
Our integral is $=-\log 2 +\text{blah}$
If we show $\text{blah}$ goes to $0$ quickly enough, we might be done.
 
Then all gets reduced to computing $\lim_{n\to\infty}n\left(\int_0^1 \frac{dx}{1+x^n}\, \mathrm{d}x-1\right)$
 
@Chris'ssisterandpals Yes.
 
but $1=\int_0^1 dx$
 
omg!
that's clever!
 
Then $\lim_{n\to\infty}n\left(\int_0^1 \frac{-x^n }{1+x^n}\, \mathrm{d}x\right)$
At this point we employ that geometric progression sum and we are doneeeeeeeee!!!!!!!!!1
:)))))))))
 
6:28 PM
@Chris'ssisterandpals You have two $dx$s.
@Chris'ssisterandpals Geometric progressions?
 
Yeah, that geometric progression sum you talked about.
 
integral_0 to 1 of 1 + x + ... + x^{n-1} dx
 
I think I can show that $\int_0^1 \frac{dx}{1+x^n}-1=-\log 2+\text{blah}$
 
Actually, it remains to compute $\lim_{n\to\infty} \sum_{k=1}^n} (-1)^k \frac{n}{k n +1}$ that is $-\ln 2$. Thus the answer is 1/2.
 
@Chris'ssisterandpals $1/2$!
@Chris'ssisterandpals Good one.
@Chris'ssisterandpals I can't follow where you get to this.
 
6:34 PM
Just a typo.
@PeterTamaroff: I interchanged the sum and the integral.
 
NBP
If a question I asked got an answer but not to my satisfaction how can I resurrect it withotu reposting? as it seems to be ignored even though I commented that the answer is not satisfactory
 
@NBP you can add a bounty to it
 
@Chris'ssisterandpals No, no, look to which message I pinged to.
 
NBP
how do I add a bounty?
 
@NBP Don't do that. Just bump it, and be patient.
 
6:36 PM
hmm, that might require 100 rep actually
which question is it?
 
NBP
bump it by commenting again? it would be my second consecutive comment
 
@NBP No, edit it.
 
NBP
it's a simple concept intuitively though the answer provided is lacking in my opinion
 
@NBP Done.
 
NBP
Thanks Peter
 
6:37 PM
@NBP give the answerer a chance to see the comment and expand on his answer
in internet terms, 13 hours is not long
 
what an utterly horrible problem, how could you possibly bring yourself to care about it?
 
NBP
Alright, sorry.
user58512, It's a simple concept from my intorudction to probability course
 
Peter, at the beginning I used $$\lim_{n\to\infty} e^{n\left(\int_0^1 \frac{dx}{1+x^n}\, \mathrm{d}x-1\right)}$$
 
NBP
if I fail to grasp it, I fail at being a maths student in my opinion, so I care.
 
@NBP, why do you even say 100 balls when you only care about the first 3? putting more balls in doesn't change anything
 
6:39 PM
@Chris'ssisterandpals Oh, well.
 
NBP
user58512, what if I asked about any ball between the first and second balls?
the ordering of the balls is insignificant to the question
the probability space being insignificant is something I can't see
 
@Chris'ssisterandpals Striclty speaking, we need to justify the integral and sum swap.
 
Yeah, this shouldn't be a problem.
 
Can you justify it? =P
 
@NBP I've commented on that post.
 
NBP
6:43 PM
Thanks Arkamis
seems clearer now
I just redefine a probability space basically
excluding all the other balls
 
@Chris'ssisterandpals I think I have a nice idea.
 
Abel's theorem? It could be. I'll do this when I put it on paper.
@PeterTamaroff: what other idea do you have?
 
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = \int\limits_1^\infty {{{n{x^{n - 1}}} \over {1 + {x^n}}}{{dx} \over {xn}}} = {{\log 2} \over n} + \int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right)} \over {n{x^2}}}} $$
 
NBP
Arkamis, why is the probability still uniform though?
 
So that.... $$n\left\{ {\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} - 1} \right\} = \log 2 + \left\{ {\int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right)} \over {{x^2}}}} - n} \right\}$$
 
6:49 PM
I don't see what you mean/why it's relevant.
But one answer is "the relative position of balls 1, 2, 3 is independent of the initial distribution of balls"
 
@PeterTamaroff: that looks nice. :-)
 
There's nothing uniform in this distribution; you're asking what the probability of an event is -- essentially a simple Bernoulli trial.
You have only two events. Event 1: the ball sits between 1 and 2; Event 2: the ball does not sit between 1 and 2
If this was uniform, both events would have probability 1/2. But they don't, so it's not uniform.
 
NBP
right, why does event 1 get the probability 1/3 though
it's because the balls are distributed randomly
 
Distributed randomly does not mean uniform
Ignore the other 97 baskets for now -- we've already agreed we can push them off a cliff without changing anything
The ways of lining up the balls are:

1 2 3
*1 3 2*
2 1 3
*2 3 1*
3 1 2
3 2 1
 
NBP
alright, so I pushed them off a cliff. can I say about the 3 remaining baskets that the balls in them are distributed randomly too?
I understand this Arkamis, I just wonder why we assign all omegas equal values
 
6:54 PM
Ah -- because each of the combinations enumerated above is equally likely
 
Athos, Porthos, @Arkamis and Dartagnan
 
NBP
right, why is that so, is my question
 
@Charlie That is actually the origin of this name ;)
@NBP Because each basket can only have one ball.
 
NBP
I know the initial ordering with the 100 baskets had equally likely arrangements, I'm just wondering why it reflects towards our case that they are also equally likely
 
@PeterTamaroff: thank you for your precious work! I need to leave right now.
 
6:55 PM
@Chris'ssisterandpals I think I have something.
@Chris'ssisterandpals Oh, noes.
Let me just tell you this last bit
 
@NBP because putting balls into baskets is independent of the number of baskets/balls, if each basket only has one ball inside it
 
We got to
$$\log 2 + \left\{ {\int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right)} \over {{x^2}}}} - \int\limits_1^\infty {{n \over {{x^2}}}} } \right\}$$
Since $\int_1^\infty x^{-2}dx=1$
 
@Arkamis I know things ;)
 
Now, for very large $x$ $\log(1+x^n)\sim \log(x^n)=n\log n$
 
6:57 PM
There's no connection between the original 100 balls and the 3 balls
None whatsoever.
 
So we're looking at
$$ \sim \log 2 + n\int\limits_1^\infty {{{\log x - 1} \over {{x^2}}}} dx$$
 
NBP
alright Arkamis
I got it now, thanks.
 
But that last integral is zero.
 
@PeterTamaroff: right. (hmmm ...)
 
I think we can make it rigorous with some Taylor expansion of $\log(1+x^n)$
Using the error term.
 

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