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11:10
Uh, I just thought of something . . . how can one prove that at least four triangles are needed to form a closed polyhedron in 3-space?
I mean, it's "intuitively obvious", but . . .
It seems L.G. came up with the closed form to one of my integrals, but no complete brilliant way is revealed though.
13
Q: Evaluating $\int_0^1 \frac{z \log ^2\left(\sqrt{z^2+1}-1\right)}{\sqrt{1-z^2}} \, dz$

Chris's sisWhat kind of real analysis tools would you employ for this integral? $$\int_0^1 \frac{z \log ^2\left(\sqrt{1+z^2}-1\right)}{\sqrt{1-z^2}} \, dz$$ EDIT: Here is a supplementary question, the cubic log version $$\int_0^1 \frac{z \log ^3\left(\sqrt{1+z^2}-1\right)}{\sqrt{1-z^2}} \, dz$$

Maybe I should change my username and put Chris's sis the artist!:-)
@robjohn did you manage to work on that series of mine I showed you yesterday?
Changed.
11:27
@pjs36 Thanks for the link, this is exactly what I was inquiring about.
@TedShifrin I will look at your recommended book, thanks.
Done.
It's weird the username didn't change though ... Ah, great! :-)
$$\int_0^1 \frac{z \log ^2\left(\sqrt{1+z^2}-1\right)}{\sqrt{1-z^2}} \, dz$$
$$=2G-\frac{13\pi^2}{48}+\frac{\pi}{2}+\frac{\ln^22}{4}+\frac{\pi\ln 2}{4}-\ln 2+2$$
I like $G$ some more for denoting the Catalan's constant.
@SohamChowdhury how do we finish it in the spirit of the art?
Finish what?
Isn't that the answer?
@SohamChowdhury Well, it's just the final answer, but not the way to go. Using some stuff like Mathematica, but I need a brilliant work by hand.
(Also, like I would know anything about these integrals. I doubt I can integrate $\sqrt{\tan x}$.) :P
@SohamChowdhury ;)
11:37
I'm waiting for the day when you meet your own four-color theorem ;)
@SohamChowdhury lolllllllll :D
I'm writing a solution to a magazine now (it's actually a proposed problem).
BBL
Hi :)
I have a question about functionals
I want to show that the continuity for the weak norm does not ensure the continuity for the strong norm. (Hint: functional of arc length)
Could I show someone what I have tried?
The functional is this: $J(y)= \int_a^b \sqrt{1+y'(x)^2} dx$, right?

I have tried the following:

If $y_1, y_2 \in V$ then:

$$|J(y_2)-J(y_1)|= \left| \int_a^b \sqrt{1+y_2'(x)^2} dx- \int_a^b \sqrt{1+y_1'(x)^2}dx \right| = \left| \int_a^b (\sqrt{1+y_2'(x)^2}- \sqrt{1+y_1'(x)^2}) dx \right|= \left| \int_a^b \frac{y_2'(x)^2-y_1'(x)^2}{\sqrt{1+y_2'(x)^2}+ \sqrt{1+y_1'(x)^2}} \right| \leq\int_a^b \frac{|y_2'(x)^2-y_1'(x)^2|}{|\sqrt{1+y_2'(x)^2}+ \sqrt{1+y_1'(x)^2}|} dx \leq \frac{1}{2} ||y_2'-y_1'||_{\infty} (b-a)$$
12:06
Anyone can help me to answer my question? I don't have any reps to make bounty this. math.stackexchange.com/questions/1236075/…
I'm lost with my books and getting lost too far to learn :-(
I think the question is a bit confusing for people to answer, but honestly, If I re-read my question I don't find any difficult words used.
12:54
@evinda: did you work out the dim, analysis for the ODE yesterday? I had to wander off
@Ridho Try on Stats.SE if you don't get answers here.
@SohamChowdhury Should I flag to moderator for migrate request to Stats.SE or just make new question in Stats.SE?
You could make a new question there and edit the questions to reflect that you've posted it on two different sites.
@SohamChowdhury I will try. Thanks for reminding me about Stats.SE
hello,everyone
13:07
@Chris'ssistheartist I'm sorry... I was preoccupied with a lot of things on and off-line, I will have to look at it later today.
@robjohn No hurry with that.
nice :-)
theres matlab code which explains what i was asking
lim=50;
syms x;y=0:0.1:lim;
f=@(x) x^2-x; g=y.^2-y;

h=plot(y,g);
for j=10:-0.1:0.1
c=0;
for i=2:j:lim
c=c+1;
b=f(i);a=j
v(c)=rectangle('Position',[i 0 a b]);axis([0 lim 0 b]);
end
pause(0.1);
for i=1:c
delete(v(i));
end
end
delete(h);
pause(1);

plot(y,g);%ezplot(f(x));
hold on;
for j=1:-0.01:0.01
c=1;ii=0;
for i=2:j:lim
if (ii>lim), break; end
ii=ii+i;
b=f(ii);a=(i+j)
v(c)=rectangle('Position',[ii 0 a b]);
axis([0 lim 0 b]);c=c+1;
end
pause(0.1);
for i=1:c-1
delete(v(i));
end
end
13:26
@Chris'ssistheartist I was looking for the series you mentioned, and I can't find it. Do you have a link to it?
r9m
r9m
@Chris'ssistheartist This integral is epitome of devilry =P I just realized where the root of the evil lies :P lol Awesome!! :-) It's unlikely anyone will notice the mischief unless they know Knuth's series and a certain Harmonic series :P
$$\sum _{n=1}^{\infty } (-1)^{n+1}\frac{1}{(2 n+1)n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\cdots+(-1)^n \frac{1}{2n+1}\right)$$
$$=\frac{7}{4}\zeta(2)-4+2\log(2)-\frac{\log^2(2)}{2}$$
@r9m lolllll :-)
r9m
r9m
@Chris'ssistheartist how did you think of arriving to the integral in the first place?! :O I'm shocked!
@r9m Tell ya one day, not here. :-)
@r9m Shocked? :D
r9m
r9m
@Chris'ssistheartist ya! I lack an exact word to the feeling I have :P the closest thing seems to be shock! =P
13:32
@r9m I feel that integral can be calculated in a very nice way! :-) The same thoughts for the supplementary integral. :-)
r9m
r9m
@Chris'ssistheartist I know it can be calculated in a very nice way !! :-) But Idk about the supplementary problem though ..
@Chris'ssistheartist when and where? please! I am dying to know what goes on in that devious brain of yours to create such twisted problems :P
I'll try and write a solution after I'm done with the logic assignment at hand :| (It's killing me)
Is a neighborhood of $x\in S$ defined as a subset of $S$ containing a closed nonzero-radius ball centered on $x$ which is entirely in $S$, or an open ball of this kind?
@r9m :D Also your brain is devious in terms of mathematics, in a very good sense I mean. :D
@r9m Logic assignment?
r9m
r9m
@Chris'ssistheartist me devious? not even close! :P
Nooooooooooo :-)
r9m
r9m
13:37
@Chris'ssistheartist I have homework problems from Math Logic course (assignment)
@r9m Ah, I see.
i had also tortuous thinking when solvin mathematic problems :p
0
Q: for any $k$,there exist set $B$ (be a permutation of $A$),such $B+A=A$

inequality Define $A+B=\{a_{i}+b_{i}|1\le i\le n\}$,where $A=\{a_{1},a_{2},\cdots,a_{n}\},B=\{b_{1},b_{2},\cdots,b_{n}\}$, Let $B$be a permutation of $A=\{-3,-2,-1,0,1,2,3\}$ I found there exsit $B$ such $A+B=A$. For eaxmple:$$A=\{-3,-2,-1,0,1,2,3\},B=\{0,1,3,-2,2,-1,-3\}$$ since $$-3,-2,-1,0...

thats rieman seies with equally wide strips
r9m
r9m
13:45
@robjohn the wiki link for Fibonacci polynomial in Jack's answer didn't have the closed form .. I just figured it's like the alternating Chebyshev 2nd poly .. :( so my approach is not so different from Jack's answer after all !! :(
r9m
r9m
14:00
@Chris'ssis when you said about generalizing Knuth's result .. did you have any premonition about what form the generalization could take?
@r9m Premonitions? :-) No, I only have some ideas that might help me to look deeper at a possible generalization ...
@SohamChowdhury Neighbourhoods are open balls.
r9m
r9m
@Chris'ssistheartist do you think we can generalize the result from catalan number to some sort of generalized catalan number identity? There are only a few things common between the catalan and its generalized version! .. I'd be interesting to see if those properties carries over to the series as well! :D
@r9m First I need to study the matter more, it's too early to say more. :-)
What are you two working on?
r9m
r9m
14:07
@Chris'ssistheartist 'kay! I was just randomly throwing some ideas :P
@JoshuaA us?
@r9m Yes.
@r9m The generalization idea has something to do with the supplementary integral of mine (I posted on main). :D
thats rieman sums with gradually larger strips
r9m
r9m
@JoshuaA we are trying to generalize the identity here .. it appeared as problem 11832 in April 15 issue of AMM.
@r9m Interesting. For what purpose?
r9m
r9m
14:12
@JoshuaA purpose? idk! just for the fun of it I guess :P ;)
@JoshuaA This is the worst kind of question, really. :-)
@r9m Oh. That's impressive xD
r9m
r9m
@Chris'ssistheartist :P I was asked once in an interview .. 'Why do you like mathematics?' :P
@r9m What answer did you give?
@r9m Ooooo, I almost have only terrible experience from interviews in terms of questions. :-))))))
@r9m Once someone gave me to solve a problem during an interview, and I said: "If you have nothing against, while I'm thinking of it, you can have fun with problem $X$" :-)
@r9m Of course, if I call them now they have no idea how to solve it not even today. :-)
r9m
r9m
@JoshuaA I was honestly dumbstruck at the moment .. :P I didn't know what to say .. had it been an informal situation I was willing to ask back 'why do people have sex?' .. but figured that'd be too rude for an interview in a prestigious institute :P
@r9m LOL. That would be an interesting one :).
@r9m What did you answer?
@r9m I'd answer simply: it defines me.
because is very funny?
maybe?
r9m
r9m
14:19
@Chris'ssistheartist that's a strong reply!!! I don't have that much self-confidence!
all what you like is about pleasure
@r9m Well, the truth is that in a crazy world mathematics may give an amazing beautiful sense to your life! Of course, I'm a lover of art and beauty, that's why I love mathematics so much.
5
I quick question guys... can you see this? math.stackexchange.com/questions/1309199/…
r9m
r9m
@Chris'ssistheartist I fumbled out some sorry excuse of a reply :p it was 3 years ago! I don't remember all that much ..
@Chris'ssistheartist That's a nice reply! :-) I like it (+1)
@r9m Most of the people that know almost nothing about mathematics probably imagine that doing mathematics is a kind of torture of one's brain, I tell you that from my exprience. I don't say it as an expert, no, not at all, but as someone that realized how amazing the beauty of mathematics can be. I'm amazed almost every day looking at its beauty and never get enough of it.
r9m
r9m
14:26
@Chris'ssistheartist :D
@Chris'ssistheartist, you says something very important: beauty... the aesthetic perception is one of the basis of intelligence... only aesthetic can move people from a not elemental necessity
of course math is beatiful :) ( as u keep a safe distance from insanity )
@Agawa001 I tiny bit of insanity may be welcome anytime! :-)
in reality @Agawa001 I dont study mathematics on university cause I had a bit of fear, not because insanity more like "I need to focus on something material"
14:32
yes , that was same with me , its like u fear your own fate :) but you cant ever avoid it , bein a math-spirited
i did computer science , and im blaming myself in dailybasis for going against stream
yeah @Agawa001, similar here, I studied biochemistry but a really like more maths
That tiny bit foregoing gives you the imbold to create integral problems like $$\int_0^1 \frac{z \log ^3\left(\sqrt{1+z^2}-1\right)}{\sqrt{1-z^2}} \, dz$$ :D
@JoshuaA Not a set containing an open ball around an element?
anyway I need to know about basic system of life :)
@SohamChowdhury What sorry? A neighbourhood is usually defined as an open ball of some radius centred on the point. $N_r(p)=\{x\in X| d(x,p)\lt r\}$
14:36
Oh. Bredon defines it as a set containing such an open ball.
maths wont let you down @Masacroso , trust me, atleast thats what i believe!
So in Bredon, he defines it $\leq r$?
No, wait.
the first definition is for metric spaces, the definition from sets is more general, is a topological definition @JoshuaA
"If X is a topological space and $x\in X$ then a set $N$ is called a neighborhood of x in X if there is an open set $U \in N$ with $x\in U$."
Actually, Armstrong defines this with closed sets, which was the cause of my confusion. I guess it doesn't matter, though, because most of the definitions don't change if you take complements of everything.
14:39
Indeed. I assumed you were referring to metric spaces, due to referencing balls.

In general topology we merely care about having all the elements in the topology explicitly - or having some basis element containing it, within the neighborhood.
I.e. if you want $U\subset X$ to be open, you can think about every element $x\in U$ being covered by tiny balls $B_i$, that are contained in $U$.
That is the metric space connection.
@SohamChowdhury maths.kisogo.com/index.php?title=Open_set it's one of the older pages, so feedback
@SohamChowdhury i just need to know whether second rieman segmentation is applicable like the first one
goddamit someone should do something about image uploading , it takes light centuries to get to server side
r9m
r9m
15:06
@Chris'ssis I just received a reply to my query from amm .. ' All acknowledgment of correct solutions is made at the time of publication for the problem's solution.' :-) seems we have to wait that long :P
@AlecTeal I like how you use phrases like "puff up". Aids the intuition, I guess -- but once you have it, you prefer writing long strings of quantifiers! :P
@BalarkaSen @AlexC Working through Bredon is slowly doing wonders for my $\epsilon$-manipulating skills. (I guessed you'd be interested, forgive me if that's not the case.)
was it a dumb question to ask ?
r9m
r9m
@JoshuaA I posted a link (comment) of my evaluation of the cubic euler sum you asked on main! lemme know what you think of it after you've checked it .. :-)
15:22
@r9m I wanna have a long summer! No hurry to that moment! The winters are not that friendly here, and I'm sick and tired of them! :-)
r9m
r9m
@Chris'ssistheartist :( .. that means I'll sit through the summer with a long face :P
r9m
r9m
BBL .. dinner time! :-)
15:36
@Chris'ssistheartist lol, i like winter , even tha harshest days ! i dont think i can process any thought of anything with blood boiling inside my head cuz of heat :D , i cant even mind my birthday
what do you think about that @Agawa001? math.stackexchange.com/questions/1309199/…
:p
@Agawa001 if you live in the city then it may be nice ... :-)
@Masacroso , how is column and row swapping goes like ?
you can change any column by other column, same for rows @Agawa001
changing column 1 by 3 and viceversa, by example
@Chris'ssistheartist i live in a village which is bein object of urbanisatin these last three years
so im gonna discover that feeling
15:46
is $\Bbb Q$ separable?
so if u swap any couple of columns ar rows , they remain same matrices ?
guten morning
I have a question I wanted to ask while reading apostol
yes @Agawa001, under any number of swaps
for stereographic projection where do points on the north pole gets mapped to
?
@Masacroso that should be either too easy question , or too hard question
:D
15:53
Could you take a look at my question?
0
Q: Proof that $f(x)=0 \forall x \in [a,b]$

evindaLemma: If $f \in C([a,b])$ and $\int_a^b f(x) h(x) dx=0 \ \forall h \in C^2([a,b])$ with $h(a)=h(b)=0$ then $f(x)=0 \ \forall x \in [a,b]$. Proof of lemma: Suppose that there is a $x_0 \in (a,b)$ such that $f(x_0) \neq 0$, for example without loss of generality we suppose that $f(x_0)>0$. Bec...

I think is like "too hard" @Agawa001, is the same problem changing matrices by complete graphs trough rows (or columns)... or something similar :p
searching isomorphisms of these graphs
If infinitely many (Greatest Common Divisor type von Mangoldt function) matrices share some common eigenvalues, what does it say about the matrices?
theres two matrices where it doesnt change anything when u swap whichever column or row u want
$\left(
\begin{array}{ccc}
0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0
\end{array}
\right)
$ or $\left(
\begin{array}{ccc}
1 & 1 & 1 \\1 & 1 & 1 \\ 1 & 1 & 1
\end{array}
\right)
$
but ...
if u want any configuration of any two column/row swaping where the matrices stay identical , thats hard
no, no @Agawa001... the matrices dont stay identical, you can transform one into another by swapping... this is what I want
ah ok
16:01
you have a matrix then you swap rows or columns all that you want and you get isomorphic matrices (under swapping)
then exist a number of really different matrices under swap transformation, so they arent isomorphic... the number of these matrices is the number of equivalence classes for the swap transformation
in the case of squared (0, 1) matrices it "maybe" possible to evaluate or something
@Agawa001 I feel myself pretty isolated during the winter. I actually am isolated here, but, well, I might say I've got used to that to a certain extent.
where are you @Chris'ssistheartist??
what is "ro"?
ok, ok
there are beautiful women in Romania :)
@Masacroso Indeed! :-)
this ia all about Romania I need to know :p
joking
@Masacroso hahaha, good! :D
16:08
being isolated from people for awhile , is good for brain , and being isolated from women is waaay better @Masacroso
and im neither misogynic nor misanthropic
r9m
r9m
@Masacroso true! indeed :P ;)
@Chris'ssistheartist btw I got all 5 problem and solution section of amm Jan to May this year from mixedmath !! :-)
@r9m AWESOME! :-)
r9m
r9m
@Chris'ssistheartist will I send them to you? :-)
@r9m Can you? Yeah, please! :-)
how old are you guys? Im 37yo so for me is "normal" evade boring people :)
r9m
r9m
16:14
@Chris'ssistheartist mixedmath is awesome! he sent me an issue of crux last year too :D
@r9m I'm sure he's awesome! :-)
r9m
r9m
@Chris'ssistheartist lol .. seems I have forgotten your email address :P (I have forgotten the username of the dummy gmail account I made to comunicate with you too :P) can you send them to me in private? :-)
@r9m May I remind you your username in private?
r9m
r9m
@Chris'ssistheartist okay! :) sure (that'd be great)
i was at work yesterday , and i was struggling to find a mathy sloution for some puzzle, my boss said : stop playing these mindteasers u have got work to do , and thats one of many reasons i avoid people who mistake math for any crossword grid :p
r9m
r9m
16:26
@Chris'ssistheartist thanks! got it :D
@r9m Did you see it?
@r9m :D
r9m
r9m
@Chris'ssistheartist ya saw it ... :) I was confusing my account name with the login id :P poor me
@r9m hehe. It happens sometimes. :-)
r9m
r9m
@Chris'ssistheartist lemme know when you get them :)
@r9m Tons of thanks!!! :-)
r9m
r9m
16:40
@Chris'ssistheartist :-) no problemo :)
@r9m 11685 is delightful (in the first paper), by Knuth again. :-)
r9m
r9m
@Chris'ssistheartist but it's well known result too :)
@r9m Then I wonder why Knuth proposed it and why AMM accepted. :-) It's possible I met it before but I don't remember.
:)
:-)
r9m
r9m
@Chris'ssistheartist 11685? or 11832?
16:54
@r9m 11685 It's just a telescoping stuff there, but I don't remember I met that before. It's a cute question though, I also posted it here some days ago or so.
r9m
r9m
@Chris'ssistheartist idk! but it sure is fun :)
@Chris'ssistheartist Your name's being shortened to "Chris's sis the".
@SohamChowdhury "Chris's sis the artist" :-)
r9m
r9m
There is a user named The Artist ..
17:09
is he the artist @r9m
r9m
r9m
@KarimMansour ? that user indeed used to come here in chat at one point in time .. idk anything else apart from that
oh
@r9m are you familiar with stereographic projections ?
an artist on crisis
r9m
r9m
@KarimMansour not exactly .. I have read about it on wiki though
because I am reading in apostol in intro to complex analysis like they say this as a way of visualizing the complex numbers
but now what about point on north pole itself how will they be projected?
all points on north poly will project onto 1 point
so its not injective maybe my understanding of this is wrong
17:14
@r9m how about Ovidiu's problem on that page? 11682
Hey guys I have a quick question for anybody that may have purchased or seen the Chinese edition of Milnor's "Morse Theory" in the US. Is it written in English with more modern typesetting or no?
@KarimMansour stereographic projections are not defined at the north pole.
It's defined everywhere else.
(I think)
r9m
r9m
@Chris'ssistheartist ya its nice :-)
17:24
How can I prove that $\frac{x}{e}+(\ln(x)-1)^3-(\ln(x)-1)^2+(\ln(x)-1)<x,\forall x\in I\subset V\subset\mathbb{R}$
?
What's I?
What's V?
I= open interval
which one?
and V=is a set
I = (0,1)?
So why is V there?
17:29
soham I= can be any open interval
This question is missing info.
It doesn't make sense.
as long as $x\in I$
is there any way to write $\sum_i {(\frac{1}{2})^i (x+\frac{1}{2}+...+(\frac{1}{2})^i)^2}$ in term of integral ?
@SohamChowdhury V=is a closeness of $\mathbb{R}$
thank you @SohamChowdhury
17:32
@Lucas I'm not sure what you mean by "closeness", @Lucas.
vicinity
thats another formulation of my last question
a vicinity of $\Bbb R$?
Again, I can't understand what you mean. :(
@SohamChowdhury we need to find an open interval I such that $x\in I\subset\mathbb{R}$
17:33
have an ideea how we can prove $$\frac{x}{e}+(\ln(x)-1)^3-(\ln(x)-1)^2+(\ln(x)-1)<x$ ?
@Lucas to start with, you can write $x=e^u$
can you give me more details ?
it can be rewritten as $v+v^2+v^3<e^{v+1}-e^v$ for instance ($v=u-1$ and $x=e^u$)
which at least looks easier to work with
u can start with $ln(x)<x $
r9m
r9m
17:40
@anon $v-v^2+v^3 < e^{v+1} - e^v$ (Lucas wrote a - sign)
It reduces to $u(u^2 - 3u + 1) < e^u$, if I haven't made any mistakes.
oh
@r9m Even Turoso approached it differently, using power series, it's nice, but it cannot be compared to the trivial way presented in the paper.
r9m
r9m
@Chris'ssistheartist which one?
r9m
r9m
17:42
@Chris'ssistheartist ah! I haven't checked all the solutions yet ,.
@r9m It's worth looking at that. ;)
r9m
r9m
@Chris'ssistheartist 'kay :-) I'll surely read through them later!
@anon how does one get from that to $(0, \infty)$?
@r9m I didn't read all of them but that one. OK!!!
iunno, thinking out loud
17:44
For zero, both sides are equal, so it's not part of the solution set.
suggesting ideas
But obviously all those sweet higher order terms in $e^u$ dominate everything else for positive u.
Oh, no.
The solution set is $\Bbb R$.
You then restrict to $u$ for which $\log u$ is defined.
Which makes it $(0,\infty)$.
@Lucas: ^
@Agawa001 what are the dots here @Agawa001?
another sum from 1 to i?
hello why if $||Tx||\leq 1$ for all $x\in E$ please
yes
geometric series
wait a min , excuse me plz im strugglin with lucas ' question
ok it gives : $\frac{x}{e}+(\ln(x)-1)^3-(\ln(x)-1)^2+(\ln(x)-1)<x^3-4*x^2+(6+1/e)*x-2 $ right ?
and $x^3-4*x^2+(6+1/e)*x-2$ is always $<x$
am i wrong somewhere ?
17:58
@anon can you help me please
@Agawa001 wait a minute
indeed but what did you led to say <x^3-4*x^2+... ?
and nothing else
it seems a easy sum @Agawa001
@Masacroso i want to write it in terms of integral
18:02
in the end you had 3 geometric series
??
oh , i forgot , thats the not easy part :p
integral respect to i?
maybe respect to x?
the sum is the integral respect to i so...
nah , if u integrate it , u do it in function of $dx$
and $i$ goes away
so what you want exactly
make the sum, after take derivative on x and you can write x as a integral
18:07
Hello@r9m
r9m
r9m
@Rememberme hey there
Are you free @r9m??
r9m
r9m
@Rememberme ya sort of ..
Well I have something to ask you If you are??
r9m
r9m
ask
18:10
Lets say we are giving an exam and for each correct answer in the exam we get 4 marks and for each wrong answer we get a -1, there are total 30 questions in the test. SO not considering the negative marks can all the numbers between 0 to 120 be achieved with the given marking system, That is with only 4's and -1's?? @r9m what are your takes on this
Given a fiber bundle $(E, \pi, M)$, the associated jet bundle $J^k(\pi)$ is of dimension $\dim E$, correct?
r9m
r9m
@Rememberme can we leave out questions? (I mean if question with 0 score allowed or not)
If you leave a question then you dont get any marks for it.... @r9m
Howdy@Balarka
r9m
r9m
@Rememberme I see .. so if someone does k of 30 questions and leaves out some m questions, ie, (30 - k - m) wrong answers .. then the net score will be 4k + 0m - (30 - k - m) where, $0 \le k+m \le 30$ right?
18:14
Yes....
r9m
r9m
so the question is what values $f(k,m) = (5k+m-30)$ take when $0 \le k + m \le 30$, where $k,m$ integers ..
@Balarka One very striking result....
In Sieve theory there is this amazing conjecture known as the Elliot-Halberstom conjecture which if assumed true then $p_{n+1}-p_{n}\leq 6$
yes.
but you should learn about sieve theory before you decide which result is striking and which is not :P
Yes thats something But isnt this result shocking to the eye....
hmm @r9m So how is the question :p
r9m
r9m
there has been improvements on the upper bound .. in fact Prof. Ram Murty will be giving a 3day lecture about that here ..
18:19
I have suddenly flown into so much of info. after entering into topology
I have to really utilize this info
So @r9m What is the way of proceeding the question after what yous aid
r9m
r9m
@Rememberme idk .. write a code and verify
Codes dont know any of 'em
@r9m It would be nice to have a few distinct approaches. Does Jack's approach leave something missing then?
r9m
r9m
@robjohn I don't think so ... Jack's answer looks complete to me (just missing a few inbetween steps .. that can be ignored) ..
@robjohn who the hell downvoted your answer????!!! :O
18:27
@robjohn, my friend (:p), say something about that, please
@BalarkaSen You were right about that knot thing (as already mentioned). Do you know what the universal cover of the hyperbolic knot complement is, or I guess its geometry?
r9m
r9m
@robjohn downvoting a perfectly fine answer is just sick :(
@r9m It happens quite frequently. People don't like you for some reason, and they downvote randomly.
r9m
r9m
@robjohn that kind of behavior is downright BS >:( I have seen this happen to chris's sis too and a few other high rep users .. just sad.
@PaulP: The universal cover of a hyperbolic link is still hyperbolic, hence the name. The point is that the theorem you used - Milnor-Svarc - demand that the action be cocompact.
18:39
@r9m Yeah, but they are entitled to their vote, and they can cast it any way they want. It may be that they are not good for the site, but I can't think of how to do much about it, even if I know who the down voter is.
okay I figured it either didnt act nicely or the geometry was weird, guess it was, the action
Well, the action by definition gives us the knot complement, which isn't compact
r9m
r9m
@robjohn you are the moderator right? can't you complain to the SE community about the situation?
(A simply connected, complete $n$-manifold with constant negative curvature is isometric to $\Bbb H^n$. Because a hyperbolic link is a link whose complement supports a complete Riemannian metric of constant negative curvature, its universal cover must be $\Bbb H^n$.)
I like fresh manifolds, with a bit of salt they are yummy :)
18:49
I want to show that the Euler equation for the functional $J(y)= \int_a^b f(x,y) \sqrt{1+y'^2}dx$ has the form:
$f_y-f_xy'-\frac{fy''}{1+y'^2}=0$

$L(x,y,y')= f(x,y) \sqrt{1+y'^2} dx$

Substituting $L_y(x,y,y')=f_y(x,y) \sqrt{1+y'^2}, \ L_{y'}(x,y,y')= f(x,y) \frac{y'}{\sqrt{1+y'^2}}$, I got the following:

$f_y(1+y'^2)-f_x y'- f y''+ \frac{f (y')^2}{(1+y'^2)}=0$

Can we get from this relation to the desired one, or have I done something wrong?
@anon Do you maybe have an idea?
Hello@KarimMansour
Ah makes sense. Not sure why I havn't buckled down and learned this stuff, I got the background needed. I guess I will start today, as my plans today got cancelled (plus flipping though hatcher it looks really fun, and the sort of thing I feel like doing now) @MikeMiller
@Masacroso ur question lacks of details, you may give examples ?
Hey @robjohn
Do you have an idea how we could solve the differential equation $y'= \sqrt{\frac{cx^2}{1-cx^2}}$ ?
what details you want? Is full on details @Agawa001 :)
well, "swap" maybe misinterpreted... but if I says "swap columns" is very hard to misunderstand what Im talking about
19:03
example maybe
how matrices stay same if their rows are swapped manytimes
matrices never stay the same... matrices are equivalents under these transformation
well thats the point
Row swapping is one of the row operations@Agawa
row swapping or/and column swapping
19:05
i know but how equivalent ?
And if i row operate one matrix to get some other matrix both of the matrices are called equivalent
and not same :p
Hi @Rememberme
I want to know the number of equivalence classes under these transformations
Think when will you call two matrices the same
19:06
for a squared (0,1)matrix of dimension N
:p
When each of the entries in the matrices are the same.....
if u operate $a$ u get $b$ when u swap $b$ u get $a$
If you row operate one matrix to get some other matrix then not every entry in the two matrices are the same
Hatcher's 3-manifolds notes @PaulPlummer? It doesn't really talk about the geometry, I think I recommended you Peter Scott's survey article before which is very nice
Yes if you do some hotch potch on one of the matrices and get another they will be called equivalent not same
19:09
@robjohn you can help me please
the sums of rows and columns are conservative but the sums, by themselves, didnt says that one matrix is isomorphic to other
Okay gtg Brain dead need to sleep
as i said its either super easy or super hard :p but what i know about isomorphism is the addition of two swapped matrices gives the swapped sum matrix isnt it ?
I dont know, LOL
lets denot $S$ swap operation , $a,b$ are matrices $\epsilon M$ where $M$ subgroup of binary square matrices , $S(a)+S(b)=S(a+b)$
i think this s definition of isomorphism
19:18
the formula is correct, but the definition of isomorphism is more broad I think
see that, on graphs we dont need to define some transformation to see if one graph is isomorphic to other, yes we define 2 functions to check if they are isomorphic... but isomorphism itself lies in the structure of the graph
the same happen with these matrices, in a very similar way
19:40
No, I was talking something a bit more pedestrian, algebraic topology. I have only picked up a little knowledge here and there and never really sat down a learned it @MikeMiller I think more serious geometry like those notes will have to wait (and you did recommend Scott's notes at one point)
(introductory algebraic topology)
hi chat
@evinda formally, you can integrate both sides of that ODE to get an expression for $y$. how to compute that integral is a different question, but one which should presumably yield to some smart $u$-substitution
19:59
@PaulPlummer: Fair enough! Should be fun

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