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3:01 PM
Go the way with the system of equations for a really nice solution (and very easy at the same time).
 
I dont get how that would help. Havent solved them but I guess I+J would end up in something easy to calculate, I-J too and then I'd do I+J-(I-J) right?
 
@UserX you get a system of equations from which you get $J$, the integral you need.
 
That sounds lengthier than calculating J elementarily though. Haven't tried the system way yet.
 
@UserX Using the system of equations makes things easier.
For instance $I+J =\int_0^1 \frac{1 + 1/x^2}{(x - 1/x)^2 + 2} \ dx$ is immediately done after making the variable change. The same thing for $I-J$.
 
What variable change? I'm using partial fractions and it's HUUUUGE
 
3:09 PM
@UserX What is the derivative of $x-1/x$? :-)
 
1/X^2+1
Wait
Lol
 
@UserX :-) You have that in numerator. Thus the integral becomes extremely easy. You make the variable change $x-1/x=y$.
Then $(1+1/x^2) dx = dy$. Fit the new integration limits, perform the elementary integration and you're done extremely fast.
@UserX Think that both $I+J$ and $I-J$ can be computed in one line each one. You really have a very fast way at hand.
:D
 
3:25 PM
That was awesome
I knew you would be the right person
 
@UserX :D
The stuff I was playing with
$$\int_0^1 \int_0^1 \int_0^1\frac{\ dx \ dy \ dz}{(1+x+y+x y ) (3
+x+y+z)(1-x-y-z+x y+x z+y z+7 x y z) }$$
 
That feels like it doesn't have a nice closed form
 
@UserX It has a very nice one.
@UserX $\displaystyle \frac{\pi^2}{36}$
 
Where do you find these?
 
@UserX I don't find them, I create them daily. :-)
 
3:39 PM
I mean you can't come up with that integral and magically find that it has that of a perfect solution
 
@UserX They usually come from research, often I'm also surprised by the closed forms of some integrals.
 
Hello all, I have a question that's probably not appropriate for a MO question
What's the most mathematically technical and pedantic way to say "Tying a knot in a rope?"
 
@MikeAkers Whatever for?
 
Mainly to poke fun at a mathy friend
 
"knot" is sufficiently technical as it is.
 
3:44 PM
well, a knot is closed isn't it?
so an overhand knot in an unclosed length of rope would not be a "Math Knot"
 
Ah, right; mathematical knots are closed.
 
@J.M.isback. nice avatar
 
@anon Thanks, it's an old elliptic function plot I had lying around. :)
 
@J.M.isback. I see the triangular lattice, but the origin seems special
@MikeAkers I think you mean tying a rope in a knot
 
ok fine :)
I was hoping for something like "deforming a high aspect ratio manifold embedded in R^3" or something like that
but i'll go with "tying a knot"
 
3:51 PM
what do you mean by "aspect ratio"?
 
@anon Yes, the origin is a multiple zero. That's why the phase coloring varies quite a lot there.
 
well, i don't know how to say it technically
but i'm trying to say that a rope is basically a really long thin cylinder
that's super bendy
 
if you just have a strand with free-hanging ends then there's no topological way to distinguish the rope being knotted or not. you'd have to fix the endpoints. also, do you want to refer to the act of knotting a rope, or the end result where the rope is knotted (by colloqial standards)?
 
the act of knotting
 
@MikeAkers That's not entirely accurate; it's a bundle of intertwined fibers. ;)
 
3:53 PM
@J.M.isback. if the origin is a multiple zero, then so are its translates. numerical artifact?
mathematically, a rope would be a function from [0,1] or S^1 into a space...
 
@J.M.isback. true
 
or maybe the picture doesn't show the full fundamental parallelogram...
 
@anon Well, it's actually more akin to the theta functions than the elliptic functions, so I think "quasi-periodic" is the accurate description.
 
gotcha
 
i.e., dividing different translations of that function yield genuinely doubly-periodic functions.
 
3:57 PM
say a rope is a path [0,1]->Dx[0,1] (where D is the unit disk in R^2) that is smooth isotopic to a line segment from (0,0) to (0,1). the act of knotting it would be an smooth isotopy to another path with the same endpoints which fixes the first endpoint (0,0) the entire time.
that probably works
 
that sounds pretty good
 
Now, that is a genuine elliptic function.
 
hexagons are bestgons
 
@anon Yes, that's part of the reason I'm obsessed with this particular family of elliptic functions. :)
 
4:02 PM
what'd you use to make the picture btw? mathematica with tweaks of standard options?
 
@anon Mathematica, yes. But I had to write a bunch of image-processing stuff myself.
There is of course an analytic proof that the function is invariant under the transformation $z \to z \exp(2 i\pi/3)$, but I found the picture way more convincing.
 
4:20 PM
how do I obtain the joint pdf of 3 random variables?
I read that using the jacobian
 
4:48 PM
please help
 
 
1 hour later…
5:57 PM
What dot products? Do you mean the following?

the unit vector a is perpendicular to n : $a \cdot n =0$

the angle between a and t is theta : $a \cdot t =|a| |t| \cos \theta=|t| \cos \theta$
 
...and since $\cos(\pi/2)=0$, point 1 follows from point 2.
 
Huy
6:13 PM
wtf is going on here
new people
 
Like who?
 
Huy
idk
where's mike
 
@DanielFischer @robjohn @Huy
Show that $r(t)=\left (\cos^2 t-\frac{1}{2}, \sin t\cos t, \sin t\right )$ is a parametrization of the curve
of intersection of the circular cylinder of radius $\frac{1}{2}$ and axis the $z$-axis with the sphere of radius $1$ and centre $\left (-\frac{1}{2}, 0, 0\right )$.

Do we have to show that r(t) satisfies the conditions $\left( x+ \frac{1}{2}\right)^2+y^2+z^2=1$ and $x^2+y^2=\frac{1}{4}$ or do we have to find the intersection and show that r(t) satisfies the latter?
 
Huy
whichever you prefer
so probably the former
 
The former is easier to do.
 
Huy
6:18 PM
no offense but those are not very manifold exercises you've been working on these days
@J.M. What does JM stand for?
 
@Huy @J.M.isback.
Substituting $x= \cos^2 t-\frac{1}{2}$ into the equation $x^2+y^2=\frac{1}{4}$ we get $y^2=\frac{1}{4}- \left( \cos^2 t-\frac{1}{2}\right)^2=\frac{1}{4}- \cos^4 t+ \cos^2 t-\frac{1}{4}= \sin^2 t \cos^2 t \Rightarrow y= \pm \cos t \sin t $. Why is the latter equal to $y= \cos t \sin t$ ? Or isn't it?
@Huy manifold?
 
@Huy They are my initials.
 
Huy
@J.M. I assumed that, which is why I asked "what do they stand for?" - but if you don't want to tell me that's fine of course
 
@evinda The idea is to replace all instances of x,y,z with the corresponding components.
 
Huy
@evinda: Why are you taking the square root?
 
6:21 PM
@Huy As you say. :)
 
Huy
Hm. I'm trying to remember someone who might be back.
 
@evinda You should end up with an expression entirely in terms of the parameter.
 
Huy
was your previous username your full name or just initials too, @J.M.?
 
Now I want to show that any point on that intersection is of the form $(\cos^2 t-\frac{1}{2}, \sin t \cos t, \sin t)$. @J.M.isback. @Huy
 
@Huy Just initials, yes.
 
Huy
6:23 PM
Ok.
 
@Huy Not yet :P
 
Huy
@Hippalectryon: I'll start doing it tomorrow, so you better do too.
 
@Huy Shouldn't I take the square root?
 
@Huy Okay, I'll see this weekend. I still have Prasalov's book to read too :D
 
Huy
@Hippalectryon: I want to study it a bit more in detail to then apply it to curved spaces too, because my prof told me it could be interesting for me with my background
do you have any knowledge of functional analysis already or none at all, @Hippa?
 
6:33 PM
@TobiasKildetoft So we can write $a$ in the form $a=A t+B n+C b$ but we cannot find the coefficients $A$ and $C$, but since $a$ is in the plane spanned by $t$ and $b$, we know that $B=0$. Is this correct?
 
@MaryStar yes
 
@Hippalectryon $$\int_0^1 \int_0^1 \int_0^1\frac{\ dx \ dy \ dz}{(1+x+y+x y ) (3 +x+y+z)(1-x-y-z+x y+x z+y z+7 x y z) }$$
 
Huy
how's it going @Khallil
 
@Huy a bit. not much.
@Chris'ssistheartist :o new forms everyday
 
@Hippalectryon because I work every day, without exception.
;)
 
6:39 PM
:D
 
@Hippalectryon $\displaystyle \frac{\pi^2}{36}$
 
That's unexpectedly nice
 
@Hippalectryon :D
 
Huy
@Hippalectryon: which of Prasalov's books are you referring to btw ?
 
@Huy Problems in Linear Algebra
 
Huy
6:43 PM
ok
anything interesting to share?
 
Well, it has some cool theorems I didn't know of. The proof are a bit too short sometimes (like a "it is straightforward" that takes you 5 minutes to understand), but overall I really like it
 
Huy
like which?
 
@Hippalectryon "straightforward" is always a relative term.
 
ugh Readcube is downloading updates ಠ_ಠ
 
Huy
wat is readcube
 
Huy
yea
ELI5
why is it useful
 
It auto adds references for your pdfs
 
Huy
wat
ELI5
 
Basically it makes it easier to organize research papers and find new ones. Unlike having 50 pdfs in one folder with weird names.
 
Huy
ic
so you're like one of those researchers
very organized
reading many papers
respect
 
6:55 PM
I'm just a student :/
With so many pdfs, I don't even know who is what anymore :P
 
Huy
ah ok
respect withdrawn
 
Huy
aren't there like many versions of this readcube? I mean alternatives
I heard about this stuff before but never bothered trying it out because I don't feel like I'd use it remotely to its full extent
 
No idea. I found readcube somehow and I liked it, so I didn't look any further.
 
Huy
and you use so many pdfs simultaneously?
takes me weeks to get through a short article
references? years
 
6:57 PM
Well I download quite many pdfs. Too many.
 
Huy
ok
 
7:17 PM
Can someone explain to me what trace property is being used in this answer? math.stackexchange.com/questions/853696/…
To get the last inequality
Robert Israel has high rep, so I assume it's true
 
@Lepidopterist I don't see any answer by R.Israel there, which one are you talking about ?
 
sorry, it was the wrong link @Hippalectryon
1
A: Convergence of a product under the trace

Robert IsraelLet $A = U \Sigma V^*$ be the singular value decomposition of $A$. Then $$|\text{tr}(A (B_k - I))| = |\text{tr}(\Sigma V^* (B_k - I) U) | \le \text{tr}(\Sigma) \|B_k - I\|$$ where $\text{tr}(\Sigma)$ is the sum of the singular values of $A$. Moreover, this is best possible (it's an equality if...

that's it there @Hippalectryon
 
Hm I don't know much about spectral norms :/
 
he doesn't explain it as if it should be an obvious property
 
Huy
@Lepidopterist: $\|AB\| \leq \|A\| \|B\|$
oh wait is it a different norm?
 
7:28 PM
i'm not sure about your question. the norm is the spectral norm
that is how i defined it in the question and he didn't specify another one
 
Huy
isn't it Cauchy Schwarz?
 
does that apply to the trace?
can you explain your thinking?
 
@bentham what is 'pdf' ?
 
probability distribution function
 
@Lepidopterist Ok
 
7:37 PM
If $(x,y,z)$ lies on the sphere $\left( x+\frac{1}{2} \right)^2+y^2+z^2=1$ then $z^2=1-\left( x+\frac{1}{2} \right)^2-y^2 \leq 1$. Therefore $|z| \leq 1$. It follows that $z$ must be of the form $z= \sin t$ for some $t$.

Substituting $z= \sin t$ into the equation $z^2+x=\frac{1}{2}$ we get $\sin^2 t+x=\frac{1}{2} \Rightarrow x= \cos^2 t-1$.

Substituting $x= \cos^2 t-\frac{1}{2}$ into the equation $x^2+y^2=\frac{1}{4}$ we get $y^2=\frac{1}{4}- \left( \cos^2 t-\frac{1}{2}\right)^2=\frac{1}{4}- \cos^4 t+ \cos^2 t-\frac{1}{4}= \sin^2 t \cos^2 t \Rightarrow y= \pm \cos t \sin t $.
 
@Huy, why do you think it's Cauchy-Schwarz?
 
Huy
@Lepidopterist: $|\operatorname{Tr}(\Sigma V^*(B_k-I)U)| = |\langle (B_k-I)U, V \Sigma \rangle| \leq \|(B_k -I)U\| \|V \Sigma\| = \|B_k-I\| \|\Sigma\| = \operatorname{Tr}(\Sigma) \|B_k - I\|$
 
@Huy i don't quite follow. the first equality expresses the trace as an inner product. what kind?
 
Huy
@Lepidopterist: $\langle A, B \rangle = \operatorname{Tr}(B^*A)$
 
isn't that the frobenius norm?
 
Huy
7:41 PM
idk what you call it, we called it Hilbert-Schmidt
and it is an inner product
the norm induced by it could be called Frobenius norm
 
i see. so the norm you are using is not the spectral norm
 
Huy
yea, because you got a trace
 
his answer was deceptive, because i asked for the spectral norm
thanks, though
 
Huy
you said or any other matrix norm
norms are equivalent anyways since you're finite dimensional
 
well maybe he should have specified...
 
Huy
7:43 PM
idk, maybe he did a different thing in his head with your spectral norm
 
equivalent, but he presented it as a tight inequality
 
Huy
but it looks a lot like Cauchy Schwarz and so does his comment "this is the best possible inequality"
 
right
you are probably right. his answer should be edited, though
thanks @Huy
 
Huy
actually, according to wiki you can write the spectral norm in a very convenient way as an inner product, @Lepidopterist
using that inner product and that $U, V$ are unitary, you should get the same with Cauchy Schwarz immediately
 
i don't see how, since you can't express the spectral norm as a trace, @Huy
 
Huy
7:47 PM
wait a second, maybe I've been too hasty
 
@Lepidopterist see (D.49) theorem of abstract harmonic analysis hewitt 2, page 704 , so $$|\text{tr}(A (B_k - I))| \leq \Vert A\Vert_{\infty} \|B_k - I\|_1 $$
 
Huy
trace and inner product are linear though
so it should be ok, no?
 
@RaziehNoori how does that relate to the spectral norm and the trace?
@Huy maybe i'm missing something, let me see
@Huy I guess I'm not sure what you mean unless you want to use some other inequality between the trace and spectral norm
@Huy What you did works because the H-S norm IS the trace of a certain matrix
 
Huy
well yeah because $\Sigma$ is as in SVD
is that not given here?
I just assumed by notation
 
The spectral norm of a matrix A is the largest singular value of A . trace norm is sum of singular values
so spectral norm\leq trace norm
 
7:58 PM
yes, @Huy. As @RaziehNoori says the spectral norm is the largest singular value
 
Huy
I don't see the problem
is my argument wrong?
 
@Huy i don't quite understand what your argument is
 
Huy
the one before with the Hilbert Schmidt norm
 
that made sense. but you implied you could relate it to the spectral norm because the trace and inner product were linear. i don't get what you mean
 
Huy
yeah my idea was wrong
I was too hasty
sorry about that
 
8:03 PM
thanks for thinking through it with me. i would have prefered the spectral norm, but the frobenius norm is probably enough for my purposes
@RaziehNoori i don't think the infinity norm is the spectral norm. it's the maximum column norm as i recall
the maximum column 1-norm, that is
 
In mathematics, a matrix norm is a natural extension of the notion of a vector norm to matrices. == DefinitionEdit == In what follows, will denote the field of real or complex numbers. Let denote the vector space containing all matrices with rows and columns with entries in . Throughout the article denotes the conjugate transpose of matrix . A matrix norm is a vector norm on . That is, if denotes the norm of the matrix , then, iff for all in and all matrices in for all matrices and in Additionally, in the case of square matrices (thus, m = n), some (but not all) matrix n...
 
yes @RaziehNoori look at wikipedia's definition of the infinity norm. it's the maximum column 1-norm
 
hewii uses my symbol
 
@RaziehNoori on what page does he define this?
 
see page 703
my phd thesis is about $\frak{E}_p(I)$
i know these spaces
 
8:17 PM
i don't see it
earlier he defines it as $\|\cdot\|_{sp}$
 
‎$$\|E_i\|_{\varphi_p}=\Big(\sum_{j=1}^{n}{|s_{j}^{i}|}^p\Big)^{1/p}$$ and‎
‎$$\|E_i\|_{\varphi_{\infty}}=sup\lbrace{s_{1}^{i},s_{2}^{i},...,s_{d_{i}}^{i}}\rbrace,$$‎
‎$$\|E_i\|_{\varphi_p}=\Big(\sum_{j=1}^{n}{|s_{j}^{i}|}^p\Big)^{1/p}$$ and‎
‎$$\|E_i\|_{\varphi_{\infty}}=sup\lbrace{s_{1}^{i},s_{2}^{i},...,s_{d_{i}}^{i}}\rbrace,$$‎
 
isn't that just the $l_\infty$ norm?
it even says these are the $l_p$ norms. that is not the spectral norm
@RaziehNoori the spectral norm is defined on page 397
 
The spectral norm of a matrix A is the largest singular value of A
 
i understand that
 
now see $\|E_i\|_{\varphi_{\infty}}$
 
8:24 PM
Do you agree that the diameter of a hexgon equals twice its side.
 
@RaziehNoori what is $E_i$?
 
i copied it, you can use $E$
$$\|E\|_{\varphi_{\infty}}=sup\lbrace{s_{1},s_{2},...,s_n\rbrace,$$‎
 
this text is unreadable. you are citing D49 to prove the inequality?
it would be nice if you made an argument instead of citing a text that is impossible to read
unless you have spent tons of time with it. he's not using any standard notation
 
excuse me d.39
in your symboles |trac XY|\leq \|X\|_{spectral norm}\|Y\|_{trace norm}
 
8:44 PM
Ok, Razieh. I will upvote your answer because I think you are probably right. I have to read it carefully but I trust you. However, if you want it to be generally understood you should use the standard notation. It's an interesting theorem, but I'm not sure I get the proof
 
8:59 PM
Hey guys, given a binary operation, does the non-existence of an inverse imply that there is no identity?
 
@Lucas no
try Z as a monoid under multiplication
 
9:23 PM
Hello someone can help me with statistics?
to get the joint pdf I have to calculate the determinant of the jacobian?
 
@TedShifrin !!!
my prof gave me 90 % in my first assignment in topology
he is very picky !!!
 
ok better ask in the site
 
Try here @Learner
:-)

 Ten fold

CrossValidated's general room for gossip, grumbles, and idle c...
 
9:41 PM
@J.M.isback. Nice to see you around again :-)
 
 
1 hour later…
10:45 PM
Artwork is everywhere here in my work today.
(some sleep is needed though)
 
does anyone hear about zalgo or zolgo text
b̢̢̯̺̜̺̹̊͑̒͂͛̑ͨ̓ͣͣ͊ͮ̆̋͗͂̾̍ȯ̴͍͙͈͕̤̫̥ͭ̈̅̍̍̆̕͟͠o̊̓̋̑̇ͪͣͭ͊͑̑̌̿̔̾ͥ͐ͩ̍͏̴̰̯͙̹̠͔̠̜͈͈͔̖̱‌​̦̟̱̱͓o̗̥͉͓͚͈̊ͣͪ̔̅ͩ͋̆̀ͩ͘͢͢o̯̦̖̭̞͊ͤͭ̆͒͒ͣ̊̌̌͌͗̓͆ͫ̄̓́ö̼͈̹̤͍̭͚̼̹͈̟̗̰̯͓̲̺̤͙̒̽ͧ̔̿̍͟͠͡o͆‌​̴̷͇͇̱͉͚̠̼̼̱ͩ͆̍̌̈́͌͌͋̓͌͜
 
11:04 PM
thanks @skillpatrol
 
Could someone of you take a look at my question:
0
Q: Signed curvatures

Mary StarLet $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ be two plane curves. Show that, if $\textbf{$\tilde{\gamma}$}$ is obtained from $\gamma$ by applying an isometry $M$ of $\mathbb{R}^2$, the signed curvatures $κ_s$ and $\tilde{κ}_s$ of $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ ar...

?
 
@robjohn Just got my computer back the other day, so... :) great to be back!
 
Sometimes I hate being on an active college campus. Music blasting in the big greenspace while I'm trying to get work done. sigh
 
@J.M.isback. How long were you without it?
 
heya @JMoravitz and @robjohn
 
11:19 PM
@TedShifrin How are things today?
 
Yucky ... it's getting hot and my cold is not going away. So I'm an unhappy smurf.
 
@robjohn More than a year, I think.
 
I don't even recognize J.M.
although I do know a mathematician with that name
heya @PVAL
 
hi
im done with my slave labor for the week
 
congrats
I'm trying to give away volunteer math tutoring to a community college, and they won't even answer my email :(
 
11:22 PM
@J.M.isback. Wow... why so long?
 
@robjohn I had to pawn it, you see.
 
ah ...
 
@TedShifrin Wow, they are missing out..
 
thanks, mr eyeglasses ... You wanna give me a testimonial? :P
 
I bet it read like a phishing attempt.
 
11:34 PM
so what math should we talk about tonight, mr eyeglasses?
Or it went to spam because of attachments ... so I re-emailed her this afternoon with no attachments.
I don't think what I wrote looks like any phishing I ever got
 
Well you wouldn't believe how many mothers moved from Sweden to Texas and were looking for tutoring for their sons or daughters by trusted professionals like me.
 
LOL
I got a few of those when I was on the faculty at UGA.
I suppose I could just drive up there and try to walk in and talk to the person, but I think that's rude. I could try snail mail, too. It's ridiculous.
 
Maybe the guy teaching the class with his masters from polynomial institute of Arkansas would feel threatened by you.
 
lol
They want to rip off grad students with cheap pay but won't accept a free established mathematician
 
@PVAL between the "please tutor my jailbait daughter" emails and the "your mailbox needs to be upgraded to the newest version or it will be locked" emails, I think I hate the ones from HonorSociety.org and Qollaboration the most.
 
11:42 PM
omg I keep getting e-mails from those scam honor society things
they are coming from my school too, so I can't block them or anything
 
@JMoravitz I hate the ones written by my university president far more than anything else.
 
@PVAL, I didn't contact the math department, although I would if I wanted to adjunct. I contacted the academic support center.
 
@morphic I doubt there are any funded grad students at a community college.
 
No, there aren't grad students at all.
 
@PVAL Oh, I guess I was just referring to my own school system. Our PhD candidates teach at our community colleges
 
11:45 PM
as adjuncts, yes
 
@TedShifrin This is SDCC?
 
San Diego Mesa College, @PVAL, part of the same system.
 
@TedShifrin I thought all the teaching grad students were considered adjuncts
 
not ones teaching in their own university, @morphic
@PVAL: It's just a two-year college.
 
Well I mean our graduate school doesn't have any teaching positions available for grad students, so...
 
11:48 PM
at most research universities, grad students TA large calculus lectures and often teach their own courses.
 
I had to teach people about springs...
 
proof of that here
 
I don't know anything about springs...
 
@PVAL Did you get to play with Slinkys in class
 
You'd rather teach about falls and winters?
 
11:49 PM
lol
 
I'd rather teach mathematics
 
ಠ_ಠ
 
don't be a snoot about applications, @PVAL
 
I've never heard it as a noun before...
I've only heard the adjective, "snooty"
 
Ironically, two of my worst evaluations about 20 years ago in a multivariable calculus class (regular) came from two chemistry majors, who both complained that I did too much physics in the class.
@JMoravitz: I endeavor to keep you on your toes :)
 
11:51 PM
Your unicity is palpable @Ted
 
ponders what the hell that means
 
@TedShifrin If I never need to hang a weight from a metal spring and throw it straight downward, I'll know what 2nd order ODE to solve.
 
You are teaching engineers and physics majors, not just pure math students.
My first quarter at Berkeley I was assigned to TA the sophomore DE class. I was upset that they didn't give me multivariable, but then, ultimately, I had a blast teaching the class and had some wonderful students.
 
At least we avoided the chapter (with the pretty insane flow chart) about what first-order ODE I need to solve if I want to detect painting forgeries.
 
Radioactive decay? :)
 
11:58 PM
Radioactive Spider-man
 
ugh, the chiropractor bruised my ribs, and it still hurts a week later :(
 
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