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8:00 PM
Then it was differential equations.
Pretty mechanical, no fun.
 
You made it sound like only the first was ok.
 
The first one, I had the vector field $$F=\left( {\frac{{ - 2z}}{{1 + {x^2} + {y^2} + {z^2}}} + \cos y,{e^x},\frac{{2x}}{{1 + {x^2} + {y^2} + {z^2}}}} \right)$$
And I had $M=\{(x,y,z):z=y^2+x^2\;,z\leqslant 4\}$
 
Oh, well there is lots of interesting DE stuff. Just not in these intro courses.
 
I had to find $$\int_M F$$
 
Another one to close off with a cap.
 
8:02 PM
Right, I closed the cap, then evaluated the integral over the disk.
It turned out to be zero.
 
What's the divergence?
 
@TedShifrin Zero divergence.
 
Hmm, really?
 
@TedShifrin ?
 
I guess I need to compute.
 
8:04 PM
@Pedro: You play soccer by any chance?
 
@DonLarynx Nope, tennis.
 
Oh, I see. It's clear.
 
@TedShifrin So, the integral over the disk turned out to be zero.
 
Chilly tennis weather here now, but I like that!
Yes, that integral is clearly $0$ by symmetry.
 
i hate multivariable integrals
so annoying
that class was NOT fun.
 
8:06 PM
Not annoying. Often important and sometimes elegant.
 
you mean you prefer single variable integrals
???
 
@TedShifrin Heh, did remember you on that one.
 
@Adam: I performed excellently in my calc 1 and 2 class. then calc 3....
 
@DonLarynx Well, that's different.
 
it went downhill?
 
8:07 PM
very downhill
 
Multivariable requires some thought and visualization.
 
@TedShifrin At any rate, I computed the integral and I had an $$\int_0^{2\pi}\cos \mu d\mu$$ term so it vanished.
But I did remember your symmetry arguments.
I just didn't give one, thought they'd prefer that.
 
Some year I'll get you to appeal to symmetry and oddness :)
You can state and prove a general elegant oddness lemma for appropriate regions in $\Bbb R^n$ :)
 
@TedShifrin Oh?
 
Sure :)
 
8:11 PM
@TedShifrin Which is?
 
Throw in a $\Bbb Z/2$ representation leaving the region invariant :)
I leave it to you to create the statement/proof. Did your prof like your Fourier essay?
 
@TedShifrin ¬¬?
@TedShifrin I am handing it in on Monday.
 
Ah.
 
@TedShifrin I will let you know. =)
 
What are you writing about?
 
8:16 PM
@Adam Fourier Theory.
 
Havent studied that.
 
thats a good one....
 
cool geometry problem for you, @Adam: Prove that a triangle with two congruent angle bisectors must be isosceles.
 
my first reaction was like: already done that, but reading through this I have only done that the median and altitude in isoceles triangle are the same so I will definitely look into this
 
8:24 PM
I just heard on the radio today that a very famous applied mathematician got hooked in high school on math when his teacher posed him that problem ...
 
@anon Yo.
 
hello
 
Hi @anon.
 
8:40 PM
Everybody: When did you start being interested in math? Did you always love math?
(As far as the cool geometry problem goes, I havent found a solution yet, but I like the problem, it has interesting structure.)
anybody there?
 
Indeed. I initially did maths to get better at physics, then was on over by it.
 
I always sucked at physics, I only understand the math part.
 
@Ethan why remove the question?
It seemed interesting.
 
@Alyosha it wasn't a question, it was an asymptotic expansion of an analytic-number-theoretic sum
 
8:55 PM
For exposition's sake or asking for a proof?
 
for experimental trophy-admiring purposes methinks. he posts sums and expansions and deletes them with some regularity in the chatroom.
 
Why remove anything from chat?
 
regret, uncertainty, insecurity, change of social strategies, etc.
 
Whoa ... let's call in the psychoanalyst!
 
I am not sure a person with so many issues would benefit from this chat
 
8:59 PM
my armchair is quite comfy
 
LOL
 
good for you, mine too
how can I remove something?
 
You can't edit/remove if you wait long.
 
hover your cursor over a message and see all the secret gadgets that appear on the left and right
over your own messages, you only get a left menu, but within a 2 min window you can edit or delete
 
pity, I thought removing the question asking why is removing a good thing to do would be rather classy...
 
9:09 PM
Sorta like the barber who shaves himself?
 
perhaps
you mean the barber who shaves only people who do not shave themselves
???
 
Perhaps. Just so long as he's not from one of the tribes down the fork in the road.
 
:-)
that reminds me of this: there is an all-powerful being. Can he create a mountain so high that not even he can see its top?
 
Is he entombed?
 
:-) what?
no, I dont think so....
 
9:15 PM
Then he should see everything.
 
but then he is not capable of creating a mountain such and such and he is not all-powerful
 
That proves it: He's entombed.
 
:-)
no I dont think so, he is capable of everything so I suppose that includes seeing through objects...
 
Is there a generating function $F(x)$ such that $[x^n]F(x)=\zeta(2n)$?
 
for any sequence $a_n$ there is a GF such that $[x^n]F(x)=a_n$; just set $F(x)=\sum a_n x^n$
very strange question
 
9:29 PM
Indeed, but is a simpler form known for $F(x)$?
 
oh, you want a nice form for it
 
That is, if $[x^n]F(x)=1 \Rightarrow F(x)=\frac{1}{1-x}$
 
try pluggin in the explicit formula for zeta(2n)
that should work
 
Euler product or sum?
 
bernoulli numbers and powers of pi
when I said explicit I meant explicit
 
9:31 PM
@Alyosha Are you looking for this?
 
It seems so, thanks!
@robjohn the proof is very pretty.
 
@Alyosha Thanks :-)
 
Any tips on the best way to integrale
$$ \frac12 \int_0^\infty \frac{t^2\mathrm{d}t}{e^t-1} $$ Numerically?
The singularities throw me off.
 
9:46 PM
@N3buchadnezzar $\frac12\Gamma(3)\zeta(3)$
 
I know =)
Does not help me that much when I have an exam in numerical analysis on monday
 
@N3buchadnezzar In actuality, there are no singularities. The function dies away like $t$ near $0$.
 
@N3buchadnezzar This one can be computed without pen and paper.
 
@Chris'ssis Does not help me that much when I have an exam in numerical analysis on monday
 
9:55 PM
@Chris'ssis you know what $\zeta(3)$ is numerically without pen and paper?
@Chris'ssis indeed; robjohn already mentioned it and N3bu responded
 
@anon I'm afraid I don't understand your point. Is there something you think I possibly miss?
 
why are you telling N3bu things he already knows, have already been said by robjohn, and obviously don't help him?
that is what my point is
 
@anon That "I know" doesn't necessarily mean he knows that formula I pointed out in the link. On the other hand, maybe my comment didn't help on his specific point (indeed, I see that "numerically"), but in case he didn't know that formula, I'm sure he's glad he knows it now.
 
@Chris'ssis Yes, Chris. Everything can be computed without pen and paper.
 
Then at the exam given only pen and paper, they surely just throw me another integral to calculate numerically
 
10:01 PM
Of course.
 
@Chris'ssis it's like you're not reading what I'm typing. oh well! :)
 
So memorizing formulas specific for this problem is a waste, I am more looking for general methods for estimating integrals numerically.
I tried Romberg, I tried Trapezoid, I tried Simpsons
I even tried changing the domain into 0.1
 
@Chris'ssis Again, Chris. You're always telling everybody things like "that is trivial", "that can be computed without pen and paper", "that is easy", "I am awesome and all my questions are awesome".
 
Hi guys, Suppose $\psi \in C^\infty(\mathbb{R}^d)$ have non empty compact support. I want to show that the family of functions $\psi(x-j)$ where $J\in \mathbb{Z}^d_+$ is bounded in $H_{-d}$. Here, $H_{-d}$ is the L2 sobolev space, ie $f \in H_{-d}$ such that $$||f||_{-d}^2 = \int |\hat{f}(\xi)|^2 (1 + |\xi|^2)^{-d}$$
 
Well Gaussian Quadrature worked, but not the adaptive one =/
 
10:03 PM
@PedroTamaroff Well, I just tell the way I see things. I think is really really fun to compute things without pen and paper. (enjoyable I mean)
 
I also want to show that the family of functions has no convergence subsequence in any $H_s$ ($s\in \mathbb{R})$. I feel like this has to do with Arzella-Ascolli theorem and Rellich's theorem
 
Can someone tell me how to protect a question? I don't see a button or menu for it that is obvious.
 
@Chris´s sis: by computing without pen and paper, you mean computing in your head?
 
@PedroTamaroff Hey, I didn't say I'm awesome! :-)
 
Is there a way to map (0,\infty) onto (-1,1) ? :p
 
10:07 PM
sine function?
 
$\sin(\infty)$ is undefined
 
You can get $(0,\infty) to $(0,1)$ with the aid of $x\mapsto{1\over x^2 + 1}$. Then shift over
 
How about $\frac{x-1}{1+x}?$
 
(brb - preparing to write up some proofs)
 
I was going to say $(4/\pi)\arctan(x)-1$
 
10:12 PM
@Adam Yes. But I try to avoid this way of saying things since it may be annoying.
 
@Chris'ssis do you have a collection of these penpaperless identities? I think it'd be useful to collect them somewhere online.
 
@anon Any identity has the potential of becoming a penpaperless identity if you like to do things mentally.
 
only a subset of identities fit inside one's field of vision, and only a subset of those can be done penpaperlessly even by those who like those methods, and only a subset of those will be known to people collectively at any given point in time
 
@TedShifrin You have a great mustache.
 
LOL, gee tanks, @Alizter.
 
10:28 PM
@anon For instance, with/without paper I'm helpless here. Compute
$$\sum_{n=1}^{\infty} \frac{\zeta(2n)}{n(2n+1)2^{2n}}$$ And there are lots of such questions! Why? I know just a tiny bit of math! :-(
 
I'm pretty quick with a lot if math, but there's very, very little I can do sans paper and pencil/pen.
 
math teaches us how to be "helpless"
 
@Chris'ssis there are lots of such questions for any value of "such," and everyone only knows just a tiny bit of X for any value of X
 
I don't agree with that at all, @badass
 
@TedShifrin why?
 
10:32 PM
I think it teaches us to use our brains, be resourceful, and not give up.
 
I agree.
 
@TedShifrin "not give up" that part I like at most. I'd say "never give up and always believe you can reach any peak".
 
But we know very little...
 
Good for you, @Chris'ssis!
Keep learning, @badass.
 
All we can do is try :-)
 
10:35 PM
Ah, the former skull.
 
aka
 
@Ted Shifrin: Do you remember the geometry problem you gave me?
 
Of course @adam
 
Well, I didnt quite prove it yet, but I did show the converse....
 
OK. That's definitely easier by symmetry. Probably need a contrapositive proof of the desired direction.
 
10:49 PM
@TedShifrin hi Theadore
 
Good evening, Charles :)
 
@TedShifrin how are you?
 
Doing well, thanks, and you?
 
@TedShifrin I'm fine, a bit tired
 
Yeah, me too. Our finals start this coming week.
 
10:52 PM
@TedShifrin :-/
 
Well, I'm giving, not taking ;)
 
@TedShifrin :-/
 
Can't make you happy :)
 
@TedShifrin very stressful, Teddy, very stressful...
 
Well, for me the stress is an MRI to check my cancer is still away.
 
10:56 PM
Il.change my name spelling to Mariglia
@TedShifrin oh my, teddy
 
Is Mariglia your true name?
 
@TedShifrin Marília
@TedShifrin your name sounds like "i adore you "
Theodore means gift from God
Theodoros
 
I have two triangles and no angle of the first triangle is the same as any angle of the second triangle.
Can I deduce that no side of the first triangle will be the same as any side of the second triangle?
 
@Adam Nope
 
I see.
 
11:11 PM
Take any 2 triangles with 6 angles and no two equal, then enlarge/shrink one until the 2 triangles have a pair of equal sides
 
Knowing that the lifetime (in hours) of an object has an exponential distribution with parameter 0.001. 6 objects are tested and the time where failures occur is written. What is the probability that any objects fail before 800 hours?
 
Map $(a,b)$ onto $(-1,1)$ or $(0,1)$ ?
 
is there any moderators in here? for the forum
 
@N3buchadnezzar What do you mean?
 
@Danny what's up?
 
11:24 PM
@N3buchadnezzar You're trying to solve theintegral?
 
Just making a short hand for usefull mapings / substitutions
 
@N3buchadnezzar Meh.
 
meh
 
no i was thinking about one thing ,but, maybe it might require a lot of work. Questions about theorems and some exercises , are often related to some book ... if people could seek bookwise about recent questions , it would be a good way to have such a structure
this forum would be a monster library ()
 
11:29 PM
@Danny Well, it is usually better if people just said "Hello, this is problem is from book X author Y chapter Z page W."
Then we would search problems easily.
 
yeah but ...
if they were able to choose where to put that question
i mean, i could "tag it"
on "Rudin"
by book titles
ofcourse this is a great deal of work
 
@Danny Ah, not bad. But some authors have maaaaaaany books. And not really useful, I think
 
@PedroTamaroff @robjohn @OldJohn HI !
 
Braces yourselves.
 
@Danny The problem is that there are A LOT of books and there would be A LOT of tags. If people want to find author X then typing it into search usually gives sufficient results.
 
11:32 PM
@PedroTamaroff Figured it out
 
@mick Hi
 
maybe there should be such "text box" where you write it down , book, title , page
 
@Danny Go search an author using the search bar
 
which could be optional
 
it should suffice
 
11:34 PM
I reformulated my question of yesterday. @robjohn @PedroTamaroff @OldJohn Feel free to answer ! :) math.stackexchange.com/questions/587458/…
 
@mick Not my area, I'm afraid
 
$x \mapsto \frac{c-d}{a-b}x + \frac{ad - cb}{a-b}$ will map $(a,b)$ onto $(c,d)$ Qwop
 
@OldJohn well at least the question makes sense now ... I think :)
 
@OldJohn is Statistics your area?
:)
 
I think Harmonic Analysis
 
11:36 PM
@DavidRobertJones Old john is a number theorist
 
@DavidRobertJones Nope - analysis
 
that too :)
 
number theory is just a hobby in my retirement
 
interesting
 
i was just suggesting. And having such "special text boxes" where one could write down book title,page, etc (when submitting a question) would give people an incentive to do so. and i people would do it aswell. But once again that was just a taught
 
11:37 PM
@robjohn are you intrested in my question ? How are you , wife , kids and cats btw ? :)
 
@Danny what about ISBN? on the other hand, is Statistics your area?
 
I like contour integration :)
 
@mick When you say "Let $M=[0,\infty)$ be a semiring." you want to say rather something like "Consider $M=[0,\infty)$ endowed with $+$, so $(M,+,\cdot)$ is a semiring"
 
@PedroTamaroff how about writing +\infty ?
 
@mick ?
 
11:39 PM
@DavidRobertJones sorry i'll do statistics next semester
 
@PedroTamaroff Let $M=[0, + \infty)$ be a semiring
k ?
 
Also, why do you give subindices to your variables? I mean, why not just write "is iso with $M[X]/(1+X+X^2+\cdots+X^{n-1})$"
@mick No, you're not declaring how it is a semiring.
 
@PedroTamaroff Will you correct that for me plz ?? I have to go now :(
Goodnight all x
byee
@PedroTamaroff any other problems with my question ?
 
"Explicit isomorphisms for $X_n$"?
 
@PedroTamaroff yes ?
 
11:41 PM
@PedroTamaroff where are you from?
 
$X$ is an indeterminate. I can't understand what you're writing.
@mick "We can give the isomorphism explicit by giving $X_3=1^{1/3}$?
 
@PedroTamaroff its a quotiënt ring
 
@DavidRobertJones Argentina.
@mick Yes, $M[X]/(1+\cdots+X^{n-1})$ is a quotient ring, that's not new news.
 
ah i should say Let M(X) be the polynomial semiring ! right ? @PedroTamaroff
 
@PedroTamaroff that's what i suspected
 
11:43 PM
@DavidRobertJones Why?
 
@PedroTamaroff because of your name. i'm from Uruguay
 
bye
 
@DavidRobertJones Ah. @PabloRotondo is from Uruguay too.
 
@PedroTamaroff interesting... does he knows statistics? :P
 
@DavidRobertJones Dunno. Doesn't @robjohn know?
I have to go now.
 
11:45 PM
@PedroTamaroff know what?
 
stat
 
@robjohn statistics
 
@DavidRobertJones I know some probability. I don't know all the methods and tests in statistics.
 
@robjohn do you know statistics?
@robjohn were you able to read my question?
 
@DavidRobertJones where?
 
11:47 PM
ROFL
 
@robjohn i posted it 31m ago
 
@OldJohn My god
 
Knowing that the lifetime (in hours) of an object has an exponential distribution with parameter 0.001. 6 objects are tested and the time where failures occur is written. What is the probability that any objects fail before 800 hours?
 
we haven't met in so long!
 
@Benja Hi!! Yes, its been a long time, indeed
 
11:48 PM
@OldJohn what you been up to lately?
 
A bit of maths - mostly analysis, and lots of hill-walking :)
And you?
 
@OldJohn there's so much stuff in AG
 
:12418381 I looked at that book once - but not too long, when I saw what followed after the first few pages!
@Benja It is huge, I believe
 
@OldJohn hah, which part did you look, from the beginning or from the middle?
Yea
 
@robjohn any idea?
 
11:50 PM
@Benja I seem to recall failing to understand even before I got to the end of the first chapter :(
 
well that's because he just chucks in many results in comm. algebra and expects you to be familiar with them
It's kinda ridiculous
 
I hope you are coping with it all OK
 
yea slowly :D
 
@Benja :)
@Benja Yeah - no problem - I have it open right now
 
ok
@OldJohn just sent you a message
 
11:52 PM
@Benja got it - will drop out of here
 
@DavidRobertJones Does the exponential distribution mean that after 800 hours an item has $1-e^{-0.8}$ chance of failing?
 
@robjohn i dont know where you get 0.8
 
@DavidRobertJones $800\times0.001$
 
@robjohn honestly I'm not sure
 
@DavidRobertJones I have to go. Sorry. If that is the meaning, then I would get the answer to be $1-e^{4.8}\doteq99.8\%$
 
11:59 PM
@robjohn thanks!
 

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