« first day (1201 days earlier)      last day (4116 days later) » 
00:00 - 19:0019:00 - 00:00

19:00
@Ethan well, there is nothing wrong in over applying
@Ethan I don't know about that.
:12218787 Are you still shooting formulas out of nowhere in their faces?
no
@robjohn where did you go to school?
@Ethan UCLA and Princeton
2
19:05
did you go to grad school?
I presumed from the two different schools that the latter was his doctoral institution
Id much rather go to a big name for grad school
Besides I would imagine its hard to keep up grades at a place like that
@robjohn If I had your mind Id' rule the world. :-) I often think my math background here is the poorest one.
@Chris'ssis you give some good answers... why do you say your background is the poorest?
(back to my work -- I'm preparing some nice proofs)
19:12
@Chris'ssis see? :-)
@robjohn :D
$$\frac{e^{-x}-e^{-1}}{x-1}+\cosh(x)=\sum_{n=0}^\infty \frac{x^n}{n!} \{ \frac{n!}{e} \}$$
Where {.} is the fractional part function
@Chris'ssis hi Chris!!!
@Ethan nice. How about this one?
Let $f:(1,\infty)\rightarrow \mathbb{R}$ be a differentiable function such that

$$f'(x)=\frac{x^2-(f(x))^2}{x^2((f(x))^2+1)}, \space x>1$$
Compute
$$\lim_{x\to\infty} f(x)$$
@Ethan I did it in one line.
its a problem not an identity though
19:19
@Ethan I'll work on your identity. I like it!
@Charlie hi The Cat!!!!!!!!! How are you doing? :-)
@Chris'ssis I'm fine :) and you?
@Charlie Not that bad. :D I was just working on some cute proofs.
@Chris'ssis wonderful :D
20:08
I wonder if @pedro is around
@Charlie I am
@PedroTamaroff oh! You really are :D
@PedroTamaroff you're quiet Any exercise?
@Charlie I was going to add something to an answer, yes.
20:20
Hello
@PedroTamaroff ah :)
@PaulRS hello
Greetings
Are groups only with binary operations
@Alizter yes, a binary operation that satisfies certain properties
@Charlie i read that they have to follow associative operations does this mean that exponention cannot be a group under positive reals?
20:28
@Ethan , what's the motivation for the identity?
@Alizter why not?
Bear with me
@Alizter lets see, define the operation a+b as $e^ {a+b}$
okay so if we have$ (\Bbb R^+_0 , \exp ) $
@Alizter Come again?
20:32
Where$a \exp b$ denotes $a^b$
Sorry it is really hard to latexinate on a tablet
Switching to computer
You're correct that that fails to be a group, but even more fundamentally than associativity, it has no identity
@Alizter Ah.
That is not usual notation! =)
The exp, you mean?
Usually $\exp a$ is $e^a$.
I have a statement in my notes that surely must be wrong. If a divides b => a divides b*x for all x in R, R a commutative ring. Surely if R is the real numbers, we can take a = 1, b = 1 and x = 1/2 and the statement fails. So the statement is incorrect, right?
20:35
i know but as there is no superscript operator:-P
@sonicboom Well, no.
You're thinking divisiblity of integers.
why does the statement fail?
I know euler used this notation before $\exp_a b = a^b$
because we get remainder 1/2
lets modify to $a \exp b = a^b$
20:36
@sonicboom As I say, you're thinking division of integers.
so $\exp$ is closed under positive reals
In a general ring (IIRC we want domains), $a\mid b$ means $b=ac$ for some $c\in R$.
@PedroTamaroff what should I be thinking of
ah
Then surely $bx=a(cx)$ is divisible by $a$.
so it can be any r in R?
20:37
@sonicboom What do you mean?
However I am reading that for an operation $\circ$ for a group
$G$ to be a group it has to be that $a\circ (b \circ c)=(a \circ b) \circ c$
My problem is that doesn't work with $\exp$ does it?
a|bx where a = 1, b = 1, x = 1/2...so we have 1 | 1/2...and it does because we can choose m = 1/2 in r - and m*a = 1/2
@Alizter the positive reals fail to be a group under your operation. They don't have associativity, nor do they have an identity, and so on...
Let us pretend that some set $S$ is closed under exponentiation. There is an identity element $e$ for example. Exponentiation has an inverse operation logorithms
however there is no asociasiativity
does that mean It cant be a group or is there another alegbraic structure it falls under?
No, you don't have an identity element. An identity element, by definition, must satisfy $I * a = a * I = a$ for all $a$
20:42
Ah
In this case, you have $a^1 = 1$ for all $a$, but $1^a = 1$
Are groups only restricted to binary operations
Check out the definition here
(The answer is yes)
Thank you people
Why is this much more difficult than it looks? $$I=\int_0^1\frac{\ln x}{\left(1+x\right)\left(1+x^{-\left(2+\sqrt{3\vphantom{\large3}}\right)}\right)‌​}dx.$$
@Chris'ssis Did you solve my other problem?
That looks terrifyingly difficult to me
20:54
@Alizter In the last period of time I've mainly worked on a new set of questions I received. I'm in delay with some proofs ...
Is there a site where you can type in latex then later reference?
I remember Alexander Gruber gave an example in meta, but I can't find it.
$$\int \frac{dx}{(1+x)(1+x^a)}$$
21:19
Significantly cut down @Ethan 's problem, but partial proof is too small for the chat: gyazo.com/c24979470d4cba0514ec489d75aca90d .
Or it could also be too large.
There should be a minus in front of the final Iverson bracket.
21:35
Very happy
Hello
21:56
@Alizter you think there's a closed form?
@Alyosha Yes
22:12
$\sum_{k \ge 0}\frac{(-1)^nx^{n+1}}{n+1}\left \lfloor \frac{n}{a} \right \rfloor$ that closed enough for you?
@Alyosha Explain
Summing geometrically, which may be a bit dodgy:

$\sum_{b \ge 0}(-x)^k \sum_{c \ge 0}(-1)^c x^{ac}=\sum_{n \ge 0} c_n x^n$

$c_n= (-1)^n\sum_{ac+b=n}1=(-1)^n \left \lfloor \frac{n}{a} +1\right \rfloor$
It's really ugly, though, so I'm sure there's a nicer way
If it wasn't indefinite it would be quite easy to use residues or Ramanujan's Master theorem.
@Alyosha Do the bounds [0, 1] help? ;)
Very much so
$a \in \mathbb{N}_0$?
Consider a finite group and the sequence (g^n)n∈N. It has infinitely many elements, so they can't all be different. Choose two equal elements with different n's, and ... now you're supposed to get the neutral element somehow, but I don't see how.
22:31
@Alyosha Indeed.
@Chris'ssis In case you forgot Compute
$$\int^1_{-1}\left(\frac{\sqrt{\frac{3-\cos(2x)\tan^2x-\cos2x+3\tan^2x}{2+2\tan^2x}+\sin^2x-x}}{2\sqrt{\cosh^2(ix)+\frac{1+\cos(\pi+2x)}{2}+x}}-\frac{\sqrt{\cos2x+2\sin^2x-x}\sqrt{\frac{1-\tan^2x+2\sin^2x+2\sin^2(x)\tan^2x+x-x\tan^2x}{1+\tan^2x}}}{2x-2\sin^2x+2\cos^2x}\right)\;\mathrm dx$$
@Alyosha I'm preparing an absolutely amazing question for my former teacher.
Any idea about proving $[ \text{n is even}]=-\left \lfloor n!\sum_{k \ge n+1}\frac{(-1)^k}{k!} \right \rfloor$?
@Chris'ssis which field?
@Alyosha analysis. The question is pure insane!
@Chris'ssis I was thinking of asking a help to solve a problem. But I lost the paper with the question and I don't remember :(
22:37
@Charlie :-). Well, I mainly love integral, series, sequences, limits and geometry. However, the latter one I didn't attend too much in the last period of time.
@Alyosha imagine this: One asks you for computing without touching integrals
$$\lim_{n\to\infty} \frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{i+j}{i^2+j^2}$$
Isn't this absolutely lovely? :-)
@Chris'ssis it wa a number theory question :)
@Chris'ssis How would you do that with integrals: turn it into a Riemann sum?
Surely you need $n^{-2}$ to give a double integral (if it is a Riemann sum, that is)?
@Chris'ssis Can one not observer that $\frac1n$ is going to get really small to the point that it is basically zero?
@Alizter how would you use that?
I don't know
I am not approaching this with rigour
22:46
@Alyosha You have all you need for a double integral.
Could you explain why $n^{-1} \rightarrow didj$, not $n^{-2} \rightarrow didj$ as $n \rightarrow \infty$?
@Alyosha well, things are there, but a bit hidden. You can put $1/n^2$ in front. Just look at it carefully.
$\frac{1}{n^2} \sum_{1\le i \le n} \sum_{1\le j \le n}\frac{\left ( \frac{i}{n} \right )+\left ( \frac{j}{n} \right )}{\left ( \frac{i}{n} \right )^2+\left ( \frac{j}{n} \right )^2}$
@Chris'ssis I never had too much practice with series but symmetry? $$\lim_{m\to\infty}\frac1n\sum^{2n}_{k=1}\frac1k$$
$\int_0^1 \frac{x+y}{x^2+y^2}dxdy$
22:52
@Alyosha Do you see now my point?
Indeed, I was a bit needy with asking for help as I've not done much with Riemann sums.
@Alyosha no pb, I also miss things especially when I'm tired.
@Alizter $m\to\infty$?
oops
thats n
@Alizter When meeting $\sum^{2n}_{k=1}\frac1k$ we may think of harmonic number that is $H_{2n}$.
@Chris'ssis No no this is your question
if we look at the sereis
wait now
IGNORE EVERYTHING
2
22:56
@Alizter ok :-)
Is there much theory about non-affine (i.e. that cannot be described by a matrix) transformations?
@Chris'ssis I was trying to turn the double summation into a double single summation
@Chris'ssis on.my birthday, a friend gave me an integral :D
@Charlie Really? What a nice friend! :-)
@Chris'ssis yes :)
23:02
@Chris'ssis I was trying to turn the double summation into a double single summation
@Chris'ssis yes, very nice :D
@Alizter interesting. Did you already do that?
@Chris'ssis Unfortunately not
@Alyosha A non-affine transformation is a not-terribly-simple function from the space to itself. Some such functions are nicer than others, but you'll have to be more specific.
my attempts look good but with further inspection there is always a flaw
on another note
why do limit laws not apply to your sum?
23:04
@Alizter I'm not sure I get your point. What laws do not apply?
@Chris'ssis EG $$\lim_{n\to\infty} \frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{i+j}{i^2+j^2}=\lim_{n\to\infty} \frac{1}{n}\times\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{i+j}{i^2+j^2‌​}=0\times\lim_{n\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{i+j}{i^2+j^2}=0$$
@MarkS. I'll post something on main tomorrow (GMT), but essentially I want to prove problems involving the ratio of the area of a convex quadrilateral to some shape constructed on that quadrilateral (for instance, join the midpoints).
@Alizter There you have an indeterminate form $0\cdot \infty$
It's easy for parallelograms, you just need to prove it for squares then apply a matrix on it. The problem with non-affine transformations is that the relationships between areas seem more complicated, with some areas of the plane ballooning out and others shrinking.
@Chris'ssis So you HAVE to evaluate the limit before you can carry out the mult?
23:08
@Alizter didn't you study the limit's rules? Do you know how many indeterminate forms one can get by $1$, $0$ and $\infty$?
@Chris'ssis Not yet. I so far have self taught most of the mathematics I know
@Alyosha Well, if you know where each point in the plane goes, you can see the effect on areas of various polygons with the shoelace theorem: artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem
@Alyosha Well, if you know where each point in the plane goes, you can see the effect on areas of various polygons with the shoelace theorem: artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem
Only the other day when I was reading about number theory I really learnt how to find gcd and lcm
@Alizter Ah, I see.
@Alizter: $1 = \lim 1 = \lim n/n = \lim n \times \lim 1/n = \lim n \times 0 = 0$
23:10
@Chris'ssis I had a vague Idea about why that wouldn't work but I was confused
The UK maths curriculum is a bit pants, self-teaching is advisable.
so yep, that doesn't work
@MarkS. I'll examine that tomorrow. That may be all I need, thanks.
Sure thing
@Chris'ssis Out of interest what was the solution to the problem?
23:14
@Alizter your integral is excellent, I think complex analysis should do it if $a$ is a positive integer, but am not proficient enough at it. Ask on main?
@Alyosha Already have :)
If Mhenni can't do it I doubt it's possible. Maybe add the limits of integration?
@Alyosha It comes from an attempt at this
@LoganM Hi. I just realized I've probably bumped into you on campus...
@FernandoMartin Did you sort out your problem?
23:21
@PedroTamaroff: the topology one?
@FernandoMartin Yiss.
not yet, but I'm sorting other stuff out now
I'm writing down my solutions for tomorrow's test
the TA contest
@FernandoMartin Ah.
So you write and then hand in? What's the deal?
@FernandoMartin Hm...
23:23
they're 12 exercises, you must solve two of them there (they don't let you bring your own paper)
I'm writing them down so I'm sure I got all the details
@FernandoMartin Ah. Neat =)
What's a good example of a bounded-above set where $\lim \sup A$ exist, but does not have a least element?
where A does not have a least element, that is
the set of all negative numbers
...lim sup of a set?
woops, I meant largest element.
23:25
@Mike Yep.
sorry
I think you just mean sup
It is defined.
How so
@PedroTamaroff: I didn't get your comment
23:26
@Mike Take an ordered set. Take a subset, $A$.
@nsanger $(-\infty, 2)$
let me rewrite that: what is a bounded-above set $A$ where $\lim \sup A$ exists, but A does not have a largest element?
@Mike Define $x$ to be an almost upper bound of $A$ if $a\leqslant x$ for all but finitely many $a\in A$.
It's okay, @PedroTamaroff, you can stop :)
Then define $\limsup A=\inf\{x:x\text{ is an almost upper bound of }\; A\}$.
@Mike Oh? Thought you were interested.
23:28
I was just trying to get a handle on @nsanger's question; I don't have much interest in order theory-type stuff. If you want to talk about this, though, go on
Didn't mean to shoot you down!
Hmm, how about $A = {x : x < 1}$? There's definitely no largest element, and $\lim \sup A = \sup A = 1$.
Yeah, that works
$A = \{ x : x < a \}$
@Pedro! Hola.
@TedShifrin Helloes =)
23:33
Hi @Ted
Hi @Mike
hey @PedroTamaroff
@FernandoMartin Yes?
I guess we should complete the circle
Hi @Fernando
In your third question sheet from linear algebra
Hi @Mike, @Ted
are you supposed to know what an isomorphism is?
23:34
@TedShifrin I will do some multivar calc problems now!
Hi, @Fernando :)
I know that you're not supposed to use determinants, since they're introduced later in the course
@FernandoMartin Hmm, part 3 is linear transformations so I'd say yes.
Determinants is 5.
4 is dual space.
23:35
Penance for me, @Pedro? :D
@TedShifrin Hehe, of course.
I appear to have lost my copy of Atiyah-MacDonald :(
I have a square matrix $A$, and I have proved that there are change of basis matrices $C, D$ such that $A=BED$, with $E$ a matrix of rank 2
@TedShifrin I have one already.
Is it ok to say that A has rank 2 since $C,D$ are isos?
23:36
@FernandoMartin What is the problem at hand?
@FernandoMartin Yes.
I don't know if that's too high level for that question sheet
ok, thanks
$A$ is just the matrix of $E$ in other bases.
So represents the same li
So has the same range.
Yes, @Fernando, if you've proved relevant theorems :)
I know it's correct! I'm just asking if people already know about matrix range being preserved by isos
@FernandoMartin You needn't talk about isos if you want, just chage of bases maybe =)
23:37
by the time they've reached the 3rd question sheet
So it is more intuitive, even so.
It is just renaming.
ok, good
thanks!
@TedShifrin I have a problem already.
@Pedro, no, not same range. That depends on the basis.
@TedShifrin What do you mean?
23:38
isn't range invariant under isomorphism?
By range I mean the "image" of the linear transformation.
Dimensions are the same, but the image of the map looks different in different bases.
@TedShifrin Well, of course.
I think the right word would be rank
not range
@FernandoMartin Yes.
23:39
But the two words are the same in Spanish
That is dimension, yes.
Ahhhhh... Rang in French.
@PedroTamaroff There's nothing wrong with saying the additive group of a ring :).
Same $\ne$ iso :)
It just means the underlying abelian group under the ring addition, with additive identity.
@DanielRust I didn't read it off like that the first time.
23:41
@Ted would you consider canonically isomorphic objects to be the same?
goes back to the corner
No, @Mike.
Why not, out of curiosity?
It depends on level and context, but equal is equal. Two subspaces of the same dimension are rarely equal, but of course are iso.
@TedShifrin Ted.
Yes @Pedro?
23:46
@TedShifrin Suppose I have a cylinder $\{x^2+y^2=1:0\leqslant z\leqslant 1\}$ topped with a sphere $\{x^2+y^2+(z-1)^2=1:z\geqslant 1\}$, and I am given this as the outwards pointing normal orientation.
True. I'm thinking more along the lines of $(A \otimes B) \otimes C \cong A \otimes (B \otimes C)$
The boundary of this surface is the circle in the $xy$ plane given by $x^2+y^2=1$.
@TedShifrin If I look at the surface from up above the $xy$ plane, should I orient the boundary counterclockwise?
Yes, it's as if you deform the surface continuously, tracking orientation, until it's the unit disk.
@TedShifrin How can I be sure that's the correct orientation?
Because you have a continuous family of surfaces with tge same boundary. Orientation is a discrete entity so it cannot change in a continuous family.
23:50
@TedShifrin ?
Ugh @typos.
@TedShifrin OK. So, I have to compute $$\int_S {\nabla \times F}dS$$
This is $$\int_{\partial S}(xdx+ydy)$$ because the $z$ terms all killed off.
Or $d\omega$ :)
And that is zero, right?
Good ol' Stokes' theorem
23:53
Yes.
@TedShifrin I wonder.,
Equivalently, you integrate the curl over any other surface with the same boundary, e.g., the disk.
Why do they write $$\int_{\partial S} F\cdot {\bf n}ds$$ instead of just $$\int_{\partial S} F$$
@Daniel: Stokes's. :)
Isn't ${\bf n}ds$ just redundant?
23:55
@Ted my mistake :D
No, @Psdro, it's WRONG.
Flux shows up with divergence, not curl.
@TedShifrin ?
@TedShifrin Oh, the $\bf n$ is Hodge's star right? I mean, Gauss is Stokes': $$\int_S d(\star \omega)=\int_{\partial S}\star \omega$$
With Stokes's and curl, you need work around the boundary, not flux. It's $\int_{\partial S} F\cdot T, ds$.
Because if $\omega$ is a one form, $d(\star \omega)={\rm div }\;F\;dx\wedge dy\wedge dz$
@TedShifrin What are you calling flux?
no, you need $1$-form. Let's be careful about dimensions here.
00:00 - 19:0019:00 - 00:00

« first day (1201 days earlier)      last day (4116 days later) »