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00:00 - 20:0020:00 - 00:00

00:00
Hmm, branch points or essential singularities wouldn't be good.
@TedShifrin In particular, I'm concerned about the CPV when the function being integrated has some logarithmic points
Right, me too.
@Peter: Countable union of countable, so yes.
@TedShifrin Know where I could read about this? Cartan's book on complex analysis, perhaps? I don't wanna ask my advisor yet in case its something stupid that he thinks I've already understood
@TedShifrin So I get that $Q(f)(v)P(f)(v)=\rm id$.
Don't know. CPV is not a particular complex variables topic. I'll have to look in my office Mon.
00:06
For if $v\neq 0$ then $Q(f)(v)P(f)(v)$ must be nonzero.
@TedShifrin I'll look in my references; let you know if I figure it out
$v$ on RHS, yes. But $P(f)$ not mono means ...
@TedShifrin Well, I did write $\rm id$.
$\text{id}(v)$
@TedShifrin Right.
Potato pohtato. =D
00:08
No, you're usually more pedantic than I.
@TedShifrin Heh, maaaaaaaaaaaaaybe.
So are we done?
@TedShifrin =/?
@TedShifrin What do you mean with "So are we done"?
00:14
There's $v$ so that $P(f)(v)=0$.
@TedShifrin No, there cannot be! (But of course you know this)
Agh, I need to scroll up. Maybe I was thinking contrapositive.
@TedShifrin Yes, yes.
That's what we did, didn't we?
No, you had $P(f)$ not mono.
this is the second time I post this question and nobody answers:
0
Q: Calculus Problem: Do I have to consider separate cases to prove this?

TwinkConsider 4 trajectories parametrized by $x(\theta)=r(\theta)\cos(\theta-n\frac{\pi}{2})$ and $y(\theta)=r(\theta)\sin(\theta-n\frac{\pi}{2})$, for $n=0,1,2,3$. I have to show that each trajectory is defined by the following differenial equation: $$\frac{d}{d\theta}r(\theta)=r(\theta).$$ I tri...

is it so difficult? :S
00:17
@TedShifrin We assumed $P\notmid m_f$ then proved $P(f)$ is mono, right?
@Twink [text](link)
no Pedro
because they don't see it
sorry
@Twink Yes, we do.
no, Ted didn't see a link I gave some minutes ago
@Twink Get used to it, this is not "Q&A express"!
No, @Pedro, I'm doing direct. $P(f)$ not mono $\implies P\mid m_f$.
00:19
@TedShifrin I already proved that.
Oh, OK. Sorry, I thought you knew I proved that.
Oh, no wait.
You said you had the converse
Why are you changing the proof now? @TedShifrin
I'm not, dammit.
We had a nice proof. $P\not\mid m_f\implies QP+Q'm_f=1$ evaluating at $f$ gives $Q(f)P(f)=\rm id$ so $P$ is mono.
Ted are you in a bad mood today or what?
00:21
@Twink Nah, cursing is nice every now and then.
Oh, ok. Same proof. So we're done. You said you had the converse when you came in.
Pedro did you see my question?
@TedShifrin Yep.
@Twink Yes.
is it difficult?
I wasted all that thinking :)
00:23
@TedShifrin Nah!
@Pedro: FYI \nmid
@Twink I don't understand your notation. You write $x(\theta)=r(\theta)\sin \cdots$. What is $r$?
I don't know
I thought it was in polar coordinates
Note that $r'=r$ gives $r(\theta)=A \exp(\theta)$
@TedShifrin Thanks.
@PedroTamaroff Was just about to say that
00:25
that's why I wrote $r(\theta)=\sqrt{x^2(\theta)+y^2(\theta)}$
@Twink I am thinking you're copying the ex. wrongly or your book has a big typo.
why?
@Twink Because what you have makes no sense.
Let me show you the original
I agree that the question doesn't make sense, as asked. I can't understand it at least.
00:27
I think it should be $x'=\dots, y'=\dots$.
I'm almost sure there aren't mistakes
it's an admission exam for a university
Ah.
Yes, the path is an exponential spiral.
Give me a sec.
ok
This is a famous problem. I think there's definitely a typo.
it's about ants
I'm sorry it's in spanish Ted
if you want I can translate it for you
00:30
I can read enough for math. :)
It should be a DE ...
so my question wasn't understandable because I didn't write the whole problem?
I thought it was enough giving the equations of the trajectories of the ants
I am saying the equations seem wrong.
Just like your tennis protégé, @Pedro
but when Pedro read the problem he understood it
so it's necessary to explain the context of the problem?
I think I'm gonna delete my question
even if it took me half an hour to type it
thanks for the book @PedroTamaroff
00:45
@Twink ;)
@Twink: The key part of the problem is in the words, not those equations. The words give the differential equation that governs the motion of the ants.
ok
@TedShifrin Suppose I have an $m\times n$ matrix $A$. What is the criteria for $A$ to admit a right inverse?
Those equations just say once you have $r(\theta)$ how you symmetrically get the paths of the four ants.
@Pedro: Think rank.
but I don't have to consider each case for each $n$ then?
00:52
@TedShifrin Thinks rank.
The differential equation is the same. Just initial conditions are different for the four ants. It's all in the words.
@Twink Do you know how to interpret the derivative in polar coords?
@Pedro: More generally, when does a function admit a right inverse?
the derivate of what? of $r$?
@TedShifrin Sorry.
00:55
it's not the tangent vector?
Onto.
@Twink Oh. What do you mean by that?
@TedShifrin Onto functions.
So it must have full rank?
I don't know
:(
I don't understand
I'm so stupid :(
Not full. Depends on $m$ and $n$.
@TedShifrin Say my matrix is $3\times 4$.
No, onto means $m$.
00:59
Then I must have rank 3, yes?
@Twink: If you haven't done physics/calculus with polar coordinates, this problem is challenging.
Yes @Pedro
@TedShifrin OK, that is what I meant by full rank.
Well, what if it has rank $n<m$?
@TedShifrin Come again?
Just saying you need $m\le n$.
01:04
@TedShifrin Yes.
Guys to draw the graph for $f(a)=\int_{-\infty}^\infty e^{-ax^2}dx$ on $(0, \infty)$ do I have to calculate that integral?
the graph of $e^{-ax^2}dx$ is like a bell
@Twink Consider differentiating and drawing tiny slopes.
differentiating $f$?
You can relate the integrals for different values of $a$ by changing variables.
I was thinking that since the integral gives the area under the curve
the graph was an exponential
because the area inceases
when $x$ increases
01:11
You can't compute the antiderivative of $e^{-ax^2}$. That's a famous fact.
how is that of chaning variables?
02:02
@TedShifrin Hey! I was wrong!
Alexandroff is arbitrary intersection of opens is open.
02:17
@PedroTamaroff
who's there?
hello?
02:36
hey @PedroTamaroff
@Twink Need anything?
I just wanted to say hello
:-(
 
1 hour later…
03:51
hey
what is a good resource for a rigorous theory of generating functions
Generatingfunctionology isn't good enough?
Hi folks
@FernandoMartin: I have that book. It is a good book. But is there a more terse and rigorous resource?
hi
04:42
A good question for you folks here from a relative noob. A brand new poster put up a poorly written question and we chatted a bit. He really wanted to know how techniques for writing better expository type mathematics that is clearer to the reader using MathJax, as he puts all of his notes in that format. I prodded him to some success and eventually took his best edit and edited that to this end. How should I interperet this.
I liked the question he was "trying" to ask.
I am in no way bothered by the resulting judgement, I may have gone a bit overboard here even making this effort to get this general question of writing clear math into the fold. I was just surprised.
Given $x \in A \land (x \notin B \lor x \notin C) \land x \in C$, can I conclude that $x \in A \land x \notin B \land x \in C$ because of disjunctive syllogism?
Because, heck, $x \in C$ is right there.....
:[
@J.W.Perry It seems like a reasonable question if your interpretation is correct, and I'd vote to reopen it if it were edited like that. I probably would have rejected the edit if it had come up to me, since it's quite substantial (I do find it funny, though, that it was rejected both for being too major and too minor...) Why not share the link to the suggested edit rejection page, and ask OP to make the edit himself?
@T.Bongers Thanks, every word you said is what I was thinking. I guess the OP just has to do it himself, and @Omega I would say clearly yes, but I really want someone to tell me I am wrong because it tlloks so clear I am surprised you are even having a fit over it. Maybe it is the beer... bleh
OOh @Omega I see something there.
@J.W.Perry Whaaaat. How can that possibly be wrong :(
No I see noting.
lol
04:56
Oh good
nothing*
Been studying for exams... goddamn, every exercise is, like, different. Different strategy etc. I'm not a creative person...
I guess the thing I saw for a second was that it was not expressly saying in the middle $(x \in C \wedge x \not \in B \vee x \in B \wedge x \not \in C)$, but that is crap.
Or is it? hrm.
Fun!
Does $(x \not \in B \vee x \not \in C) \Rightarrow (x \in B \vee x \in C)$?
Seriously, clearly $x \in A \wedge x \in C$, but can we conclude that $x \not \in B$?
Ah rats, that is the disjunctive syllogism. Next question...
05:16
:]
yes
because $x \notin B \lor x \notin C$
Man I am glad I showed up
It is a tautology
Well, I'm taking a break from set theory. Now I'll take a look at cardinality. Seems pretty simple...... number of elements in a set........ hmm....
Lol, just realized that is still set theory
:[
Goddammit.
and since $x \in C$
we must have $x \notin B$
Exactly!
very cool
05:18
Thanks :D
05:52
i guess it is true that for any martin-loef random binary sequence then there are only a fifinite number of prefixes that represent primes?
since "the n'th prime number" always has a short enough encoding even when n is random
but there is a paradox i am confused about
heuristically for a uniformly selected real, the expected number of binary prefixes that are prime is infinite.
yet martin-loef reals have measure 1 in the unit interval right?
is that a question yet? gustavo anyone what's the deal?
Do you know where can I find solved exercises about determining elementary set stuff? Like this question I made: math.stackexchange.com/questions/507529/…
I tried searching online, but, eh... I don't really know what terms to look for. Sometimes I end up with exercises that are clearly many levels above mine..
:(
@Omega I think this is a fine place your question got an answer
at that level i guess there might be some automated theorem-proving stuff that is better than wolfram alpha?
@DanBrumleve: No I mean, I want to find exercises online so I can solve them
For practice
And of course, it would be awesome if they came with their solution / how-to so I can check if I did the right thing
why don't you write a program to generate random formulas
the program can find the solution too
try perl or python
That sounds like a great idea. I'd do it if my exam wasn't tomorrow.
:D
06:03
ok then go to sleep
Nah, I need more exercises :(
once i had to memorize a whole page of shakespeare and i did it but it all happened in my sleep
write the perl program in your head and dream
My head is a slow interpreter
mr infinity try omega^2
And doesn't have multiple threads
06:06
ok i'll make up one
One what? A formula randomizer?
Ok @Omega, I broke out some old notebooks. This is a good one. Let $P \Rightarrow Q$ be the statement. If $P is false$, and $Q$ is true, is the statement true of false?
no just a good formula give me a minute
Uh.... it is true.
very good
heehee
06:07
:(
STOP PLAYING WITH ME D:
sorry i don't know of any actually
still wondering about my previous q
Anyway, I heard you mention something about wolfram alpha automated something. What was that?
I'm not playing man, I am reading some ancient crap I wrote. For example, "Implication is false only when antecedent is true and consequent is false" straight from a dusty box labeled 2008 hehe.
page 2
@DanBrumleve, meanwhile, can you still recite that shakespeare page?
I would in no way ever try to torture you in fear of being tortured myself.
Ah hehe I do. Will have to dig.
06:11
0.233333... has 2, 23, 23, 233, 2333, 23333, maybe more as prime prefixes. does the subset of [0,1) with an infinite number of prime prefixes have measure 1?
my mistress with a monster was in love
near to her close and ... it goes on from there
@DanBrumleve: I don't know.
(edited)
heuristically there are an infinite number of expected primes in the seuquence
but it can't by martin-loef random, contradiction
@Omega maybe coq.inria.fr ? i haven't tried it
i mean is it a real paradox or do i have something wrong? i guess it might be related to the arithmetical hierarchy stuff they talk about in the wiki page?
06:32
@omega Regarding set equivalence problems, I doubt I have any that you do not already have yourself. Best of luck with your test.
06:49
@J.W.Perry: Thanks!
 
4 hours later…
11:15
hi
 
4 hours later…
15:33
The first of the sequences from these troll formulas are series coeffcients of a Lambert W function. I am wondering if the other two have Lambert W expressions too:
$$\exp \left(\lim_{s\to 1} \, \zeta (s) \sum _{k=1}^{1 n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}\right) = \text{If}\left[n=0,1,\frac{n^{1 n}}{\frac{(1 n)!}{1^n}}\right]$$

$$\exp \left(\lim_{s\to 1} \, \zeta (s) \sum _{k=1}^{2 n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}\right) = \text{If}\left[n=0,1,\frac{n^{2 n}}{\frac{(2 n)!}{(1\ 2)^n}}\right]$$

$$\exp \left(\lim_{s\to 1} \, \zeta (s) \sum _{k=1}^{3 n} \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}\right) = \text{If}\left[n=0,1,\frac{n^{3 n}}{\frac{(3 n)!}{(2\ 3)^n}}\right]$$
16:11
Greetings
hihii
@Chris'ssis Hello.
@Potato Hi. I saw your comment and I wa thinking you wanna see my approach.
16:26
@Chris'ssis I do, thank you.
@Potato and @Chris'ssis I would like to see it too, thanks!
Why is that? I just use the Dirichlet eta function formula here en.wikipedia.org/wiki/Dirichlet_eta_function
Q.E.D.
@Potato and @brenna I have only one thing to ask you: don't post my solution anywhere on site.
@Chris'ssis Thanks for the solution. Why do you not want to share it?
@robjohn please delete my solution above, the whole texts. Thank you!
@Potato I need no one recongnition. I do math only for love and pleasure.
You have an $e^y-1$, but the wikipedia uses $e^y+1$. What accounts for the difference?
@Chris'ssis You can post anonymously if you wish. It seems strange that you wish to destroy your interesting solution.
16:35
@Potato oh, yes. You're right. Then we use the product between zeta and gamma function here mathworld.wolfram.com/GammaFunction.html
@Chris'ssis Yes, that seems correct.
Thank you.
@Potato I did it in a hurry and missed that sign.
@Potato welcome any time.
that's a nice way to do it, although i'm not familiar with the dirichlet eta function. thanks!
@Chris'ssis got it
@robjohn thank you.
16:47
I'm still curious why you want your solutions to be destroyed.
They are quite helpful.
@Potato Thank you.
17:05
Hi guys
@Potato: I think that is correct.
I have a small question, If I want to find the volume $y = ln(x)$ from 1 to e, using the shell method around the x axis
I have to write the function in terms of x right?
so it becomes $x=e^y$?
what is it with all the deletions?
Is that right?
and does it mean the integral for the volume is given by:
$V=2*pi*\int_0^1 e^y * y~dy$
?
do you mean the area under the curve y=ln(x) from x=1 to x=e rotated about the x-axis?
if so, your shells have height e-e^y for y=0 to y=1 no?
because you draw a line segment from (e^y,y) to (e,1) and rotate this line segment about the x-axis to give the shell
17:19
Well, yes, I'm trying to find the volume so it would be the area under
also use \pi for pi and don't bother using astericks for multiplication
@anon okay then, but why is e-e^y?
read the rest of my comments about the line segment
I still don't understand that part, about it being e-e^y
have you drawn any pictures?
draw y=ln(x) from x=1 to x=e
then draw horizontal lines under the curve
17:22
Yes, done that.
then ask yourself "what are the length of these lines?"
the lengths of the lines is the displacement in x-coordinates
Hmmm... oh.
the distance between (e^y,y) [which is on the curve] and (e,y) [which is the right endpoint of the horizontal line containing (e^y,y)] is e-e^y
Okay then.
so wait, e^y is getting the horizontal distance between ln(x) and the y axis correct?
yes
17:23
Okay then, that makes sense, thank you.
so my integral becomes:
$V=2\pi\int_0^1 y(e-e^y)~dy$ correct?
yes
Thanks!
Would I be able to integrate that with integration by parts?
the ye^y part yeah
the ye part you can do directly
 
1 hour later…
18:55
@anon Yao yao.
Yo.
@anon What are you up to today?
@JayeshBadwaik Good. I like my steak bloody.
What are you upto?
18:58
@PedroTamaroff Well, I guess I just got told by a 14-year old copy-pasting verbatim from a flawed stillborn MathWorld article.
I am writing a draft of an answer to that question, also writing up an awesome original computation of the discriminant of cyclotomic fields I've patched together, and enjoying the weather.
@anon Sad.
@anon I'd like to see that computation! Can you tell me about it?
@PedroTamaroff nah, kid knows not what he do. I think it's hilarious.
@anon Heh! Could you tell me about cyclotomic fields and their discriminants?
19:06
Hello everyone! Anyone remember me? ;-)
@amWhy Hey.
@amWhy Hello.
@PedroTamaroff Hay!
@JayeshBadwaik Yellow! Oops, hello!
@anon I am reading "Introductory Real Analysis" by Kolmogov and Fomin and sipping my lovely shiny coffee. Now reading about the Minkowski functional and stuff.
I use a product formula for resultant to relate the discriminant of $x^n-1$ (which can be easily be computed directly) to the discriminants of cyclotomic polynomials $\Phi_n$ plus some extra baggage (pairwise resultant factors that always yield prime powers). Then I perform Moebius inversion to compute the discriminants exactly.
19:08
@anon Cawl.
But what are cyclotomic fields?
I know some of those words!
@PedroTamaroff ${\Bbb Q}(\zeta_n)$
@FernandoMartin LOL.
@anon Ah.
you and your "shiny" coffee
@anon You'd love it if you tried it.
19:09
I don't like coffee :(
@amWhy :-)
which reminds me, someone should invent "white chocolate" hot cocoa
@anon Hmmmm. Would be too greasy, IMO.
Damn it, used 3 acronyms in too little time. This chat is getting to me.
Anyone up for some topology problem?
I have to prove that a $T_0$ topological group is $T_2$
19:20
@FernandoMartin Hmm...
Munkres defines a topological group as a $T_1$ space that makes composition and taking inverses continuous, but apparently it makes sense to drop $T_1$. $T_1\implies T_2$ is easy but I don't see how to do it with $T_0$.
@FernandoMartin $T_1\implies T_2$ assuming it is a topological group, right?
Yes
Wouldn't make much more sense if not, haha
@FernandoMartin $T_0$ iff every singleton is closed, aye?
@FernandoMartin =)
No, that's $T_1$
$T_0$ is that points are topologically distinguishable
19:22
@FernandoMartin So the closure of $\{x\}$ is distinct from that of $\{y\}$ if $x\neq y$; right?
I can't prove that! I think I need a break.
I'm pretty sure that holds.
@FernandoMartin Dude, use the def. of closure.
Ok
Time for a break then
Yes, it's obvious
19:32
@FernandoMartin Cool.
I'd like to prove that points are closed using that it's $T_0$
or just that $\{e\}$ is closed, since left multiplication by any element of the group is an homeomorphism
@FernandoMartin Is it raining over there?
Nope
is it over there?
@FernandoMartin Yep, quite strongly.
Ok, Bourbaki says that $T_2$ iff the identity is closed. Haha.
According to this, $T_0\iff$ Tychonoff!
for topological groups, that is
19:55
Awesome music to study to.
OK!
Think I got it!
:)
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