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18:20
Does the IVP $$e^x \dfrac{dy}{dx} = e^{y^2} \sin x - \dfrac{1}{y^2+1}, \: y(0) = 0$$ have an unique solution in a nbhd of origin?
If $f(x,y) = \dfrac{dy}{dx}$, do I have to determine analyicity of both $\partial f/ \partial x$ and $\partial f/ \partial y$ in a nbhd of origin or just the second?
I thought both partial derivatives (along with $f$) had to analytic, but it seems I'm wrong
Dear god
This is awful
@BalarkaSen what is?
@Lozansky I'm afraid I haven't dealt with systems this nonlinear.
Maybe someone else can help. Have you tried posting in MSE?
@BalarkaSen No, not yet
I don't think existence-uniqueness says anything about the linearity
Yeah but this is not autonomous is it
You can't just invoke Picard-Lipschitz
18:31
No, but it's first order
Okay this seems to answer my question
Oh, you could use Picard-Lipschitz for things of the form $y' = f(x, y)$ right
I had forgotten
Yeah
You need Lipschitzness in the $y$ factor or something
So the continuity/analyticity of $\partial f/ \partial x$ is irrelevant
18:34
Dang it, I assumed it was relevant
Anyway, thanks for the help
Thanks for reminding me about that version of P-L
No problem :)
I'm studying for my complex analysis final. Any didacts here have any common mistakes to watch out for? Cases where students often dont realize the hypotheses for a thm aren't fulfilled?
Hi all.
Hello thenotyou
18:42
:)
I wanted to generate a random subset of the set {1,...,n} with probability of each subset proportional to (n choose k). I figured that in order for this to work, $P(A)$ must be (n choose k) over (2n choose n)
Trying to figure out an algorithm out of this expression is driving me nutts!!
What's wrong with what you just wrote
That it is not an algorithm
Ooh I see
I dont want to bother to check if it's normalized, just gonna assume you're right. Let's think
the expression signals that we generate a subset of {1,...,2n}, and then somehow using it we figure a subset of elements of {1...n}
It is normalized, trust me
First thing I always check is how hard it is to find the generalized inverse of the cdf
Doesn't seem to be trivial here
Or cdf/cmf I mean
18:52
I don't think we even need the CDF here.
Quite often in probability, if you look at the probability each "object" must be yielded with, you can come up with an algorithm
For example, if you want a subset of {1,..n} proportional to the number if its subsets, P(A) must be (2^k / 2^n) (the size of the set is k) after normalization
And that clearly signals that what you do is generate a random string of length n over the alphabet {0,1,2} and then if you yield the string (e1,...,en) you take as a subset {i | ei in {0,1}}
You get what I mean?
Yeah but I inverse transform sampling is a great algorithm if you can solve FX^inverse, but you usually cant
Golden rule of sampling
Ah I see your point
You know that algorithm?
I will read about it
But, I just had an idea.
(n choose k) over (2n choose n). What if we randomly sample a subset of {1,...,2n} (assume we can do that), and then take the first k elements that are in the range {1,...n}, if any
WP doesnt mention the discrete case, but it's just thus
If F is only really monotonic, choose the min of each subinterval where its coinstant
anyway
19:02
sanity check: an infinite bounded set in R^n has a limit point
Does B-W work in R^n?
Oh you're not assuming connected
@GFauxPas what?
the open disc in R^n is connected
path-connected
You didn't say an open set
what?
Leaky Nun
14:02
sanity check: an infinite bounded set in R^n has a limit point
19:11
I don't get you
I thought you were asking if that were true
yes
I don't get what you're talking about open set
What about S1\{irrationals}
Sn i.mean
N-+
N-1
every point in S^(n-1) is a limit point
Even if you remove all points with irrational coords?
19:13
yes
Well then, carry on
what?
I'm saying ignore me
ok
ah, proof by compactness
How?
I mean, how do you know the set is compact
19:44
@GFauxPas I mean by compactness of the closed ball which contains the infinite set in question
Ah, nice
19:58
I wrote down that's impossible for a complex power series to converge in $D(a,r)$ and its closure except at exactly one point on the boundary $|z-a|=r$. , why is it impossible?
in my lecture notes
maybe i wrote it down wrong
or maybe i meant
oh i see what I meant
I assume you want it to be a power series around z=a with radius of convergence r
I meant that it converges absolutely in $D(a,r)$ and converges on $\mathbb C\setminus D(a,r)$ perhaps conditionally, and it diverges at exactly one point $z$ with $|z-a| = r$
I wrote down that that's impossible
but converges on, say, the rest of $|z-a| = r$
so that it's radius of convergence is $r$ because r is the supremum of $\{r \ge 0: \sum a_n z^n \text{ converges for } |z| < r\}$
okay, that makes sense
20:23
@Alessandro I finished reading the presentation you linked
Thank you very much for that
It's beautiful
I haven't :P
I had a group theory exam yesterday and I'll have a Galois theory one on Thursday, I'll finish reading it afterward
Good luck!
I'm sure you'll ace both of them
How was group theory?
Oh for some reason I read yesterday as tomorrow
ma nama closed timelike jeff
@Dami it went very well I think, it was fairly easy
20:32
Alessandro did you start topology yet
I'm still studying for my final tonight
I already had topology last year
Ayo @Akiv
There's no geometry at all this semester... luckily I'll have differential geometry next semester
oh okay
20:35
Ayup
are you in my class or am I confusing you with someone else?
hiya Akiva
Not only am I in your class, I'm standing right behind you.
oh snap
the guy with the knife or the guy with the gun?
20:37
Not that I know, are you in Italy?
lol no, theres someone in my class that hangs out here but i forgot his handle
Not me then :P
(Apologies for Dami who is slightly red-green colorblind if I'm not mistaken)
I guess @GFauxPas was referring to me.
oh hi Antonios
Oh apparently it's a Simpsons reference
your avis are both black and white and both of you have "A"
so its your fault
Hmm.
Well anyway, what's up?
I knew your name wasnt Alessandro but I figured it was some internet handle you made up lol
i ddint even start the top. final ive been too busy studying for my final tonight, i guess it will be tomorrow and thursday i do it
did you?
Yeah, I've just about finished.
20:42
is it tough?
Not particularly. A little bit tedious.
What flavour of topology?
It's basically elementary Algebraic topology and a little bit of point set business.
mostly algebraic
and - e;f,b
@AkivaWeinberger Thank you for that information. I'll send @Daminark a red-green dichromatic thonk
and some knot theory tangentially
anyway, I am going to get back to work. The sooner I'm done with this the sooner I can start on the massive pile of self-studying I've unfortunately assigned myself.
just to show an application of Van Kampen's thm
which was like 3 classes
Later, all. Good luck @GFauxPas.
thanks\
you too
These are all equivalent, right?
$G$ is simply connected $\iff$ $\pi_1(G) = 0$ $\iff$ every circle in $G$ can be filled in continuously into a disk $\iff$ every closed path is h.e. to the trivial path
Re: edit, yes
k thanks for confirming :)
Re: Akiva's response, yes
20:50
Re: normies
Amendation to my thanks Re: Balark's confirmation to Akiva's answer: thx
21:06
@Balarka thonk you for the red monochromatic thonk
rip "every closed path is h.e. to the trivial path" 2017-2017
@Daminark finest communist thonk
Re: Re: Re: Re: Please remove me from this mailing list
Hm, there's a terminological flubberduck that's happening on my end
So I call the dynamical system given by the linear ODE $\mathbf{x}' = A\mathbf{x}$ hyperbolic at the fixed point at the origin if $A$ has no eigenvalues of zero real part.
I hope I get unofficial extra credit in this course because I was the one who caught his minor algebra/notation mistakes
21:14
I am seeing that some places define a matrix $A$ on $\Bbb R^n$ with fixed point origin hyperbolic if $A$ has no eigenvalue of modulus $1$
I guess the point is if you write down the flow for my ODE, that's $\varphi_t(x) = e^{tA}x$
So $e^{tA}$ should be a hyperbolic matrix in the latter sense
If $\lambda$ is an eigenvalue of $A$, $e^{t\lambda}$ is an eigenvalue of $e^{tA}$, right?
should be, yeah
by power series
$Av=\lambda v\implies A^k v=\lambda^k v\implies f(A)v=f(\lambda)v$
So that's the point, I guess
I'm always a little paranoid about that latter implication, tbh
21:20
The converse is true too, surely. If $\lambda$ is an eigenvalue of $e^A$, $\log(\lambda)$ is an eigenvalue of $A$
Yeah me too
convention question: do we consider a power series a laurent series?
It should be a little bit of work. I can severely beat it up and make it happen by giving $A$ the Jordan normal form
Balarka, my gut instinct tells me it should be easier than that
Yeah I don't know an easier proof
I'm open to suggestions
What's the question?
21:22
@who?
What you're trying to prove.
nothing yet, I just want to know if i can feed a power series into any theorem that eats a laurent series. It just has all the coefficients on the negative terms = 0 so I dont see why not
Ya the laurent series about a regular point agrees with the power series
I thought you and Balarka were working on the same thing
21:24
only in the cosmic sense of we're both doing math
everything is math
I'm listening to breakcore bruh
math sucks
Or more appropriately, everything is Water, so we're all always doing the same thing -Thales of Miletus
@KevinDriscoll everything is quick maths
21:26
You say The Beatles-I say Neutral Milk Hotel

You say Led Zeppelin-I say Animal Collective

You say Pink Floyd-I say Godspeed You Black Emperor

You say Metallica-I say Kanye

You say Queen-I say Radiohead

You say The Who-i say Indie

You say Classics-I scream Alternative!!!

You say jhonn lenin-i punch you in the face

92% of teenagers have turned to dadrock.If you are part of the 8% that still listen to real music, copy and paste this message to another 5 topics. Dont let the spirit of JEEEESUUUS CHRII-iii-IIIST die.
Found from the depths of 4chan/mu/
I am BACK!
I have to decide at what point i should take a break before the exam, if I study straight for another 3 and a half hours until the exam I might be mentally fatigued
@Semiclassical I see you're speechless at this beautiful piece of poetic exposition on modern music
Sure, let's go with that.
21:35
I AM DONE GRADING! YAY!
Radiohead's been around for a long time though
time for whiskey
and I don't care that it is 1:30 in the afternoon, damnit
@Xander congrats!
By whiskey, you ofc mean working on solera aging optimization problems
@Lozansky Try this version: You say The Beatles-I say Neutral Milk Hotel

You say Led Zeppelin-I say Neutral Milk Hotel

You say Pink Floyd-I say Neutral Milk Hotel

You say Metallica-I say Neutral Milk Hotel

You say Queen-I say Neutral Milk Hotel

You say Aerosmith-I say Neutral Milk Hotel

You say Kiss-I say Neutral Milk Hotel

You say The Who-I say Neutral Milk Hotel

You say Tupac-I say Neutral Milk Hotel

You say Rock-I say Neutral Milk Hotel

You say Classics-I scream Neutral Milk Hotel
21:38
no more
We can only handle so much glory in one day
Aw I wanted to post a version with v a p o r w a v e on the right hand side
Tomorrow we'll recharge
It feels like half my complex course was "cauchy's integral theorem: variants and applications"
and the other half is, "what happens when you can't use cauchy's theorem? actually you can still use Cauchy's theorem"
21:45
Half of complex analysis is basically Cauchy's theorem plus log not being globally defined
but by Cauchy's thm, log can be locally defined
etc
@Daminark By whiskey, I mean a bottle of 25 year old single malt
is there a compact notation for $B(a,r) \setminus \{a\}$?
compact in the colloquial sense of the word
my book just adds $'$, $B'(a,r)$, but I try to avoid primes because they can mean infinity different things
@GFauxPas I don't know that there is a universally accepted notation for the punctured disk / ball
I was afraid of that
21:56
I've seen both $B'$ and $B^{\ast}$
I kind of like $B^{\ast}(x,\rho)$, personally
but, as always in mathematics, define your terms clearly, then do what you like ;)
It's reminscent of the notation $S^*$ being an algebra with a non-invertible element removed
I'm removing one thing and keeping the outside
not saying there's any meaningful connection
just saying it reminds me of it
And so the take-home final is done, and so is the semester.
22:14
grats
I have an exam at 8:00 and I think I need to stop studying for at least an hour before the exam so i dont get mentally tired
I seem to be assuming this in my notes, is it true or me being sloppy?:
$f$ holomorphic with $f(a) = 0$, $a$ a zero of $f$ of order $n$ in some $D(a,r)$
then
$g=1/f$ is meromorphic in some $D(a,r)\setminus \{a\}$ with $a$ a pole of $g$ of order $n$?
the order of the zero becomes the order of the pole when you take the reciprocal?
I mean, it seems reasonable
@XanderHenderson not familiar with that but woot nonetheless
22:51
When integrating when is substitution not appropriate?
@Trey when your boss is watching
really
@Trey There's no general procedure for deciding if a substitution will simplify the calculation or not
you can always do a substitution, the question is knowing when it will be helpful
unfortunately the answer youre gonna get is "you get better at recognizing it with practice"
23:15
@Daminark You have never heard of single malt scotch?
You live a deprived life :(
konnichiwa :)
I feel like starting at any morbid sad boi song and going through the rabbit hole of youtube recommendations is a good approach to knowing a lot of different musical genres which I am interested in
Like I started with breakcore and I am finding lots of generative ambient musics now
Pick a time frame.
Yeah I don't drink alcohol so I don't know much along those lines
Demonark is still under-age :P
23:28
I need to find the area for the region bounded by y=2cosx and y = -sinX I have the graph but I need to find the x value when they intersect how can I solve for x given 2cosx = -sinx
And I cant use graphing calculators
Hi professor
@MATHASKER: Write that equation with one trig function only and use inverse trig.
hi skull
Hi @TedShifrin
hi @Balarka
Who drinks alcohol? Pfeh. Do yellow bugpowder
23:29
Not humorous.
There
That's better
what do you mean by one trig function @TedShifrin?
@MATHASKER: Can you rewrite that equation with one trig function only? Not two.
hi demonic @Alessandro
Are you still listening to Bowie? @BalarkaSen
I think I have listened to most major works by him
23:32
like substituting one of the trig Identities for -sinx or 2cosx? @TedShifrin
I have moved on
I see.
Simpler than that, @MATHASKER.
What function involves both of those?
cant sin be rewrote to be same as cosine and cosine to be as sine @TedShifrin
There is an identity, yes, $\sin^2 x + \cos^2 x = 1$. But that'll make for more challenging algebra.
I'm not going to just tell you the answer. Are there any other trig functions?
23:38
yes but I don't know any trig function that involves both of sine and cosine @TedShifrin
Seriously?
oh
nvm tangent
Well, that only took 10 minutes :P
So you want two consecutive $x$ values satisfying what equation?
Also, cotangent :-)
wait how can I re write the equation in tangent form of 2 cos x and sin x
23:43
What is the formula for tan?
sin/cos
So, from your equation, what is $\sin x/\cos x$?
its tangent but i have $2cosx = -sinx$ so that means I can just divide the cosx over and leave the two?
You know how to do basic algebra. Come on.
oh ya Idk what im sayin so it would be 2 = tanx
23:47
Almost, but not quite.
and then I can see which measures in radians of tangent equals two?
2 = -tanx
So tan x = ?
Right, so one x will be $\tan^{-1}(-2)$ (or $\arctan(-2)$), which you can find with a calculator or leave that way. What's the next x?
23:50
No.
wait but I think I need exact value of that
cuz I have to take integral from 0 to that value where the two graphs intersect
From 0? I thought it was between two consecutive intersection points.
oh wait ya its from two consecutive points I had my graph only on the first quadrant so i got confused
They don't intersect in the first quadrant.
ya but why is it not +2pik
23:53
Because it's not.
i meant the 4th
I only have 1st and fourth so i got confused
You'd better graph more.
I did that cuz like on the question it says [0,pi]
Huh? That's first and second quadrants.
What is the EXACT question?
Area for the region bounded by y = 2cosX and y = -sinX on x is the element [0,pi]
23:56
So on part of the interval one curve is on top and on part of the interval the other curve is on top. They want the total geometric area between the curves on the whole interval.
That's like doing $\int_0^\pi |f(x)-g(x)|dx$.
Howdy @anon @Lucas
howdy
wait this is the picture right
I thought it was asking for the area of the shaded
But you also need the part to the right of the intersection, too.
But, as I wrote above, which curve is "above" switches.
sup homies
23:59
jus maffs
hey @EricSilva. How go the apps?
I've finished my apps for SMALL brown and berkeley
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