Mathematics - 2018-05-09 (page 1 of 4)

so the residue of $\coth(\pi s/2)$ is $2/\pi$ at every pole, so you have the residue of the laplacand (lol) is $\displaystyle \frac{2}{\pi}\dfrac{1}{1-4n^2}$

Semiclassical

12:11 AM

nice

Akiva Weinberger

I feel like we should be able to use Gaussian elimination somehow

Semiclassical

@AkivaWeinberger you can write that as $f(z)=i+5i\dfrac{z-i}{z+3i}$

which is...nice?

mercio

I think that long ago I had a geometric procedure to get the fixpoints of a Mobius transform given 3 paris of inputs-outputs

:|

Akiva Weinberger

Möbius transformations are just linear transformations on $\Bbb C^2$, quotiented out by $[x,y]\sim[\lambda x,\lambda y]$, $\lambda\in\Bbb C$, right?

mercio

I found that $-i$ is sent to $-4i$ because of the circle going through $i,-2,2$

Semiclassical

12:13 AM

nice

GFauxPas

Akiva yes, in fact we have that any mobius transform is equivalent (not uniquely) to $\begin{bmatrix} A & B \\ C & D \end{bmatrix}$ on , I think, $\mathbb CP^2$?

mercio

oooh i know !!

it conserves right angles

and that's how we can find the image of infinity

because the unit circle is orthogonal to the real axis

GFauxPas

or maybe $\mathbb C P^1$

mercio

and so, their image must also be orthogonal

GFauxPas

well we kinda knew that because mobius transformations are conformal, no?

mercio

12:17 AM

and on the image of the real axis you find the images of $0$ and $\infty$

I still don't see how $2i$ is the second fixed point

geogebra doesn't have "circle orthgonal to this circle and that also goes through those 2 points"

GFauxPas

anyway semi, you come out with $\dfrac 2 \pi \displaystyle \sum_{n = -\infty}^{+\infty} \frac{e^{2 i n}}{1 - 4n^2}$ I think

what does it mean that a circle is orthogonal to another circle

Akiva Weinberger

Two curves are orthogonal if the tangent lines are orthogonal

at the place where they intersect

GFauxPas

ah

mercio

oh, the second fixed point has to be on the circle containing $i,-1,-2$

so it has to be $2i$

GFauxPas

$\dfrac 2 \pi \displaystyle \sum_{n = -\infty}^{+\infty} \frac{e^{2 i n t}}{1 - 4n^2}$ rather, it needs to be a function of $t$

something like that

mercio

12:29 AM

imgur.com/a/T7k6HQ5

Joe Shmo

does anybody know of a good proof to Poincarè's Lemma?

GFauxPas

red is the image of blue there mercio?

mercio

no, red is the image of red

and blue is the image of blue

GFauxPas

sorry im not sure how to read this image :(

mercio

I'm not certain why the image of circle I A A' sohuld be the same circle

:s

it was making sense while i was drawing it

GFauxPas

12:34 AM

well that can't be right, because if a Mobius map has 3 distinct fixed points on $\mathbb C \cup \{ \infty \}$, then that's the identity map

mercio

the image of circle IAB is IA'B'

those are the red circles

GFauxPas

so a whole circle cant map to the same cicle

mercio

it can

GFauxPas

then there are fixed points all along the circle ....

mercio

not necessarily

maybe all the points are moved along the circle except 2

GFauxPas

12:36 AM

oh, rotation

uhh hmmm

right

mercio

the image of the real axis has to be acircle going through A' and B' that is also orthogonal to the big red circle

so that gives you the blue circle

and with it you get the images of $0$ and $\infty$

but i can't make sense fo the green circle anymore

I think it only works because the derivative at the fixed points is real

and you can tell it is real because the imaginary axis is conserved

so near $i$ there is no rotation of the angles

so the image of the green circle must e tangent ot the green circle at $i$

GFauxPas

aanyway, I think I should just set up cross ratios

mercio

and also has to contain A'

GFauxPas

it gets really ugly though

mercio

which forces it t obe green circle itself

GFauxPas

12:40 AM

though you can eliminate 2 of the parameters with the 3 equations in 4 unknowns

mercio

but geometry is pretty

throws circles to everybody

GFauxPas

except the exam is timed :P

mercio

a hwell

that's a powerful argument

GFauxPas

still though, $(z,-1,i,1) = (w,-2,i,2)$ isn't fun

in terms of cross ratios

mercio

o..o

GFauxPas

12:43 AM

I tried and it got miserable

mercio

you can also do uit with determinants

GFauxPas

please share

mercio

$\det \begin{pmatrix} zw & z & w & 1 \\ 2 & -1 & -2 & 1 \\ 2 & 1 & 2 & 1 \\ -1 & i & i & 1 \end{pmatrix} = 0$

GFauxPas

what is this creature

mercio

you just put $(zw,z,w,1)$ for each point

you develop it and do lots of computations and it literally tells you the equation

GFauxPas

12:47 AM

but where are the coefficients

mercio

also I wrote $w$ for the image of $z$ and not the other way around

the coefficients are the $3$ by $3$ minors

imagine what you et if you develop the first row

$zw$ times something + $z$ times something + etc = 0

the somethings are the coefficients

:|

GFauxPas

hmm

mercio

I prefer doing the thing with all the circles

Semiclassical

by my lights, the fastest route is to write $f(z)=i+(z-i)/(cz+d)$ then require $f(1)=2$, $f(-1)=-2$ and solve for $c,d$

it's not great but it's not terrible

you get $c+d = (1-i)/(2-i)$ and $d-c=(-1-i)/(-2-i)$

GFauxPas

doable

my summation is wrong because I want to have $\pi$ somewhere in the exponential so I can write it without $i$ using Euler's formula

meh

GFauxPas

1:40 AM

bleh, I think I'm done studying for today

there's always tomorrow

I'll leave off with the question I gave up on:

Let $f \in L^1(\mathbb R^n), p >1, q \in (1..p)$. Suppose:

$\forall t > 0, |\{x \in \mathbb R^n: |f(x)|>0\}| \le (1+t)^{-p}$. Then prove $f \in L^q$.

off to play vidjagames/cry/have dinner

Secret

4:43 AM

in The h Bar, 6 mins ago, by Secret

> Define $\mathcal{O}$ to be a megalomaniac controprositute such that the glomeralogy of the silverton is given by the following esperatorationism:

Gibberish at its finest

skull

indeed

Secret

> Let $\tau$ be a disteranion colmeralogy, defined as following:

$$\tau = \frac{sarxiv}{\infty} \aleph_{\omega} \frac{us}{t}\to \leftarrow$$

We first wrote the chain of katombo, given us follows:

$$\to \leftarrow \leftarrow \infty \to \fork {rsv}{usdvt} \defork stuv$$

Apply the Operator L to the chain gives:

$$L (\to \leftarrow \leftarrow \infty \to \fork {rsv}{usdvt} \defork stuv) = \circ udv$$

The next step is to solve for the roots of the above. We achieve that by acting the pseudomsticular operator $\mho\Omega$

$$\mho (\circ udv) \Omega = \int^{H^1(dis)}_{\Bbb{ESDV}} rdf(\mu)d\mu$$

which give us the following sucurrent relational:

$$\int odv = e^{\int ods T_{\mu}^{a^{sdvt}}}$$

Expanding gives:

$$\int odv = \text{############## @%#%####} [][][][] I(x,y)\aleph_{s}\omega^{\omega} $$

Dividing both sides by zero thus yield the required results:

$$\frac{\int odv }{0 Px} = \text{Hail Topology}$$

$\square$

Twink

5:02 AM

Secret

that sounds like a very annoying recursive relation to evaluate

$$1000 + \frac{(x-1000)}{7} + 2000 + \frac{(x-1000 - \frac{(x-1000)}{7})}{7} + 3000 + \frac{(x-1000 - \frac{(x-1000)}{7} - 2000 - \frac{(1000 - \frac{(x-1000)}{7})}{7})}{7} + \cdots$$

Tobias Kildetoft

@Secret But we are also told that each brother gets the same amount

So those first two terms are equal

Secret

hmm... so that means we can find x, the amount of land being split

But I think the hardest part about that question is to solve for the number of brothers one need to solve an equation of the form $\sum_{k=1}^{n} f(k) = nx$ where $x$ is found by equating the first two terms

skull

i think it is the sum of the first two terms...

...so each pair of terms are equal.

Tobias Kildetoft

@skull Right, I used the term "term" loosely there to actually refer to two terms.

skull

5:16 AM

kk

:-)

looks like a jee question

Secret

not surprising as they love nested functions for some reaosns

skull

and they love to test for how fast you can come up with an answer

Semiclassical

i get why they do that, i guess

lots of people apply, but much fewer accepted

so you need a pretty high threshold in order to winnow it down

skull

some say, it's the most competitive exam; second place goes to the med school admissions test

Famous Michael Wang

6:27 AM

I totally agree

A J

@BalarkaSen Kindly keep the language family-friendly. Thank you.

skull

sorry @AJ i forgot about that when i starred it

great satire, nonetheless :-)

Madhuchhanda Mandal

7:10 AM

@Twink 36000 and 6?

philmcole

8:19 AM

Hi, I want to find critical points of $f(x,y) = x\cdot\sin\left(y\right)+ax^2+by^2$. After computing $D_xf$ and setting it zero I obtained $\sin(y)+2ax=0$ and $x\cos(y)+2by=0$. How would you approach this non-linear system of equations and solve it?

philmcole

9:18 AM

@Gibbs Hi, why did you change my question?

mercio

9:30 AM

I think Secret broke

Tobias Kildetoft

@mercio How can you tell the broken version from the usual nonsense?

mercio

it's nonsense squared

Balarka Sen

@AJ Oh yeah sorry about that.

philmcole

If my Hessian matrix of a critical point is the zero matrix, what can I say about this point? Is it a maximum, minimum or saddle point?

mercio

you can say nothing of any sort except that the point hates you

Balarka Sen

9:44 AM

lol

philmcole

Oh nice :D @mercio T

Isn't there another criterium or so?

mercio

are you looking at your map $f$ at $(0,0)$ ?

philmcole

yeah

mercio

are you sure that the hessian is the zero matrix ?

philmcole

I mean yeah pretty sure, the function is $f(x,y)=x^3-y^3$

mercio

9:56 AM

oh your function changed then

philmcole

yeah sorry I didn't get much out of the first one

Alessandro Codenotti

Is there a notion equivalent to the field of fractions for semirings?

Tobias Kildetoft

@AlessandroCodenotti Hmm, no idea actually

Does the construction even always work like it should?

mercio

well then I suppose you have to use the secret forbidden technique of looking at the signs of your function evaluated at a bunch of points close to $(0,0)$ in hope of it giving you ideas about what happens

Alessandro Codenotti

@Tobias I'm not sure. I'm interested in a very nice case actually, namely the semiring of ideals of a Dedekind domain

philmcole

9:59 AM

Okay. So in my case is it actually a saddle point at zero? Since it doesn't look like a saddle because it goes down in two adjacent directions and up in the two other ones @mercio

Alessandro Codenotti

I wanted to know if fractionals ideals can be introduced via a field of fractions like construction

mercio

yup

philmcole

okay thanks :)

Tobias Kildetoft

@AlessandroCodenotti Hmm, is there any obstruction to just extending the addition to make it a ring?

mercio

don't you have $x+x = x$ in that semi-ring ?

Tobias Kildetoft

10:01 AM

(Also, is that semiring even calcellative?)

@mercio Ahh, right, so doing Grothendieck group stuff would kill everything

mercio

I'm not sure about ring of fractions but you can have fractional ideals

Alessandro Codenotti

Cancellative means $a+b=a+c\implies b=c$, right?

Tobias Kildetoft

@AlessandroCodenotti I meant for the multiplication

Alessandro Codenotti

I don't think it holds for the addition either

Tobias Kildetoft

Right, it does not hold for addition at all

Alessandro Codenotti

10:05 AM

Pick $b,c$ coprime with $a$. I'm not sure about the multiplication

Tobias Kildetoft

But actually, it seems like it should hold for the multiplication by unique factorization.

mercio

yeah

Alessandro Codenotti

Oh, right, that makes sense

mercio

as far as multiplication is concerned, ideals form a free monoid

Tobias Kildetoft

in which case, you can do Grothendieck group of the multiplicative monoid to get a "semifield of fractions"

mercio

10:06 AM

those are called fractional ideals I'm pretty sure

Tobias Kildetoft

assuming this plays well with the addition, which my guess would be that it should.

mercio

you can also describe them as finitely generated $O$-submodules of $K$ I think

Alessandro Codenotti

I'll look up what the Grothendieck group is later and see if it works, thanka for your help! @Tobias @mercio

@mercio That's the usual definition

mercio

:|

well then you can easily define $+$ and $\times$ on that set

Alessandro Codenotti

Actually if everything goes according to the plan that Grothendieck group should actually be the ideal group of the number field we got the Dedekind domain from?

(Assuming the Dedeking domain was the ring of integers of a number field)

mercio

10:14 AM

en.wikipedia.org/wiki/Fractional_ideal should be relevant

Mark S.

11:18 AM

I have a terminology question (topology context if it matters): If we have two maps with the same domain, we can combine them to get $(f,g)$ with the same domain and the product codomain. And if we have two maps with arbitrary domains we can make $f\times g$ from the product of the domains to the product of the codomains. Are both of these called "product functions"? How should I refer to these two different operations?

Tobias Kildetoft

@MarkS. Hmm, so the first one the obtained by composing the second with the inclusion of the diagonal. So maybe something like diagonal product?

Mark S.

That sounds good to me. My one concern is that the Wikipedia page for product (in an arbitrary category) suggests that it's fairly standard to call the first one the product morphism. But if topologists generally call the second one "the product of the maps" then I don't want to go against that

Tobias Kildetoft

Hmm, not actually sure there

Mark S.

Ok, thanks for your thoughts. I'll try to check a selection of topology books later.

skull

11:40 AM

\o semic

philmcole

12:04 PM

Why is the parametrization of an ellipse $(a\cos(t),b\sin(t))$? How can I see it from the equation $x^2/a^2 + y^2/b^2 = 1$?

Mark S.

@philmcole are you comfortable with the parametrization of a unit circle?

user8469759

1:45 PM

Hi guys, if $A = [0,a]\times[0,b]$ and $(X,Y)$ is a random variable in $A$ with distribution $p_{X,Y}(x,y) = \frac{1}{ba}$ shouldn't result $\longbar{(X-\mu_X)(Y - \mu_Y)} = 0$?

I've done the computation by myself and it is seems to be true

other than it's obvious to me

Semiclassical

@user8469759 I presume you mean $\overline{(X-\mu_X)(Y-\mu_Y)}$

yes

kk

is it right?

I think so. The most obvious case occurs if you do a linear transformation to map $A$ to $[-1,1]^2$

user8469759

1:55 PM

ok because I've implemented a naive algorithm in C++ and it doesn't seem to work

so I was wondering wheather or not I got something wrong

Semiclassical

the point is really that there are four possibilities: $X$ and $Y$ both exceed the mean, only one of the two does (in two ways), or neither do

user8469759

@Semiclassical sorry?

Semiclassical

first and last make the product positive, middle two make it negative

user8469759

point of what? I got lost

Semiclassical

Point of $\overline{(X-\mu_X)(Y-\mu_Y)}=0$

Take the example of $a=b=2$.

Then $\mu_X=\mu_Y=1$, and we're left to compute $\overline{(X-1)(Y-1)}$

A quarter of the time, both X and Y will exceed 1. A quarter of the time, X will and Y won't, etc.

user8469759

2:00 PM

ah I see

Semiclassical

first case means that the product is positive, second and third that it's negative, and the last case would be it being positive agian

so it shouldn't be too surprising if the expected product comes out as zero

A simpler case for testing purposes, though, might be the uniform distribution on $[-1,1]^2$

in that case, $\mu_X=\mu_Y=0$

user8469759

isn't something like this supposed to implement the covariance?

Semiclassical

yeah

user8469759

for (auto i = 0; i < intensity.rows; ++i) {
		for (auto j = 0; j < intensity.cols; ++j) {
			w = intensity.at<float>(i, j);
			sigmax2 += (i-mean[1])*(i-mean[1])*w;
			sigmaxy += (i-mean[1])*(j-mean[0])*w;
			sigmay2 += (j-mean[0])*(j-mean[0])*w;
		}
	}

Semiclassical

oh, your code

dunno. i'm not so familiar with c++

user8469759

2:03 PM

yes, the relevant part

doesn't matter what language

just this line

sigmaxy += (i-mean[1])*(j-mean[0])*w;

assuming $w$ is always the same, like $\frac{1}{ab}$

Semiclassical

How many samples are you taking?

And how far off of zero is it?

user8469759

$0 \leq i, j \leq 4$

Semiclassical

That's not a lot of samples.

user8469759

no, but it's completely wrong the result

Semiclassical

actually, is this actually a random sample here? Seems like you're just plugging in various values of i,j

user8469759

2:07 PM

that's because it's supposed to compute...

Semiclassical

in which case appealing to probability theory is superfluous

user8469759

$$\overline{(I-\mu_I)(J - \mu_J)} = \sum_{0 \leq i < 5} \sum_{0 \leq j < 5} (i - \mu_I)(j - \mu_J) \frac{1}{25}$$

Semiclassical

I see.

user8469759

basically I'm implementing that formula, in the specific case I proposed

Semiclassical

well, if you're doing integer $i$ satisfying $0\leq i<5$, then that's really the discrete uniform distribution on the events $0,1,2,3,4$ with mean value $2$

in which case one can substitute $i'=i-2$, $j'=j-2$ to get the sum as $\displaystyle\sum_{i'=-2}^2 \sum_{j'=-2}^{2} i' j'/25$

which should definitely vanish, since positive products occur as often as negative products

Lozansky

2:18 PM

Is standard notation $\int_{\Omega(x_1)}\int_{\Omega(x_2)}...\int_{\Omega(x_n)} f(x_1,x_2,...,x_n) dx_n...dx_2 dx_1$ or $\int_{\Omega(x_1)}\int_{\Omega(x_2)}...\int_{\Omega(x_n)} f(x_1,x_2,...,x_n) dx_1 dx_2... dx_n$?

I wanna say the former

Semiclassical

ugh, i always hate worrying about that ordering

Droleulb

why does it matter?

it depends what's simpler in the context

i.e. makes calculation clearer

Lozansky

Yes but surely there is a convention?

Droleulb

yeah you just write $d^nx$?

Lozansky

That does not take ordering into account

Semiclassical

2:21 PM

that's standard in physics

Lozansky

I know

Semiclassical

it's less standard outside of physics as far as I know

Droleulb

oh shit is my physics showing???

Semiclassical

I take the issue to be that your inner limits can be functions of the outer limits

since otherwise you wouldn't expect the integration order to matter regardless

Lozansky

@Semiclassical Sorry they are not functions, it is just bad notation. What I mean is for example $\int_{x_1=a}^{b}$

The problem is knowing which integration limit applies to which variable

Semiclassical

2:26 PM

well, if you’re denoting them explicitly using x_1=a_1, x_2=a_2 etc then I again think it doesn’t matter

It’s when you don’t that I’m not sure

Lozansky

So there is no standard notation? :/

Semiclassical

for instance $\int_0^1\int_0^2 f(x,y)\,dx\, dy$

Lozansky

Yea

vanaghka

regarding the answer in math.stackexchange.com/questions/360888/…

Semiclassical

Yeah, I’m not sure. Standard seems to be that you order the limits of integration opposite of the integration variables

vanaghka

2:30 PM

if $|H| =15$, why does the homomorphism $\theta$ map a 60 element group to a 24 ?element group

Semiclassical

So that you can insert parentheses between the integral signs / differentials and get the same result

Lozansky

So $0,1$ is $y$ and $0,2$ for $x$?

Semiclassical

Right, since inserting parens gives $\int_0^1(\int_0^2 f_, dx)dy$

I sorta hate that tbh but it does seem typical

Lozansky

I thought that was the convention, but I have seen numerous examples of the opposite (order of limits of integration = order of integration variables)

Semiclassical

weird

i'll note that if you write integrals as $\int dx\,f(x)$, as is often true in physics

then you can perhaps read it as $\int_0^1 dx \int_0^2 dy\,f(x,y) = \int_0^1 dx (\int_0^2 dy\,f(x,y))$

but you really wouldn't see that either in physics

it'd be $\int_0^1 \int_0^2 dx\,dy$

Lozansky

2:38 PM

Ugh I do not like that notation

Semiclassical

in which case the notation is really pretty ambiguous

@Lozansky which one specificaly?

$\int dx f(x)$

ah

Like... why?

why don't we define Lebesgue integration using the Lebesgue measure on the area under curve?

Lozansky

2:40 PM

Guess it's better than the naming conventions for atomic states

Semiclassical

I think I started seeing it in QM a lot, since there you commonly want expressions like like $f(p)=\langle p|f\rangle = \int_{-\infty}^\infty \langle p |x\rangle\langle x| f\rangle \,dx$

Lozansky

Yeah, Griffith's employs it quite a lot

Semiclassical

what you'd then want to extract from that is the identity expansion $1=\int_{-\infty}^\infty |x\rangle \langle x |\,dx$

the problem being that, if you want to apply it, you have to shove stuff between $\langle x |$ and $dx$

Lozansky

Isn't $|x \rangle \langle x|$ an operator?

Semiclassical

it is. but so is $1$, in the sense that $1f(x)=f(x)$

Lozansky

2:43 PM

I suppose

Semiclassical

really it's not $1$ there but $\text{id}$

GFauxPas

Hi friends, just so you all know I don't like infinite products

Semiclassical

i.e. $\text{id}{f}=f$

so for that reason I think people start writing it as $\text{id}=\int_{-\infty}^\infty dx|x\rangle \langle x|$

so in that case it's more visually tidy when applying it to the right, as you'd often do

otoh, you then have it looking a bit icky when acting from the left

so it's not really a great solution

Lozansky

We should just scrap QM and forget it ever existed

Sha Vuklia

3:24 PM

guys, is there an efficient way to determine the commutator subgroup of this group:

Semiclassical

:S

Sha Vuklia

I determined the center, which makes things a little bit easier

Balarka Sen

instant death

Sha Vuklia

lol is that directed at my group?

Balarka Sen

yes

Sha Vuklia

3:27 PM

I already determined all the inverse elements, the order of the elements, generators, some subgroups, some left/right cosets, some quotient groups

but not sure if I can use that

my only idea now is to just start with an element, and then calculate the commutator with all the other elements

(skippings the center)

mercio

oO

Semiclassical

@BalarkaSen I'm not sure why that makes me think of this: youtube.com/watch?v=WMMWwqwRggQ

mercio

I would try to find a better description of your group by now

@ShaVuklia

like maybe it is Z/24Z in disguise

then it would be easy to find the commutator subgroup

well it's not Z/24Z but maybe you get the idea

Abcd

If $g(\sin 2t) = \sin t + \cos t $ $\forall t \in [-\pi/4, \pi/4]$, can I find $g(\cos 2x)$ (x not t) ?

mercio

can you find $g(0.8)$ ?

why would you say that

Sha Vuklia

3:36 PM

@mercio yea I got a list with possibilities (groups of order 24), but a lot of them are with the semidirect product, so I'm not sure that's going to help me a lot

and it's not simple things like C24 or D24, or C2xC2xC2, so I think it must be a semidirect product

mercio

maybe you can just throw a computer at the group then ?

Abcd

@mercio =$\sin (\arcsin (0.8)/2) + \cos(\arcsin (0.8)/2)$

mercio

yep

Sha Vuklia

@mercio you mean to write a code?

mercio

either that or use a CAS that can do it

Sha Vuklia

3:39 PM

never heard of CAS

mercio

computer algebra system

like maple

Abcd

@mercio then how to find the range of $g(\cos 2x)$

Sha Vuklia

ah I've never worked with that

but I guess programming it wouldn't be too crazy

mercio

the range ?

Abcd

@mercio yes?

mercio

3:41 PM

programming it is certainly good practice

MUH

Hi everyone,
I am looking for a small study group for studying anything related to Algebraic geometry, algebraic topology, commutative algebra, functional analysis, complex analysis, geometry and topology. If anyone is interested, please let me know. We can also look for expository work afterwards.

mercio

maybe you should start with the range of $g$ then ?

GFauxPas

en.wikipedia.org/wiki/… why does the image of $\infty$ get absorbed into the constant?

oh I guess it makes that term be constant, intuitively

wait, no, I don't see it after all :(

Abcd

@mercio its $[-\sqrt2 , \sqrt 2]$

mercio

probably

Abcd

3:51 PM