Mathematics - 2017-07-31 (page 2 of 5)

skullpatrol

06:00

Define define

Typhon

@mick we can examine forms of numbers. I'm not looking at a particular number. I'm looking at arbitrary forms. It's fun and I enjoy it. So what? I've got nothing better to do.

@mick -_-

it trivializes lots of number theory and solves several unsolved conjectures by proving them

to me, that indicates that it is likely not true

shrugs

who knows though?

my curiosity though is whether a Riemann Hypothesis would be true in various algebraic integer rings.

skullpatrol

It is a beauty.

Twink

I'm feeling bad

Typhon

@Semiclassical fair enough. For me it is legendre's conjecture and the collatz conjecture. I like to play with numerical programs doing stuff with them. Interestingly enough. If we combine inverse operations with the regular collatz conjecture operations we can prove 27 goes to 1 (by induction) in 4 steps. Just four steps

in fact

Twink

like dizzy

Typhon

06:03

the highest I've gotten out of the first 1000 nontrivial numbers is a whopping 35

(if at any point I manage to get a constant value that indicates I've found the mappings that prove the statement and I just have to view the list)

skullpatrol

The collatz's simplicity is extremely attractive.

Typhon

^^^

Semiclassical

That it is.

skullpatrol

1, 2, 3

Typhon

it's complex not in its complexity of knowledge but rather in very clever design

whoever conjectured it was quite brilliant

I know. Collatz conjectured it.

:p

Twink

06:06

:(

Typhon

Plus, when you're sitting though an incredibly boring ballgame you can do the calculations of various steps.

mick

No number needs more than 4 n^n steps in collatz ( conjecture by tommy1729 )

Semiclassical

Some discussion and sources here: hsm.stackexchange.com/q/3587/107

Typhon

just don't do what I did and think 3n + 1 can map to odd numbers. XD

skullpatrol

Go to a doctor @Twink

mick

06:07

Do you believe that 4 n^n conjecture ?

skullpatrol

No.

mick

WoW you do not

Why

skullpatrol

It sounds like crank math.

Twink

it's not necessary

Typhon

@mick I don't know of it. I believe the Collatz Conjecture is worth trying to prove because when I examined it I easily saw that all numbers of the form 4q + r have the conjecture true.

except when r = 3

that tells me that it is highly likely to be true

divergence would to me require that there at least be a few more tricky cases

(I know that sounds weird but intuition is rarely not weird)

Twink

06:11

@skullpatrol read it

skullpatrol

A lot of very smart people have spent a very long time on it.

mick

The problem with collatz is you can only investigate a mod b , for SOME b. But never all b < a + 1.

Thus modular aritmetica fails as a proof

And No other way is known

Typhon

huh?

that sentence made no sense

mick

For instance proving collatz for all prime twins has not been done

Typhon

why would you investigate all b < a + 1

mick

06:16

Sigh

Typhon

a mod b is the remainder when a is divided by b or it is the equivalency class of elements equivalent to a mod b.

mick

If it holds for all a mod b , it holds for all integers !!

Typhon

all a

in mod b

not the other way around

you want to test the a's

not the b's

mick

Ah yes

Typhon

XD

mick

06:17

I switched srr

Typhon

it's ok

mick

But the argument remains

Typhon

actually you can investigate them all for a finite b

doesn't mean the investigation is fruitful

but you can investigate nonetheless

for instance, I am looking at 12b + a

I can write out every a

and do the operations

(tbh, I think a sufficiently advanced way of doing 2-tuple operations on the remainder-quotient forms would be useful)

I might write a library for doing operations with those forms

would be useful for its own sake

division and stuff could return whether the operation is "decidable" (i.e. whether i have to assume something about the quotient to proceed)

@mick think of it this way. I'm not just investigating the conjecture. I'm looking at the function itself and its properties. Defining and proving various relationships in the general case might lead to a proof, but that doesn't make the relationships not useful for their own sake.

and if there is a weakness of modular arithmetic then perhaps modular arithmetic needs to be thrown away in the pursuit of a better way of representing things (in this context anyways)?

im gonna head out

skullpatrol

06:43

Cya

How I feel about the collatz conjecture now :P

mick

06:59

Lol

Secret

google.com.au/…

Now, perhaps someone can start to figure out how to make a molecule with non Hausdorff topology, then it will be interesting

ray

@Typhon Yes, the substitution of z = -y makes it easier to see. Thanks.

Liad

08:15

@SteamyRoot @TobiasKildetoft Hi !

Tobias Kildetoft

@Liad Hi

Liad

Are you familiar with this course : http://shnaton.huji.ac.il/index.php/NewSyl/80519/2/2018/ ?
Any chance you know a good book for this subjects? @TobiasKildetoft

Daminark

Hey there!

Tobias Kildetoft

@Liad math.ku.dk/noter/filer/matematik.htm The notes by Christian Berg are quite good (and in English)

Liad

There are some PDFs there, I guess you are talking about this :math.ku.dk/noter/filer/koman-12.pdf

Tobias Kildetoft

08:21

yes

Liad

Alright , thank you. last one :shnaton.huji.ac.il/index.php/NewSyl/80517/2/2018

Daminark

Ooh measure theory is nifty

Liad

:P you know a good book ?

Tobias Kildetoft

Sorry, not familiar with a good book in that

Daminark

My professor hadn't used a book for that, but Stein-Shakarchi and Rudin are both good

I had also used a bit of Evans but that was more for the GMT stuff like Vitali/Besicovitch and Denjoy topology

Liad

08:27

Alright , thank you both .

Alessandro Codenotti

08:38

Hi chat

Daminark

Hey @Alessandro!

Alessandro Codenotti

how are you?

Daminark

Doing alright, how about you?

SteamyRoot

@Daminark Stein-Shakarchi's complex analysis - good?

Daminark

Oh I was talking about measure theory

SteamyRoot

08:49

Oh okay

Carry on then

Daminark

I dunno much about Stein's complex though my classmates like it a lot

We're using Titchmarsh officially, and I'm supplementing with Narasimhan a bit

Alessandro Codenotti

@Daminark Quite well

I'm on holiday after all

SteamyRoot

We've had a postdoc guest-lecturing CA at our campus last term, and he switched textbooks to Stein-Shakarchi...

It has a few nice exercises, and that's the only positive thing I have to say about it :P

Lucas Henrique

morning.

Daminark

Lol, what do you like for CA then?

Hey @Lucas!

SteamyRoot

08:54

Greene & Krantz

Daminark

@Alessandro 'tis true

Lucas Henrique

I had a terrible night

Daminark

@Steamy never heard of that one

SteamyRoot

It's definitely not the most well-known book

Lucas Henrique

couldn't sleep because I was thinking of a system of diophantine equations

SteamyRoot

08:56

But, in my experience, it's a much better "first introduction" to CA than SS

Also, the proofs are actually, like, rigorous.

Daminark

SS is handwavy? I didn't anticipate that

meh

Lucas Henrique

even if against my will, my brain tried to see if there was a general solution to a system of quadratic and linear equations on 5 variables

Daminark

Oh lord

Lucas Henrique

couldn't rest 1 hour... I see that today's class is gonna be long.

(fun fact - I found some restrictions to the problem with quadatric residues and modular reciprocity)

Perturbative

09:10

Hi chat

user84215

Hello.

user84215

I hope there exists no mafia among the users of MSE.

Daminark

Hey @Perturbative

@aminliverpool a mafia soldier is after you right now! :O

Perturbative

Yo! @Daminark

How's things going?

Alessandro Codenotti

hi @Perturbative

Perturbative

09:23

Hey! @AlessandroCodenotti

@AlessandroCodenotti I haven't had much time to go over Kunen's stuff in the Set Theory group :(

Daminark

Everything's aight, how about you?

Perturbative

Aight over here too, just spent 40 mins on a proof, only to find out a simpler proof for it was trivial

Alessandro Codenotti

no problem, it's not like you have to if you're busy

Perturbative

I'd have more time if I didn't have to do these awful lab-reports for my Physics course

Lucas Henrique

hello @Perturbative :)

user84215

09:38

I failed a physics lab course twice because of lab-reports.

Perturbative

Howdy @LucasHenrique :)

user462339

10:03

Hello. I had read that the base $\Delta\subset \Phi$ is determined by the choice of the Borel $B$. How do I see that this is the case. I see that in the Lie algebra setting $\mathfrak{g} = \mathfrak{h} \bigoplus_{\alpha\in \Phi} \mathfrak{g}_{\alpha}$, so if anything I feel I could see that the choice of the Borel could fix the positive roots, rather than fixing the base.

Kirill

@user2457324 donno, ehrlich gesagt! I had to go out that time. Hope you've got luck!

user84215

If your answer to a question on MSE is very similar to (almost the same as) the previous one/ones, then it will be deleted by moderators, downvoted by others, or ...?

Kirill

@aminliverpool I don't know.

@aminliverpool you can try :)

Martin Sleziak

10:22

@aminliverpool I am not sure whether this is the situation you have in mind, but this was discussed recently on meta: Getting beaten to the correct answer

You can find a few related posts in the comments:

Some older posts which seem related (maybe you can find more): About posting identical answers to a question, Courtesy/Etiquette: Writing an Almost-Identical Answer just After the Previous. and Should I delete my answer if it is identical to others. — Martin Sleziak Jul 26 at 16:18

Kirill

if there is a sequence $\varphi_n$ of functions that converge uniformly to $f$, shows $\Vert f - \varphi \Vert_{\infty}$ the biggest difference between two functions or shows it the biggest value of the difference function? Is that the same?

user84215

@MartinSleziak Thanks

user462339

@Kirill I don't understand your question.

What does 'shows ...' mean?

You have $\{\varphi_n\}_{n=1}^\infty$ converging uniformly to $f$ with respect to $L^\infty$?

Kirill

@user462339 lets talk simply about the functions in $\mathbb{R}$. What does the norm mean on the paper? Is that the greatest distance between the two function on the intervall, or how can I interprete this visually?

@user462339 I donno the $L^{\infty}$ notation. But the rest is true.

user462339

Oh, I understand your question now.

Kirill

10:36

I think I know the $\Vert \cdot \Vert_{\infty}$ as $max_{\ldots} \vert \varphi_n(x) - \ f(x) \vert$, still cannot imagine the maximum of the difference of two functions

user462339

So you are considering $\|f(x)-\varphi(x)\|_\infty$, and so in the measure theoretic sense, you have that this is the inf number $C$ such that $|f(x)-\varphi(x)|\leq C$ a.e.

Well draw two functions $f,g:\Bbb R\to \Bbb R$ on $\Bbb R^2$ and find the maximum separation of $y$.

Kirill

@user462339 so, the biggest distance in $y$?

user462339

Consider on $L^\infty[0,1]$ $f(x)=x$ and $g(x)=-x$

Kirill

in this case - wehn there is a sequence that converge - means the norm the greatest absolute value of every $\varphi_i - f$? Is it one norm-value for every function, or is there a value for every $x$ from the domain? Sorry for vague formulations.

@user462339 I do not know the $L^{\infty}$ notation

user462339

You haven't worked with $L^p$ spaces?

Kirill

10:45

@user462339 I am sure I had, but without naming them that way

user462339

There is one norm for all of the specified domain, not one for each value.

Kirill

@user462339 so, we give every function $\varphi_i - f$ a value $\Vert \varphi_i - f \Vert_{\infty}$, like we usually do when we use norms? And you mean, all these different functions build a space that you call $L^{\text{something}}$, right?

user462339

Yes, you take functions satisfying convergence of specific $p$-norms, and obtain $L^p$ space.

And yes, we give a value to each $\varphi_i -f$ via the $L^\infty$ norm.

(Assuming it belongs to $L^\infty$)

(meaning it is an essentially bounded measurable function)

Kirill

@user462339 pretty clear about that point now. Can we make the $-x$ and $x$ example to the end? I am watching precisly these two on $[0,1]$.

user462339

So you can find the value of $(f(x)-g(x))$ at each $x\in [0,1]$, very easily for $f(x)=x$ and $g(x)=-x$.

Kirill

10:52

@user462339 yes, the difference spans from 0 to 2 units.

so, the norm will be two?

user462339

So $|f(x)-g(x)|\leq 2$ in $[0,1]$?

Kirill

sure

user462339

Indeed, although it's uninteresting, since we don't have to care about the a.e. part of the definition.

Kirill

and then you take the $\inf\{f(x) - g(x) \mid x \in [0,1]\}$?

user462339

No, it's the inf of $C$ such that $|f(x)-g(x)|\leq C$ for a.e. $x\in [0,1]$.

Kirill

10:55

oh, my mistake, right

@user462339 ok, I got the point. Thank you.

user462339

Not a problem. I hope I was clear.

Kirill

@user462339 sure, I mention that I have to express things clear

I am really wondering how dynamic mathematics are. We talk about sequences, limits, series. There is really motion in these ideas, not a static character!

user84215

Do MSE users earn more reputations and gain privileges to become the moderators of this site?

Secret

11:20

moderators are elected each year

Tobias Kildetoft

11:34

@Secret No, they are elected when new ones are needed

Secret

oops sorry

Alessandro Codenotti

@perturbative I think you're right on your retracts question, it does seem to be immediate from the definitions

Kirill

11:54

question: if you have a sequence of step functions defined on $[a,b]$ that converge to somewhere, do they have to use the same partition of $[a,b]$? Is it possible ot build sequences of step functions with different partitions?

Perturbative

@AlessandroCodenotti Thanks! Just wanted to verify that, seemed to easy to be an exercise'

I'm off to a Physics prac now, cheers everyone!

Alessandro Codenotti

good luck

Kirill

@Perturbative good luck!

Alessandro Codenotti

I don't understand what you mean @Kirill

Kirill

@AlessandroCodenotti the definition of a regulated function says, that some function $f$ is a regulated function, if there is sequence of step functions that converges uniformly to $f$. Ok, so I am asking myself, how this sequence should look like. To build a step function on $[a,b]$, we make a partition on $[a,b]$ with $a=x_0<x_1<\ldots<x_n=b$. Then we say that the step function should have a value $c_i$ on $(x_{i-1}, x_i), 1 \le i \le n$. @AlessandroCodenotti

So, again, should the sequence use the same partition of $[a,b]$, or are there sequences that involve different partitions of $[a,b]$?

shorter: how we build a sequence of step functions?

Alessandro Codenotti

12:05

they can use different partitions

They have to, how would you get a sequence of step functions converging to, say, $x^2$ on $[-1,1]$ otherwise?

Kirill

@AlessandroCodenotti I was sure that partition is held, so that only the values of the functions converge to the given function.

Alessandro Codenotti

If the partition is fixed there aren't many regulated functions

try it with an easy example, $f(x)=x$ on $[0,1]$, fix a partition and see how well the function can be approximated by step functions without changing the partition

Kirill

@AlessandroCodenotti is it possible to find a prescript for step functions, according to your example with $x^2$? Or is that a theoretical approach that says that such a sequence exists, without an idea how to define certain terms?

@AlessandroCodenotti I can imagine this one: for a given intervall $[a,b]$ the $\varphi_i$-s term of the sequence $\varphi_n$ cuts the intervall in $i$ pieces and take the $min$ ob the function $f$ on each piece. I have ordered a value, according to the values of $f$. Is that a proper prescript?

Alessandro Codenotti

$f_n=\sum\limits_{i=0}^{n-1} \left(\frac{i}{n}\right)^2\chi_{[\frac{i}{n},\frac{i+1}{n}]}$ should converge to $x^2$ on $[0,1]$

@Kirill yep, that's the same as the function I wrote

Kirill

@AlessandroCodenotti wow. What does the chi mean in the formula?

Alessandro Codenotti

12:18

$\chi_{A}(x)$, for $A\subseteq\Bbb R$ is called a characteristic function, its value is $1$ if $x\in A$ and $0$ if $x\not\in A$

Kirill

@AlessandroCodenotti super! I have seen it only as a characteristic polynomial till today

Alessandro Codenotti

the step functions are those that can be written as a finite linear combination of characteristic functions of intervals, that is in the form $\sum\limits_{i=0}^n a_i\chi_{A_i}$ with $a_i\in\Bbb R$ and $A_i$ intervals that partition the set on which you want to define the step function

Semiclassical

Is $A_i$ a single interval?

Alessandro Codenotti

yep

Akiva Weinberger

@Typhon What's this a reply to?

Semiclassical

12:26

That bothers me a little, then. It'd seem to exclude a periodic step function

Alessandro Codenotti

we were working on a bounded interval

Semiclassical

Ah, okay

Alessandro Codenotti

yeah the finite linear combintions thing doesn't seem to work very well over the whole of $\Bbb R$

Semiclassical

And anyways if you've got a periodic function all you need to do is specify $f(x)$ on the fundamental period (not sure that's the right terminology but w/e)

Kirill

@AlessandroCodenotti I have heard of linear ocmbinations only according to vectors, so, are there also linear combinations of functions? Or, are functions $\sim$ vectors here?

Alessandro Codenotti

12:36

Consider the set of all functions $[a,b]\to[a,b]$ with $(f+g)(x):=f(x)+g(x)$ and $(\alpha f)(x):=\alpha f(x)$ for $\alpha\in\Bbb R$, this is a vector space

Kirill

@AlessandroCodenotti ok. But characteristic functions are colinear, can they be used as a basis?

Akiva Weinberger

@LucasHenrique By the way, I assumed $n$ was positive for this; if not, $(1+\sqrt{-3})^3=-8$ provides a counterexample

Alessandro Codenotti

No, a basis for that space is going to be an atrocious mess

Akiva Weinberger

but I believe that $1-\sqrt{-3}$ and $1+\sqrt{-3}$, raised to a mutiple of $3$ power, are the only such counterexamples

Kirill

@AlessandroCodenotti wordhippo.com/what-is/another-word-for/atrocious%20mess.html

Alessandro Codenotti

12:41

Extremely ugly

Secret

Actually, is this the correct in the nutshell on how bayanesian works?

in The h Bar, 11 mins ago, by Secret

Meanwhile Bayesianist don't know or don't have the probability distribution of all events they are interested in. They first assign a prior probaility to some given question, and then each trial of an experiment or other incoming evidences will serve to update the probability distribution, so if all the evidences and experiments are not crappy and of high quality, then eventually the probability distribution will converge towards that given by the frequentists

Kirill

@AlessandroCodenotti ok, so that is a extremly ugly infinite basis

Alessandro Codenotti

yeah, what I mean is that there is no nice description of it

Secret

btw, why we cannot define a linear combination with infinite terms?

Julius

Hi. What would be some analogues to initials and final topologies? I am looking for examples in mathematics of something smallest and largest giving some property

Kirill

12:45

@AlessandroCodenotti but we can still name the sum of scaled characteristic functions as a "linear combination"?

Alessandro Codenotti

@Secret Because the basic structure of a vector space isn't enough to make sense of an infinite sum

Secret

Ah I see

Alessandro Codenotti

@Kirill why not? A linear combination is just a sum of vectors with coefficents

Kirill

@AlessandroCodenotti oh true. Somehow I thought they should be basis vectors for that. That was false.

Akiva Weinberger

@Typhon If you were referring to the proof I did for Lucas that $(a+b\sqrt n)^k$ isn't an integer—that was the absolute value on $\Bbb R$, not the norm

(And you meant $x^2-by^2$, not $x^2+by^2$)

Kirill

12:53

can we look at the integral as at the function with three arguments $(a,b,c)$, where $a$ is an integrand, $b,c$ lower and upper bounds?

or does it makes sence only for single variables?

Secret

One problem of that when generalising to multiple variables is that b and c becomes sets (as you are integrating over some hypervolume). One might be able to get around that by defining $\int_{-}(-)d\mu : L^{something} \times \Bbb{F}^n \mapsto L^{something}$, I think...?

and if your integration domain is some manifold, it gets even tricker to define it even though the integral operator is really a linear map

Kirill

13:16

@Secret hard to follow, I do not know manifolds are hypervolumes. As I see the integral of a step function, it is the function itself, defined on a class of functions. I thought it was not bad to add $b,c$ as arguments. But in multivariable thing we have to show, if we integrate on $t$, or $x$, or another nice letter

Secret

Actually, I am not 100% certain whether hypervolumes in integration domains must be manifolds, Danu knows differential geometry better than me.

Multivariable integrals you also need to worry about how to parametrise the integration domain, and IIRC, if you are doing exterior algebra, then you need to worry about the differential elements as they multiply in certain ways that don't commute. If the integration domain is specified, the image of the integral operator should no longer be a class of functions since you have specified some boundary conditions

Put it simply, in 1D, you only need to worry about b,c, that is your integration domain is an interval (b,c)

But in higher dimensions, specifying the integration domain is the same as specifying te boundary conditions

Kirill

@Secret thanks for explanation

inside the integral test of convergence we say that the series $\sum_{k=1}^{\infty}f(k)$ converge iff the integral $\int_{1}^{\infty}f(x)\mathrm{d}x$ exists. Is that essential to start at $k=1$? What happens, if I choose $k=0$, or $k=4$?

is it necessary for $f$ to be monotonically decreasing, and only at the intervall $[1, \infty)$?

Secret

13:32

I don't know real analysis to answer that rigorously, but based on my intuition, I would imagine everything should be fine for monotonic $f$ (because imagine plotting the area you are trying to integrate, then moving k to the left or right should not matter as long that new f(k) don't blow up. however, things can get complicated if f is only a continuous function, cause if the k don't match the lower limit of the integral, then it might end up blowing up because the
limit failed to converge due to some "cancellations"

Manolis Lyviakis

can some1 help ? math.stackexchange.com/questions/2377690/…

Kirill

@Secret but there is a small space to play, isn't it? If $f$ is defined on $[0,\infty)$, monotonically decreasing, then "$\sum_{k=0}^{\infty}f(k)$ converge" $\Leftrightarrow \int_{0}^{\infty}f(x)\mathrm{d}x$ exists" should also work?

Secret

I think if f is monotonic, you will probably be fine, but if f has all sorts of turning points, and has positive and negative components, then its convergence may be sensitive to the domain of integration (i.e. where you put k)

Kirill

@Secret thats clear. So, seems like the monotony is essential, not the interval itself

Secret

But remember, I am not a real analysis student, so please double verify what I said youself as all of that is just my intuition and basic calculus

Semiclassical

13:42

hi @MikeMiller

Mike Miller

hiya

Semiclassical

There was something I wanted to ask you, but not I can't remember it. :/

something something Morse homology, I think

Farhad Rouhbakhsh

math.stackexchange.com/questions/2377669/…

I'd appreciate it if anyone could help. 10 hours thinking on this problem...

Farhad Rouhbakhsh

14:05

The question is false

Semiclassical

@FarhadRouhbakhsh The hint I'd give, I guess, is to consider the triangles ABC and ACD.

From the tangency conditions, you know quite a bit about these triangles right off the bat.

The Raiders of Las Vegas

14:27

hi @EricSilva

Eric Silva

Hi

The Raiders of Las Vegas

How's it going?

Kirill

15:01

why the center of the series is called the center of the series?

Semiclassical

Do you mean in the sense of a power series, e.g. $a$ is the center of the series $\sum_{n=0}^\infty a_n (z-a)^n$?

Kirill

@Semiclassical exactly. I read everywhere that $a$ is called so, but have no idea what does it mean. It has something to do with disks I guess.

Akiva Weinberger

Center of the interval it converges in

or the disk it converges in if this is in $\Bbb C$

Kirill

15:16

@AkivaWeinberger how does it look like visually? All $x$ outside the disk seem to be chaotic, but being inside the disk they tend to some point? Like a vacuum cleaner?

Semiclassical

One way to see the significance of $a$ is that the only term in the power series I wrote which is nonzero at $z=a$ is the constant term.

And the closer you are to $z=a$, the smaller that $(z-a)^k$ will be. So the power series can be understood as telling you not only $f(a)$ but also how $f(z)$ behaves near $z=a$.

(So long as the series is convergent. If you've got $\sum_{n=0}^\infty n!(z-a)^n$, for instance, then the radius of convergence is zero and the only point where this series makes sense is $z=a$ itself.)

Kirill

I do not understand the concept.

Semiclassical

That's an empty statement. What don't you get about it?

Kirill

@Semiclassical where we are (plane, $\mathbb{C}$, etc.), what we are doing, with what purpose. Is the radius of converges a visual disk on the plane, is it a methaphorical name? What does it mean, that the series has a radius, how can we measure convergence?

Semiclassical

Well, let's focus on the case where the center is zero. Then a power series will be of the form $f(z)=\sum_{n=0}^\infty a_n z^n$.

Kirill

15:28

Disks have radius, series is a sequence of the partial sums. What is the radius of the sequence of the partial sums?

@Semiclassical ok

Semiclassical

Now, when $z=0$ that's just $f(0)=a_0+a_1(0)^1+a_2(0)^2+\cdots =a_0$. So $f(0)$ is always well-defined.

Kirill

@Semiclassical so, we take the series as a function?

is $z \in \mathbb{N}$?

Semiclassical

No. $z\in \mathbb{C}$.

Felix.C

I have a bit of an embarrassing question, why is it again that we have $r dr d \phi$ when we substitute $dxdy$ with polar coordinates? Actually when having $x=r cos \phi$ and $y = r sin \phi$ I get $dxdy = cos^2 \phi dr d \phi$ or $dxdy = -sin^2 \phi dr d \phi$ Depending if I'm doing $dx/dr$ and $dy/d \phi$ or vice versa...

Typhon

@AkivaWeinberger Your starred post. "Don't let society tell you what the norm is"

Semiclassical

15:30

However, the tricky thing is that this series is not necessarily convergent for all $z$.

@felix Jacobian.

Felix.C

@Semiclassical I don't see how it follows from Jacobian

Manolis Lyviakis

What do we call when we have a limit of a sequence and we compute the limit as a function ?

Semiclassical

Then you're not doing the Jacobian right. It will follow immediately from that if you've computed the partial derivatives correctly.

Typhon

r dr dt is the area relationship isn't it?

Kirill

@Semiclassical why do we take series as a function? I thought series is just s big thing that exists.

Manolis Lyviakis

15:33

say $\frac{n}{e^n}$ we compute the $\frac{x}{e^x}$ ?

Typhon

doesn't r dr dt represent the area of the "area" we are computing as the individual riemann sum rectangles in three dimensional space?

Kirill

@Semiclassical and, by the convergence you mean, "an infinite sum has a finite value", right?

Semiclassical

@Kirill If one is interested in a disk of convergence, then one is interested in actual values of $z$.

Felix.C

@Semiclassical
I have this Jacobian:
https://wikimedia.org/api/rest_v1/media/math/render/svg/032138d29af7b8bc7aa38fa7b63e4112385f9274
If you mutliply the diagonals you get different values...

Manolis Lyviakis

how do we call this thing?

Semiclassical

15:33

No, you don't. @Felix.C

Manolis Lyviakis

How do we call the procedure working on functions to get results in sequences?

Semiclassical

The determinant of a matrix \begin{pmatrix} a & b \\ c & d\end{pmatrix} is $ad-bc$.

Kirill

@ManolisLyviakis don't understand you. Call it work.

Semiclassical

So the determinant of that is $(\cos \phi)(r\cos \phi)-(-r\sin \phi)(\sin \phi)=r(\cos^2\phi+\sin^2\phi)$.

And that last factor had better be familiar.

Felix.C

@Semiclassical I'm not asking about the determinant!!! I'd like to know why we we substitute $dxdy$ with $dxdy = r \phi dr d \phi$

Manolis Lyviakis

15:35

there is a theorem that connects functions with limits of sequence

Semiclassical

What you're asking is precisely the Jacobian (determinant).

Typhon

@Felix.C he just told you

user8469759

hi guys

Kirill

@ManolisLyviakis I don't know what you mean.

Semiclassical

$dx\,dy = \left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,dudv$

That || is precisely the determinant of the Jacobian matrix.

Typhon

15:37

Semiclassical

Yeah, that's a good picture.

user8469759

if $x_0 = \left(\prod_{j=0}^{+\infty}(1+2^{-2j}) \right)^{-1}$ and $x_{j+1} \leq(1+2^{-2j})x_j + 2^{-j}$ is it true that $x_{j+1} < x_j$?

Typhon

A laymen's answer.

Semiclassical

Right. It's obviously heuristic, but it captures the right intuition.

BAYMAX

Lewis Carrol :)

Typhon

15:38

@Semiclassical I think using this "Jacobian" is overdoing it. I've never heard of it and I know double integrals just fine.

Alessandro Codenotti

@Typhon how do you do change of variables apart from the usual ones then?

Semiclassical

If you're only interested in polar coordinates, then the above picture is fine as heuristic.

Typhon

@AlessandroCodenotti what do you mean?

you don't do other coordinate systems...

Semiclassical

But if you want to do more general changes of variables, then the Jacobian formula I gave earlier is indispensable

HAHAHAHAHAHAHA

SteamyRoot

^

Semiclassical

15:40

Oh thats funny.

Alessandro Codenotti

How do you deal with a double integral where a smart change of coordinates is needed?

Typhon

(well, most people don't)

Kirill

@user8469759 that is such a scarry formula

Typhon

@AlessandroCodenotti You do a normal u-substitution.

changing coordinates is pretty much in a weird way just another way of doing substitution. Or at least, I remember the professor doing something to show it's equivalent.

Semiclassical

I actually agree with that, if it's understood properly.

Felix.C

15:42

Ok thanks a lot guys, my actual, perhaps a bit strange question was, why isn't it the same if I multiply the partial derivatives of x and y with respect to r and $\phi$ not the same if I'm doing vice versa...e.g. $(dx/dr)*(dy/d \phi) \ne (dy/dr)*(dx/d \phi)$ this confused me in the first place...

Semiclassical

Suppose we have $x=x(u,v),y=y(u,v)$. (Ted would probably object to this notation, and he'd probably be right, but oh well)

Typhon

u-substitution is a method in solving single integrals. Assuming the integral isn't sufficiently advanced to not be in a table (somewhere), I doubt that most u-substitutions or other basic methods of solving integrals will fail.

Semiclassical

Then $dx=\frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv$ and $dy=\frac{\partial y}{\partial u}du+\frac{\partial y}{\partial v}dv$

Typhon

@Semiclassical yes, but when doing a double integral you only need to integrate by either x or y once while treating the other as a constant. Hence, the only difficulty is if there are arbitrary constants in the wrong place.

for instance

user8469759

@Kirill is it?

Alessandro Codenotti

15:44

Like $\displaystyle\int_A \frac{y}{x^4}\mathrm{d}x\mathrm{d}y$ where $A$ is the region bounded between $y=x^2$, $y=2x^2$, $y=x$ and $y=2x$ is a nice integral which becomes much nicer after a change of coordinates

Semiclassical

^

Typhon

$x^y = yx^{y-1}$ isn't that bad

to integrate for x

Semiclassical

That's not really the problem here. The problem is the range of integration.

Typhon

y would be a little trickier to do without looking up the formula

@Semiclassical oooh

XD

Alessandro Codenotti

@Semiclassical I mostly think about change of coordinates as a way to turn the domain of integration into a rectangle or a circle tbh

Semiclassical

15:46

At least, I think that's where the problem is. I'm not really trying to visualize the domain of integration.

@AlessandroCodenotti Same.

Typhon

aren't most of the 16 standard coordinate systems based around parabolas and various hyperbolic/regular trig functions anyways?

and most of those regions would be easy to represent but I might not be thinking that through.

Semiclassical

Anyways, if one plugs in the above differentials into $dx dy$, one formally has $$dx\,dy=(\frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv)(\frac{\partial y}{\partial u}du+\frac{\partial y}{\partial v}dv)$$

Typhon

uuuh

ok

Semiclassical

If one 'multiplies' those out, one gets four terms with $du\,du$, $du\,dv$, $dv\,du$, and $dv\,dv$ respectively.

Felix.C

@Semiclassical Ok thanks a lot for your help. I confused derivatives with infinitesimals..

Kirill

15:49

@user8469759 sure.

Typhon

@Semiclassical ook

Semiclassical

And now I'm trying to figure out how to justify the next step, at least heuristically.

Typhon

...

Semiclassical

To do it right, one needs the concept of a wedge product.

Alessandro Codenotti

something something differential forms and other arcane magic

Semiclassical

15:50

Right. And something something oriented area.

Alessandro Codenotti

the smacks would be flying if Ted were around

Semiclassical

Basically, you interpret $dxdy$ as $dx\wedge dy$ where $\wedge$ is the so-called wedge product.

The key property of which is that, if $a,b$ are 1-forms, then $b\wedge a=-a \wedge b$.

That's basically intended to capture the orientation of your area element, with dy dx having an opposite orientation as dx dy

If you take that for granted, then you immediately conclude that $du\wedge du=dv\wedge dv=0$ (an antisymmetric product of equal objects is zero by symmetry)

user8469759

@Kirill but any help whether or not it is monotonic?

Secret

I wonder how area preserving transformations are like. I am guessing that will require some jacobain like term with determinant 1?

Felix.C

@Semiclassical I though again about what you said..Shouldnt it be
$dx= [ \frac{\partial x}{\partial u}du \quad \frac{\partial x}{\partial v}dv]$ and $dy= [ \frac{\partial y}{\partial u}du \quad \frac{\partial y}{\partial v}dv]$
With dx and dy being row vectors?

Semiclassical

15:56

And similarly $dv\wedge du=-du\wedge dv$. Hence $dx\wedge dy$ simplifies to $$dx\wedge dy = \left(\frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv\right)\left(\frac{\partial y}{\partial u}du+\frac{\partial y}{\partial v}dv\right)=\left(\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right)\,du\wedge dv$$

That term in front, though, is the determinant of a 2-by-2 matrix. That's the Jacobian.

@Felix.C No, not really.

$dx$ is not a vector.

Or, rather, whatever $dx,dy$ are, they shouldn't be different kinds of objects than $du,dv$.

But again this really all comes back to the Jacobian.

Secret

because, inspired from homeomorphisms, if given some generic definite integral (I have no idea how will the indefinite integral look other than sheer messiness)

$$\int_S f(x)dx$$

then suppose I can find some are preserving transformation $T$ such that

$$\int_S f(x) dx = T\left(\int_S g(x)dx\right)$$
where g is something simple like a square, then in theory all definite integrals will be computable

Semiclassical

The above can probably be made rigorous, but for me I just use it as a convenient mnemonic for how the area element transforms under changes of variables.

Felix.C

@Semiclassical I'm a bit overstrained with all this generosity but at least if we have our previous x such that $x:\mathbb{R}^2 \to \mathbb{R}$ then its a row vector

with $(x,y) \in \mathbb{R}^2$

Semiclassical

$x$ is not R^2 to R^2.

Felix.C

Sry made some edits

Semiclassical

15:59

Hence whatever dx is, it's not a vector. It's a scalar.

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