Mathematics - 2019-05-22 (page 1 of 2)

7:24 AM

This is very dumb but I don't see how the translation operator, translating a vector by a constant vector c is linear. T(x + y) = x + y + c which is not the same as T(x) + T(y) = x + y + 2c

It's not a linear transformation: it doesn't even preserve the origin! It's an "affine transformation"

Exactly, then why is the translation operator in Quantum mechanics called linear ?!

I don't know quantum mechanics

@Semiclassical might be able to give you an answer

Yea it's been eating my head since morning. Could you help @Semiclassical

From what I see by looking it up it acts on a wavefunction $\psi(r)$ by sending it to $\psi(r + x)$

That's a perfectly fine linear transformation on the Hilbert space of wavefunctions or whatever

It's not translating by a vector in $\Bbb R^n$ like you said

7:35 AM

How's that a linear transformation on the Hilbert space

$T_x(\psi + \phi)(r) = (\psi + \phi)(r + x) = \psi(r + x) + \phi(r + x) = T_x(\psi)(r) + T_x(\phi)(r)$

i.e., $T_x(\psi + \phi) = T_x(\psi) + T_x(\phi)$ as wavefunctions. That's the same thing as saying $T_x$ is a linear transformation on the space of wavefunctions

Oh I see.

Well it will take me getting used to this function space.

Q: Non-isolated singularity points

I am thinking about a complex function where the singularities is a dense set

something like $\frac{1}{1-[\Bbb{Q}]}$ where $[\Bbb{Q}]$ is the rational indicator functionfunction

i wonder if all the rational points become essential singularities due to the density

5

I am having trouble understanding non-isolated singularity points. An isolated singularity point I do kind of understand, it is when: a point $z_0$ is said to be isolated if $z_0$ is a singular point and has a neighborhood throughout which $f$ is analytic except at $z_0$. For example, why would $...

complex-analysis

10:09 AM

To be added: (infinite, finite) x (bound, unbounded) 2x2 table

10:31 AM

Hi guys

I need help

I have math exam tomorrow

I want to know values of

11:11 AM

@Alessandro hey

quick tell me, what's a good notation for the first coordinate projection $X \times Y \to X$?

i have $\pi, p, \pi_1$ all in use

the first two are projections and the third is the fundamental group :3

Oh there's a chat! Nice

Is the chat TEXed?

I'm on mobile and It seems as if it isn't.

tinyurl.com/cfqcvpc

PM 2Ring

11:41 AM

@Simone We don't have automatic MathJax in chat, but there are scripts that can be used. The bookmarklets kind of work on mobile. See meta.stackexchange.com/a/220976/334566

Thank you Balarka 😊

And PM 2Ring

PM 2Ring

No worries. There's currently some kind of technical problem affecting the Stack Exchange chat network. It's been pretty flaky for several hours. Hopefully, it will be back to normal in the next hour or two, when business hours commence on the east coast of the USA...

11:56 AM

Nice! I'm using the Brave browser and the Chrome extension works like a charm

$(x_i +1)^2$

@Alessandro I ended up using $\text{proj}$ because I'm a hipster fuck

@Balarka why not $\pi_X$

Hi @Balarka, saw your message only now

$\pi^1$?

12:12 PM

What's the difference between Mathematics Stack exchange an Mathoverflow?

the first is mathunderflow

@ÍgjøgnumMeg didn't cross my mind

good suggestion

@AlessandroCodenotti lol

Well, you see, I was projecting to $\widetilde{X}$

so $\pi_{\widetilde{X}}$ is uh

@Simone Mathoverflow is primarily for research level mathematics, while Mathstack is for mathematics of any level (and if it is deemed worthy it will be migrated to Mathoverflow)

@ÍgjøgnumMeg Ah, so it's under the same umbrella.

Hey @Alessandro

Can I post questions in here that I feel to embarrassed to ask on the main forum? XD

12:22 PM

Of course

Unless they are "Where do babies come from", I feel that'd be considered off topic

Hi @ÍgjøgnumMeg

How are you?

@ÍgjøgnumMeg ah... so question num. 2 then:

@Alessandro I'm good, working and preparing for the masters as usual! How are you?

I did my seminar talk yesterday, went pretty well, so now I have more time for the other courses

When are you going to move to Germany again?

The absolute value of a complex number $z=x+iy$ is defined as $\sqrt{x^2+y^2}$. Hence, when evaluating the absolute value of $x+i$ I get the number $\sqrt{x^2 +1}$; but the answer to the problem says it's actually just $x^2 +1$. Why?

Where's the error?

12:30 PM

@Alessandro cool :) Moving probably start of October

@Simone you're right, it should be under a square root

mmh, I probably should ask this on the forum.
The full problem asks me to show that we can choose $log(x+i)$ to be $$log(x+i)=log(1+x^2)+i(\frac{pi}{2} - arctanx)$$
So I'm trying to find the polar coordinates (absolute value and an argument $\theta$) of $x+i$ to then apply the $log$ function on it

err

Flagged

nitsua60

@ÍgjøgnumMeg Thank you.

wtf lol. Probably not the place mate

Damn it it doubled everything in edit

nitsua60

I'm going to remove that second half as uninteligible, given the flag-removal of the first half.

12:45 PM

@nitsua60 hey can you fix my question above too? It looks like I cannot edit it any further. It doubled all the formulas.

@Simone I don't see anything duplicated

ah I reloaded the page an it fixed the issue.

Hi I have question

@ÍgjøgnumMeg @ÉricoMeloSilva

In fourier series

Half range questions

@Simone

@Secret @skullpetrol

Generaly question is written to detemine that it is half range series question?

@Adam @AlessandroCodenotti @Albas @Astyx @AkivaWeinberger

How to determine that it a half range series question?

@Mast @Mathphile @MaryStar

Mast

@Pie Stop asking random people. You're not a vampire, so don't act like one.

@Pie pinging everyone in the chat is bad practice, if you want to ask a question just post it and somebody will look at it if they decide to, if not just post it on main

Abhas Kumar Sinha

1:04 PM

Knock Knock let the Devil in...

A question, what's the ending of mathematics will look like?

Seriously

[hop]

@AlessandroCodenotti

@Pie If I remember right, that happens when the series has only sines or only cosines

and then you only need half the period to determine the function everywhere

but please don't ping everyone like that

user131753

@BalarkaSen I will be glad to receive them. My email id is user170039@gmail.com.

@AkivaWeinberger yes i have two question can tell which one is half range and which is not?

imgur.com/jBXmAVq

yo yo yo yo yo

imgur.com/RhkhtWG

FuzzyPixelz

1:18 PM

I know this must've been asked before but I couldn't find it, math.stackexchange.com/questions/3235652/…

@AkivaWeinberger

yo Mr. P

@FuzzyPixelz i have looking for similar question but could not find any

Hey chat.

@Ultradark i need help

1:21 PM

I don't know f-series

@Pie OK so $f$ is defined from 0 to pi, right?

Yes

But the period of $f$ is $2\pi$

(Do you have LaTeX on?)

I do not know

It turns the dollar sign stuff into math symbols

1:22 PM

How can i do that?

There's a link on the top right^>

tinyurl.com/cfqcvpc

It doesn't really matter that much right now

So the period of f is 2pi, but we only have it defined from 0 to pi

so we only know it on half of its period

Mast

@AbhasKumarSinha It's limit will likely go towards infinity, so who knows?

Can you tell me which link is of fourier series and which link is of half range?

@AkivaWeinberger is this half range?

@Pie Yes, for the reason I gave

The second one defines it for 0≤x≤2pi

so that one's a full range

because it's defined for the entire period.

They're both Fourier series

but the first one is half range.

1:39 PM

Ok you are right @AkivaWeinberger

@AkivaWeinberger what about this question?

imgur.com/D0QlhxQ

@Pie Type "https://i.imgur.com/D0QlhxQ.jpg"

Ok

skullpetrol

@pie pinging everybody in the room will get you suspended

@skullpetrol ok i will not do it.

skullpetrol

np

:-)

1:46 PM

@skullpetrol can you help me?

skullpetrol

I have to leave soon, sorry.

@skullpetrol np

Lol

can anybody help me solve this equation

@AkivaWeinberger can tell me is fourier or half range?

If you type or copy-past "https://i.imgur.com/D0QlhxQ.jpg" it shows up as an image

1:49 PM

@Ultradark which equation??

$\exp\Big(\frac{n}{\ln(\pi(x))}\Big)=\pi(x)$ for $n=1,2,3,...$

@Pie What range is $f$ defined on?

Range is shown in right of this image

@AkivaWeinberger

Yes

I know

I know the answer to the question I asked

You tell me

Ok so is it half range or not?

1:50 PM

What range is f defined on @Pie

@AkivaWeinberger

It is 0 to pie

OK

And how big is the period?

Since range sums upto pie/3 + pie/3 + pie/3= pie

2pi, right?

So you only have half of the period

(The period is 2pi, since sin(x) and cos(x) have period 2pi)

Also -

another hint should have been that the series it gives you is all cosines

Half-range questions are either all sines or all cosines

Since our function is defined from [0,pi], we need to find some way to define it on the other half of the period ([-pi,0]).

If we use cosines, it will be extended in such a way that you get an even function

If we use sines, it will be an odd function

I recommend drawing a picture of this

Om

Ok

@AkivaWeinberger

user131753

2:10 PM

Let $X$ be any nonempty set and $\sim$ be any equivalence relation on $X$. Then are the following true:

user131753

(1) If $x=y$ then $x\sim y$.

user131753

(2) If $x=y$ then $y\sim x$.

user131753

(3) If $x=y$ and $y=z$ then $x\sim z$.

user131753

Basically, I think that all the three properties follows if we can prove (1) because if $x=y$ then since $y=x$, by (1) we would have $y\sim x$ proving (2). (3) will follow similarly.

user131753

This question arised from an attempt to characterize equality on a set $X$ as the intersection of all equivalence relations on $X$.

user131753

2:18 PM

I don't know whether this question is too much trivial. But I have yet not seen any formal proof of the following statement : "Let $X$ be any nonempty set and $∼$ be any equivalence relation on $X$. If $x=y$ then $x\sim y$."

What's your definition of an equivalence relation?

user131753

@AlessandroCodenotti A given binary relation $\sim$ on a set $X$ is said to be an equivalence relation if and only if it is reflexive, symmetric and transitive.

what does reflexive mean?

user131753

@AlessandroCodenotti The pair $(x,x)\in \sim$ for all $x\in X$.

user131753

Of course this would mean that $x=x$ implies that $x\sim x$. In other words the statement that this proves is $\forall x(x=x\to x\sim x)$ but we want to prove that $\forall x\forall y(x=y\to x\sim y)$

2:26 PM

Sure, but you can substitute equal terms for equal terms

user131753

@AlessandroCodenotti So you are using the substitution axiom of equality in FOL right?

2:39 PM

That is definitely a new person, not going to classify as RHV yet as other users have already put the situation under control it seems...

(comment on many many posts above)

In other news:

> C -2.5353672500000002 -1.9143250000000003 -0.5807385400000000
C -3.4331741299999998 -1.3244286800000000 -1.4594762299999999
C -3.6485676800000002 0.0734728100000000 -1.4738058999999999
C -2.9689624299999999 0.9078326800000001 -0.5942069900000000
C -2.0858929200000000 0.3286240400000000 0.3378783500000000
C -1.8445799400000003 -1.0963522200000000 0.3417561400000000
C -0.8438543100000000 -1.3752198200000001 1.3561451400000000
C -0.5670178500000000 -0.1418068400000000 2.0628359299999999

probably the weirdness bunch of data I ever seen with so many 000000 and 999999s

@user170039 yes

user131753

@AlessandroCodenotti I completely forgot about it. Silly me.

user131753

Anyway, thanks @AlessandroCodenotti.

noob

3:05 PM

https://stackoverflow.com/questions/56250062/how-to-calculate-the-double-integration-in-r

Anyone please help me with this question.

user131753

But I think that to prove the implication for transitivity the inference rule an use of MP seems to be necessary. But that would mean that for logics for which MP fails we wouldn't be able to prove the result. Also in set theories without Axiom of Extensionality the desired result will not hold. Am I right @AlessandroCodenotti?

MP?

user131753

@AlessandroCodenotti Modus Ponens.

ONN's Autistic Reporter II

Ah, I see

I don't know, I'm not familiar with weak logics and related questions

I like FOL and working in strong set theories :P

Adam

►

user131753

3:11 PM

@AlessandroCodenotti A logic without MP is not necessarily weaker than FOL.

@AlessandroCodenotti I wonder if there's a such thing as a "strong independent theory"

Adam

always compliment police officers on how shiny their costume is

this has been an important social skill I have learned

user131753

Wait, @AlessandroCodenotti, when you were talking about substitution axiom, what axiom precisely were you talking about?

I don't have a precise formulation in mind, but in all presentations of FOL I've ever seen there's an axiom whose intuitive interpretation is "you can substitute equal terms for equal terms"

user131753

@AlessandroCodenotti A precise formulation would help in this case because I am trying to understand whether a proof of the statement which I mentioned at the outset depends really on the equality axioms or the FOL axioms (without equality axioms).

user131753

3:21 PM

This would allow in some cases to define an "equality like" relation for set theories for which we don't have the Axiom of Extensionality.

I don't know, looks like it's really needed though

And set theories without extensionality are very weird

set theory is weird*

3:42 PM

Can someone give an intuitive explanation why $\mathcal{O}(x^2)-\mathcal{O}(x^2)=\mathcal{O}(x^2)$. The context is Taylor polynomials, so when $x\to 0$. I've seen a proof of this, but intuitively I don't understand it.

set theories without extensionality -> urelements

Conjecture: Anything is not weird enough if it has a positive description

Counterexample: Axiom of choice

4:15 PM

my head hurts

both from my new pair of glasses of 8D & 9D and this chat

does anyone here get demotivated often that they aren't good enough at math and leave it for a few days then come back to doing math?

kinda like a motivation battery

GFauxPas

@schn write out the inequalities that define big O for two functions f and g. Then add the inequalities.

Oh, you want intuition.

Well $\vert K_1x^2 - K_2 x^2 \vert = \vert K_1-K_2 \vert x^2$

Define $K_3 = \vert K_1 - K_2 \vert$

@GFauxPas sort of like (big infinity)-(small infinity)~ infinity?

@schn: The minus is irrelevant (for example, the thing you are subtracting could be negative). When you add two things that are of the order of $x^2$, of course the sum is the same (or possibly smaller). For example, $3x^2-x^2=2x^2$. You could have $x^2+(x^3-x^2)=x^3$, which is still $\mathscr O(x^2)$.

@GFauxPas: You only know $|f(x)|\le K_1 x^2$ and $|g(x)|\le K_2 x^2$, so that won't be a valid proof, of course.

4:31 PM

very informally, it means( I might be wrong here), the growth of the function is still dominated by order $2$ of $x$

@SubhasisBiswas: He's talking about $x\to 0$, so "growth" is perhaps the wrong intuition :P

@TedShifrin ahhhh

then it is like $x \approx \sin x $

when $x \to 0$.

amirit

Right. Indeed, $\sin x - x = \mathscr O(x^3)$.

hi chat

Heya @Semiclassic

4:37 PM

about 7 hours until my flight

Oh, how exciting!~

yep

You'll have a great time.

yeah, should be cool

Talk all prepared? :P

4:42 PM

slides, yes

mentally...we'll see

GFauxPas

@TedShifrin he said he saw the proof but was looking for some intuition

I understand.

GFauxPas

K

I wish I was more suited for this chat. I don't know much of mathematics. I try. I fail. Get demotivated. Don't study for days. Then go all in.

any cure for this cycle? I am just asking for some help

It's not a law that everyone must love mathematics unwaveringly :P

4:49 PM

hi @Ted

hi demonic @Alessandro

well, I love it. But you know, small arguments keep happening all the time in almost every relationships.

that's how I like to see it.

Q: Complex monic polynomial $f(z)$ with $|f(z)|\leq 1$ for $|z|\leq 1.$

So math is acting passive-aggressively to you?

N. Maneesh

Suppose there is $a_k\neq 0$. So, $f(z)=z^n+a_k z^k$. Consider a point $z=1$, It doesnot implies $|f(z)|\leq 1$. Hence $a_k=0, \forall k$. So, $f(z)=z^n$

0

Let $f(z)=z^{n}+a_{n-1}z^{n-1}+\cdot\cdot\cdot+a_{0}$ be a complex polynomial such that $|f(z)|\leq 1$ for $|z|\leq 1.$ I have to prove that $f(z)=z^{n}.$ I tried it as As $|f(z)|\leq 1$ for $|z|\leq 1$ we must have coefficient $a_{0},a_{1}\cdot\cdot\cdot a_{n}$ to be zero because by triangul...

complex-analysis polynomials complex-numbers

@TedShifrin kind of xD

N. Maneesh

4:51 PM

Is my proof correct?

No @N.Maneesh

Consider $z^n + z^3 - z^4$.

@GFauxPas @TedShifrin Thanks for the replies. Now, why is it we're only interested when $x\to 0$? When we do a taylor approximation cantered at x=0, aren't we interested in all the values of our approximation, even those not near 0?

Yes, @schn, but the key thing is the size of the error for small $x$.

Why though? Why is that more important than the error for other values not close to x?

Indeed, one thing a lot of texts don't emphasize is this: if $P$ is a polynomial of degree $\le n$ and $f(x)-P(x)=\mathscr O(x^{n+1})$, then $P$ is the (unique) Taylor polynomial of degree $n$ of $f$ at $0$.

4:55 PM

@TedShifrin So if we were to center the taylor polynomial at x=5, we'd only be interested for x close to 5?

Yes. Then we'd talk about $\mathscr O((x-5)^k)$.

Of course, Taylor polynomials can give you interesting information away from $x=0$, too. That's the point of the remainder formulas.