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00:00 - 17:0017:00 - 00:00

12:07 AM
jk it's $76/201$
which is approximately $.378$
 
12:49 AM
Hello one and all! Is anyone here familiar with planar prolate spheroidal coordinates? I am reading a book on dynamics and the author states
If we introduce planar prolate spheroidal coordinates $(R, \sigma)$ based on the distance parameter $b$, then, in terms of the Cartesian coordinates $(x, z)$ and also of the plane polars $(r , \theta)$, we have the defining relations
$$r\sin \theta=x=\pm R^2−b^2 \sin\sigma, r\cos\theta=z=R\cos\sigma$$
I am having a tough time visualising what this is?
 
Is there any short way to check the type of sigularity?
1
Q: Check the singularity of the function at $z = 0$

StrugglerConsider the function $f(z) = Sin\left(\frac{1}{cos(1/z)}\right)$, the point $z = 0$ a removale singularity a pole an essesntial singularity a non isolated singularity Since $Cos(\frac{1}{z})$ = $1- \frac{1}{2z^2}+\frac{1}{4!z^4} - ..........$ $$ = (1-y), where\ \ y=\frac{1}{2z^2}+\frac{1}{4!...

Here, He has done the problem using Laurentz series.
Is there any aliter way?
By Finding the limit at zero. Right?
$\lim_{z\to 0}\sin(\frac{1}{\cos(\frac{1}{z})})\neq 0$ or $\infty$
So, Given singularity is essential.
Am I correct?
 
 
6 hours later…
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.
 
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
Q: Non-isolated singularity points

Q.matinI 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 $...

 
 
2 hours later…
10:09 AM
To be added: (infinite, finite) x (bound, unbounded) 2x2 table
 
Pie
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.
 
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
 
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
 
@ÍgjøgnumMeg Thank you.
 
wtf lol. Probably not the place mate
Damn it it doubled everything in edit
 
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.
 
Pie
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
 
@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
 
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
 
@BalarkaSen I will be glad to receive them. My email id is user170039@gmail.com.
 
Pie
@AkivaWeinberger yes i have two question can tell which one is half range and which is not?
 
yo yo yo yo yo
 
1:18 PM
I know this must've been asked before but I couldn't find it, math.stackexchange.com/questions/3235652/…
 
Pie
@AkivaWeinberger
 
yo Mr. P
 
Pie
@FuzzyPixelz i have looking for similar question but could not find any
 
Hey chat.
 
Pie
@Ultradark i need help
 
1:21 PM
I don't know f-series
 
@Pie OK so $f$ is defined from 0 to pi, right?
 
Pie
Yes
 
But the period of $f$ is $2\pi$
(Do you have LaTeX on?)
 
Pie
I do not know
 
It turns the dollar sign stuff into math symbols
 
Pie
1:22 PM
How can i do that?
 
There's a link on the top right^>
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
 
@AbhasKumarSinha It's limit will likely go towards infinity, so who knows?
 
Pie
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.
 
Pie
1:39 PM
Ok you are right @AkivaWeinberger
@AkivaWeinberger what about this question?
 
@Pie Type "https://i.imgur.com/D0QlhxQ.jpg"
 
Pie
Ok
 
@pie pinging everybody in the room will get you suspended
 
Pie
@skullpetrol ok i will not do it.
 
np
:-)
 
Pie
1:46 PM
@skullpetrol can you help me?
 
I have to leave soon, sorry.
 
Pie
@skullpetrol np
Lol
 
can anybody help me solve this equation
 
Pie
@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
 
Pie
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?
 
Pie
Range is shown in right of this image
@AkivaWeinberger
 
Yes
I know
I know the answer to the question I asked
You tell me
 
Pie
Ok so is it half range or not?
 
1:50 PM
What range is f defined on @Pie
 
Pie
@AkivaWeinberger
It is 0 to pie
 
OK
And how big is the period?
 
Pie
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
 
Pie
Om
Ok
@AkivaWeinberger
 
2:10 PM
Let $X$ be any nonempty set and $\sim$ be any equivalence relation on $X$. Then are the following true:
(1) If $x=y$ then $x\sim y$.
(2) If $x=y$ then $y\sim x$.
(3) If $x=y$ and $y=z$ then $x\sim z$.
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.
This question arised from an attempt to characterize equality on a set $X$ as the intersection of all equivalence relations on $X$.
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?
 
@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?
 
@AlessandroCodenotti The pair $(x,x)\in \sim$ for all $x\in X$.
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
 
@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
 
@AlessandroCodenotti I completely forgot about it. Silly me.
Anyway, thanks @AlessandroCodenotti.
 
3:05 PM
https://stackoverflow.com/questions/56250062/how-to-calculate-the-double-integration-in-r

Anyone please help me with this question.
 
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?
 
@AlessandroCodenotti Modus Ponens.
 
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
 
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"
 
always compliment police officers on how shiny their costume is
this has been an important social skill I have learned
 
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"
 
@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).
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.
 
3:23 PM
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
 
@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!~
 
You'll have a great time.
 
yeah, should be cool
 
Talk all prepared? :P
 
4:42 PM
slides, yes
mentally...we'll see
 
@TedShifrin he said he saw the proof but was looking for some intuition
 
I understand.
 
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.
 
So math is acting passive-aggressively to you?
 
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
Q: Complex monic polynomial $f(z)$ with $|f(z)|\leq 1$ for $|z|\leq 1.$

neelkanthLet $f(z)=z^{n}+a_{n-1}z^{n-1}+\cdot\cdot\cdot+a_{0}$ 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...

 
@TedShifrin kind of xD
 
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.
 
if a function is Taylor expandable at $x=a$ (satisfies the necessary conditions for an expansion) then in a certain neighbourhood of $a$, the function can be approximated as you wish
 
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