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Socrates
Socrates
00:00
@SylentNyte better computers
Sylent Nyte
Sylent Nyte
@Socrates yea ive already b4een running the script for an hour
mick
mick
00:12
Hi all
Latest question
1
Q: Conjecture about $A(z) = \lim b^{[n]} ( c^{[n]} (z) ) $

mickLet $b(z),c(z)$ be analytic on the strictly positive reals. Let $^{[*]}$ denote composition. Conjecture : If $A(z) = \lim b^{[n]} ( c^{[n]} (z) ) $ Such that 1) the limit ( $A(z)$ ) exists for all strictly positive real $z$. 2) the sequence $a_n(z) = b^{[n]} ( c^{[n] } (z) ) $ is bounded i...

complex-analysis limits inequality convergence dynamical-systems
Mary Star
Mary Star
@StellaBiderman We have that $\sum_{n=1}^{\infty}\frac{1}{(1+x^4)^n}=-\frac{1}{x^4}$, right? So, for $x\neq 0$ the series converges, and from the Weierstrass criterion, it follows that $\displaystyle{\sum_{n=0}^{\infty}\frac{\sin (x)}{(1+x^4)^n}}$ converges uniformly. But what about $x=0$ ?
Socrates
Socrates
can there be a proof about proofability, without actually proofing something?
mick
mick
00:39
Yes
When the number of tools for a potential proof have a known countable structure , we can sometimes do it
But USUALLY we show undecidable in zfc or such , although that might be equivalent to the above sometimes
Hope those mean something to you
@robjohn @Ramanujan my last question may be at your level
mick
mick
01:09
What does a closed question imply ? No more possible to comment / answer ?
mick
mick
01:26
Nevermind
1
Q: Conjecture about $A(z) = \lim b^{[n]} ( c^{[n]} (z) ) $

mickLet $b(z),c(z)$ be analytic on the strictly positive reals. Let $^{[*]}$ denote composition. Conjecture : If $A(z) = \lim b^{[n]} ( c^{[n]} (z) ) $ Such that 1) the limit ( $A(z)$ ) exists for all strictly positive real $z$. 2) the sequence $a_n(z) = b^{[n]} ( c^{[n] } (z) ) $ is bounded i...

complex-analysis limits inequality convergence dynamical-systems
Goodnight !
Simple Art
Simple Art
@mick Or good morning!
Can differential equations have non-continuous solutions? if so, could someone give an example?
mick
mick
Yes , poles and singularities ?
Also , not sure if you accept it as a differential equation but look at the " fabius function " @SimpleArt
Simple Art
Simple Art
01:41
Thanks
Simple Art
Simple Art
01:58
If your feeling a little stressed today...
user image
 
3 hours later…
Secret
Secret
04:48
Last night dream has a horror element: There's an invisible gate where if your numerical representation is not coprime with the gate, you will be diced into cubes in midair, each cube being rotated, and then diced again into oblivion. Interesting while in the dream I am aware of this, no actual scene of the slicing is ever shown in the dream's events
3
DHMO
DHMO
@Secret I'll be 1 then.
Adeek
Adeek
05:02
@BalarkaSen can you check this when you come back math.stackexchange.com/questions/2096939/…
@MikeMiller if you could check it aswell it would be nice.
 
3 hours later…
Balarka Sen
Balarka Sen
07:36
@Adeek You're looking at the map $I \times S^{n-1} \to B^n \to Y$, where the first map is the quotient map you defined. This is continuous because both $I \times S^{n-1} \to B^n$ and $B^n \to Y$ are continuous. That's the nullhomotopy.
I have no idea what $X$, or $F$ is, or why it's relevant.
Ramanujan
Ramanujan
08:07
is he taking left,right hand derivative correctly?
Alessandro Codenotti
Alessandro Codenotti
No, look at the first comment
Hi @Balarka
Balarka Sen
Balarka Sen
Hi @Alessandro
Ramanujan
Ramanujan
@AlessandroCodenotti when should we use that type for differentiation ?
I only know that it can be used in Lagrange mean value
Alessandro Codenotti
Alessandro Codenotti
What do you mean with "that type"?
Ramanujan
Ramanujan
If F(x) = sin x then ?
Alessandro Codenotti
Alessandro Codenotti
08:22
I don't really like that definition of differentiable since it doesn't generalize well, differentiable should mean approximable by a linear function
I don't understand what's the question @Ramanujan
Ramanujan
Ramanujan
$F'(x) = \frac{\sin x - \sin 0}{x-0}$ ?
@AlessandroCodenotti
Alessandro Codenotti
Alessandro Codenotti
$F'(0)=\lim\limits_{x\to 0}\frac{\sin x-\sin 0}{x-0}$
Ramanujan
Ramanujan
=1?
Alessandro Codenotti
Alessandro Codenotti
If the limit exists (which it does)
Yes
Ramanujan
Ramanujan
But f'(x) = cos x
Alessandro Codenotti
Alessandro Codenotti
08:27
That's $F'(0)$
And it agrees with the fact that cos0=1
Ramanujan
Ramanujan
ok,thanks
Animesh Ashish
Animesh Ashish
08:39
Total no. Of local maxima and local minima of f(x) =
cap
cap
Hi, is $f(X)$ always a random variable if $X$ is a discrete random variable?
My book says yes, but what if $f$ isn't measurable?
Animesh Ashish
Animesh Ashish
Few questions I couldnt solve :'(
imgur.com/a/aZFwF
Help me , please?
@AlessandroCodenotti
Alessandro Codenotti
Alessandro Codenotti
My phone has troubles loading that image, can you type here the question you're having problems with?
aquire
aquire
09:05
can i prove equilateral triangles have the same angles by rotating them?
Animesh Ashish
Animesh Ashish
09:16
@AlessandroCodenotti : Do something , Im dead without you :(
Socrates
Socrates
09:37
@AnimeshAshish it would certainly help if you upload it, so that we can read. otherwise type it
Ramanujan
Ramanujan
09:52
user image
aquire
aquire
10:24
for question number 61 what did you tried @Ramanujan
Secret
Secret
How on earth does one visualise an object with continuum many degrees of freedom?
Alessandro Codenotti
Alessandro Codenotti
You visualize an object with $n$ degrees of freedom (whatever that means) and then let $n=2^{\aleph_0}$
Secret
Secret
NB reading some introductory quantum field theory btw
::roll eyes:: It wonders me how can physicists contain a $2^{\aleph_0}$ sized object in their finite brains
Ramanujan
Ramanujan
10:47
@aquire that's not my question
aquire
aquire
ok what's yours?
Ramanujan
Ramanujan
1 hour ago, by Socrates
@AnimeshAshish it would certainly help if you upload it, so that we can read. otherwise type it
aquire
aquire
ah ok @Ramanujan
Ramanujan
Ramanujan
@aquire btw I got a=1 , how to get b?
?
aquire
aquire
can't a = -1 @Ramanujan
i got $$a^2=1$$
how did you get a=1?
Ramanujan
Ramanujan
11:01
@aquire so "a" can't be negative
aquire
aquire
why?
$$$$
Ramanujan
Ramanujan
1
Q: Why sqrt(4) isn't equall to-2?

Stepo Possible Duplicate: Square roots — positive and negative $\sqrt{4} = -2$. WolframAlpha says "false"! Now lets take a deeper look to my idea. Well...we know that, $$2^2 = 4 \iff \sqrt{4} = 2$$ $(-2)^2 = 4$ so why can't $\sqrt{4}$ be equal to $-2$? I'm a little bit confused // Than...

arithmetic radicals
Check accepted answer for this
Fargle
Fargle
@aquire In his problem, if $a$ were negative the limit would grow without bound.
Ramanujan
Ramanujan
@Fargle and if "b" was also infinity,than a can/can't be negative?
Fargle
Fargle
@Ramanujan b can't be infinity.
Ramanujan
Ramanujan
11:08
If there was no "a"(-ax term) then b will be?
Fargle
Fargle
If a were 0, b would have to be infinity for this to make sense, because $\sqrt{x^2 - x + 1}$ grows without bound.
But it would be more accurate to say that the equality couldn't be true.
Ramanujan
Ramanujan
Ok,now how to find "b" ?
Fargle
Fargle
To find b, you can use the fact that $\sqrt{x^2 - x + 1} < x$ for large $x$, so that the limit without $b$ would be negative.
Since taking away $b$ would give $0$, $b$ has to equal the value of that limit, so $b$ is negative.
By process of elimination, (B) is the right answer.
Ramanujan
Ramanujan
@Fargle by taking "x" out of sqrt LHS=RHS ( I may be wrong so correct me)
Fargle
Fargle
@Ramanujan How I got the inequality: We know that $\sqrt{x^2} = x$ for positive $x$, so if we take away a little bit inside the square root, it'll be less than $x$.
Ramanujan
Ramanujan
11:16
$x\sqrt{1-1/x + 1/x^2 } < x$
$x=\infty$ 1/x =0
Balarka Sen
Balarka Sen
Hi @Fargle
Ramanujan
Ramanujan
Wrong?
Fargle
Fargle
I'm very nervous about "plugging in" infinity.
Hello @Balarka.
Ramanujan
Ramanujan
OK,so infinity/infinity doesn't make sense
But how to exactly get solution?
Fargle
Fargle
That I'm not sure of, it requires more advanced techniques than I remember how to use.
Ramanujan
Ramanujan
11:36
math.stackexchange.com/questions/2097263/…
I think angle ABC is needed to calculate angle BAC
Or not?
Mary Star
Mary Star
I want to check the pointwise and uniform convergence $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{nx-n^2}}$, $x\in (0,1)$.

Do we use the Weierstrass criterion?

We have that $\left |\frac{1}{nx-n^2}\right |=\frac{1}{n^2-nx}=\frac{1}{n(n-x)}$.
By what function could we bound it?
Balarka Sen
Balarka Sen
@Fargle What's new?
Astyx
Astyx
Do you know about normal convergence @MaryStar ?
Socrates
Socrates
what is sgn(x)?
Astyx
Astyx
The sign of $x$ ? @Socrates
Socrates
Socrates
11:53
then I'd say f(x)=|2sgn(2x)|+2 has a removable discontinuity at x=0
Mary Star
Mary Star
@Astyx I looked the definition now in wikipedia. Do we have to show this one first?
Astyx
Astyx
You don't have to
Mary Star
Mary Star
12:04
@Astyx WHat else could we do?
Astyx
Astyx
Well, what methods do you know for proving convergence ?
Mary Star
Mary Star
The Leibniz criterion, but we don't need it here. The Direct comparison test, the Ratio test, the Root test. Maybe there are also some others that I don't remember now. @Astyx
Kasmir Khaan
Kasmir Khaan
12:25
@MaryStar hi have you done wierstrass test?
Mary Star
Mary Star
Yes, but by which function can we bound $\frac{1}{n(n-x)}$ ?
For example we have that $\frac{1}{n(n-x)}\leq \frac{1}{n}$, right? But the series of $\frac{1}{n}$ doesn't converge.
Kasmir Khaan
Kasmir Khaan
Am thinking about it hold on :)
hmm
when you take derivative of $\frac{1}{n(n-x)}$
you get -( nx-n^2)*n = 1/( n-x)
for x in (0,1) its maximum when x = 0
you could compare this serie with -1/n^2 for wierstrass
Studentmath
Studentmath
13:26
So, I'm asked to show that every irreducible polynomial over $Z_p$ of degree $\neq 3$ doesn't divide $x^{p^3}-x$. But one of degree 1 obviously could, couldn't it?
Or am I missing something here..
Alessandro Codenotti
Alessandro Codenotti
Doesn't every polynomial of degree 1 divide it actually?
Studentmath
Studentmath
13:51
Yeah, necessarily does
Alessandro Codenotti
Alessandro Codenotti
14:05
I suppose they mean of degree 2 or higher then
Studentmath
Studentmath
Probably, thanks!
Astyx
Astyx
14:43
@MaryStar You can bound it by ${1\over n(n-1)} = {1\over n-1} - {1\over n}$ of which the series is easily computable
aquire
aquire
15:15
@Ramanujan
Mary Star
Mary Star
@Astyx But the original series begins from $n=1$, but the series for ${1\over n-1} - {1\over n}$ cannot begin from $n=1$, right?
Astyx
Astyx
That doesn't matter here
Try to understand why
Mary Star
Mary Star
15:31
We have that $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{nx-n^2}=\frac{1}{x-1}+\sum_{n=2}^{\infty}\frac{1}{nx-n^2}}$.
From the Weierstrass criterion we have that $\sum_{n=2}^{\infty}\frac{1}{nx-n^2}$ converges uniformly. But what about the whole series?
Astyx
Astyx
What do you think ?
aquire
aquire
@Ramanujan Hint: $(\sqrt{\alpha}-\beta)\times (\sqrt{\alpha}-\beta)=\alpha-\beta^2$
Ramanujan
Ramanujan
@aquire for finding "b" ?
aquire
aquire
correction: 2nd - should be +
yep
my internet connection is bad bear with me.
make use of this:
$$(\sqrt{\alpha}-\beta)\times (\sqrt{\alpha}+\beta)=\alpha-\beta^2$$
Ramanujan
Ramanujan
Let me try
Mary Star
Mary Star
15:43
To find this without calculating the derivative can we justufy this as follows?
$n^2-nx>n^2-2n^2=-n^2$
And so:
$\left |\frac{1}{nx-n^2}\right |=\frac{1}{n^2-nx}<-\frac{1}{n^2}$
Or is it better to justify it using the derivative?
@Astyx I don't really know...
Astyx
Astyx
Using derivative seems a bit overkill for me
Just say that $n^2-xn\sim n^2$ and $\sum {1\over n^2}$ converges
Thus by series comparison $\sum{1\over n(n-x)}$ converges
So you have pointwise convergence
Ramanujan
Ramanujan
@aquire iam not getting :( so first step is I need to multiply and divide by $(\sqrt{x^2 - x +1}+ ax + b) $ ?
Mary Star
Mary Star
Ahh. So, we cannot use it as I did wieth "<" so that we can use the Weierstrass test for check for uniform convergence? @Astyx
Astyx
Astyx
Now evaluate the difference between the general term and the limit :
$$\begin{align}\left|\sum_{n=1}^{+\infty}{1\over n(n-x)} - \sum_{n=1}^{N}{1\over n(n-x)}\right| &= \left|\sum_{n=N+1}^{+\infty}{1\over n(n-x)}\right| \\
&\le \sum_{n=N+1}^{+\infty}{1\over n(n-1)} \text{for $N\ge 1$}\\
&={1\over N}\\
&\to 0 \end{align}$$
@MaryStar not sure what you mean by that
So you have uniform convergence on $]0, 1[$
Mary Star
Mary Star
16:01
@Astyx We have that $n^2-xn\sim n^2$ and $\sum {1\over n^2}$ converges. Do we have then that ${1\over n(n-x)}\leq {1\over n^2}$, so that we can use the series comparison test?
aquire
aquire
16:12
@Ramanujan yep
Ramanujan
Ramanujan
What I got is$ b+\sqrt x =1$
@aquire
aquire
aquire
have you considered the limit @Ramanujan?
Ramanujan
Ramanujan
No
Astyx
Astyx
16:38
@MaryStar Sure, however you have to point out that the terms are positive
Owatch
Owatch
Hey.
I'm reading my book here about signals, and it states at one point that: $e^{i\omega} + 2 + e^{-i \omega}$ is equal to: $2 + 2cos(\omega)$.
How is this the case?
aquire
aquire
have you got up to this point @Ramanujan?
$$
\begin{align}
0=&\lim_{x\to\infty}\left({\sqrt{x^2-x+1}-ax-b}\right)\\
=&\lim_{x\to\infty}\left[\left({\sqrt{x^2-x+1}-ax-b}\left)\times{{\sqrt{x^2-x+1}+ax+b}\over{\sqrt{x^2-x+1}+ax+b}}\right]\\
=&\lim_{x\to\infty}\left({(1-a^2)x^2-(1+2ab)x+1-b^2}\over{\sqrt{x^2-x+1}+ax+b}\right)\\
\end{align}
$$
Semiclassical
Semiclassical
@Owatch Euler's formula $e^{i\omega}=\cos \omega+i \sin \omega$.
Owatch
Owatch
Yeah FML. @Semiclassical
I was reading the one for $sine$
It's right below it on my formula sheet.
Semiclassical
Semiclassical
Happens.
Owatch
Owatch
16:49
Thought I
had caught a mistake. :(
euclid
euclid
@MaryStar hi. do you know primitive roots?
Balarka Sen
Balarka Sen
Hi @MikeMiller.
Owatch
Owatch
lol.
Mike Miller
Mike Miller
Hi @Balarka.
Balarka Sen
Balarka Sen
I feel like I should write down random things on the internet about differential forms while simultaneously revisiting it. I dunno where.
I don't wanna TeX it. Maybe I'll open up a chat and start writing garbage.
Mike Miller
Mike Miller
16:56
I only got an A on M5 :( It didn't like my speed.
If you like. Whatever gets you inspired.
Balarka Sen
Balarka Sen
Oh. Wasn't the S-streak already broken?
I redid some of the training maps
Mike Miller
Mike Miller
Yeah, but I'd still like as many as I can get. No partkcular reason.
Balarka Sen
Balarka Sen
Ah, gotcha
Mike Miller
Mike Miller
I suspect you play less aggressively than you need to get an s-rank.
My advisor told me to think about other things while I write. So at less I've got that now.
Socrates
Socrates
ah, i see now
Alessandro Codenotti
Alessandro Codenotti
17:01
One day or another I'll ask you to teach me about differential forms @Balarka
Balarka Sen
Balarka Sen
@Alessandro All the more reason I should write stuff up
Mary Star
Mary Star
@euclid Hello!! Tell me your question. Maybe I will have more time later to answer you.
Balarka Sen
Balarka Sen
Thanks for the extra motivation, BTW
aquire
aquire
@Ramanujan
$$
\begin{align}
0=&\lim_{x\to\infty}\left({\sqrt{x^2-x+1}-ax-b}\right)\\=&\lim_{x\to\infty}\left[\left({\sqrt{x^2-x+1}-ax-b}\right)\times\frac{\left({\sqrt{x^2-x+1}+ax+b}\right)}{\left({\sqrt{x^2-x+1}+ax+b}\right)}\right]\\=&\lim_{x\to\infty}{\frac{(1-a^2)x^2-(1+2ab)x+1-b^2}{\sqrt{x^2-x+1}+ax+b}}
\end{align}
$$
euclid
euclid
@MaryStar i dont know how to find primitive root of $p^2$ and other powers of $p$ also primitive roots of $2p^2$
Alessandro Codenotti
Alessandro Codenotti
17:12
Ted sent me his book to read about them
Balarka Sen
Balarka Sen
I like his exposition much better than G-P's, yes
If you want a push, let me know. You actually don't need much of the previous material in G-P to get introduced to chapter 4.
Adeek
Adeek
@BalarkaSen I want to ask you a question
In topology okay ?
Balarka Sen
Balarka Sen
k
Adeek
Adeek
@BalarkaSen Consider the following theorem a map $f : S^{n - 1} \rightarrow Y$ is null-homotopic iff there exists a extension $\phi : B^n \rightarrow Y$.
I understand the backward direction is obvious.
Suppose we go for forward direction
Balarka Sen
Balarka Sen
sure
aquire
aquire
17:22
@Ramanujan I think you have got this. Now you should see clearly that $a^2=1$.
Since we have argued that $a\ne -1$, $a\text{ should be }1$.
$$\begin{align}\lim_{x\to\infty}\frac{-(1+2b)+\frac{1-b^2}{x}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}+1+\frac{b}{x}}\end{align}$$
After that you just have to apply the limit and solve for b.
arctic tern
arctic tern
@euclid there is no general formula. for a specific modulus, you will just have to test values until you find a primitive root. have you typed "how to find primitive roots" into a search engine yet?
Adeek
Adeek
As $f : S^{n - 1} \rightarrow Y$ is null homotopic so there exists a homotopy $F : I \times S^{n - 1} \rightarrow Y$. Define a new function $\phi : B^n \rightarrow Y$ as follows $\phi(x) = F (1 - ||x||,\frac{x}{||x||})$.
adjusted
Now consider the map $q : I \times S^{n - 1} \rightarrow B^n$ defined as $q(t,x) = (1 - t)x$. We proved that this map is a quotient map. Now, this $\phi$ will be continous iff $\phi \circ q$ is continous.
why is $\phi \circ q$ continous ?
euclid
euclid
@arctictern i know there are theorems for it. you deleted our room?
Adeek
Adeek
The above is by univerisal property of quotient space.
arctic tern
arctic tern
@euclid no, I didn't delete it. it's still there.
Balarka Sen
Balarka Sen
17:28
All of this sounds way too hard, @Adeek. A nullhomotopy $F : I \times S^{n-1} \to Y$ is constant along $\{1\} \times S^{n-1}$, so universal property already says that it factors through a map $I \times S^{n-1}/\{1\} \times S^{n-1} \cong B^n \to Y$, doesn't it?
I don't get why you're doing the extra computation
Sophie
Sophie
I made the world's most inelegant Machin-like formula
Astyx
Astyx
Do you consider this as an achievement ?
Sophie
Sophie
it does what I wanted it to do, so that's nice
Adeek
Adeek
how do we get that it factors through a map $I \times S^{n - 1} / \{1\} \times S^{n - 1} \cong B^n$ ?
@BalarkaSen
arctic tern
arctic tern
identify points in fiber
Adeek
Adeek
17:32
ohh
oh ok I agree with you.
that is way better thanks @BalarkaSen
Balarka Sen
Balarka Sen
no probs
aquire
aquire
17:54
1+1=2 happens only for linear systems, right? @BalarkaSen
Balarka Sen
Balarka Sen
that question sounds like nonsense to me
aquire
aquire
or else "=" defined only to linear systems?
is "+" only a linear operation?
it came to me seeing that 1+2+3+... converges to -1/12 @BalarkaSen
Semiclassical
Semiclassical
1+2+3+... quite certainly does not converge to -1/12. There are reasons to assign a value of -1/12 to that series, but convergence isn't one of them.
aquire
aquire
ok @Semiclassical then there should be other values we can "assign" for 1+1 right?
other than 2
Semiclassical
Semiclassical
not really. if the series converges, then there's no reason to assign it another value.
Jonathan Krill
Jonathan Krill
18:02
hey guys, someone knows what this identity is called or where i can find information on it
user image
Astyx
Astyx
1+1 isn't a series anyway is it ?
Semiclassical
Semiclassical
@Astyx Eh, partial sums of 1,1,0,0,0,...
aquire
aquire
finite series @Astyx
Astyx
Astyx
You mean sums
But that's going backwards
Semiclassical
Semiclassical
I'm okay calling 1+1 a series on the grounds of it being 1+1+0+0+...
Astyx
Astyx
18:03
The only reason 1+1 = 2 is because we define it as such
Semiclassical
Semiclassical
...if you mean that that's the definition of 2, sure.
but there's not going to be a way to say that the value of 1+1 is 1+1+1, for instance.
Astyx
Astyx
I mean that there is no other intrinsic deeper meaning to this equality
arctic tern
arctic tern
@JonathanKrill doesn't need a name, it's just the fact that if a is a root of f(x) then f(x)=(x-a)g(x) for some g(x). hence if {a,b,...} are all the zeros (counted with multiplicity) then f(x)=(x-a)(x-b)...
Semiclassical
Semiclassical
Sure. But there's no need to pull anything in about the value one assigns to a divergent series; that has nothing to do with that.
Astyx
Astyx
I agree
Jonathan Krill
Jonathan Krill
18:06
@arctictern thanks i look into it
Semiclassical
Semiclassical
@JonathanKrill In what context are you seeing that identity? If we're talking about polynomials over $\Bbb C$, then that's false since $2^p-2\neq 0$.
arctic tern
arctic tern
any ring of characteristic p
Semiclassical
Semiclassical
Fair enough.
Jonathan Krill
Jonathan Krill
its the field Z_p[x], p is prime
arctic tern
arctic tern
the equivalent over C would be prod (x-zeta) where zeta ranges over roots of unity
(note in Z_p, all of 1,2,...p-1 are roots of unity)
aquire
aquire
18:08
@Semiclassical may be physicists use that -1/12 value due to some property of the system where the first "1" is different from the next "1"
Astyx
Astyx
Is that true for any ring of characteristic $p$, $p$ being any integer ? I thought it had to be prime
arctic tern
arctic tern
yes p prime
Astyx
Astyx
Okay
Semiclassical
Semiclassical
18:24
@aquire Mathematically, what's going on is that for positive $s>1$ one can define the Riemann zeta function as $\zeta(s)=\sum_{k=1}^\infty k^{-s}$.
This definition only makes sense if it converges, so literally speaking one can't use it outside of this domain. In particular, it doesn't converge for $s=-1$.
However, the Riemann zeta function can be 'analytically continued' into the complex plane and from there to values with $s<1$.
It's that function which can be said to equal -1/12 at $s=-1$.
When stuff like 1+2+3+... shows up in a physics calculation (and it does, in the context of stuff like QED), one should understand that as a 'shorthand' expression for $\zeta(-1)$.
This is also sensible from a physics perspective. When you do a calculation of the Casimir effect, for instance, a naive computation of its magnitude would lead to a divergent calculation that is nonsense on the face of it.
Krijn
Krijn
I must say, I have never seen a non-prime non-zero characteristic
I have worked with $\mathbb{Z}/n$ of course
But never written down $char = n$ or anything
Semiclassical
Semiclassical
The trick is that the 'true' calculation is a bit more subtle, requiring one to modify the calculation in a way to make things actually sensible. When you take that into account, it's $\zeta(s)$ that matters rather than $\sum_{k=1}^\infty k^{-s}$.
To the extent that one uses a value of -1/12 there, it's with the understanding that if you did do things more carefully then you'd be able to get -1/12 in an honest way. It's a shortcut, is how I think of it.
Socrates
Socrates
18:50
if we have the representation of basisvectors with respect to another basis. can we use this representation to build a basischanging matrix?
Semiclassical
Semiclassical
Certainly.
Alessandro Codenotti
Alessandro Codenotti
Do you know how to write a transformation from the canonical basis to yours?
Socrates
Socrates
the basis(s) are not given tho, i might add.
Semiclassical
Semiclassical
What's the actual question?
PVAL-inactive
PVAL-inactive
19:08
@Mike @Ted Ever figure out that Klein bottle thing? I think I have a way to get an embedding using the isometric immersion. Call this immersion $f$. Let $\phi$ be a diffeomorphism $K \to K$ sending the intersection points to points with distinct last coordinate in the image of the immersion. Let $F_1: K \to \Bbb R^5$ with $F_1(x)=$(f(x),0)$ and $F_2(x)=(0,f(\phi(x))$ Then $F=F_1+ \epsilon F_2$ should be an isometric embedding (for small enough epsilon).
Mike Miller
Mike Miller
Ah that should work.
Mats Granvik
Mats Granvik
So it is not possible to locate the zeta zeros by integration of the zeta zeros counting function. I thought it was, until I received an answer to a related question on Math Overflow. (Assuming the Riemann hypothesis.)
PVAL-inactive
PVAL-inactive
@Mike Okay I'm not sure what I said is quite suffcient to guarantee no self intersections.
Mike Miller
Mike Miller
Seems like the right idea.
PVAL-inactive
PVAL-inactive
I still think like a linear combination with two such embeddings should work.
There's an embedding you get into $\Bbb R^8$ by the trivial version of this trick.
Mike Miller
Mike Miller
19:28
@PVAL I'm just thinking about the fact that embeddings are generic.
math is love
math is love
19:41
Hello everyone.
Let {a _ n} be a sequence of positive numbers converging to zero .prove that the series summation a_ n z^n is uniformly convergent on { z:|z|<=1} and on { z:|z|>=€} for some €>0
Semiclassical
Semiclassical
@mathislove FYI, you can use mathjax in chat if you use what's linked in the room description.
Balarka Sen
Balarka Sen
If I want a coordinate free definition of the exterior derivative, what would it be? Let's say, $\omega$ is a $k$-form on an $n$-fold $M$ and I want to find out what $d\omega(v)$ is, where $v \in T_p M$. Is there a nice expression for it? (I know how to write it in local coordinates)
Semiclassical
Semiclassical
@MatsGranvik Which question was that?
Balarka Sen
Balarka Sen
Sorry, I meant, $v$ is a $(k+1)$-tensor on $T_p M$.
So $d\omega(v_1, \cdots, v_{k+1})$
Alessandro Codenotti
Alessandro Codenotti
Is there a nice and direct way to prove that every family of compact sets of $\Bbb R^n$ with the finite intersection property has nonempty intersection?
Mike Miller
Mike Miller
19:53
@BalarkaSen There's one that's a generalized version of the one for 1-forms: $d\alpha(X,Y) = X\alpha(Y)-Y\alpha(X)-\alpha([X,Y])$.
Balarka Sen
Balarka Sen
What's the finite intersection property you have in mind?
@MikeMiller Hmm
Semiclassical
Semiclassical
@MikeMiller That's rather cute, due to the commutator showing up in there.
Alessandro Codenotti
Alessandro Codenotti
every finite collection of sets from this family has nonempty intersection
Balarka Sen
Balarka Sen
Ideas: If it's a countable collection you can enumerate them as K1, K2, ..., and take iterative intersection $P_n = \cap_{i = 1}^n K_i$ to get a decreasing sequence of compact sets. If it doesn't stabilize then pick a sequence of points $x_k \in P_{k+1} - P_k$; which has no convergent subsequence cuz otherwise the limit point would be contained in $P_{\infty} = \emptyset$.
Alessandro Codenotti
Alessandro Codenotti
I had a topology exercise earlier asking whether $(\Bbb R^n,\tau)$ with $\tau$= "sets whose complement is a compact set in the euclidean topology" is a compact topological space
and the answer is yes, but being compact is equivalent to "every family of closed sets with the finite intersection property has nonempty intersection"
and in this topology the closed sets are precisely the compact sets of the euclidean topology, so I was trying to write a more direct proof
Mike Miller
Mike Miller
20:03
this is actually an application of a theorem of logic!
Compactness theorem
In mathematical logic, the compactness theorem states that a set of first-order sentences has a model if and only if every finite subset of it has a model. This theorem is an important tool in model theory, as it provides a useful method for constructing models of any set of sentences that is finitely consistent. The compactness theorem for the propositional calculus is a consequence of Tychonoff's theorem (which says that the product of compact spaces is compact) applied to compact Stone spaces; hence, the theorem's name. Likewise, it is analogous to the finite intersection property characterization...
Semiclassical
Semiclassical
Godel proved the countable case in 1930. Neat.
Balarka Sen
Balarka Sen
The countable one
my proof-idea for the countable version is ok, though, right?
Mike Miller
Mike Miller
yes
Alessandro Codenotti
Alessandro Codenotti
it looks right to me
@MikeMiller that was unexpected
Balarka Sen
Balarka Sen
cool. no idea how to do it for uncountable; guess that's why the wiki thing is interesting
Semiclassical
Semiclassical
20:07
I wonder how Godel proved that.
Oh, it says how---as an application of his completeness theorem.
Mike Miller
Mike Miller
@BalarkaSen Manifolds are sigma-compact, hence Lindelof, hence every open cover has a countable subcover. Apply to the complements of the compact sets to write the intersection of the compact sets as a countable intersection.
Balarka Sen
Balarka Sen
Ahh, I completely forgot that fact
That makes sense
Andrew Thompson
Andrew Thompson
20:32
Wow, this is the first time I've heard the word "Lindelof" since my first topology course.
Mike Miller
Mike Miller
That's good to hear.
Anderson Felipe Viveiros
Anderson Felipe Viveiros
Is every connected Riemann surface path connected?
It is clear to me that every Riemann surface is locally path connected
Semiclassical
Semiclassical
Path connected = connected for manifolds, doesn't it?
Anderson Felipe Viveiros
Anderson Felipe Viveiros
But I can't imagine an example in which the Riemann surface is not globally path connected...
@Semiclassical Hm, really? I didn't know...
Is easy to prove?
Semiclassical
Semiclassical
I'll be honest, i'm relying on google-fu for that claim.
But said google-fu yields this MSE question: math.stackexchange.com/questions/1145293/…
Kasmir Khaan
Kasmir Khaan
20:41
hi chat , can anyone give me some tips on how to get better at visualize surfaces in 3 d ?
am doing tripple integrals and i cant see the boundries
Semiclassical
Semiclassical
Do lots of examples.
Then do more examples.
Anderson Felipe Viveiros
Anderson Felipe Viveiros
Thank you!
Kasmir Khaan
Kasmir Khaan
am doing alot but feels like each problem is new
also am bad at drawing
Alessandro Codenotti
Alessandro Codenotti
I think locally euclidean is enough to get "connected$\implies$path connected" but I'm not 100% sure
PVAL-inactive
PVAL-inactive
Read the answer.
@Mike Alright let $\phi$ take each pair of intersection points take one to a point which is positive on all of the shared coordinate and the other which is negative on all of the shared coordinates. Then the map $F_1+\epsilon F_2$ should be an embedding for small enough epsilon.
I don't know if theres a tranversality way to do it, because that requires local perturbations.
Mike Miller
Mike Miller
20:55
@PVAL-inactive Is it still liable to be isometric?
PVAL-inactive
PVAL-inactive
Yes the pullback metric is the sum of two constant curvature metrics on K which are all the same up to a constant multiple. So the pullback is a constant multiple of the unique flat metric on K
@Mike So in general the pullback metric should be a constant multiple of a normalized flat metric
Mike Miller
Mike Miller
K.
PVAL-inactive
PVAL-inactive
If you want the normalized one multiply the whole thing by a constant.
Actually I dont know
The pullback metric is more annoying than that maybe
You probably need DF_1 and DF_2 to be orthogonal to each other.
@Mike Maybe if you ask on MO, RB will tell you the answer.
If anyone would know how to do this it'd have to be him.
Vrouvrou
Vrouvrou
21:16
Hello, what is the relation between $L^{\infty}$ and Holder space $C^{1,\alpha}$
Socrates
Socrates
21:49
Ok, I managed to find a basischangig matrix(in fact it's reverse too). How can I now find for a given $_B[f]_B$ the $_{B'}[f]_{B'}$? And what does this even mean? Is $_B[f]_B$ the left endomorphism, that translates $K^n$ into $V$?
Adrien
Adrien
22:07
Hi, I have a question about model categories. One of the axioms for model categories is that the distinguished classes of morphisms are closed under retracts. Is it equivalent to require that they be closed under sections (the dual notion to retracts)?
Ted Shifrin
Ted Shifrin
22:33
@PVAL @MikeM: I just got home and haven't been following all your discussion, but perturbing to stay in the isometric category is very subtle. I'm not sure I understand what PVAL proposed earlier.
Sophie
Sophie
how do you prove that $\displaystyle\sum_{n=0}^\infty \frac{(2n)!!}{(2n+1)!!}x^n=\frac{\arcsin(\sqrt{x})}{\sqrt{x(1-x)}}$?
PVAL-inactive
PVAL-inactive
@TedShifrin I was trying to do things with linear combinations of immersions thinking that the induced pullback metric would be a sum of the two metrics.
I think you need some sort of orthogonality for that to work.
It does give you an isometric embedding from $K$ into $\Bbb R^8$.
I'm confident in that.
Semiclassical
Semiclassical
hi chat.
@Sophie Probably not too helpful, but if you replace $x\mapsto x^2$ and multiply both sides by $x$ then that becomes $\displaystyle \sum_{n=0}^\infty \frac{(2n)!!}{(2n+1)!!}x^{2n+1}=\frac{\arcsin x}{\sqrt{1-x^2}}$
One thing that's possible to try then is the substitution $x=\sin\theta$, which renders the RHS as $\theta \sec\theta$.
Problem is, you'd need to do something clever with the infinite sum of $(\sin \theta)^{2n+1}$.
And I'm not seeing it :/
Sophie
Sophie
22:50
I've found some alternative representations so if I know closed forms for $\sum \frac{x^n}{{2n+1 \choose n}}$ or $\sum \frac{x^n}{{2n \choose n}}$ then I can probably find what I'm looking for
Semiclassical
Semiclassical
Neat.
Ted Shifrin
Ted Shifrin
@PVAL: You absolutely need orthogonality :)
Akiva Weinberger
Akiva Weinberger
Hello
Oh, wait, you responded
Ted Shifrin
Ted Shifrin
I answered it ... I even pinged with your real name.
Semiclassical
Semiclassical
Hmm, one thought following the above. If I move the factor of $\cos\theta$ to the other side, then that becomes $$\theta = \sum_{n=0}^\infty \frac{(2n)!!}{(2n+1)!!}(\sin\theta)^{2n+1}\cos \theta$$
Akiva Weinberger
Akiva Weinberger
22:54
Ah. Sorry, it didn't work for some reason
Sophie
Sophie
@Semiclassical maybe taking the derivative of this is interesting
Semiclassical
Semiclassical
Which if nothing else is weirdly awesome.
Akiva Weinberger
Akiva Weinberger
Is it really just algebra? Because it looks like it would end up as a very nasty quartic equation
Ted Shifrin
Ted Shifrin
The Dandelin proof is super cool, though, if you haven't done it.
Yeah, it's just algebra. If you want to send me what you've done, I'll look at it.
Semiclassical
Semiclassical
Dandelin spheres?
Ted Shifrin
Ted Shifrin
22:56
#13 is also bizarre but interesting. It shows up in some interesting differential geometry (rolling an ellipse and looking at the locus of the focus, then making a surface of revolution).
Right @Semiclassic
Semiclassical
Semiclassical
Balls rolling on surfaces can do some weird stuff.
Even in 1D you get fun pictures like this: math.ucr.edu/home/baez/rolling
Ted Shifrin
Ted Shifrin
DogAteMy: If you got to quartic, you didn't do algebra smartly.
Semiclassical
Semiclassical
(I mean that the surface that one is rolling on is 1D)
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