Mathematics - 2025-01-30

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Xander Henderson

01:10

@Thorgott Well, that rules out anything interesting. :P

leslie townes

01:38

i feel like it ought to be possible to do something like this just fiddling with geodesic coordinates around a point that contains both points of A and points of A complement. but intuitively i find myself needing to know 'how' the relevant piece of A sits inside even a ball in these coordinates.

of course the fractal guy would say that :)

Ben Steffan

oh, does Xander not have Hausdorff dimension 3 like the rest of us? :/

leslie townes

anyway the question was closed (as expected). it would be nice if the OP amended the question. i think a lot of people interpret closure as some kind of "this is not wanted" message instead of what it actually is. despite all of the documentation around what closure is

Xander Henderson

02:01

@BenSteffan G-d, I hope not. Lungs work better in a slightly lower dimension.

@leslietownes what question?

leslie townes

02:52

the most recently linked question in this chat

Q: A special diffeomorphism

Let $A$ be an open subset of $(M,g)$ closed riemannian manifold. Given $\varepsilon>0$, is it true that there exists a diffeomorphism $\phi:M\to M$ such that $$ \phi^{-1}(A) = Vol(M)-\varepsilon \qquad? $$

differential-geometry

5 hours later…

Thorgott

08:16

@leslietownes yeah, that's essentially my boundary worry

2 hours later…

Monty

09:50

anyone free to chat about KD-trees?

psie

$\mathbb Z$ is closed in $\mathbb R$ because $\mathbb{R}\setminus\mathbb{Z}=\bigcup_{n\in\mathbb{Z}}(n,n+1)$. Now, the sets $(n,n+1)$ regarded as subsets of $\mathbb R^2$ are not open. So is $\mathbb Z$ closed in $\mathbb R^2$?

Ben Steffan

You should figure this out on your own.

3 hours later…

one potato two potato

13:04

It seems Basse book on Einstien manifolds is telling the truth but most of the proofs say "follows easily".

I just checked one formula and the full calculation requires one full page.

Soumik Mukherjee

@psie The complement of $\Bbb{Z}\times\{0\}$ in $\Bbb{R}^2$ is very much different from the complement of $\Bbb{Z}$ in $\Bbb{R}$. So the sets $(n,n+1)$ are not of much help in here.

Ben Steffan

13:35

@onepotatotwopotato physics.jpeg

one potato two potato

my motivation for differential geometry is in physics, unfortunately.

psie

14:31

@SoumikMukherjee yes, so I guess arguing from saying that the complement is open is more cumbersome than simply saying $\mathbb Z$ has no limit points, and thus is vacuously closed.

Binky

$(1+i)^{2z}=n$

How can I solve it in the complex field?

Sine of the Time

@Binky how did you define $w^z$?

where is this exercise from?

Soumik Mukherjee

14:51

@psie it's not cumbersome, prove that the complement is open by using open balls

Sine of the Time

do you have to show the complement is open?

Soumik Mukherjee

It's just one of the ways

Thorgott

the point is that it should be evident that Z is closed

Binky

@SineoftheTime $e^{zln(w)}$

Thorgott

you have to move on from trivialities at some point

Sine of the Time

14:59

@Binky ok, where are you stuck?

@SoumikMukherjee yes; I was wondering if psie is required to show it's closed using the definition

Binky

@SineoftheTime at the start

Pizza

@Binky maybe you have to write $(1+i)^2 = 2i$

Sine of the Time

@Binky huh?

Pizza

so $(2i)^z = n$

Binky

@SineoftheTime Im blocked at the pizza point

Sine of the Time

15:12

just use the definition

Binky

$e^{z\ln{2i}}=n$

Sine of the Time

@Pizza you have to be careful since the rules of exponents does not always hold in $\Bbb C$

Binky

...

Pizza

@SineoftheTime I haven't practiced these types of exercises yet :(

sorry if I said something wrong

Sine of the Time

keep in mind that $a^{bc}$ is not always equal to $(a^b)^c$

Binky

15:16

$e^{2z\ln{1+i}}=n$

Sine of the Time

$n$ is a natural number?

Binky

Yes

$2z\ln(1+i)=\ln(n)$

$z=\frac{\ln(n)}{\ln((1+i)^2)}$

Pizza

@SineoftheTime How do I understand if I could do it now or not?

Binky

$z=\frac{\ln(n)}{\ln(2i)}$

👍

Sine of the Time

@Pizza when $a>0$ and $a,b,c$ are real

Pizza

15:28

@SineoftheTime yes but only if I put | 1 + i | i can do this

that is, if I only write 1 + i, I cannot tell if it is > 0 because it only applies to real numbers

Sine of the Time

yeah that's why when you have $w^z$ you usually rewrite it as $\exp(z\operatorname{Log}(w)))$ and you specify which branch of the log are you using

Pizza

@SineoftheTime Does the uppercase in 'Log' clearly indicate the principal argument of the logarithm?

Sine of the Time

I used it to indicate the complex log

Complex logarithm

In mathematics, a complex logarithm is a generalization of the natural logarithm to nonzero complex numbers. The term refers to one of the following, which are strongly related: A complex logarithm of a nonzero complex number z {\displaystyle z} , defined to be any complex number w {\displaystyle w} for which e w = z {\displaystyle e^{w}=z} . Such a number w...

ILikeMathematics

$$\underbrace{f(x + h) - f(x)}_{\Delta f(x)} + o(h) = d f(x)[h] =f'(x) d x[h] = f'(x)\underbrace{((x + h)- x)}_{\Delta x}$$

Pizza

$\operatorname{Log}(w) = \ln|w| + i \arg(w)$

yes ?

Sine of the Time

15:36

yes

ILikeMathematics

What does the $[h]$ mean?

Binky

19 mins ago, by Binky

$z=\frac{\ln(n)}{\ln(2i)}$

@SineoftheTime So it's correct, right?

Z.G.

16:30

@SohamSaha Heya!

Soham Saha

@Z.G. Hi!

Z.G.

Oh you are into logic

Soham Saha

Anything interesting

Z.G.

Cool

Will see you around!

Soham Saha

Where are you from btw?

Z.G.

16:36

Haha... This is a public place

Soham Saha

@Z.G. yep. You can ping me anytime in here

@Z.G. oh leave it then

Z.G.

@SohamSaha check aops

Soham Saha

What are you interested in? Like which topics

Z.G.

Algebra, number theory and I hope to learn some logic before uni.

Soham Saha

@Z.G. Now aops is rate limiting my messages

Z.G.

16:42

Yeah that's annoying.

Soham Saha

AOPS chats are private I guess?

Z.G.

Yeah I guess so

So you're in 12th grade this year or took a gap year? (If you don't mind)

Soham Saha

This year

Z.G.

Hmm, you did stuff like PSS by Engel or no?

Soham Saha

Just had a look. Most are too much time taking

Did you give jee mains this year?

Z.G.

16:49

@SohamSaha That is so.

@SohamSaha I'm in 11th this year.

Soham Saha

Gave RMO?

Z.G.

But I don't think I'll sit for that, I mean I suck at chemistry.

Soham Saha

Z.G.

@SohamSaha haha, I don't like sitting for exams unless absolutely necessary.

Soham Saha

Free pass to ISI/CMI if you qualify INMO, after going through IOQM and RMO

Sorry, gotta go now. Again, ping here anytime if needed.

Bye

Z.G.

16:53

The entrance route is fine, how did INMO go btw?

@SohamSaha Good Night

Soham Saha

@Z.G. Good night

Sine of the Time

17:32

@Binky I did not check your computation

DaPlumer

18:05

Someone needs to put the Riemann hypothesis up as a question, see if anyone solves it:)

Ben Steffan

...

Xander Henderson

@DaPlumer No.

1 hour later…

hbghlyj

19:22

Question 1.2. Polynomial $P(x) \in \mathbb{Z}[x]$ takes values $\pm 1$ at three different integer points. Prove that it has no integer zeros.
[Hint : Divisibility condition of polynomials over $\mathbb{Z}$]
My work: $P(x)^2-1$ has three integer zeros, hence $P(x)^2-1=(x-a)(x-b)(x-c)Q(x)$ for some $a,b,c \in \mathbb{Z}$ and $Q(x) \in \mathbb{Z}[x]$. Now, $P(x)^2-1=(P(x)-1)(P(x)+1)=(x-a)(x-b)(x-c)Q(x)$. Now, $P(x)-1$ and $P(x)+1$ are coprime, then what can I do?

Soumik Mukherjee

@hbghlyj if it had an integer zero (say at k) then -1=(k-a)(k-b)(k-c)Q(k)

then at least two of the factors among (k-a),(k-b),(k-c) will be same(+1 or -1), a contradiction

hbghlyj

I see. Thanks.

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