Mathematics - 2022-12-09

geocalc33

02:40

anyone here?

how do i find my motivation

uses command-F to find motivation

one potato two potato

04:40

peep

nickbros123

06:57

Chat gpt is at best a decent search engine

PM 2Ring

08:11

@imbAF You can use the inverse transformation method, but instead of using [0, 1) for the uniform sample interval use [cdf(tau_min), cdf(tau_max)), as explained in en.wikipedia.org/wiki/…

Joe Willyam

I have a question about notation. Let $A$ a ring and $a \in A$, what is $A_a$?

Lukas Heger

08:41

@JoeWillyam it's the localization at the set $S=a^\Bbb N$

one potato two potato

0

Q: Show that ${1\over\pi}\iint_{\Bbb D}|\psi'_\alpha|^2\ dxdy =1$

If $\psi_\alpha(z) = (\alpha-z)/(1-\bar{\alpha}z)$ for $|\alpha|<1$, prove that $${1\over \pi}\iint_{\Bbb D}|\psi'_\alpha|^2\ dxdy = 1.$$ [Hint: The integral can be evaluated without a calculation.] Direct calculation shows that $\psi'_\alpha(z) = (|\alpha|^2-1)/(1-\bar{\alpha}z)^2$. Note that ...

Hmm...

Xander Henderson

11:53

@onepotatotwopotato poop

Magnus Alexander

12:43

Hello everyone! Maybe someone has an idea: figure out whether series converge or not:
$\sum\limits_{n=2}^{\infty} \dfrac{(2+5\cdot (-1)^{\dfrac{n(n-1)}{2}})^n}{n\cdot \sqrt{\ln{n}}} \cdot \dfrac{1}{7^n}$

geocalc33

13:38

i've run out of bounties for a question. how can i maximize the odds for an answer?

Asked 1 year, 3 months ago
Modified 1 month ago
Viewed 149 times
Upvotes 2

Koro

13:51

Can anyone please tell me how one can find the harmonic conjugate of a real valued function u(x,y) without integration?

I know one way to do so without integration but I don’t know why it works.

one potato two potato

14:02

2 days ago, by one potato two potato

If $u(x,y)$ is a harmonic function on a disk $x^2+y^2<R^2$ then a function
$$f(z) = 2u\left({z\over 2},{z\over 2i}\right)-u(0,0)$$
is holomorphic on $|z|<R$ with real part $u(x,y)$. I know how to solve this but it uses some series expansion (double power series comes out) and is quite nasty. Is there any simple way to prove this? Maybe using Schwartz/Poisson integral formula.

@Koro This?

Koro

14:25

Yes. @onepotatotwopotato

I don’t understand why it is reasonable to put z/2 instead of x in u(x,y).

It’s also mentioned in section 1.2 of Ahlfors’.

But I don’t understand why it is allowed to put z/2 in place of x.

@geocalc33 Did you try posting it on mathstack overflow?

nickbros123

Disc? Only time I've seen discs is in rotational dynamics problems

Rockstar ff- op

Hello!! I need help with proof.

Koro

Here is what I think: Suppose you want to find harmonic conjugate of u(x,y). Put x= (z+ z’)/2, y= (z-z’)/(2i).

Xander Henderson

Just ask. Don't ask to ask.

Koro

This gives a function of z and z’. We want analyticity so del u over del z’ should be 0.

one potato two potato

14:40

@Koro I just read that part. Ahlfors only stated that part formally. But the identity actually holds for complex $x$ and $y$ so it's allowed to put $z/2$ in $x$ and $z/2i$ in $y$.

Koro

@onepotatotwopotato for complex x and y? Why?

one potato two potato

@Koro Yes that needs verification and that's the part where double summation comes out. I can give you a reference: Invitation to complex analysis by Ralph P. Boas section 19B.

19B to 19D I should say

nickbros123

14:58

Most of you here are profs or high level experts in the field. What can I do to make Spivak not feel like one of the most annoying and frustrating books

Koro

15:25

@onepotatotwopotato thanks a lot :). I’ll check it out.

@nickbros123 if you don’t understand something, just ask.

I think that people here will surely help you out. Don’t be discouraged in case you don’t understand something in the book.

Goku

Why do I feel like something bad is happening right now

Rockstar ff- op

16:03

It's impossible to write $(3z)^n=(x^2-y^2)-(a^2-b^2)$ if $(3z)^n=(r^2+s^2)-(p^2+q^2)$ & vice versa for integers & if possible then one of $(x^2-y^2)$ or $(a^2-b^2)$ can never be a perfect power & in the same way one of $(r^2+s^2)$ or $(p^2+q^2)$ can never be a perfect power, where $n$ is any positive odd integer & $x,y,a,b,r,s,p,q$ are all distinct from each other. How to prove or disprove this?

For e.g.,

$3^3=14^2-13^2$

here, $14^2=50^2-48^2$ similarly $13^2=85^2-84^2$

So, $3^3$ is expressible in the first way but it's not possible to express

$14^2$ as $r^2+s^2$

So in this case both situations can't exist simultaneously.

Now, the essence of the question lies for $(3z)^n$ for any positive integer $z$ & any positive odd integer $n$.

The question is,

Is this true for all $(3z)^n$?

I can just guess that it's true;-)

one potato two potato

16:24

57

Q: The Complexity of "The Baby Shark Song".

This question is just for fun. I hope it's received in the same goofy spirit in which I wrote it. I just had the pleasure of reading Knuth's "The Complexity of Songs" and I thought it'd be hilarious if someone could do an analysis of the complexity of an infinite version of "The Baby Shark Song" ...

asymptotics recreational-mathematics information-theory music-theory

doo doo doo doo doo doo

ILikeMathematics

17:41

@copper.hat It is at least monotonic increasing then and at most strictly monotonic increasing (in the case of $f(x) = x^3$), right?

nickbros123

18:09

Is Chen combinatorics good

Rockstar ff- op

It's impossible to write $(3z)^n=(x^2-y^2)-(a^2-b^2)$ if $(3z)^n=(r^2+s^2)-(p^2+q^2)$ & vice versa for integers & if possible then one of $(x^2-y^2)$ or $(a^2-b^2)$ can never be a perfect power & in the same way one of $(r^2+s^2)$ or $(p^2+q^2)$ can never be a perfect power, where $n$ is any positive odd integer & $x,y,a,b,r,s,p,q$ are all distinct from each other. How to prove or disprove this?
For e.g.,

$3^3=14^2-13^2$

here, $14^2=50^2-48^2$ similarly $13^2=85^2-84^2$

So, $3^3$ is expressible in the first way but it's not possible to express

user193319

18:50

I want to prove that $4a^2+1$ and $2a+1$ are relatively prime for any $a \in \Bbb{Z}$. Is the following valid? $gcd(4a^2+1,2a+1) = gcd(2a+1, 4a^2 + 1 \mod 2a+1) = gcd(2a+1,2)$. Since the only positive divisors of $2$ are $1$ and $2$, and $2a+1$ is odd (meaning it isn't divisible by $2$), it follows that $gcd(2a+1,2) = 1$, thereby proving $4a^2+1$ and $2a+1$ relatively prime...?

Hmm...the book I am working through says that $gcd(a,b) = gcd(b, a \mod b)$ holds if $a$ and $b$ are positive integers...$2a+1$ is not necessarily positive...

Hmm...this seems to be giving the same sort of reasoning euclid.colorado.edu/~roymd/m2001/sols10.pdf

See the second page...So, maybe what I am trying to do works...I just don't see how to get around the fact that $2a+1$ might be negative.

Karl Kroningfeld

19:09

I think the method is fine. The division algorithm steps can be used to write $1$ as a linear combination of $4a^2+1$ and $2a+1$. Thus $4a^2+1$ and $2a+1$ must be relatively prime.

leslie townes

19:27

user: you may only caught up in whether something might be negative because your interpretation of "a mod b" requires that, not because it affects the result.

you can replace a with a plus or minus some multiple of b and the value of gcd(a,b) will not change. even if you happen to subtract so many multiples of b that the result is negative.

user193319

21:57

I'm trying to find a pair of integers $a,b$ that does not have a greatest common divisor, and that it is the only pair of such integers. Would $a=b=0$ be the right pair?

Thorgott

22:08

what is your definition of greatest common divisor

I would say that the greatest common divisor of $0$ and $0$ is $0$, but your mileage may vary

leslie townes

yeah, that question is a stress test of the particulars of how you have defined gcd.

in particular, the 'why' (0,0) does not have a gcd (if indeed it doesn't, and i agree that this is probably the pair to focus on) will depend on the particulars of the definition.

geocalc33

$$g_{jk}(\theta)=
\int_X
\frac{\partial \log p(x,\theta)}{\partial \theta_j}
\frac{\partial \log p(x,\theta)}{\partial \theta_k}
p(x,\theta) \, dx.$$

can someone clarify those subscripts?

where $p(x,\theta)$ is any normalised probability distribution with parameter $\theta$

$g$ is the Fisher metric

$j,k$ index the coordinates

If $p$ is a 1-parameter distribution then I think you can just drop the subscripts

Jakobian

23:16

well yeah. $\theta$ is probably from some manifold, that's why it has multiple parameters

um, partial derivatives

you know what I mean

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