« first day (4510 days earlier)      last day (518 days later) » 

2:40 AM
anyone here?
how do i find my motivation
uses command-F to find motivation
 
 
2 hours later…
4:40 AM
peep
 
 
2 hours later…
6:57 AM
Chat gpt is at best a decent search engine
 
 
1 hour later…
8:11 AM
@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/…
 
I have a question about notation. Let $A$ a ring and $a \in A$, what is $A_a$?
 
8:41 AM
@JoeWillyam it's the localization at the set $S=a^\Bbb N$
 
0
Q: Show that ${1\over\pi}\iint_{\Bbb D}|\psi'_\alpha|^2\ dxdy =1$

one potato two potato 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...
 
 
3 hours later…
11:53 AM
@onepotatotwopotato poop
 
12:43 PM
Hello everyone! Maybe someone has an idea: figure out whether series converge or not:
$\sum\limits_{n=2}^{\infty} \dfrac{(2+5\cdot (-1)^{\dfrac{n(n-1)}{2}})^n}{n\cdot \sqrt{\ln{n}}} \cdot \dfrac{1}{7^n}$
 
1:38 PM
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
 
1:51 PM
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.
 
2:02 PM
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?
 
2:25 PM
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?
 
Disc? Only time I've seen discs is in rotational dynamics problems
 
Hello!! I need help with proof.
 
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).
 
Just ask. Don't ask to ask.
 
This gives a function of z and z’. We want analyticity so del u over del z’ should be 0.
 
2:40 PM
@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 $y$.
 
@onepotatotwopotato for complex x and y? Why?
@onepotatotwopotato for complex x and y? Why?
 
@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
 
2:58 PM
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
 
3:25 PM
@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.
 
Why do I feel like something bad is happening right now
 
4:03 PM
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;-)
 
4:24 PM
57
Q: The Complexity of "The Baby Shark Song".

ShaunThis 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" ...

doo doo doo doo doo doo
 
 
1 hour later…
5:41 PM
@copper.hat It is at least monotonic increasing then and at most strictly monotonic increasing (in the case of $f(x) = x^3$), right?
 
6:09 PM
Is Chen combinatorics good
 
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
 
6:50 PM
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.
 
7:09 PM
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.
 
7:27 PM
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.
 
 
2 hours later…
9:57 PM
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?
 
10:08 PM
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
 
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.
 
$$g_{jk}(\theta)=
\int_X
\frac{\partial \log p(x,\theta)}{\partial \theta_j}
\frac{\partial \log p(x,\theta)}{\partial \theta_k}
p(x,\theta) \, dx.$$
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
 
11:16 PM
well yeah. $\theta$ is probably from some manifold, that's why it has multiple parameters
um, partial derivatives
you know what I mean
 

« first day (4510 days earlier)      last day (518 days later) »