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3:08 AM
Ask not when politics is a stage show. Ask when is politics not a stage show.
 
2 hours later…
4:43 AM
@copper.hat yeah definitely seems like a stage show. The unseriousness of it all is a bit hilarious
 
1 hour later…
6:06 AM
Id like to redact my above statement
yeah, it was pretty bad. i am curious how an audience member was killed and another injured. one report said that the bullet that clipped Trump hit someone behind.
I did not realize that people had been killed when I wrote my comment above.
6:34 AM
thankfully it didn't turn into a mass shooting as in Las Vegas.
6:47 AM
He shot into a crowd of 50,000 people.
 
3 hours later…
10:17 AM
@copper.hat My wife and I went through the same thought processes.
10:38 AM
In the probability generating function, when someone writes $E(t^X)$, where $X$ is a random variable, what is $t^X$? It can't be a function of a random variable, since $t$ isn't fixed.
11:22 AM
@psie the random variable $t^X$ for some fixed $t$
yeah, it wasn't harder than that, thanks :)
the probability generating function is the function $t\mapsto E(t^X)$
perhaps that's whats confusing to you
yeah, now that you write it it's much clearer
11:57 AM
0
A: Compact sets of Moore plane

José Carlos SantosI will denote the Moore plane by $M$ and the $x$-axis (that is, $\Bbb R\times\{0\}$) by $X$. If $p\in\Bbb R^2$ and $r>0$, then $D_r(p)$ is the open disk centered at $p$ with radius $r$. Let $K$ be a compact subset of $M$. Then $K$ is also compact with respect to the usual topology on $\Bbb R\time...

this answer that characterizes the compact subsets of the Moore plane got downvoted for no reason
12:29 PM
@copper.hat There were a lot of shots fired.
What really gets me about the whole thing is that a bunch of secret services officers were risking their lives to protect Trump, but rather than just GTFOing, he kept popping his head out of the scrum and raising a fist, putting those people at further risk.
12:58 PM
NY Times suggests that an AR-15 was used: nytimes.com/live/2024/07/13/us/biden-trump-election/… .
1:12 PM
Silly question, but Abel's theorem says that the limit from the left at $x=1$ of $\sum_{n=0}^\infty c_n x^n$ exists if $\sum_{n=0}^\infty c_n$ converges and $|x|<1$. Now, I am looking at probability generating functions, specifically derivatives of such, where one has $$\sum_{n=k}^\infty n(n-1)\cdots (n-k+1)t^{n-k}P(X=n).$$I'm confused as how to apply the theorem here, since it doesn't appear to be of the form $\sum_{n=0}^\infty c_nx^n$.
1:50 PM
@psie first you can make a substitution $m = n-k$ so that $\sum_{m=0}^\infty (m+k)...(m+1)t^m P(X = m+k)$ is your series
then you see that this is $c_m = (m+k)...(m+1)P(X = m+k)$ where $t$ and $x$ are just dummy variables
ah ok, thanks, this helped.
 
3 hours later…
5:00 PM
@XanderHenderson it is rather puzzling.
where can I learn about Vector spaces over finite fields, Ive checked sheldon axler, hoffman, and they dont seem to talk about them in much detail
im looking for introductory level stuff, get my feet into the water
5:23 PM
what about them?
A finite geometry is any geometric system that has only a finite number of points. The familiar Euclidean geometry is not finite, because a Euclidean line contains infinitely many points. A geometry based on the graphics displayed on a computer screen, where the pixels are considered to be the points, would be a finite geometry. While there are many systems that could be called finite geometries, attention is mostly paid to the finite projective and affine spaces because of their regularity and simplicity. Other significant types of finite geometry are finite Möbius or inversive planes and Laguerre...
are you looking for a book on something like this
5:39 PM
no no
I just wanted to know what fundamentally changes (if any) when studying LA over finite fields (or general fields not R or C)
kinda found the answer here: math.stackexchange.com/q/2865696/1118236
5:58 PM
@nickbros123 theory of quadratic forms is interesting even for $\mathbb{R}$
e.g. Minkowski's space
pretty much all of basic linear algebra does not care about the ground field
quadratic forms are a good example of something that does care
about Moore plane though, I think its interesting to replace $\{(x, 0)\}\cup B((x, r), r)$ with $\{(x, 0)\}\cup U_{x, r}$ for some open sets $U_{x, r}\subseteq \mathbb{R}\times (0, \infty)$ such that $(x, 0)\in \overline{U_x}$ and $U_{x, r}\subseteq U_{x, s}$ for $r < s$.
It shouldn't matter for most properties of the Moore plane, but should change how things converge to points $(x, 0)$, in particular it would change what the compact sets are
6:16 PM
hmm... it does need that $\bigcap_{r > 0} U_{x, r} = \emptyset$
and not sure if its enough to assert complete regularity
In the argument it matters that $U_{x, r}$ are convex, but also that the balls "converge" to $(x, 0)$ in a particular way
Hello everyone!! can anyone please explain how the probabilities can be calculated here for picking any random ball ??
In the argument you take some set $U_{x, r}$ and define $f(x, 0) = 0$, $f(p) = 1$ for $p\in \partial U_{x, r}\setminus \{(x, 0)\}$ (boundary in $\mathbb{R}^2$), and extend this linearly to $U_{x, r}$ and let it be $1$ otherwise. Then you need to assert this is continuous.
@SonuGupta Isn't that what you are supposed to be figuring out?
I am just confused about the many possible permutations of the picking of ball due to special balls
so for $\varepsilon > 0$ there has to exist $0 < s < r$ such that $f[U_{x, s}]\subseteq B((x, 0), \varepsilon)$
6:28 PM
@XanderHenderson yes and I really tried hard but couldnt get any answer so far. Even I tried to understand the editorials but couldnt understand them. Thats why I'm asking here
I wonder what would be some easy conditions for $U_{x, r}$ to have so that this property is satisfied
@copper.hat could you help? This is, at heart, about convex sets
@SonuGupta What you have posted appears (to me) to be a competition problem. I am not really interested in working a competition problem for someone else.
Also, it appears to me that it is a coding problem, not really a math problem. It seems that you are supposed to simulate it.
@XanderHenderson No, its over. Believe me. I am just curious to understand the concept behind it which is probability.
@XanderHenderson The main issue I'm facing here is that due to the special balls in n set of balls we can have many possible combination of the balls which would lead to different expected values for each of them ?? Then how can I uniquely determine it as it is asked in the question.
would it make sense to say polar coordinates are specified in $\mathbb{R}\times\mathbb{R}/2\pi$ or $\mathbb{R}/360$ for degrees
i guess more precisely $[0,\infty)\times[0,\infty)/2\pi$
6:55 PM
I'm studying PGFs and it is claimed that if two random variables have the same PGF, then they have the same distribution. The author of my book refers to the literature for this result and remarks this follows from the "uniqueness theorem" for power series. I have been trying to look up this theorem and found two theorems in regards to this, but to a great extent different proofs. This makes me think these theorems are not the same, but I can't spot what the difference between them is.
The first theorem is from baby Rudin. It reads as follows:
> 8.5 Theorem Suppose the series $\sum a_n x^n$ and $\sum b_n x^n$ converge in the segment $S=(-R,R)$. Let $E$ be the set of all $x\in S$ at which $$\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty b_n x^n.$$ If $E$ has a limit point in $S$, then $a_n=b_n$ for $n=0,1,2,\ldots$. Hence the above equation holds for all $x\in S$.
The second theorem is from these lecture notes.
> Corollary 10.23. If two power series $$\sum_{n=0}^{\infty} a_n(x-c)^n, \quad \sum_{n=0}^{\infty} b_n(x-c)^n$$ have nonzero-radius of convergence and are equal in some neighborhood of $c$, then $a_n=b_n$ for every $n=0,1,2, \ldots$.
The corollary is much easier to prove it seems, so it makes me think it is a different statement from the theorem in Rudin, but I can't see what the difference is.
@Jakobian Sure, if I can, I will. Where does the question start?
@psie In one theorem, both series on defined on a set which has a limit point. In the other, they are defined on a neighborhood (i.e. a set containing an open set).
@XanderHenderson ah ok, subtle difference, but I think I see what you are saying. $E$ in the corollary is a neighborhood around $c$ (i.e. a set containing an open set), whereas $E$ in Rudin's theorem is simply a set which has a limit point. Grazie!
 
2 hours later…
8:54 PM
@copper.hat So consider $X = \mathbb{R}\times [0, \infty)$. For each $x\in\mathbb{R}$ you have a convex set $U_{x, r}\subseteq \mathbb{R}\times (0, \infty)$ such that $(x, 0)\in \overline{U_{x, r}}$ and for $0 < s < r$ you have $U_{x, s}\subseteq U_{x, r}$. Define $f:\overline{U_{x, r}}\to [0, 1]$ to be $1$ on $\partial U_{x, r}\setminus\{(x, 0)\}$ and $f(x, 0) = 0$, and extend linearly. What would be some simple conditions on $U_{x, r}$ to make it so that for each $\varepsilon > 0$ there exists
$0 < s < r$ such that $f[U_{x, s}]\subseteq B((x, 0), \varepsilon)$?
Example of such family is $U_{x, r} = B((x, r), r)$
By extend linearly, I mean that on the segment from $(x, 0)$ to a point $p$ on the boundary you are defining it linearly. Also assume $U_{x, r}$ are non-empty, open and bounded
so $f((1-t)(x, 0)+tp) = t$ for $t\in [0, 1]$
 
2 hours later…
11:16 PM
Hi everyone, I need to prove $f(x)=e^x$ convex without using the derivative test.
I got stuck at
$\lambda e^{(x_1-x_2)}e^{\lambda(x_2-x_1)}+(1-\lambda) e^{\lambda(x_2-x_1)} \geq 1$?
I need to prove the above inequality. Is there any hope with this?
$\lambda \in [0,1]$
Interestingly, numerical attempts show that this inequality always holds
Given $x_2\geq x_1$ and $x_1,x_2\in \mathbb{R}$.
4
Q: How to prove the convexity of the exponential function?

zeynepHow to prove that the convexity of exponential function? It is not allowed to use second derivative of $e^x$.

11:33 PM
I saw this post but I didn’t fully understand the answer.
the first one?
Yes
I think OP meant that proving convexity is equivalent to proving the last inequality
I feel the answer is missing details. Mine is more lucid if I could prove the above inequality.
I can’t type my full answer because I’m typing from my phone.
what are you allowed to use?
11:41 PM
I don’t understand your question.
what tools can you use?
Tools what?
you mean typing?
no
you said you can't use derivatives
I need to use the definition which is the one stated in the answer you’ve linked
I feel the answer is not complete.

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