« first day (5185 days earlier)      last day (24 days later) » 

12:53 AM
Anyone know any nice open problems in high dimensional manifolds theory?
 
2 hours later…
3:20 AM
$(\sum_{i=1}^{n}|x_i|^p)^{1/p}\leq \sum_{i=1}^n |x_i|$ I know can be deduced from Jensen inequality and convexity of $x \mapsto x^p$ for positive integer $p$. is there a simpler way to deduce this?
4:05 AM
How many terabytes would a computer need to deal with computations on a 800 trillion decimal digits number?
That's an estimate on how large the number might become with my process that I feel may find the Riemann Zeta zero with value not 1/2
Provided it exists
4:20 AM
anyone studied hensel lemma in neukirch book
i have doubts in its proof
$\LaTeX$
any one explain why $q(x)\in O(x)$
 
3 hours later…
 
2 hours later…
9:15 AM
@ZacU. to store the number alone would require more than 300 terabytes!
9:33 AM
@nickbros123 Am I missing something or this is a consequence of the subadditivity of $x\mapsto x^{1/p}$ ?
9:48 AM
I'm doing the 2nd problem
I don't understand what they mean by the last two sentences
"The first bar is in exactly half of the compositions,...and so on"
Why?
10:42 AM
Hi everyone. On the following question I think I proved the impossibility of having two perfectly negatively correlated $\rho=-1$ Geometric Brownian Motion series. I hope you could take a look and tell me if you find any mistakes on the proof. Thanks beforehand.
11:10 AM
Suppose in the above theorem b) does not hold. What's the conclusion then of the theorem? Only that $f_x$ and $f^y$ are measurable for a.e. $x$ and for a.e. $y$ respectively?
@psie no, also that $x\mapsto \int f_x$ and $y\mapsto \int f^y$ are measurable
@Jakobian ah ok, to define the integral of $f_x$ and $f^y$, they need not be integrable, right?
@psie we're not having the whole "integrable vs integral exists" discussion again
@TheEmptyStringPhotographer goodness!
I guess the displayed equation also holds in case $f\geq 0$.
The integral with respect to the complete measure turns into an iterated integral of the incomplete measures.
11:28 AM
In the last para "and two sections with this property coincide on an open nbd of x" it's not clear to me that how they arrived at this consequence from axiom 1
Can someone help?
Hi
I have a question but when I have to find the absolute extremes on a circumference and I parametrize, if $\theta \in [0,2\pi]$ , then in the end I will also have to check the value of the function at $0, π/2 , π , 3π/2, 2π$?
That is, manually
Because while reviewing the exercises I noticed that here: Calculate the function definition set: $f(x, y) = \arcsin(x^2-y-1)$ and calculate the absolute extrema in the circle with center (0, −1) and and unit radius.
I can't find that red dot on the left
$f(\theta) = \cos^2(\theta) - \sin(\theta) \quad f'(\theta) =\cos(\theta) (-2\sin(\theta) - 1)$
Setting the first derivative = 0, I find i $\theta = \pi/2, 3\pi/2, 7\pi/6, 11\pi/6$
So doing the calculations I find the absolute minimum at $\pi/2$ and the absolute maximum at $3\pi/2$ , if I were to also consider $0, \pi$ then I also find the missing ones
So I was thinking that maybe in domains like squares, triangles, rectangles etc., i consider the vertices, while in the circle I have to consider $0, \pi/2, \pi , 3\pi/2, 2\pi$?
12:04 PM
$\sum^{\infty}_{n=0} \frac{x^{n^2}}{n!}$
I'm in difficulty, I don't know if I should make a substitution for x^n²
I did it like this:
$\lim_{n\to\infty} \left|\frac{1}{(n+1)n!} \cdot n!{} \right| =\lim_{n\to\infty} \frac{1}{n} \to 0$
@Pizza you don't have to check at all these points
only at the end points, i.e. $\theta=0,2\pi$
@psie ah yes, of course. That part is Tonelli's theorem and its really important
@SineoftheTime But where does that point in $\pi$ come from then?
@Gian'sPizzeria and how would you do that anyway
I mean, it seems to me that I did everything right...
12:09 PM
@Jakobian $x^n=y$
your $y$ depends on $n$...
@Pizza it's not clear here
what is $f$ ?
and how you parametrize the circle? Polar coordinates in $(0,-1)$?
Yes
@Gian'sPizzeria if you're trying to use a theorem for power series, then you didn't do it correctly
12:13 PM
I used $x = \cos(\theta), y = -1 + \sin(\theta)$
So $g(x,y)=x^2-y-1$
Yes then I only considered the argument
theorem about taking quotients $a_{n+1}/a_n$ for power series doesn't work here since your series $\sum a_nx^n$ is such that $a_n = 0$ infinitely many times
$g(\cos \theta, -1+\sin \theta)=\cos^2 \theta-\sin \theta:=h(\theta)$ for $\theta \in [0,2\pi]$
however - you can use this theorem as for regular series
12:15 PM
@SineoftheTime yes
but then you don't get a quotient of $(1/(n+1)!)/(1/n!)$ but you get a quotient $(x^{(n+1)^2}/(n+1)!)/(x^{n^2}/n!)$
you have to be precise with the notation, writing $f(\theta)$ is confusing since $f$ is the original function
Right
so now: $h'(\theta)=-2\cos \theta \sin \theta-\cos \theta=-\cos \theta(2\sin \theta +1)$
Yes
12:18 PM
sorry, typo
@Jakobian In the theorem it is written that $a_n≠0$
and now you find the values you wrote before
This additional thing is not there
Yes
$\frac{|x|^{(n+1)^2}/(n+1)!}{|x|^{n^2}/n!}$
@Gian'sPizzeria it isn't, so you can't use this theorem
12:19 PM
@Pizza so your doubt is if we have to check manually for $x=\pi$?
not as a theorem about calculating radius of power series $\sum a_n x^n$
you need to use a version for regular series $\sum a_n$
@SineoftheTime More than anything, where does that absolute maximum come from?
if you are to use this type of theorem at all
@Pizza what do you mean?
In that image of Wolfram there is that absolute maximum at π
12:21 PM
Once you find those values, you have to substitute in the original function to see where is the max
Yes
@Pizza what does it mean at $\pi$?
the plot you sent is with coordinates $x,y$, isn't it?
The point on the left
the alternative here is to use the formula $\lim_n \sqrt[n]{a_n} = \lim_n (1/n!)^{1/n^2}$ and say, Stirling approximation $n!\sim (n/e)^n \sqrt{2\pi n}$ to show $\lim_n (1/n!)^{1/n^2} = \lim_n \sqrt[n]{n/e} \cdot (2\pi n)^{1/2n^2} = 1$
@Jakobian But therefore the limit of $a_n$ that goes to infinity must not be 0
12:23 PM
g(π/2) = -1 , g(3π/2) = 1, g(0) = 1, g(2π) = 1
@Gian'sPizzeria no buts. What are you objecting to
g(7π/6) and g(11π/6) = 5/4
But > 1, so I don't have to consider it
@Gian'sPizzeria don't give me random links, give me words
and your justification for why you think something is or isn't true
and WHAT is that you think is or isn't true
@Pizza I don't understand what you're trying to say
how does the graph you sent imply that at $\theta=\pi$ you have a critical point?
I saw that to use this theorem an must not be 0 for large n, does this mean that the limit of n that tends to infinity of an must not be 0?
12:27 PM
$a_n$ being non-zero for large $n$ means that there exists $N$ such that $a_n\neq 0$ for $n\geq N$
this doesn't imply that $\lim_{n\to\infty} a_n \neq 0$
and its not actually relevant to our discussion in the slightest
there is plenty of sequences of non-zero numbers convergent to $0$, $a_n = 1/n$ being an example
I don't understand why the theorem cannot be used
the one about power series $\sum a_n x^n$?
@SineoftheTime No, it's not like that, it's just that I had seen that since at $0$ , $3π/2$ I had found the maximum that coincided with that image
And even at π/2 I found the minimum
Because $a_n = 0$ whenever $n$ is not a square
12:31 PM
Then that point that remained was π
and there is infinite amount of natural numbers which aren't squares
for example, $m^2+1$ is never a square
@Pizza what is the meaning of the image? What did you write on wolfram?
@SineoftheTime maximize arcsin(x^2-y-1) on the domain x^2+(y+1)^2 = 1
12:35 PM
@Jakobian but an≠0
@Gian'sPizzeria no buts
don't give me random links, give me words
and your justification for why you think something is or isn't true
and WHAT is that you think is or isn't true
When 1/n! = 0???
$a_n$ is not $1/n!$???
@Pizza ok, now I understood
@Jakobian Why?
12:41 PM
@Gian'sPizzeria it just isn't, what do you expect me to say
@SineoftheTime What did I do wrong?
Example: x^n/n an=1/n, why doesn't this apply here?
In order to treat the series $\sum \frac{x^{n^2}}{n!}$ as a power series $\sum a_nx^n$ we need to define $a_n = \begin{cases} \frac{1}{m!} & \text{if }n = m^2 \\ 0 & \text{otherwise} \end{cases}$
@Pizza you did it correctly
but there's a problem
that is $\arcsin (x^2-y-1)$ is not defined on the whole circle
No, so the values 5/4, are excluded
12:45 PM
$\sum b_n x^{n^2}$ isn't a power series (in that form)
in order to treat it like a power series one needs to introduce coefficient $a_n$ like above i.e. introduce the zeros
this exercise is subtle @pizza
You actually don't have to maximize on that circle, but on the circle intersected with the domain of $f(x,y)$
I did not notice that, my bad
I didn't understand, that is, did I get the exercise procedure wrong?
partially, yes
In the part where I use parameterization?
The point is: you don't have to maximize $f$ on the circle, but on the circle intersected with the domain of the function $f$
12:52 PM
How should I solve it?
Let's start by finding the domain of $f(x,y)$
$|x^2-y-1| \leq 1$
you have: $-1 \le x^2-y-1\le 1$ i.e. $x^2-y \ge 0$ and $x^2-y\le 2$
Yes
now, draw the region $x^2-y \le 2$
12:55 PM
@Jakobian And what if I wanted to treat it without it being a power series?
But I had done this step in the exercise
then you should have noticed that not all point in the circle satisfy $x^2-y\le 2$
No, it would be those 5/4 that I had found
do you see that?
In fact I excluded them
@SineoftheTime yes
12:56 PM
ok, then the parametrization is wrong. $\theta \in [0,\pi]$
and to see the other piece, you have to parametrize the parabola
first you parametrize the upper circumference, then the parabola under $x=-1$. Is it clear?
But couldn't Just i calculate the function at the intersection points between the circle and the parabola?
@SoumikMukherjee let $U, V$ be as in axiom $1$. If $s$ is a section with the properties listed, then there exists neighbourhood $U_0$ of $x$ such that $s:U_0\to V$ from continuity of $s$. Since $\pi\circ s$ is an identity map, we can take inverse of $\pi$ restricted to $V$ and compose it to show $s$ must be obtained as inverse of $\pi$ restricted to $V$ in a neighbourhood of $x$.
yes, because these are end points
but it's not sufficient
you have to parametrize, use the first derivative test and then see what happens at the end points
note that here you need two parametrizations
@Gian'sPizzeria do I have to repeat myself? What are you asking me.
45 mins ago, by Jakobian
$\frac{|x|^{(n+1)^2}/(n+1)!}{|x|^{n^2}/n!}$
its all written above there
@Pizza is it clear?
1:07 PM
Maybe yes, but is it right that I can only derive the argument from arcsin?
Anyway I'll try it again, I'll let you know, thanks for the help
@Jakobian I think I'll proceed like this, thank you very much
@Pizza what does it mean?
@Pizza try this: maximize the function over $D=\{\text{circle}\}\cap \{\text{domain of the function}\}$ and I'm sure you'll find the correct result
@Jakobian I understand, thank you very much
@SineoftheTime When we parameterized , I only considered the argument when I calculated the derivative
@SineoftheTime ok, I try
yes, you can do that
then to find the max of the function you have to plug in in $\arcsin(\dots)$
1:11 PM
Ah, I had entered the values in $g(\theta)$, not arcsin
you are asked to maximize f, not g
you introduce g to simlplify
I have to go, let me know if you can solve it
Yes, I also have to go and buy some books that I need, as soon as I get back I'll do the exercise
 
1 hour later…
2:14 PM
If there are 3 million mathematicians on a boat, how old is the captain?
2:25 PM
Regarding a recent question that I posed here, it just occurred to me that folks who find the statement $0\not\in\left\{x\in\Bbb R:\frac1x>1\right\}$ unacceptable perhaps shouldn't accept $\forall x{\in}\mathbb R{\setminus}\{0\}\;\frac1x\ne5$ either, since it merely abbreviates $\forall x{\in}\mathbb R\;(x\ne0\implies\frac1x\ne5)$ and asserts, in particular, the illegal statement $0\ne0\implies\frac10\ne5.$
@TheEmptyStringPhotographer yes
2:42 PM
@ryang I disagree that it abbreviates this
3:02 PM
@Jakobian 1. $\forall x{\in}S\,P(x)\overset{\text{ def}}\iff \forall x\,\big(x\in S\to P(x)\big)\quad\quad$ 2. $(A∧B) → C \;\:\equiv\:\; A → (B→C).$
After going thru a bit of Searcoid-metric spaces, I can conclude this books is pretty solid imo
It has a lot of equivalent definitions, which I think is great, especially for paranoid people like me
@ryang that is the formal way, sure. But writing $\forall_{x\in S} P(x)$ when $P$ involves a function $f$ say, should be interpreted as abbreviating $\forall_{x\in S} f$ is well-defined and $P(x)$
sorry not $f$ is well-defined... you know, say $f$ explcitly depends on variable $x$, then it should be interpreted as $x\in \text{dom}(f)$ and $P(x)$
@SineoftheTime i think were eventually talking about the same thing
was that a weird use of the word "eventually"?
we end up with the same kind of issue about how do we interpret $f(x)$ when $x$ is not in the domain of $f$
@nickbros123 yes, to me at least
i felt it too
it was actually unnecessary
3:20 PM
@Jakobian The whole context of that definition is that that abbreviation is informal; anyhow, i'm not following why you think $\forall x{\in}\mathbb R{\setminus}\{0\}\;\frac1x\ne5$ doesn't abbreviate $\forall x{\in}\mathbb R\;(x\ne0\implies\frac1x\ne5).$ In any case—and I'm possibly misunderstanding you—but ultimately aren't you selectively making concessions (for $\forall x{\in}\mathbb R{\setminus}\{0\}\;\frac1x\ne5$ but not for $0\not\in\left\{x\in\Bbb R:\frac1x>1\right\}\;).$
4:08 PM
I have a confusion about defining probability measures. Suppose our sample space consists of infinite sequences with values in $\{0,1\}$ and our set of events will be generated by $A_i=\{i\text{th coordinate is }1\}$ where we define $P(A_i)=1/2$ for any $i\in\mathbb{N}$. How does one actually show countable additivity for such a function? I cannot see how this follows from just our constant function definition.
I want to minimise (c*r)^r over positive integers r and 0 < c < 1. Can I always just minimise it over the reals and round to the nearest integer?
@Simd No, not exactly, but it gives you good places to look if you can apply some other theory (e.g. monotonicity).
Though it is also worth noting that for $c \in (0,1)$, the expression $c^{r^2}$ is decreasing in $r$, and can be made arbitrarily close to zero by choosing $r$ big enough. So you are not going to have an easy time finding a minimum.
4:27 PM
@XanderHenderson why $c^{r^2}$?
@Jakobian Oh, I misread the question as $(c^r)^r$.
Nevermind.
In the case of $(cr)^r$, go back to my original statement: you can't just minimize over the reals and hope that it will work by rounding, but if you minimize over the reals and then apply a monotonicity argument, you can get there.
is c fixed here?
@XanderHenderson it's not c^(r^2) though
@Simd Yes, read to the end of the chat. This has already been pointed out.
@nickbros123 yes
4:38 PM
Typing (c*r)^r is somewhat confusing, as no one bothers with multiplication symbols most of the time. Your expression is (cr)^r---the extra symbol threw me.
It's (c*r)^r
@XanderHenderson understood
@Simd Yes. I know.
8 mins ago, by Xander Henderson
In the case of $(cr)^r$, go back to my original statement: you can't just minimize over the reals and hope that it will work by rounding, but if you minimize over the reals and then apply a monotonicity argument, you can get there.
common practice in programming i suppse
but then id expect ** instead of ^
But I can't find a counter example to just rounding to the nearest integer which is frustrating
@nickbros123 Yeah, but bad practice in mathematics, where $\ast$ is often used for other things, e.g. convolution.
4:42 PM
I am very grateful if anyone can find a counterexample
Some very quick experimentation with GeoGebra seems to give a counterexample with $c=0.1474$. In that case, minimizing over the reals gives $r_{\text{min}} \approx 2.496$, which rounds down to $2$. But $(0.1474\cdot 2)^2 \approx 0.0869$, while $(0.1474\cdot 3)^3 \approx 0.0865$.
I would take the derivative of $(cr)^r$
@Jakobian I have assumed that @Simd knows how to minimize the expression over the reals---they are asking if minimizing over the reals and then rounding will always give the minimum over $\mathbb{N}$.
its equal to $(cr)^r\cdot (\ln(cr)+1)$ which sign depends on sign of $\ln(cr)+1$, that is if $cr > 1/e$ or $cr < 1/e$
Like, you know that $(cr)^r$ will be decreasing on the interval $(0,1/(c\mathrm{e}))$, and increasing after that.
4:52 PM
from this you can tell that minimizer of $(cr)^r$ will either by floor or ceiling of $r = 1/(ce)$
So the minimum over $\mathbb{R}$ will be at $1/(c\mathrm{e})$. The minimum over $\mathbb{N}$ is then what you get be either rounding that up or down, depending on which gives the smaller result.
(which is the monotonicity argument I was suggesting above).
But you can't just round $1/(c\mathrm{e})$ to nearest integer and expect it to always work out.
I think I found a counter example
5 mins ago, by Xander Henderson
Some very quick experimentation with GeoGebra seems to give a counterexample with $c=0.1474$. In that case, minimizing over the reals gives $r_{\text{min}} \approx 2.496$, which rounds down to $2$. But $(0.1474\cdot 2)^2 \approx 0.0869$, while $(0.1474\cdot 3)^3 \approx 0.0865$.
@XanderHenderson oh cool!
Thank you
That was much harder than I was expecting!
Germany vs. Netherlands soccer game today I'm watching
5:15 PM
@ModularMindset who are you rooting for?
5:27 PM
hi
@ephe If you just say $P(A_i) = 1/2$ you did not define the probability for all events
What you probably want to say is that $P(A_i) = 1/2$ and the events $A_i$ must be independent.
Even here we don't give explicitly the values of $P$ for all events, so the question should be not "Prove that $P$ is $\sigma$-additive", but "Prove that there exists a unique $\sigma$-additive $P$ satisfying these conditions"

And I guess in this case you need to give a complete description of events and their probabilities. If I'm not mistaken, any event here will be a countable union of sets of the following
@ModularMindset I'm watching you watching the soccer game
Greetings.
5:43 PM
@Almanzoris hi
@Jakobian did you get drunk last night? I saw you posting something about getting really drunk. Maybe you were sarcastic.
sarcast-hic!**&%
@copper.hat you too? :)
unfortunately not
5:46 PM
@psie yes. I drank more than half a bottle of whiskey
Went from sober to quasi-sober.
@Jakobian cool. You know, drinking and driving is illegal in many countries. Hope you didn't drive.
@psie I never learned to drive
used to be a tv public service ad in Ireland when I was growing up: "If you drink, don't drive, but if you do, just two will do."
Now you can hardly have food cooked with wine & drive.
Doesn't alcohol evaporate when cooked with wine
5:49 PM
I was thinking the same.
@copper.hat risotto is my favorite
i'm not a foodie, but i enjoy eating :-)
I don't actually enjoy eating that much
my treat dinner atm is a masala dosa
i enjoy eating with friends when i don't have to work shortly afterwards
I should get around to making dosas
I keep being reminded that that's a thing, think "hmm, I should make that" and then forget about it
5:52 PM
@Jakobian have you tried fasting?
Ted Shifrin described my approach to food preparation as "assembly, not cooking" ?
He is correct.
I was completely unaware of that masala dosa is something. But it looks good.
@psie yes. But it makes me weirdly irritated
besides it makes me weak
I don't know, its not my favorite thing to do
ok, I guess your reaction is not so unusual.
Are there chess players here?
6:05 PM
there's at least 3 and maybe more
Nice! Would be cool to play a game.
6:20 PM
$I=\int_0^{\infty} \frac{sin(x)}{x}$
how is it done?
@Gian'sPizzeria well, some solutions are given here <-- that's a link to a specific answer, where you need to know about Fubini's theorem.
I think it can also be done with the Feynman's trick
there was a user in the answer who used it
@psie Thanks very much
Bml
Bml
6:37 PM
Hi everyone, sorry to bother you, could you kindly explain why this edit was rejected? The post does not meet Math SE guidelines at all, since it is not written in English, nor in MathJax syntax. Where is the error?
@Bml it's not really an editors job to completely transform a question
changing the language its written is a major thing
Bml
Bml
@BenSteffan I understand, thank you :-)
6:55 PM
@Almanzoris in chat we often sends links to join a game, if someone is around you can ask
@Gian'sPizzeria the standard method in analysis 2 is to use the differentiation under integral sign
@Bml long time no see :)
Bml
Bml
@SineoftheTime Yes... how are you?
good, what about you?
Here's an exercise. Compute $\int \chi_D \, d(\mu\times\nu)$ where $\mu$ is Lebesgue measure on $[0,1]$ and $\nu$ counting measure on $[0,1]$ and $D$ the diagonal in $[0,1]\times[0,1]$. I have already computed iterated integrals, and they disagree. Is it possible to use Fubini's theorem to conclude that $\chi_D\notin L^1$ and hence the integral is infinite? What confuses me is that in Fubini's theorem we require the measure spaces to be $\sigma$-finite, and one of them isn't here.
I don't think this approach works.
@SineoftheTime 👍
7:16 PM
@SineoftheTime Sure! Thanks for the information.
7:40 PM
All of you have probably seen various images of Mandelbrot set about a billion times already, right? But I still want to share this result with you, because I think it is interesting since it is displayed in console window (which means no graphics, only text). And the source code of the program responsible for generating that result is only 151 bytes long, so it nicely shows how mathematics can enable a simple algoritm to generate a complex result!
@user430580 nice
thank you!
BTW here is the code in C which generated this result (I am not the author of the code, and I found it on the Internet with a note that the author is unknown):
main(n){float r,i,R,I,b;for(i=-1;i<1;i+=.01538,puts(""))for(r=-2;I=i,(R=r)<1;
r+=.00448,putchar(n+31))for(n=0;b=I*I,26>n++&&R*R+b<4;I=2*R*I+i,R=R*R-b+r);}
8:31 PM
@psie how do you define the product
@Jakobian $\mu\times\nu$ is only defined on the algebra of finite disjoint union of rectangles, namely by $\mu\times\nu(E)=\sum_1^n\mu(A_i)\nu(B_i)$. We can extend this by Caratheodory's extension theorem to the product sigma algebra $\mathcal M\otimes\mathcal N$ and get a unique measure provided $\mu,\nu$ are $\sigma$-finite.
But I guess what you are after is some kind of definition in terms of infimum of a set.
E.g. $$\mu\times\nu(E)=\inf\left\{\sum_1^\infty\mu(A_i)\nu(B_i):\text{ something something}\right\}.$$
Non-uniqueness is what I'm getting at
@Jakobian and? This very exercise is meant to demonstrate that Fubini-Tonelli's theorem fails when the spaces are not $\sigma$-finite, but I don't see how to compute that integral of $\chi_D$ with respect to the product measure.
@psie and what? There is no reason to expect a unique value
But sure, maybe we can show its infinite anyway
In mathematics, given two measurable spaces and measures on them, one can obtain a product measurable space and a product measure on that space. Conceptually, this is similar to defining the Cartesian product of sets and the product topology of two topological spaces, except that there can be many natural choices for the product measure. Let ( X 1 , Σ 1 ) {\displaystyle (X_{1},\Sigma _{1})} and...
According to wikipedia there is a minimal product measure, I'd try with that
8:51 PM
@Jakobian yeah, I believe that's the case. I think a descriptive argument would work kind of. Something like...the "smallest" cover of $D$ of rectangles would be $\{\{x\}\times\{x\}:x\in[0,1]\}$. Since $[0,1]$ is uncountable, we get infinity immediately, or?
@psie huh..
@Jakobian yes?
@psie I'm replacing "something something" here by "$\{E_n\}_1^\infty$ a sequence of finite disjoint union of rectangles which cover $E$". Or equivalently, simply "$\{A_i\times B_i\}_1^\infty$ a sequence of rectangles which cover $E$".
I think for the minimal product measure this is actually 0
But for maximal its infinity
So that its not unique
@Jakobian ok, that was trickier than I thought...hmm.
@Jakobian I don't understand the definition of the maximal product measure in the Wikipedia article you linked. Do you understand it?
There doesn't seem to be a precise definition, or a formula, like for the minimal product measure.
9:56 PM
@psie $(\mu\times \nu)_{\text{max}}(A) = \inf\{ \sum \mu(A_n)\nu(B_n) : A\subseteq \bigcup_n A_n\times B_n\}$
while $(\mu\times \nu)_{\text{min}}(A) = \sup_{S\subseteq A, (\mu\times\nu)_{\text{max}}(S) < \infty} (\mu\times \nu)_{\text{max}}(S)$, from below
10:26 PM
@Jakobian I think I misunderstood Folland. When he writes $\mu\times\nu$ he has a specific object in mind, namely the restriction of the outer measure induced by the premeasure we start with on the algebra of finite disjoint union of rectangles: $\pi(E)=\sum_1^n\mu(A_i)\nu(B_i)$. This is well defined as one can show. Absent $\sigma$-finiteness, it is true that there might be other measures that also extend the premeasure. But when he asks about $\mu\times\nu(D)$, it's a specific measure I think.
10:37 PM
Since the measure spaces in the problem are not $\sigma$-finite, there might be other measures such that the measure of every measurable rectangle is the product of the measures of its sides. But when he writes $\mu\times\nu$, its the restriction of the outer measure induced by the premeasure.
@psie no because $D$ is not a union of such rectangles
ah sorry
you mean the measure induced by the Caratheodory criterion
sure, that's most likely
yeah
above its $(\mu\times \nu)_{\text{max}}$
yeah, I believe that's the one
11:03 PM
I really wanna find out about the possible procedures which might exist in working on zeta zeroes
The original way I described a possible paper of the algorithm I built was "on the uselessness of writing leading zeroes"
11:35 PM
@Jakobian probably my last question on this, but is it true that \begin{align} \mu\times \nu(D) &=\inf\left\{\sum_1^\infty \mu\times \nu(E_n) : D\subset \bigcup_1^\infty E_n,\{E_n\}_1^\infty\subset\mathcal{A} \right\}\\ &=\inf\left\{ \sum_1^\infty \mu(A_n)\nu(B_n) : D\subset \bigcup_1^\infty A_n\times B_n\right\},\end{align}where $\mathcal A$ is the finite disjoint union of rectangles?
The first set seems more natural to me as the definition of $\mu\times\nu$ and I don't see how it equals the second.
Drinking a cup of coffee at 10 pm was not a great idea :(
@SineoftheTime yeah, that is not a good idea. Why did you do that? :)
Coffee is for mornings or afternoons.
I was having a bad headache and wanted to study a bit more. I said to myself: I'm so tired that I'll sleep anyways, the coffee won't make any effect
I see. Well, I've been there, done that. Next time you know better :)
11:55 PM
@Jakobian the first set seems more natural to me due to the above, i.e. every premeasure induces an outer measure according to equation (1.12), but you see, the sets that cover our set are from the algebra, and these are finite disjoint unions of rectangles.

« first day (5185 days earlier)      last day (24 days later) »