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5:03 PM
6 mins ago, by Rudi_Birnbaum
OK: this one point in the 1-point compactification is it topologically equivalent to a point in a compact set?
take {0,1,1/2,1/3,1/4,...}
this is a compact set
now take away 1
it's still compact
but if you take away 0 instead, then it's not compact, and its one-point compactification is isomorphic to the original set
where by "isomorphic" i mean "homeomorphic"
 
@LeakyNun OK "homeomorphic"!
 
rudi, someone finally countered your downvote on some of my answer, bringing its score to 0 o..o
 
Please help me in solving this question.
 
@mercio What?
@mercio What downvote are you referring to?
 
@mercio can you help me solving this question ..
 
5:09 PM
where do those questions come from
 
My teacher asked.
 
@TobiasKildetoft, how are we supposed to write it?
 
I am not sure I want to do anything with that question
 
@mercio I don' understand what you are refering to here???
 
5:12 PM
rudi, this one math.stackexchange.com/questions/2704399/…−1-−-i-and-1-i-evaluate/2704731#2704731
(though maybe it wasn't you, you can never really be sure with votes)
rip formatting
 
@mercio Oh that one, its possible it was me. I didn't like it since after reading one didn't get any closer to $i\pi$.
@mercio I voted it now up, and it was working, so most probably it wasn't me who downvoted it ...
 
also can someone tell(/chat to) me why infinitesimals are such a controversial topic?
 
@blue_eyed_... Not sure what you mean. But there is no $r$ such that the $n$'th term of the Fobonacci sequence is $r^n$
@MoreAnonymous Well, they are not actually controversial. They just often get treated way too imprecisely.
 
@mercio So currently you got three upvotes! Seems its good for those (upvotes) to link stuff occasionally from the chat ... (especially if your "prominent").
 
@Identicon is the a power? Are you missing brackets?
 
5:21 PM
@TobiasKildetoft how so?
an explicit example?
 
@MoreAnonymous Of what? Of how they are treated too imprecisely?
 
@mercio I know roughly about the scope of MSE, though I tend more to the people which also want Answer/Question.
 
yea ... how are they treated imprecisely
?
 
infinitesimals used to be controversial before we got the rigorous epsilon-delta definitions
 
@mercio infinitesimals have no place in the world of epsilons and deltas. They live in another world
@MoreAnonymous I don't have any concrete example handy actually
 
5:24 PM
@TobiasKildetoft u might be interested in some recent proof of mine :P
0
Q: A rough proof for infinitesimals?

More AnonymousI discovered the following relation for arbitrary $d_r$: $$ \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{d_r}{r^s}}_{\text{removable singularity}} \times \int_0^\...

 
maybe i should link to all my answers
 
lol
using a bot?
 
@MoreAnonymous No idea what all of that is supposed to prove. You don't "prove" infinitesimals.
 
@mercio I rencently made a bounty and was not aware that I'll lose the points when no answer pops up. That completely killed all my motivation to work hard on thorough answers on MSE. So I am out of that buisness.
 
I see I'll edit my post to make it clearer
 
5:28 PM
ah yeah, bounties do that
 
@Rudi_Birnbaum One should think of the bounty as a way to pay for advertisement of the question
 
you "pay" for the advertisement
I used to be a bounty hunter back in the day
 
Yeah and the loss was in NO absolutley NO relation to the work I invested in the points.
@mercio It SEEMS to be logical but so would be the contrary.
 
yeah it's a design choice
 
@TobiasKildetoft, then how do we find it's nth term?
 
5:32 PM
It could help other newbies to have some warning on the bounty bottons or letting something like that pop up. You also cannot assume everyone reads all the descriptions.
I think I am not the first one.
 
it is unlikely this will be changed
i think it's probably SE-wide design
 
@blue_eyed_... You can look up the formula by searching for it.
 
This way bounties are put on questions the asker believes to be answerable in a satisfactory manner, if one was refunded the points for not awarded bounties you could keep giving visibility to unanswerable questions, which benefits nobody
 
why do the separable elements form an intermediate extension?
 
@MikeMiller (almost) no problems with the rule itself. The big problem is about (them?) not making it more visible.
 
5:37 PM
yeah that's what i was referring to not going to change
 
@mike That its only announced on Andromeda? OK.
 
5:48 PM
how mathematician are selected for field metal?
 
@TobiasKildetoft I edited the question can u lemme know if it makes sense now?
1
Q: A rough proof for infinitesimals?

More AnonymousI discovered the following relation for arbitrary $d_r$: $$ \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{d_r}{r^s}}_{\text{removable singularity}} \times \int_0^\...

In fact if it doesnt make sense to anyone please lemme know?
 
@MoreAnonymous whats that cross $\times$ in your first equation?
 
oh maybe I should have just used \cdot
 
Does $\lim_{x \to 0}\dfrac{\sin (1/x)}{\sin(1/x)}$ exist?
Thoughts?
 
it normal multiplied
 
5:53 PM
@MoreAnonymous or parantheses? And I guess you should say something about the domains and the type of functions (integrable,differentiable,bound,...) ...
 
I wrote the integral has to abosultely converge
 
@Abcd why wouldn't it?
 
@Holo sin(1/x) is zero for many points near $x \to 0$
 
@Rudi_Birnbaum anything else I can fix?
 
@Abcd the definition of limit that I know talks only on the domain: for all epsilon>0 there exists delta>0 such that for all x in the domain if 0<|x-a|<delta implies |f(x)-L|<epsilon
So it doesn't matter
 
5:57 PM
@Holo its not possible to define a neighborhood in this case man
sin 1/x is 0 for infinitely many points around x = 0
 
I just wrote the definition I am using
 
0
A: What is $\lim_{x \to 0}\frac{\sin\left(\frac 1x\right)}{\sin \left(\frac 1 x\right)}$ ? Does it exist?

leonbloyFirst, notice that the function is defined for all reals except for $x=0$, and the points where the denominator is zero: $$\sin(1/x)=0 \iff 1/x= k\pi \iff x = \frac{1}{k\pi}$$ for integer $k$. As you correctly guessed, that the function is not defined at $x=0$ does not matter for the limit. B...

 
@MoreAnonymous not sure that I can speak for anyone else except me, but I would put the integral behind the limit or alternatively put just () around the limit and remove the dot. Then I was confused by $d_r$ I first thought its the $r$th derivative. Why not using $a_k$?
 
@Rudi_Birnbaum don't make me replace $d_r$ in the whole document (that will take very long) :(
 
@Abcd look at the comments of gimusi's answer, it depends on the convention of he definition, look at Mike Earnest's comment there for more precise possible definition
 
6:01 PM
@MoreAnonymous then don't.
 
but does it make sense besides that?
 
@MoreAnonymous I discovered that sometimes one self thinks a question "What does ... mean?" makes perfect sense and must be well understood by anyone else. However I had to learn that this question almost never was greeted with enthousiasm at MSE. I am afraid this is another case of that.
 
sigh . ...
:(
 
@MoreAnonymous Lets assume the equation is correct, then people will ask why should it be wrong.
 
k ... well if its right then hurray we can count infinitesimals for certain functions!
hurray
!
$d_r$ is the number of times the $r$'th infinitesimal appears at f(rh)h
where h tends to 0
$h \to 0$
 
6:09 PM
@MoreAnonymous I see its a rhetorical question ... (might confuse people...)
 
is anyone took part in ISEF Intel Science fair?
 
@MoreAnonymous So you want to know if $d_r$ can be considered as infinitesimals? But you define $d_r$ to be constants. So how can they be infinitesimals at the same time?
 
ah ... because they are localised at a specific point in the integral
It's under what does this formula mean
the picture
specifically what I mean by $d_r$ can be considered as an infinitesimal is can it be thought of the number of infinitesimals there are
to be counted
 
Do $d_r$ depend on $s$ somehow?
 
Nope
its an arbitrary constant
 
6:19 PM
The only thing I can contribute is that the limit of sums of derivatives are connected via the Bernoulli numbers with $\zeta$. Maybe you have some similar thing here.
 
link?
 
Euler-Maclaurin fomula and Bernoulli numbers, (wiki)
 
can I take $d_r = 10 ^{10^r}$ ?
 
I think you'll get a LHS = RHS = infinity
 
6:38 PM
@MoreAnonymous $n$ does not depend on $k$?
 
nope ...
See the standard case set all $d_r =1$ and get Reimannian integration
I think that should help for integration
and intuition
 
@MoreAnonymous I have edited the equation. The dot was odd (to me). But its just a suggestion ...
 
its k i approved :)
what did u tink of the proof btw?
@Rudi_Birnbaum
 
@MoreAnonymous I am still not sure what those $d_r$ are supposed to be. You say they should be constants, but also want to consider them as infinitesimals. But then they would be constants in what field?
 
I still try to understand it
The two two independent limites somehow feel odd.
 
6:46 PM
@TobiasKildetoft so if u look at the L.H.S ... try to ask urself where is $d_r$ localised? (The answer well according to me ... at f(rh))
 
@MoreAnonymous What does it mean for something to be localized?
 
@MoreAnonymous and again summing up function values at "supporting points" (?) connects with summing up derivatives in the Bernoulli numbers by Euler-Maclaurin.
 
What sort of object is $d_r$ anyway?
 
its a number
 
@MoreAnonymous What sort of a number?
 
6:47 PM
real
 
Then it does not make sense to speak of it being localized. And it is not an infinitesimal (since no reals are)
 
but it counts the number of infinitesimals at $f(rh)$
@Rudi_Birnbaum Im not differentiating anything
 
Your talking about infinitesimals? A derivative is the quotient of two "infinitesimals".
 
yes ... But I saw ur formula and Im pretty sure mine is different
@TobiasKildetoft $d_r =$ number of infinitesimals at $f(\frac{rk}{n})$
 
@MoreAnonymous no doubt they are not the same. Its just to suggest its no too big surprise that when doing stuff like that you get the zeta function and infintesimals involved.
 
6:52 PM
ah ohk
 
(maybe)
 
@Rudi_Birnbaum also which part don't u get (maybe I can help)
 
@I don't see how the $d_k$ in the diagram relate to those in the equation, e.g.
because $d_k = f(hk)h$ in the pic.
 
I thinki I'll add an example so it becomes clearer how my formula works?
 
@MoreAnonymous Wait, what sort of function is $f$ since there are infinitesimals "at" its values?
 
6:58 PM
an example is e^(-x)
 
@MoreAnonymous There are no infinitesimals anywhere if you just consider functions from the reals to the reals
infinitesimals are elements of larger fields than that.
 
@MoreAnonymous maybe its enough to make the diagram consistent with the formula. The $d_k$ in the digram is it the areas of the rectangles?
 
@TobiasKildetoft I see ... not really familiar with the rigour ...
Im currently writing an example
@Rudi_Birnbaum yes
but it will only work if $h \to 0$
 
@MoreAnonymous Maybe you should start by writing what you mean by the term "infinitesimal"
 
@MoreAnonymous so $d_k = f(kh)h$, no?
 
7:01 PM
@Rudi_Birnbaum yes
 
So if all $d_r$ are $1$, what happens?
 
u get Riemann integral of sum formula?
 
No you dont
 
d_r = 1 is the zeta function
yes u do from 0 to infinit
infinty
 
no you dont
 
7:04 PM
okay how would u write integrate o to infinity in riemannian sum of stips
*strips
?
 
Oh sorry you do!
 
MAGIC!
 
Lets say $k=10$ and you do the limit in $n$ what happens?
This $k$ is somehow very strangely hanging round in that expression
 
wont work
k must go to infinityu
 
that must be nonsense then
 
7:11 PM
@TobiasKildetoft I even put an example
 
can you interchange the limites?
 
@MoreAnonymous An example of what and where?
 
I edited the post
right above questions
(subtitles)
2
Q: A rough proof for infinitesimals?

More AnonymousI discovered the following relation for arbitrary $d_r$: $$ \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{d_r}{r^s}}_{\text{removable singularity}} \int_0^\infty...

*sub headings
 
You say the expression does not exist for any $k$ except $\infty$? That must be nonsense
 
@Rudi_Birnbaum for certain functions u will be able to interchange the limites
 
7:13 PM
I still have no idea how one could in any reasonable way consider some specific real numbers as infinitesimals.
 
Does the figure I've (terribly) drawn make sense?
(Have u looked at it?)
 
@MoreAnonymous What do you mean by "wont work"?
 
I use a property in the derivation of that $x \to rx$ to satisfy this equation there are only 3 numbers $\infinty$, $-\infty$ and $0$
@Rudi_Birnbaum
and I dont change their limits
 
These $\lim$s dont make sense $\frac{k}{n}$ goes to $1$ right?
 
@TobiasKildetoft have u seen the terribly drawn figure of a function ... Would u accept the $d_r$ as the number of infiniestimals at $ d_r f(kh)$
 
7:19 PM
@MoreAnonymous I still have no idea what "number of infinitesimals at [some number]" means.
 
as essentially I add: $d_1 f(h)h + d_2 f(2h)h + d_3 f(3h) h + \dots$
@Rudi_Birnbaum depends which limit do u take first?
Lets say I mapping a function + infinitesimals $\to$ the infinitesimals: Let the function be $e^{-x}$. At $0 \to e^{-h} $ at $1.5 \to e^{-(1.5 + h )}$ at $2h \to e^{-3h}$
So I have a mapping
 
That made no sense to me. Not least because you still have not defined what you mean by infinitesimal.
(also not really sure how you mean that arrow to be read)
 
@MoreAnonymous You claim $$\lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n f(\frac{k}{n}r)\frac{k}{n} $$ is the Riemann integral. I claim this is not true.
 
how so?
\int_0^\infty f(x) dx = \lim_{k \to \infty} \int_0^k f(x) dx
$$ \int_0^\infty f(x) dx = \lim_{k \to \infty} \int_0^k f(x) dx
$$
now we seperate it into $k -0/n$ strips
\lim_{k \to 0} \lim_{n \to 0} \sum_{r=0}^n f(\frac{kr}{n}) \frac{k}{n} $$
 
Hello. Can someone help me prove that if $X$ is a non-trivial linear topological space over $F = R$ or $C$. Then the topology of $X$ can never be discrete. I know that discrete topology means all subsets of $X$ are open but how does that help?
 
7:28 PM
$$\lim_{k \to 0} \lim_{n \to 0} \sum_{r=0}^n f(\frac{kr}{n}) \frac{k}{n}$$ fixed @MoreAnonymous
 
@TobiasKildetoft Does the Riemann limit of a sum use the definition of infinitesimal?
 
You meant to infinity?
 
@MoreAnonymous No. Limits do not involve infinitesimals usually
 
@MoreAnonymous I still dont see how $\int_0^k f(x)dx = $ anything like your sum.
 
I see in that case i think it is only a matter of interpretation ... I think $d_r$ can be interpretted that way ...
@Rudi_Birnbaum what is $\sum 1/r^s$
which side by the way (LHS or RHS)
?
 
7:30 PM
Show me how the integral is resolved to the Riemann sum
 
@Rudi_Birnbaum
sure
$d_r =1$
 
@MoreAnonymous An infinitesimal is by definition some "thing" which is positive but which is smaller than $1/n$ for any natural number $n$. Your $d_r$ are real numbers, which do not satisfy this.
 
yes ... but the count the number of infinitismals
at any r
 
$\int_0^k f(x) dx = \sum_ ...$
 
*they cpoutn
yes .. .
 
7:32 PM
@MoreAnonymous There are no infinitesimals in anything you have written so far.
 
go on
 
@TobiasKildetoft did u read the proof
 
@MoreAnonymous No.
 
?
sigh ...
 
$\int_0^k f(x) dx = \lim \sum_ ...$ please continue here ...
 
7:33 PM
on it
 
OK I'll start for you:
 
$\int_0^k f(x) dx= \lim_{k \to \infty} \limt_{n \to \infty} f(r \frac{(k-0)}{n}) kn $
is this correct?
(accroding to u (@Rudi_Birnbaum)
 
No. That is not how you define a Riemann integral!
You only have one limit in the correct formulation.
 
why not?
oh yea
theres a sum
:P
$\int_0^k f(x) dx= \lim_{k \to \infty} \limt_{n \to \infty} \sum_{r =1}^n f(r \frac{(k-0)}{n}) kn $
 
No, its not two limites, its only one !
 
7:36 PM
There are two limits because Im taking k to infinty at the end of this
the riemann integral is between arbitrary points b and a
but what if $b \to \infty$
 
Again: $\int_0^k f(x) dx = $ MIND $k$ is some finite number here
 
Then you just take a limit of a sequence of numbers (which each happen to be obtained as an integral)
 
Just write it down the / a Riemann sum limit for that very integral.
 
In that case $\limt_{n \to \infty} \sum_{r =1}^n f(r \frac{(k-0)}{n}) kn $
 
Then you'll see it does not work like you do it ....
 
7:39 PM
Yea ... but I never made the claim it works for finite $k$ ...
I said $k \to \infty$
 
This is not the point,
 
so then what is ur point?
 
The point is that your construction does not correspond to any Riemann sum whatsoever.
And that you will see when you write it down for finite $k$.
 
K? then are u saying $\int_0^\infty f(x) dx $ cannot be done via a riemann sum?
 
7:41 PM
so it can?
how
@TobiasKildetoft I think i'll be putting a bounty on it after 2 days hopefully that will provide ebough motivation for u to go through the derivation also ... $d_r$ seems to count the number $k/n$ s at $f(rk/n)$
 
@MoreAnonymous I have no interest in going through a wall of integrals, bounty or no bounty
And the last thing you just wrote still makes no sense at all
 
why so?
 
This is the Riemann sum: $$\int_0^k f(x) dx = \lim_{n\to\infty} \sum_{m=0}^n f(k \frac{m}{n})1/n$$
 
And then when you go with $k\to\infty$ you get $$\lim_{k\to\infty} \lim_{n\to\infty} \sum_{m=0}^n f(k \frac{m}{n})1/n$$
 
7:49 PM
@TobiasKildetoft why do u say it makes no sense?
 
@MoreAnonymous: Now I do not see how that connects with your expression?
 
@MoreAnonymous Because I have no idea what "the number of [some real number] at [some other real number]" means
 
@Rudi_Birnbaum see step 2
 
It is (and must be) $1/n$
 
of pdf
can u provide a reference?
@Rudi_Birnbaum
 
7:51 PM
Just think for a short while...
 
no
In mathematics, a Riemann sum is a certain kind of approximation of an integral by a finite sum. It is named after nineteenth century German mathematician Bernhard Riemann. One very common application is approximating the area of functions or lines on a graph, but also the length of curves and other approximations. The sum is calculated by dividing the region up into shapes (rectangles, trapezoids, parabolas, or cubics) that together form a region that is similar to the region being measured, then calculating the area for each of these shapes, and finally adding all of these small areas together...
 
$1/n$ is the with of the slices.
The $k$-th rectangle is $1/n f(k/n)$
Isn't it?
 
no
its k/n
u have a line
 
It is $k/n$
 
that starts from $a$ and goes to $b$
 
7:54 PM
Lets imagine we fix $k$, OK.
 
u divide it into $k$ parts
 
for a start.
 
what is the length of each part
?
$b-a/k$
 
Now we cut $[0,k]$ into $n$ pieces, OK?'
 
You have each part is $km/n$ running on $m$, so $k(m+1)/n-km/n=k/n$
 
7:55 PM
yes well done k /n
 
So you have $\lim_{k\to\infty} \lim_{n\to\infty} \sum_{m=0}^n f(k \frac{m}{n})k/n$
 
yes that looks good
 
what was the difference when I wrote it?
 
The sum is for finite $k$
$$\lim_{k\to\infty} \left(\lim_{n\to\infty} \sum_{m=0}^n f(k \frac{m}{n})k/n\right)$$
 
yea ... but u take k to infinity at the end
 
7:59 PM
But after summation
 

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