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12:06 AM
I don't know it.
 
Reviewed. I give it an $\sqrt{81}$ out of $10$.
 
Hi there !
 
12:21 AM
@SimplyBeautifulArt Definitions, theorems and proofs, nothing really new under the mathematical sun.
What about you ?
 
Well, I invite you to join me in my personalized chat room: chat.stackexchange.com/rooms/51337/…
 
12:34 AM
hi chat
 
1:42 AM
wow its empty here. I don't think I've ever seen that
Can anyone here explain the meaning of Semantic consequence? Also give the definition if it exists. (for instance $X\subseteq Y := \forall a (a\in X \rightarrow a\in Y)$ would be considered a definition of $\subseteq$)
 
@user400188 lol
 
I tried wikipedia but it was not very reliable. I gave the definition once in chat and (to the people that actually knew it) was quite clearly wrong.
 
1:58 AM
Hey @mikem
 
@TheGreatDuck Hi! Where you been?
 
actually don't worry
 
at home
where ive been all day
studying and such
 
okay i seriously suck at rearranging and substitution for integrals =/
difficult as heck
 
2:05 AM
@WDUK whats the problem
 
its just hard to work out how to rearrange it so i can work them out properly
im trying to do x/(1+2x^2) dx
 
Oh, have you tried ....
 
what comes to mind
 
first thoight was perhaps try logarithmic intergration. But I'm not sure if it will work.
 
The first step is to try substitution almost always
 
2:10 AM
well applying u = 1+2x^2 i got 4x du
but thats not close to what the orignal one was
 
yes it is
just multiply by a constant
 
how so ? isn't the original one technically (x)dx
 
du/dx = 4x
 
like i re-arranged to (1/1+2x^2) * x dx so shouldn't i be trying to get x dx not 4x dx?
 
not 4x du
 
2:12 AM
yeh but du = 4x dx right
 
so x dx = 1/4 du
 
Wait! I have the answer!
x = 4x/4
 
ohh so i just had to isolate x
 
So we get...
 
yes @WDUK
 
2:13 AM
(1/4)(1/(1+2x^2))(4x dx)
The 1/4 is just a constant, and we have everything we need
 
these steps are not obvious to me like it is for simpler ones lol its quite messy
so if x = 1/4 du then isn't it (x)(1/u)(du)
no wait that can't be right
ok question, why is the 1/4 written outside rather than inside?
 
Hi @Ali.
 
@MikeMiller I only pinged you because I thought you were around
But now that you are, how are things
 
They're ok
 
are you still working on that homological algebra construction?
 
2:28 AM
I have that more or less figured out. I'm reading stuff that's more on that front now since I got curious but it's not the stuff I should be doing for my research now.
 
nice
 
okay i'm confused, i got cos x/(3-sinx) dx and i got ln(3-sinx) + c but the answer seems to negative not positive
not sure where the negative natural log came from =/
 
if you diff -sin x you get -cos x not cos x
hey @arctictern
 
hello
 
do you know much about zeta functions of groups?
 
2:31 AM
right but shouldnt i be removing the negative since i need to get positive cos x?
 
I used to know more. Marcus du Sautoy wrote some cool stuff on them.
 
@arctictern Yes I am reading his paper now
 
Being rational functions of p^-s, nilpotence class, yadda yadda
cool
 
I am trying to work out how to count p-groups of order p^n with k generators
 
sounds hard. you mean estimate?
 
2:34 AM
Well marcus defines f(n, p, c, d) to count p-groups of order p^n with nil class at most c and at most d generators
 
aha
is coclass particularly relevant? I remember that being a thing for p-groups.
 
So counting 2-groups with k generators should just be f(n, 2, n-1, k)-f(n, 2, n-1, k-1) right?
 
"with k generators" means there exists a generating set of size k?
 
I would assume that would be f(n,2,n-1,k) then
 
2:36 AM
The thing is I really don't care about the classes
but if it is at most d generators then doesn't it count less than d generators as well
 
yes. if a group has a generating set of size <d, then such a set can be expanded to a generating set of size d (assuming not d is not bigger than p^n ofc), so these groups are counted as well
 
So if I want groups with minimal generating sets of size k then I need that difference?
essentially counting how many p-groups have rank k
 
@AliCaglayan yes
 
but I am not sure if I can say anything more meaningful now
The results given in the paper require some pretty heavy machinary
In the paper they prove that the series
$$
\zeta_{c, d, p}(s)=\sum_{n=0}^\infty f(n, p, c, d)p^{-ns}
$$
can be written as
$$
\zeta_{c, d, p}(s)=\sum_{N \triangleleft \widehat{(F_{c,d})_p}}\left|\widehat{(F_{c,d})_p}:N\right|^{-s}\left|\mathfrak G_p:\operatorname{Stab}_{\mathfrak G_p}\left(N\right)\right|^{-1}
$$
where $\widehat{(F_{c,d})_p}$ is the pro-$p$ completion of the free $p$-group of class $c$ on $d$ generators and $\mathfrak G_p=\operatorname{Aut}\widehat{(F_{c,d})_p}$.
 
presumably that's a combination of orbit-stabilizer and "presentations mean groups equal free groups mod normal subgroups of relations"
 
2:42 AM
I am not really sure how to use this to say that a random 2-group probably has rank 3 or 4
 
iunno
 
he has some more work to compute that function using p-adic integrals and whatnot
but this is completely beyond me at this point
 
I found the p-adic stuff easier to understand
(IIRC anyway)
lots of geometric series and splitting the domain up into balls
 
I wonder if people have considered a coarser counting function
 
well, there's the different one with coclass
 
2:51 AM
what's the rank of a p-group
 
bounding wrt number of generators doesn't really seem natural
 
size of the minimal generating set
 
i can't imagine any way a random 2-group is rank 3 or 4
 
(actually groupprops says rank of a p-group is the maximal value among the sizes of the minimal generating sets of abelian subgroups)
 
for example 2-groups with n=9
the most numerous rank is 5
I was trying to investigate this behaviour essentially
You could also consider that the rank of a p-group is the rank of its abelianization
modding out the Frattini subgroup and whatnot
 
2:57 AM
interesting
 
So the most common rank among p-groups should be the most common rank among groups as p-groups dominate groups
or more specifically 2-groups
 
Finite groups are weird.
Finiteness just does not seem like a natural property to study groups under.
 
finite groups are just not-so-free groups
 
@PVAL-inactive sure it does. you can do combinatorics.
dividing by the number of elements in a group is helpful
 
finite groups have some good links in number theory
like dimensions of irreps of monster groups being in j invariant expansions and other crazy things
 
3:04 AM
what number theory problem does this solve?
 
"Why does 196884 = 196883 + 1?"
 
classic number theory
 
I know the automorphisms of curves are forced to be finite, but I'm not sure I have good other reasons for studying finite groups .I don't know how any besides say S_n,A_n,D_n,Z/n, and linear groups of the latter appear in mathematics other than as automorphism groups of curves or equiv. isometry groups of surfaces.
 
Well it depends on your motivation really
If you are a geometer you won't really care much for finite groups
 
why do we study automorphisms of curves?
 
3:08 AM
only a few ever come up and are interesting enough to study
 
I wouldn't be surprised if the sort of moonshine stuff really comes out of the monster group being a hurwitz automorphism group of some curve.
 
I think you can use elliptic curves to count nilpotent groups
That's an interesting link
 
why do we count nilpotent groups?
why do we study math?
 
I guess to better fit your frame I could have reworded it as: We can count elliptic curves by counting nilpotent groups
Groups are natural structures in algebra not just because we said so as well
take for example Hom of pretty much anything
 
I am certainly willing to say groups are a good object to study.
 
3:15 AM
finite groups can give insight into not-so-finite groups
we have lots of examples of this in math
like I was reading about how ordinals can be used as "infinity to say things about the finite"
 
I don't think very much that I know about groups is very focused on the underlying set.
Most of it is focused of the behavior of the operation.
 
hence upto isomorphism
 
So while putting some condition with respect to the operation seems natural to me, putting some condition on the set seems kind of weird.
 
It becomes hard to talk about groups without naming some of the things they describe
so elements are really necessery
but if you want to standardise elements just take some representation
 
I'm not sure I agree.
 
3:22 AM
I am not sure i understand you
 
Well heres the thing.
The most "obvious" facts about finite groups
like most finite groups are 2-groups.
Are completely intractable
Sort of like the most "obvious" facts about say the digits of pi.
 
Well I wouldn't say that is an obvious fact
 
I think that's probably because the condition of finiteness is quite a strange one and really makes it hard to prove anything else.
 
@PVAL-inactive what would you count as a tractable fact about something?
 
@AliCaglayan something that humans have hopes to proof (or better yet have proven)
 
3:26 AM
Well that's the thing
I would say that that is an obvious fact about finite groups
I would say thats more of a not-so-obvious
The obvious ones are things that group theorists use a lot
And they definitely have been proven
but most finite groups being something is more of a combinatorial question on algebraic structures
Which are perfectly natural questions to ask
as opposed to facts about the digits of pi
*my first I would should be I wouldn't sorry
Stepping back I should point out that being natural is completely subjective. There is no reason to study anything in math apart from the fact that some things make you feel tinglier than others.
 
The tractibility of questions in a field is a very good reason to study (or not study) something.
 
anyway its 3:33 so I better be off
see ya @PVAL-inactive @arctictern
 
3:58 AM
is there a different syntax for writing something like ln|4 - x| when hand writing it ? since often it will look like 14 - x which is not ideal
 
4:10 AM
stop writing l for 1 and write bigger absolute value bars
 
4:27 AM
is there an integer n such that the fractional part of 1.5^n lies at most 0.03 away from 0.54?
 
Can some one check my workings here, want to know if i am applying the logic correctly here:
 
@WDUK du/dx = -4x
 
ah ! damn ! thanks for spotting that!
does that mean i can't do -1/4 anymore because of the x?
 
not sure what "do -1/4 before the integral" means
 
before the intergral
because it would have to be -1/4x then the rest
 
4:38 AM
@WDUK this makes no sense to me
 
i don't know how to write the fancy S shape in chat, but i would have (1/4x)S 1/sqrt(u) du
 
you can't bring the x outside
and learn to type latex
$\displaystyle \int \frac 1 {\sqrt{1-4x^2}} \ \mathrm dx$
 
thats what i mean, so my logic is wrong in the way im doing this
 
yes it is
 
hmm
i don't see what i am missing, gah!
 
4:44 AM
can you see latex?
 
yup
 
you need another substitution
 
my book hasn't taught me about using more than one yet, i feel like its telling me i can do it without a secondary subsitution
 
i didn't say secondary substitution
erase all your work and use another substitution
 
oh so im picking the wrong thing as my current substitution
okay
 
4:46 AM
yes
 
i'm thinking to include the sqrt(1-4x^2) so i have 1/u but not sure i know how to find the derivative of that
 
use another substitution
 
what else is there to substitute =/
 
trigonometric substitution
 
isn't that 1+x^2 not 1+4x^2
4sin^-1 x?
 
4:55 AM
let x=(1/2)sin u
u = 2arcsin x
 
hm i think that pointed me in the right direction
 
nice
 
thanks for the nudge in the right direction
 
 
1 hour later…
6:19 AM
@Secret zousun
 
good morning indeed (although afternoon here)
 
@Secret nei hai bingor sikui?
 
In HK or in Aust?
 
Caption: the distribution of the fractional parts of (1.5)^n from n=0 to 10000
@Secret australia
 
Sydney NSW
 
6:21 AM
nei yiga hai hongkong?
 
nope, still in syd. No return hk plans this year as PhD start in a month
But my clock says 17:22
 
right, it's 17:22 now
nei hai hong kong duk junghok?
 
yup
 
:o
do you think it is possible that i know you in real life
 
depends... A fb account can often be a good first step to check, cause I can recognise some of my friends even if their profile pics is not them
 
6:28 AM
I'm trying to generate the image for 100000 and 1000000 but they're taking very long
there are quite a few holes in the picture above
but no zeroes
 
The distribution looks quite random as far I can see from the pics. I am not sure what it can tell us
 
but far from uniform...
 
It strongly reminds of any baseline of an NMR spectrum
note those ocassional spikes of twice the length
 
from PIL import Image

p3 = 1
p2 = 1
freq = [0]*1024

for i in range(-9,100000):
	p3 *= 3
	p2 *= 2
	freq[(p3%p2)>>i if i>0 else (p3%p2)<<(-i)] += 1
	if i%1000 == 0: print(i)

max = 0
for f in freq:
	if f > max:
		max = f

pixels_out = []
for f in freq:
	pixels_out += [(0,0,0)]*f+[(255,255,255)]*(max-f)

img = Image.new("RGB", (max, 1024))
img.putdata(pixels_out)
img.save("1p5_1e5.png")
this is my current code
do you know how I could optimize?
 
I am not sure, see if you can find more effiicient algorithms online to implement the mod function % cause I suspect that and the for loop is where it is slowest

Have you try to instead write the input variable as a linear array instead of entry by entry. For me, at least in matlab, putting everything in array generally makes it faster
 
6:39 AM
the input variable?
 
for example, instead of computing one by one, compute the first 10 using a 1x10 array and then for loop that
 
but I only need them once
 
so for each iteration of the loop, you update the previous array
That is, for every 10 numbers where you calculate the mod, map them to the pixels, then on the next iteration the same thing is done again, and reusing the same 1x10 array. That might help save memory, I think...
Instead of computing them all at once in some big array and then plot them
 
I just realized that I'm stupid
from PIL import Image

p3 = 1
p2 = 0
freq = [0]*1024

for i in range(-9,100000):
	p3 *= 3
	p2 = p2*2+1
	freq[(p3&p2)>>i if i>0 else (p3&p2)<<(-i)] += 1
	if i%1000 == 0: print(i)

max = 0
for f in freq:
	if f > max:
		max = f

pixels_out = []
for f in freq:
	pixels_out += [(0,0,0)]*f+[(255,255,255)]*(max-f)

img = Image.new("RGB", (max, 1024))
img.putdata(pixels_out)
img.save("1p5_1e5.png")
this ran under 10 seconds
Caption: the distribution of the fractional parts of (1.5)^n from n=0 to 100,000
 
What mathematical operation is &, is it just the usual "and" logic symbol?
 
6:45 AM
bitwise and
Caption: the distribution of the fractional parts of (1.5)^n from n=0 to 1,000,000
 
Uh you go from
p2 = 1
p2 *= 2

which will give 2, 4, 8, 16 ,32 ,64 ...
To
p2 = 0
p2 = p2*2+1
which will give 1, 3, 5, 7, 9, 11, 13 ...
?
 
no, 1,3,7,15,31,...
powers of 2 - 1
 
ok nvm
but then you are computing the fractonal part of 3^n/(2^n-1) instead of 3^n/2^n=(1.5)^n?
 
27 mod 8 = 11011 mod 1000 = 11 = 3
27 and 7 = 11011 and 111 = 11 = 3
 
Ah I see, yeah, I think using binary is faster, so that might be why this program is more efficient than the previous one
 
6:57 AM
indeed
@Secret do you think that the fractional part of 1.5^n is dense in [0,1]?
or easier, is 1 a limit point of 1.5^n?
 
1.5^0=1, so 1 is definintely a limit point and also continuous I guess as n approach 0
 
@Secret no, the fractional part of 1.5^0 is 0
 
oops, forgot it is the fractional part not the whole function
 
7:14 AM
@DHMO Isn't asking whether $a^n$ has $1$ as a limit point equivalent to the same question for the sequence $n\log a$ and $0$?
 
@MartinSleziak yes, but i'm asking for the fractional part of 1.5^n
 
And if $\log a$ is irrational, then $\{n\log a\}$ is dens in $[0,1]$, so it will get close to $0$ arbitrary often.
@DHMO Yes and I take $a=1.5$. I'll try to rephrase it more precisely.
 
@MartinSleziak how?
 
Ok, let me think. I am not sure whether this helps or not now.
 
I am currently trying to approach this problem via continued fractions. The continued fraction of 1.5 is 1+1/2
So I just need to n power it and see if the fractional part powered can contain .1
 
7:18 AM
If we know that $n\log a$ gets arbitrarily close to an integer, it only says that $a^n$ get arbitrarily close to $e^{\mathbb Z}$. So this probably does not help. :-(
I found them by searching for $a^n$, dense, fractional in Approach0. You can try also some other reasonable search queries.
Comment by Gerry Myerson - he did some reasearch in uniform distribution theory, so he is knowledgeable in this area:
It's an open question whether, for example, the fractional part of $(3/2)^n$ is dense in the interval (and this example is related to Waring's problem). — Gerry Myerson Oct 3 '16 at 12:19
Of course, even if the question whether $(3/2)^n$ is dense is open, the question whether $1$ is a limit point might still be answerable. I am not sure.
 
The binomial expansion of $1.5^n$ is $\sum_{i \in S}^n{}^nC_i \left(\frac{1}{2}\right)^i$ Anything about the fractional part is encoded in $\frac{1}{2}$ I think
I don't know how to take binomial expansions if i is real however
 
I might have misunderstood @DHMO's question, but I suppose he only works with $n\in\mathbb N$.
 
yes, $n \in \Bbb N$
 
If $\sum_{i =0}^n{}^nC_i \left(\frac{1}{2}\right)^i$ has a closed form that is not $1.5^n$ then it might be able to be tackled
 
Ok, so at least we know that density is an open problem. @DHMO Perhaps this would be interesting enough to ask on the main site...?
 
7:32 AM
@MartinSleziak would you help me to ask it?
 
@DHMO What should be the problem. I am sure an experienced user like you knows how to post a question.
I would be certainly willing to add a bounty, if it goes some time without an answer.
 
I'm the opposite of an experienced user
 
3.3k reputation points...? And not an experienced user?
 
0 rep in math
 
But enough experience on chemistry.SE.
 
7:36 AM
experience is never enough
 
I think that if you mention in the post that the question whether (3/2)^n is dense in (0,1) is an open problem, that would count towards including context. It would be good to find some source, too.
This pdf file math.boku.ac.at/udt/unsolvedproblems.pdf lists both questions - whether (3/2)^n is dense in [0,1] and whether it is uniformly distributed - as conjectures, but does not mention the sour.ce
 
alright
 
@DHMO If you ask the question, feel free to ping me and add a link. If I will be able to suggest some improvements of the post, I will say so.
If found the book [SP, p. 2–149] referenced in the pdf linked above.
Basically it is a brief summary of know results about uniform distribution sequences of the form $\lambda\theta^n$.
I am not sure to which extent it is related to your question - it seems more closely related to the question whether the sequence is dense or uniformly distributed (equidistributed).
But I will copy that paragraph here anyway. It is from the book O. Strauch – Å . Porubský: Distribution of Sequences: A Sampler.
2 17 Exponential sequences
NOTES: J.F. Koksma (1935) proved that the sequence $\lambda\theta^n \bmod 1$ with $\lambda\ne0$ fixed is u.d. for almost all real $\theta>1$. If we take $\lambda = 1$ then we get that the sequence $\theta^n\bmod 1$ is u.d. for almost all real numbers $\theta > 1$. However, no explicit example of a real number $\theta$ is known for which this sequence is u.d.
If $\theta > 1$ is fixed then H. Weyl (1916) proved that the sequence $\lambda\theta^n$ is u.d. for almost all real $\lambda$. A D. Pollington (1983) proved that the Hausdorff dimension of the set of all $\lambda\in\mathbb R$ for which the sequence $\lambda\theta^n \bmod 1$ is nowhere dense is $\ge\frac12$.
Then it contains the exact refences to the papers by Koksm, Weyl and Pollington.
And also it mentions this MathWorld article among references: mathworld.wolfram.com/PowerFractionalParts.html
@DHMO Since the above link contains some information about the sequence $(3/2)^n \bmod 1$, it might be worth having a look and probably it could also be useful to add the link to the question, if you decide to post it on the main site.
Wow, I would not expect that a question which you asked casually here in chat and which on the first looks seemed like it is not going to be difficult is connected to so many open problems and deep mathematics.
 
8:02 AM
0
Q: Is $1$ a limit point of the fractional part of $1.5^n$?

DHMOIt is an open problem whether the fractional part of $\left(\dfrac32\right)^n$ is dense in $[0...1]$. The problem is: is $1$ a limit point of the above sequence? An equivalent formulation is: $\forall \epsilon > 0: \exists n \in \Bbb N: 1 - \{1.5^n\} < \epsilon$ where $\{x\}$ denotes the fracti...

@MartinSleziak
 
Thanks.
If you want to add some reference to the fact that it is open problem, you can add the pdf I linked above or Strauch-Porubský 2.17.1 (or both).
 
I may add them later.
 
I would probably consider adding this link mathworld.wolfram.com/PowerFractionalParts.html to the question, since it mentions some know results about the sequence of fractional parts of of $(3/2)^n$.
 
Find the solution to $\sin x = 0$

I gave the solution set $\{x : x = n\pi, \forall n \in \mathbb{Z}\}$

Is the logic inside my solution set correct ?
 
you gave the solution without the logic.
 
8:09 AM
@DHMO Here is the relevant page from the book: i.stack.imgur.com/f0NRW.png The following two pages are about the same sequences. (Some further results and references.) If you want to see them, too, I can add images of those two pages here too.
 
@DHMO No I mean to say that did I use the quantifier $\forall$ properly ? to answer the question. It is compulsory to use $\forall$ to answer this question.
 
I have also added some tags to your question - of course, if you think that some of them are not suitable, feel free to retag the question.
 
no, it wasn't used properly
 
@DHMO Why ?
@DHMO I did get a feeling that this is not correct but I can't figure out any other way.
 
@MartinSleziak I have added two links
@A---B the solution set is {n pi | n in N}
I mean Z
you can also say forall n in Z: sin(n pi) = 0
 
8:18 AM
@DHMO I have replaced "good luck finding this problem here" by a more detailed reference. (I guess it sounds a bit better this way.)
 
I don't see how what he wrote is wrong, but $\forall$ is not necessary there in his version.

Or perhaps a better way to write it is $x = \{n\pi, n\in \mathbb{Z}\}$
 
@MartinSleziak thanks
 
Hi. This chat has very different people from what it used to have five years ago.
 
except you, @JasperLoy
 
That's normal, people come and go
 
8:24 AM
Does robjohn still come here?
It seems Chris's Sis has stopped coming here.
 
@Secret nei wuimmwui seung daap ngo tiu muntai?
 
Oh, his post is 4 days ago in this room.
 
@JasperLoy That user now has a different username: Then.
 
@DHMO "ngo tiu muntai" I cannot fully interpret the cantonese behind this phrase. Please elaborate
 
@DHMO I don't recall talking much to your username which may have changed several times.
@MartinSleziak Ah yes, I saw Then in ELU room and figured immediately it was her because of the introduction.
 
8:28 AM
@JasperLoy Is Chris sister her?
 
@JasperLoy Yeah, I think the topology guys and 0celo7 have end up push her over the critical point or something. Which is too sad cause she, like simpleart, know something about symmetry of integrands, and hence helping to explore $\mathcal{C}(\mathbb{R})$
 
@MartinSleziak I have not met him or her personally, but from our conversations all these years I am going to assume it is her.
 
You cannot be sure about gender with anonymous usernames. IIRC originally the username used to be simply Chris. So I thought he or she changed it as if their sister started posting - just as a whimsy.
 
@MartinSleziak Actually, you can't be sure about non-anonymous names either, lol.
 
Since then, one of the big changes to this chat room is that integral related discussions are less revolted by the maths users here
 
8:31 AM
@Secret "my question"
 
I no longer have an account on MSE, because I don't really like the atmosphere here.
I also figured that some people are best ignored forever and not confronted at all, whatever they do on this site.
 
@JasperLoy Well, if somebody has link to his official website at the university where they work, it is quite likely that the information is not fake. Especially if the answers on the site shows that the user is knowledgeable of the area in which the real life person with that name published. (Of course, you cannot be really sure about anything. It is possible that I am simply a bot.)
Do I remember correctly, that you were the "singing guy", Jasper? Some user posted here in chat. a few youtube clips of him singing some songs
 
@MartinSleziak Yes, since you mentioned, here is a video for you:
 
@DHMO Well, I don't get anything useful from that binomial expansion for the first few n, . It is true that (1/2)^3 gives 0.125 but the other powers of 1/2 and the binomial coefficients screwed it and not giving me .1 afterwards
 
This is my latest masterpiece, and I will try not to delete my channel again, lol.
 
8:35 AM
One reason I am interested in the global geometry of $\mathcal{C}(\mathbb{R})$ is because, if one understood that geometry completely, one can intuitively guess a lot of integrals (which must be closed in this space)
since integrals are just linear functionals in this space
 
@Secret Do you mean the continuous functions on the real numbers?
 
yes.
 
@Secret By the way, are you Victoria? I like Victoria's Secret lingerie, lol.
 
obviously not
From the 5 days ago of discussions here, $\mathcal{C}(\mathbb{R}) \subset \mathbb{R}^{\mathbb{R}}$ is a huge place with size $\beth_2$
 
Actually, integral computation is very much a part of mathematics, not any less than topology.
I find the comments on Chris's Sis too much, so I often try to defend her in this chat.
But since they are all gone now, no point in discussing this.
 
8:40 AM
@JasperLoy yeah, and I do find her work interesting given how most people fear integrals and people like her who love integrals are rare
Then is actually still on the main site as far to recent January, so she is not completely gone, but chat, yes
 
There are lots of temperamental people in this chat, perhaps myself included, but I am no longer part of this chat, lol.
 
Meanwhile I am the generator of weirdness in this chat
 
A guy just entered and left. Maybe someone flagged one of my messages again.
 
and judging from how the people behave here, it seemed some of them are being slowly assimilated by my weirdness, as they started to understand when I ask them about some maths things
 
I cannot recall. Are you from Germany @Secret?
 
8:43 AM
nope, I am living in aust
 
Too many users and names confuse me.
@Secret Wait, are you Alex?
 
@JasperLoy Nope, my online username is always secret (or secret314 or secret ultraviolet)
 
He has too many accounts, lol.
 
and I am pretty sure if you compare (whoever this) alex is and my messages, you will see a lot of difference
 
Here comes a Frenchman.
 
8:46 AM
From my experience on the internet in my real life and other user's feedback, I have a rather distinctive personality signature. It is quite hard to confuse me with someone else
 
I do remember talking a lot to a Secret long ago.
 
@DanielFischer sorry, it was :Hi, does there exist a holomorphic function on a neighborhood of $0$ such that $\vert f^{(n)}(0)\vert \ge (n!)^2$ for all $n\in J$ where $J$ is an infinite subset of $\Bbb{N}?$ I know the answer is no if we replace $J$ by $\Bbb{N}.$
 
OK, I am out of this chat. I don't feel too at home in this room now that most of my comrades are gone.
 
ok
 
@DHMO As promised, I will add a bounty to your question. But the bounty only can be added if a question is at least two days old. If I forget, feel free to ping me here in chat.
BTW I am glad that you have created account on this site.
 

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