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12:00 AM
There was a day when I was more than half the comments, Balarka.
 
okay, now I enabled it. thanks for pointing that out for me, :)
 
I figured you weren't able to read LaTeX
So, what's $\delta_2(f)$? :)
 
Excellante
So $\delta_2$ is a homomorphism $\Bbb Z \to \Bbb Z$ which maps generator to generator
i.e, $1$ to $1$
What must $\delta_2$ be?
 
Hi handsome ppl
 
12:03 AM
uh, zero?
 
I have come to chew bubble gum and insert action here. And I've got a pretty good amount of bubble gum actually, sick
5
 
Well, no, let's work it out. $f : \Bbb Z \to \Bbb Z$ is a homomorphism such that $f(1) = 1$.
What's $f(n)$?
For any integer $n$
@Daminark lol that made me crack up a bit. good modification
I mean, good commodification, capitalist pigdog
 
man, this sounds easy but I'm probably over thinking it
n?
 
Yes, can you prove it?
 
Every function actually maps to 7
 
12:06 AM
STOP
 
since it's injective?
 
@user78103 Just keep it simple. Being a homomorphism means $f(n + m) = f(n) + f(m)$, right?
So write $f(n) = f(1 + 1 + \cdots + 1)$ ($n$ many $1$'s there)
And...?
 
fundamental group?
 
What does the fundamental group have to do with anything?
 
i thought it because of were adding parts, but I'm still not seeing the answer
 
12:11 AM
No, just think, you have $f(1+...+1)$
 
carpet cleaner
 
Then what can you do with it for sure? What do we know about f?
 
so if f is just n
I'm lost
 
f is a homomorphism, right?
 
yes, and that means it has an inverse
 
12:14 AM
Uh... Not quite...
 
isn't there a surjection?
 
not quite, $f$ only has inverse if it is a bijection
 
@user78103 it isn't a homeomorphism :)
 
homeomorphism is a bijection...
but if this is abstract algebra...only automorphism, isomorphism, and homomorphisms come into play...at least that's what I remembered
 
i thought they were homomorphisms, i have only half read chat (i read Dami ask and you said yes...)
 
12:16 AM
Is the set of all probability distributions with finite support considered a closed set?
 
so a homomorphism structure preserving map
 
@user78103 let's start small. can you prove that f(2) = 2?
 
by proving that the inverse of f leads to 2?
 
Anybody?
 
can you use the property that "f is a homomorphism" and "f(1)=1" to prove that f(2)=2? @user78103
f itself, nothing to do with its inverse
 
12:18 AM
@DaenerysDracarys I think so? You can add a function with arbitrarily small norm to it and go outside the set, so that sounds like the set should be closed
 
@KevinDriscoll if you can go outside then it isn't closed
 
maybe replace the number 1 with the number 2... and see if the definition holds?
 
with induction?
 
@user78103 you're over-complicating everything; what does homomorphism mean?
 
I don't remember using induction on this :/
 
12:19 AM
@usukidoll you need induction, but not now
 
@LeakyNun I mean closed in the sense of contains all its limit points, not closed in the sense of "closed under operations"
 
@LeakyNun I presumed we were talking about a topologically closed set. Not closed under some operation like pointwise addition
 
@KevinDriscoll what you said.
 
I don't want to bother you guys too much with this, is there an article that you recommend I read before I come back with question?
 
@user78103 do you know the definition of homomorphism?
 
12:20 AM
Can you explain what you mean by an arbitarily small norm, maybe provide an example @KevinDriscoll ?
 
what is the question??? like does it involve rings and subrings?
 
no, you have everything, maybe just write down the definitions and try to work at our hints? @user78103
 
@DaenerysDracarys Well, what space are you considering your probability distributions as a set in?
 
I suppose in $\mathbb{R}^{n}$
 
let's see then, a homomorphism is a structure-preserving map
 
12:22 AM
@usukidoll, i think the question is $f: \mathbb{Z} \to \mathbb{Z}$ a homomorphism such that $f(1) = 1$, what is $f(n)$ for any $n$
 
@user78103 what does "structure-preserving" mean?
 
That doesn't make sense, the functions are defined on $\mathbb{R}^n$ but the 'live in' an infinite dimensional function space. THat is, each function is a point in an infinite dimensional space. @DaenerysDracarys
 
note that both the domain and co-domain of $f$ use an additive structure
 
You're right @KevinDriscoll I just mean $\Omega$ is a set of $n$ elements.
$\Omega$ is its domain.
 
that a summation in the domain reflects in the image?
 
12:25 AM
in terms of equations? $f(a + b) = ???$
 
f(a+b) = (a+b)
 
@user78103 try not to use big words and just write down the meaning
@user78103 no that isn't right
 
f(a+b)=f(a)+f(b)
 
right
 
@user78103 substitute a=b=1 and what do you get?
 
12:26 AM
f(2) = 2
 
good
just to make sure you understand: could you write down the steps?
 
@DaenerysDracarys Okay, I'm out of my depth here, because I'm not sure what functions space is the right one for probability distributions. But consider, for example, starting with a probability distribution with finite support, say $\text{supp}(f) = [-1,1]$
 
f(a) = a
f(b) = b
f(a+b) = f(a) + f(b)
if a=b=1
 
why f(a)=a?
 
or since it's a a homomorphism
f(a+b) = f(a) + f(b)
 
12:29 AM
You'd have to make it discrete too, though @KevinDriscoll
 
@user78103 could you write down a coherent and step-by-step equation starting with f(2) and ending with 2?
 
if a = b = 1
 
@DaenerysDracarys So if the case of a finite sample space is what you care about, I'm not actually sure what finite support means there
I guess then its just that $P(A) = 0$ for some $A \in \Omega$
 
I'll start you off in the right direction $f(a+b) = f(a) + f(b)$ and $a = b = 1$
 
Meh, it's all right. I think I have enough to go on. i might be able to hand wave my way through showing I can use the Krein-Milman theorem
 
12:31 AM
@mdave16 the user already wrote these two equations twice
 
i know, i felt he was going off track :/
 
I'll just consider that for homework today since I should know this. Thank you for all the help @LeakyNun @mdave16 and @BalarkaSen
 
I have a fever
And there's only one cure
 
@user78103, wait!!! we can do this
 
@DaenerysDracarys Ok, in the discrete case, then the relevant space has to be sequences that sum to $1$. But then you can make a function of compact support no longer have compact support by adding an arbitrarily small probability to all of the outcomes which were originally 0 probability and take away a corresponding arbitrarily small ammount from the non-zero probability outcomes.
 
12:35 AM
MORE COWBELL
 
try writing the first equation out on paper, and then substitute all the $a$s and $b$s with $1$
 
I'll wait then
 
Thanks @KevinDriscoll that sounds good.
 
That new sequence will be arbitarily close to your original sequence in the l1 norm
 
and then type it to us
 
12:36 AM
Seroiusly though I do have a fever, I'm not just making Christopher Walken references.
 
So an arbitrarily small change takes you outside the set, so the set must be closed.
 
I actually had to think for a minute there, because I couldn't remember Christopher Walken's name.
Thanks @KevinDriscoll !!
 
Thats not exactly a proof
 
WEll, I think it's sufficient for my purposes.
 
But its a kind of intuitive sketch that could be turned into a proof i think, if I knew all the appropriate technicalities
 
12:37 AM
but @LeakyNun said to start with f(2)
 
If I magically get better during the night, I'll try to tighten it up, but it's really only a side thing I need in order to apply that theorem
 
@user78103 I mean start by finding the value of f(2)
from the two properties that are given
1. f(a+b) = f(a)+f(b) for all a and b
2. f(1) = 1
 
3. if a = b = 1, then f(1+1) = f(1) + f(1)
which is f(2) = 1+1 -> f(2) = 2
 
bingo
now can you find f(3)?
 
with induction?
 
12:40 AM
without induction, just pure calculation at this point
the induction comes later
 
f(2+1) = f(2) + f(1) -> f(3) = 2 + 1 -> f(3) = 3
 
can you find f(4)?
 
awesome, now, if you had to guess $f(n)$?
@LeakyNun, i notice you go to imperial, did you go to WIMP last year?
 
@mdave16 I only started university a month ago
 
f(3+1) = f(3) + f(1) -> f(4) = 3 + 1 -> f(4) = 4
 
12:42 AM
unfortunate, but make sure you go to the WIMP talks, they are pretty good
 
so f(n+1) = f(n) + f(1)
 
@mdave16 what is WIMP?
 
from there, f( $\Bb Z$ + 1) = f($\Bb Z$) + f(1)
 
Weakly Interacting Massive Particle
Its a dark matter candidate
 
Joint Warwick IMPerial conference, every year there are a series of talks held at both campusses, aimed at undergrad, with plenty of opportunity to present your own research/interests. see this @LeakyNun
 
12:45 AM
that's bad notation: $\Bbb Z$ is the set of all integers @user78103
@mdave16 thanks
 
testing: $\Bbb Z$
f( $\Bbb Z$ + 1) = f( $\Bbb Z$) + f(1)
 
you can also use \mathbb
still poor notation, try $n$ instead
 
correct notation would be $\forall n \in \Bbb Z : f(n+1) = f(n) + f(1)$
 
If all I have is the explicit equation of a tangent plane in the form of z-z0=fx(x0,y0)...how am I suppose to find vectors in that plane? There's no x^ and y^ and z^
 
so @LeakyNun , I'm still lost on how to use that from the mapping of $\Bbb Z$ to $\Bbb Z$?
 
12:56 AM
@user78103 so we have established that if $f:\Bbb Z \to \Bbb Z$ is a homomorphism with $f(1)=1$, then $f(n)=n$ for all $n \in \Bbb Z$?
 
yes
 
@mdave16 Three whole extra letters? Never!
 
@user78103 now, what is $\delta_2(f)$?
 
You ask too much!
 
what about 2 and a half?
 
1:02 AM
@user78103 on a sidenote: would it be a better idea to deal with group theory before you go to algebraic topology?
 
@user78103 In my opinion you should learn some algebra before algebraic topology.
 
i'm nothing if not reasonable
 
@BalarkaSen SNIPED
 
you wrong sniped; doesn't count
 
:c
 
1:03 AM
missed headshot bruh
 
still trying to figure out the controls
 
pew pew
the controls are smoooth
i can just do a 360 noscope
 
i also agree, with the edit of "some" to ""
 
@LeakyNun I'll look into group theory then since I'm struggling here
 
@user78103 what is this for?
 
1:05 AM
I recommend Bob Ash's guide
 
@LeakyNun just wanted to learn algebraic topology, after reading some books on topology
 
@user78103 hmm
you definitely need algebra for that :)
 
@BalarkaSen , do you mean like elementary algebra? like f(x) = mx + b?
thanks for the recommendation, @mdave16
 
i gave a wrong link, but that one should be just as good, you can find my not-so-secret source
 
@user78103 No, abstract algebra
 
1:09 AM
No, you need to know concepts from abstract algebra, like groups, rings and modules
sniped
 
look at me 360 noscoping these n00bs
 
technically i sniped all yall with a link to an abst alg book
 
I'll look into that then
@LeakyNun is the answer to that question just e?
 
sure
 
1:14 AM
so how would that make the the kernel of $\delta_2 $ equal zero?
 
Because $\delta_2$ will be the isomorphism
 
I just figured out that you can prove the formula for the geometric series using étale cohomology, which I think is pretty funny. But now I have to look through all the proofs to see if it's not circular ><
 
since a kernel means where it fails to be injective, $\delta_2 $ fails to be injective when it's zero?
 
No. Kernel 0 means it is injective.
 
yea. I"ll have to look into group theory and abstract algebraic. Thanks again, @BalarkaSen
 
1:28 AM
Is there an easier way to determine the arc length than taking the derivitaves of x(t) and y(t)?
I did that and got it right. But it was a pain in the ass.
arc length = sqrt(x'^2+y'^2)
The computer generated answer makes it seem like there could be
 
using another parameterisation, else no (at least that i know of)
 
My answer looked like this "sqrt(((exp(-t/2)*(sin(6t)-12cos(6t))/2)^2)+((exp(-t/2)*(12sin(6t)+cos(6t))/2)^2‌​))"
compared to this: "1*sqrt(36.25)*exp(-0.5*t)"
 
i'm confused as to why you put 1* at the front
 
I didn't. That is the computer generated answer.
 
oh
well, there are tricks, e.g. you can pull out the exp stuff, and trig identities along with $\sin^2 + \cos^2 = 1$ usually help
i typically try to do as much as i can manually before resorting to CAS to do everything
 
1:33 AM
The computers simplification is one hell of a jump though :/
 
Hi all
 
yo.
 
Need French help !!
 
@Balarka better yet, 2Ï€ noscoping
 
i only hablo espanol
 
oiu this is certainly the correct chat
 
wii wii baghuet
 
Math chat is notorious for people asking for French homework help
 
On p. 103
 
i'm fairly sure google has a translate via picture thing, and that should be enough... i think
 
1:35 AM
Not homework.
 
You could probably use MTURK
 
@10Replies, anyway, back to the rearranging, another trick i use is to fix a value for the trig functions, and then use the shorthand $s$ and $c$, actually helps quite a bit.
 
that sounds fun
 
after pulling out the exponentials, and forgetting about writing a sqrt, we have $(\sin(6t)-12\cos(6t))/2)^2+(12sin(6t)+cos(6t))/2)^2‌$
also pull out a quarter, then $ (s - 12c)^2 + (12s + c)^2$
which becomes $s^2 - 144c^2 -24sc + 144s^2 + 24sc + c^2 = 1 + 144(s^2 - c^2)$
which makes me think i copied down your thing wrong...
 
1:51 AM
thanks, that is useful
 
It uses some awful ancient formatting thing
 
also i made a sign error, the thing i wrote boils down to 145, which becomes 36.25 after you put the $1/4$ back in
(s - 12c)^2 + (12s + c)^2 = s^2 + 144c^2 + 144s^2 + s^2 = 145
i'm not sure, but i think i killed chat
i feel responsible for this silence
 
Well, I was hoping for some shortcut, I guess the computer is just really good at simplification.
 
i think i used 4 lines on paper, 4 isn't that many
anyway, i feel the simplification issue is one you get better at the more you practice, but the computer will be better, usuaally
I'm going to go to sleep, it's awfully late here
 
night
 
 
1 hour later…
3:20 AM
I'm surprised not one person knows
I wouldn't expect that from a site that claims to answer so many questions
 
3:33 AM
[Random]
Suddenly so many h bar people
 
in The h Bar, 1 min ago, by Semiclassical
@0celo7 I need an expert opinion: Is the latest message in the Math chat basically just baiting?
 
which latest message?
 
@Secret not a lot to choose from (but no not you)
 
Why do you need an expert opinion? You are one of the experts @Semiclassical
 
lol I knew it was not me, but I thought that message referred is somehow related to index manipulation
 
3:35 AM
hah
 
@RayOfHope We don't know what you're talking about.
 
in The h Bar, 3 mins ago, by Slereah
I think math people don't like indices because they don't actually have to calculate quantities
 
Everybody was out watching The Walking Dead
 
Slereah's message makes the nature of the "last message" very confusing
 
I want to watch Overdrive soon.
And also Blade Runner 2049.
 
3:37 AM
While gut feeling suggest what the message has to be, I still cannot rule out it might had something to do with index manipulations
I think I am index biased
 
Let me know how Blade Runner 2049 is. I loved the original.
 
O btw, there's is where I am in that coding:
def read_template2(filename,index):

#Obtain information from template<num>.com
with open('template'+str(filename)+'.com','rU') as f, open('template.ligands,com','rU') as g:
line = f.readlines()
line2 = g.readlines()

#Obtain torsion scanning information
scan = [int(i) for i in line[9].split()]
steps = range(scan[0]+1)

#Extract parameters to prepare inputfiles
#filename,L1,L2,L3,charge_spin = lineno:[1,4,5,6,12],start slicing:[6,4,4,4,0]
params = [line[i][j:-1] for i, j in zip([1,4,5,6,12],[6,4,4,4,0])]
It's tedious to see what to modify, but its getting there, slowly...
 
Sir Cumference likes big pis and he cannot lie
@SirCumference
Baby got PI
 
[Random]
3
 
4:21 AM
1,1+1,1+1+1,1+1+1+1,...
1+1+1...=w,w+1,w+1+1,w+1+1+1,...
w+1+1+1...=w+w
w,w+w,w+w+w,w+w+w+w,...
w+w+w...=w^2
w,ww,www,...
www...=w^w
w^w,w^ww,w^www,...
w^www...=w^w^w
w^w^w,w^w^ww,w^w^www,...
w^w^www...=w^w^w^w
w,w^w,w^w^w,...
w^w^w^...=e0
e0,e0^w,e0^w^w,e0^w^w^w,...
e0^w^w^w^...=e0^e0=^w2 w
e0,e0^e0,e0^e0^e0,...
e0^e0^e0^...=e1=^w^2 w
e1,e1^e0,e1^e0^e0,...
e1^e0^e0^e0^...=e1^e1=^w^2 2 w
e1,e1^e1,e1^e1^e1,...
e1^e1^e1^...=e2 = ^w^3 w
e0,e1,e2,e3,...
e0^e1^e2^e3^...=ew = ^w^w w
ew,ew2,ew^2,ew^w,...
 
4:38 AM
I was asked to show that there's always an irrational number in the interval $(a, b)$
I've shown that there's always $q \in \mathbb{Q}$ such that $\sqrt{2}q \in (a, b)$. How do I account for the case $q=0$? Cause my proof is useless if $q=0$.
 
@studrayght5 if $0\in (a,b)$ is in there then $q=0$ is a possibility. But then you can find some $\epsilon>0$ rational so that $\epsilon\sqrt 2\in (a,b)$.
 
if $0\in(a,b)$ just use your proof with $(0,b)$. that can't have $\sqrt{2}\cdot0$.
 
Or do that
 
Pick the interval (0,1). Then $\sqrt{2}q$ can be made arbitrarily close to zero for any $q \in (0,1)$ e.g. $q=\{n\in \Bbb{Z} : \frac{1}{n}\}$
 
that's poor wording
 
4:47 AM
Thanks guys!
@anon Is that because if $0 \in (a, b)$ then anything in $(0, b)$ is in $(a, b)$?
 
or just use the set $1+\Bbb Q \sqrt2$
 
I used anon's idea! God knows how he knew I had already done the proof for $(0, b)$. :)
 
5:10 AM
@Secret should that be $n \in \mathbb{N}$?
 
Isn't $\Bbb{Q}$ include the negative rationals as well?
 
But if $q$ is negative, then $q \not \in (0,1)$? My knowledge is very basic, obviously, so I'm probably misunderstanding something.
 
5:29 AM
Ah yes, my bad. Forgot I am picking q from (0,1)
 
Intriguing question: (@Secret you may just want to use intuition)
Is the set $\Bbb Q^2 \cup (\Bbb R \setminus \Bbb Q)^2$ connected in the usual topology of $\Bbb R^2$?
 
Isn't that union basically all of $\Bbb{R}^2$?
since rationals are closed under products, while irrationals not necessary, thus having the plane of irrationals and the plane of rationals they should union to form all of the real plane?
 
$\Bbb Q^2 = \{(a,b) \mid a,b \in \Bbb Q\}$
 
5:44 AM
@Secret Nope, it contains points whose coordinates are either both rational or irrational, no mixes.
 
$(\Bbb R \setminus \Bbb Q)^2 = \{(a,b) \mid a,b \in \Bbb R \setminus \Bbb Q\}$
 
Hi. I hope I'm not butting in.
 
@Ravi no worries
 
ah yes, forgot the cross terms...
So... just $\Bbb{Q}^2$ alone and all (a,b) will be discrete under the usual topology of $\Bbb{R}^2$ (since there will be gaps in the form of irrationals,irrationals and irrationals, rationals.
 
hardly discrete
they are dense
 
5:48 AM
Ah right, (have to stop thinking them as dots in plane distributed like a fractal with varying scaling factor)
Meanwhile, $\Bbb{I}^2$ is also dense, and there are gaps in the form of (irrational,rationals) and (rationals,rationals)
So putting them together, we are missing the points (rational,irrational) and (irrational,rational). This means:
Given a pair (q,r) with fixed q, there are gaps for r irrational. Given a pair (r,q) with fixed q, there are gaps for r irrational.

Similarly, for fixed r, (q,r) and (r,q) both have gaps in the form of rationals

hmm... what do these gaps look like (other than they will be dense horizontal or vertical lines) ...
 
that's a cool exercise
 
It seems like there are countably many connected components (take $\Bbb{R}^2$ as the backdrop, there are horizontal and vertical slits cutting across at irrational positions, but whenever these slits met, they don't cut there and thus you have an island. Meanwhile, there are horizontal and vertical slits cutting across at rational positions, but whenever these slits met, they form an island. Therefore the result is kinda like a dense version of the following:
(Imagine making countable copies of this unit, paste them together in all directions (and whenever two q or r slits cross, make an island), scale it nonlinearly at every point, then we should end up with $\Bbb{Q}^2 \cup \Bbb{I}^2$)
 
lol
 
6:04 AM
The Stern–Brocot tree and Calkin–Wilf tree and thomae function suggests $\Bbb{Q}$ is some kind of "semi-fractal (more technical term to be found soon)" which has nonlinear self similarity
This is in contrast to a fractal where the scaling factor is constant.
I wish I can somehow prove or disprove this intuition, though...
Conjecture: $\Bbb{Q}$ is a multifractal system
A multifractal system is a generalization of a fractal system in which a single exponent (the fractal dimension) is not enough to describe its dynamics; instead, a continuous spectrum of exponents (the so-called singularity spectrum) is needed. Multifractal systems are common in nature. They include the length of coastlines, fully developed turbulence, stock market time series, real world scenes, the Sun’s magnetic field time series, heartbeat dynamics, human gait, and natural luminosity time series. Models have been proposed in various contexts ranging from turbulence in fluid dynamics to internet...
 
6:37 AM
Is it true that not all parametric equations can be implicitized?
 
How to check $\sum_{0}^{\infty} \frac{x}{1+nx}$ is uniformily convergent to $0$ or not in [0,1]?
I tried to find the $sup|f_n(x)|$, not able to find
I tried to do using M-test
@MartinSleziak @MatheiBoulomenos @0celo7
@ಠ_ಠ
Please help me
@SimplyBeautifulArt@studrayght5
@TedShifrin
@Yashas
 
7:11 AM
@ManeeshNarayanan this is a warning: pinging individual users just to get them to prioritize helping you with your problem is not acceptable
11
 
7:23 AM
Can anybody post anyting about math.stackexchange.com/questions/2485557/…
 
7:37 AM
Is there an analogue of monotone Turing machines/process machines for the lambda calculus? i.e. producing monotonic output?
 
@ManeeshNarayanan for x = 1 isn't that the classic harmonic series?
 
@tripleee Sorry!!! I won't repeat. Actually, no one is relpying :'(. I felt bad. I don't know much about the rules and regulations of the chat. i won't repeat again. Forgive me. I am apologizing to all.
@studrayght5 yes
then interval must be (0,1)
it point wise converges to 0.
 
@ManeeshNarayanan I don't think it does. $\frac{x}{1+nx}\le \frac{1}{n}$ for $x \in [0,1]$ (and so for also $x \in (0, 1)$).
 
$\frac{x}{1+nx}\le \frac{1}{n}$???
@studrayght5
 
7:54 AM
@ManeeshNarayanan Yes.
 
@studrayght5 need not true
 
It's true, for $n \in \mathbb{N}$ and $x \in [0,1]$.
 
$\frac{x}{1+nx}\le \frac{1}{nx}$ if $x>1$ we could say without any confusion
say x=1/2 it results contradictory to your argument. Am I right?
@studrayght5
 
8:21 AM
@ManeeshNarayanan Sorry, I should have said $\frac{x}{1+nx} \le \frac{1}{n+1}$.
 
8:52 AM
@studrayght5 $1+nx<1+n$ am I right?
 
9:06 AM
Should latex be working in chat?
It doesn't for me
 
@JackM for me, it is not working
 
@JackM See link in chat description top right.
 
@TastyRomeo I have followed the instruction, i have bookmarked. still, not working :'(
 
Weird, the bookmark works fine for me.
 
9:09 AM
@ShaVuklia Good JP meme
Accurate on multiple levels
 
forgot about that
been years since I've used the chatroom here
got the greasemonkey script running
 
If you have an old script running it may indeed not work anymore, because mathjax libraries were moved to a different server a few months back
Which also changed the url needed to access them
 
* facepalm * if you must star a [random] don't star one that is going to smack readers with a giagantic block of text
 
9:26 AM
Maybe don't post gigantic blocks of texts in the first place?
 
9:59 AM
Hi, off topic. Does the Brouwer fixed point theorem hold in infinite dimensional Euclidean space?
 
user84215
Hi.
 
user84215
@TheSubstitute Yes.
 
user84215
Every continuous function $f: D \to D$ has a fixed point, where $D$ is a compact convex subset of a Banach space.
 
10:45 AM
Hi.
 
user84215
Hi.
 
$am^3+m(2a-h)+k=0$ I have $m_1m_2= -1$ ($m_1,m_2,m_3$ are roots) Then isn't $m_3= \dfrac{k}{a}$ using Vieta's formula?
or is it $-\dfrac{k}{a}$
Someone please verify. I am slightly unsure.
Never mind. I got it. There was an answer on this site which was wrong and therefore I got confused.
 
11:09 AM
@balarka Hahahah very nais. Did you know that Bosnians make memes about Indian TV series? Like, they make a trivial scene very dramatic, with the music and the zooming in on the faces. We have one meme where someone asks "hey sup", and then you get 2 min or zooming in and out, and music and slow motion, and then at the end the person answers "not much".
Or there is one where a woman accidentally breaks an egg on the ground, and they like replay that a million times, from different angles and with different speeds. Does that sound familiar?? Or are you so used to your own culture, that you're not even aware of the meme content:d
 
[Philosophy] Define mathematically: Unknown unknown
 
user84215
11:54 AM
The second week of the General Topology Course will start at 9:30 GMT on Tuesday, October 24, 2017 in this room.
 
12:18 PM
Having proven that there's always an irrational number $q$ in $(a, b)$, how do I show that one can find infinitely many irrationals in $(a,b)$? I mean I get the idea: we apply this to $a$, $q$ and $q$, $b$ then there exist $q'$ and $q''$ such that $a < q'< q < q'' < b$ and so on. But this... feels faulty!
 
Suppose there are only finitely many, say n, then the above inequality will eventually stop at the nth iteration, thus a contradiction to the denseness of rationals and irrationals
 
[Random] How to do a simple knot on a sphere:
 
@secret wonderful, thanks!
 
12:38 PM
A continuous game is a mathematical generalization, used in game theory. It extends the notion of a discrete game, where the players choose from a finite set of pure strategies. The continuous game concepts allows games to include more general sets of pure strategies, which may be uncountably infinite. In general, a game with uncountably infinite strategy sets will not necessarily have a Nash equilibrium solution. If, however, the strategy sets are required to be compact and the utility functions continuous, then a Nash equilibrium will be guaranteed; this is by Glicksberg's generalization of the...
fancy a game of continuum chess :P
 

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