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00:01
in some sense yeah i guess
00:14
@loch can you prove that [X,K(A,n)] = H^n(X;A)?
No i rmb the gist is to prove that it satisfies the axioms of a cohomolog theory
you what
some theorem tells you that a cohomology theory (e.g. singular cohomology) is uniquely determined by some axioms

so you reduce your question to checking that [-, K(A,n)] satisfies those axioms
i should probably say CW complexes
actually
im not sur eif thats how people prove it because i cant find it on google
00:32
@loch I was about to cite the exact same website lol
it's a small world under google
can we compute negative cohomology
$H^{-1}(X;\Bbb Z) = [X,K(\Bbb Z,-1)] = [X,\Omega^2 S^1]$
$\Omega S^1$ is just $\Bbb Z$
$\Omega \Bbb Z = \Bbb Z$
hmm
i think this tells you that these are 0
[these have to be pointed maps]
but you do see negative stuff if you deal with generalised cohomology theories i think
oh $\Omega \Bbb Z = \ast$ then
> The proof of lemma 2 is standard, which we will omit
@LeakyNun what
00:46
@RyanUnger look at the link lol
lol
oh god
is this just formally
like what is $\pi_{-1}$
I think so
Is there a technique to solve this system of ODEs exactly: $x^\prime = ax^2 + bxy, y^\prime = cy^2 + dxy$?
I thought to divide the equations but I can't find a way to separate variables
01:02
it's a system of nonlinear odes, so absent something clever I wouldn't expect much
 
2 hours later…
03:09
I am having trouble making these two RHS's equal: i.imgur.com/YAcPVFN.png
Made some progress on it, but keep getting stuck.
03:23
Hello, any idea how 1 < ⌈ 2x+1⌉ < 6 is equivalent to 1 < 2x+1 ≤ 5 ?
ceiling function
definition of ceiling function is ⌈ x ⌉ = n - 1 < x ≤ n but I'm not sure how to apply it to 1 < ⌈ 2x+1⌉ < 6
to get 1 < 2x+1 ≤ 5
04:05
@MatheinBoulomenos Hodge-Arakelov conflicts in time with symplectic topology ;_;
04:20
What is uniform convergence? I came across definitions in books and internet but they all are opaque to me. Any intuitive defintion without using formality?
@Hmmm Possible value of ceil(2x + 1) are 2,3,4,5. If ceil(2x + 1) = 5 then 4 < 2x + 1 <= 5; Similarly for otherside. You got that?
04:42
but why would the possible values be 2,3,4,5. There could be 5.9 still less than 6
oh because of ceil :)
 
3 hours later…
07:40
I guess my integrals are really rusty, but how is the integral of e^-log(x) equal to log(x) + c (for x>0) (according to wolfram alpha)? There is never negative area, so no value of c could accomodate that
0
Q: $[X,K(G,n)] = H^n(X;G)$ for non-CW-complex X?

Kenny LauIt is a standard fact that if $X$ is a path-connected CW-complex, then: $$[X,K(G,n)] = H^n(X;G)$$ where: $G$ is an abelian group; $n>1$ is an integer; $K(G,n)$ is the Eilenberg-Maclane space; $[X,K(G,n)]$ is the set of homotopy classes of maps $X \to K(G,n)$; $H^n(X;G)$ is the $n$th singular co...

@BalarkaSen stupid question ^
08:31
@BalarkaSen lol it has been answered already
09:07
test
09:53
@LeakyNun, Hi, how are you? can you help me to find an example for a cochain map $ψ: A• → B•$ that exists :
$ψ^i:A^i→B^i$ is an bijective (one-to-one correspondence) map (for i>=0) but $ψ∗ :H^k(A∙)→H^k(B∙)$ is not a bijective map (for k>=0)?
I don't think it exists
10:24
Can someone help?
@LeakyNun, I maen by "exsits" : that the map which I look for "Satisfies"
I know
@Nobodyrecognizeable split the integral at 0
@LeakyNun but the first integral gives $\infty$.
you must have done it wrong then
remember that |x| = -x for x<=0
@LeakyNun, but How I can find like this map ? I can I take the map : #f: S1→D2#
10:39
@LeakyNun thanks.
Hi chat.
@LeakyNun is it possible to do this by just knowing basic properties of delta functions?
I'm sure it is
@LeakyNun do you have any references I can learn to do this?
you should be able to do this
using change of variables
and the sifting property
10:53
@LeakyNun thanks. Will have a look at that.
Hello, i know that if A and B are compact then there exists $(a,b)\in A\times B, d(a,b) = d(A,B)$ I want to find an example where this is not true if A is compact and B closed
I put A=[1,2] and $B=]-\infty,0[ $ in $\mathbb{R}^*$
is it correct ?
here B is closed but not bounded then it is not compact right?
and d(A,B)=1 but d(a,b)>1
is it true ?
what is the radius of convergence of $\sum_{k=0}^{\infty} (^kx)$?
@Mathphile What does your notation mean?
$^kx$ is tetration
In mathematics, tetration (or hyper-4) is iterated, or repeated, exponentiation. It is the next hyperoperation after exponentiation, but before pentation. The word was coined by Reuben Louis Goodstein from tetra- (four) and iteration. Tetration is used for the notation of very large numbers. The notation n a {\displaystyle {^{n}a}} means a a...
$^2x=x^x$
$^3x=x^{x^x}$
$^4x=x^{x^{x^x}}$
yeah, got it
11:07
and so on
well, for $x \geq 1$ it obviously diverges
my guess is that it would be $(e^{-e}, 1)$
as $^{\infty}x$ converges for $x \in (e^{-e}, e^{1/e})$
@LeakyNun have you an idea about my question
@PolineSandra Leaky seems to be AFK.
@LucasHenrique what do you think?
11:12
what it means, is it "off line"?
@Mathphile, TBH I'm not very acquainted with hyperoperations so I can't guess it.
@PolineSandra basically, yes.
$\Heart$
damnit
$\heartsuit$ \heartsuit
11:30
25 mins ago, by Mathphile
what is the radius of convergence of $\sum_{k=0}^{\infty} (^kx)$?
@LeakyNun any idea?
@Mathphile I don't think it ever converges unless x=0. If $x\geq 1$ it clearly doesn't converge. if $x \in (0,1]$, then $^{k+1}x>^{k}x$ which implies that $^kx$ doesn't converge to $0$
@MatheinBoulomenos it doesn't converge for x=0, fight me
okay so it just doesn't converge lol
is tetration even defined for base $0$?
@MatheinBoulomenos lol
0^0 = 1
11:35
well, convergence does happen
hold on
yeah
or maybe it's just really slow
for which x?
for x=.1, it seems to stabilize at 0.399
but might be floating point errors and the series growing really really slowly
Can you post the program?
maybe I was wrong about that inequality. Let's see
oh wait, sum of those
ok, right, that doesn't converge
11:37
yeah, the point is, $^kx$ will converge to something nonzero (if it converges)
{let i = .1; for(let j = 0; j < 10000; ++j) {i = .1 ** i; console.log(j, i)} }
this is js for individual iterations
at least for .1, it doesn't look like the sum of iterations will converge
11:55
24 mins ago, by MatheinBoulomenos
@Mathphile I don't think it ever converges unless x=0. If $x\geq 1$ it clearly doesn't converge. if $x \in (0,1]$, then $^{k+1}x>^{k}x$ which implies that $^kx$ doesn't converge to $0$
this isn't right when $0 \lt x \lt 1$
@MatheinBoulomenos take $x=0.5$
$^1(0.5) \lt ^2(0.5) \gt ^3(0.5)$
the power tower seems to oscillate
within certain intervals, they alternate
in others, they keep growing
only value in which nothing happens seems to be 1
when $x \gt 1$, they will obviously keep growing
certain intervals that are already within 0,1, that is
12:01
@towc do you have an example of an interval within 0,1 where it keeps growing?
yeah hold on, I take that back
I thought I saw them
one thing was that below a certain threshold, they seem to converge to multiple values (<0.065, around)
or maybe it just converges much more slowly
for $e^{-e} \lt x \lt e^{1/e}$ all values of $x$ converge for $^{\infty}x$
i don't think there is a radius of convergence
but the limit $^\infty x$ is positive, according to wiki
but the sum of those won't be
@MatheinBoulomenos yes that's the reason the sum won't have a radius of convergence
12:10
right, but if $^\infty x \neq 0$, then $\sum_{k}^\infty {}^k x$ doesn't converge
well, there's more steps
@MatheinBoulomenos exactly
but we're agreeing
why are there more steps? If $a_n$ doesn't converge to $0$, then $\sum_{k}^\infty a_k$ doesn't converge
well, is $2^\infty = 0$?
12:14
even $\sum_{k=0}^{\infty} (^k \left( \frac{1}{x} \right))$ should not converge for the same reason
someone can help me on topology ?
@towc no, but also $\sum_{k}^\infty 2^k$ doesn't converge
sorry, $1/2^\infty$
I think I'm making a mess of all mathematical notation
are we increasing k or x or that sum?
do you mean $1/^{\infty}2$?
I assumed k for whatever reason, but it's x, right?
@Mathphile nope
12:18
yes $1/2^{\infty}=0$
well then
12:36
i think $\sum_{k=0}^{\infty} (-1)^k(^kx)$ might converge for $0 \lt x \lt 1$ though
so we are increasing k?
yes in each term k is increasing
where I can read a material for solving this question: find an example for a cochain map $ψ: A• → B•$ that exists :
$ψ^i:A^i→B^i$ is an bijective (one-to-one correspondence) map (for i>=0) but $ψ∗ :H^k(A∙)→H^k(B∙)$ is not a bijective map (for k>=0)?
@MatheinBoulomenos do you think the new sum will converge?
@Mathphile no
12:44
okay lets take $x=0.5$
$(-1)^k ({}^kx)$ doesn't converge, does it?
wait a minute
what would be $^0(0.5)$?
is $^0(0.5)=1$?
lets evaluate $\sum_{k=1}^{\infty} (-1)^k(^k(0.5))$ for simplicity
$\sum_{k=1}^{\infty} (-1)^k(^k(0.5))=-0.5+0.707-0.612+0.654-0.635+0.643-0.640+0.641........$
lets group the terms into pairs of two
13:02
it doesn't converge. A necessary condition would be that $\lim_{k \to \infty}(-1)^k ({}^k(0.5)) =0$.
Eg: $-0.5+707=0.207$, $-0.602+0.654=0.042$, $−0.635+0.643=0.008$, $−0.640+0.641=0.001$
$\sum_{k=1}^{\infty} (-1)^k(^k(0.5))=0.207+0.042+0.008+0.001...$
it does seem like it will converge
again, might converge to two values, and one of them might be 0
probably a better way to phrase it
@MatheinBoulomenos can you prove that $lim_{k \to \infty} {^k(0.5)}-^{k-1}(0.5) \ne 0$?
if $\lim_{k \to \infty}{}^k(0.5)=x$ exists, then it is a solution of the equation $x=(0.5)^x$, so $x\neq 0$
the $(-1)^k$ is inconsequential for the limit of the sequence: if $a_n \to a$, then $|a_n| \to |a|$, so the limit of $(-1)^k ({}^k(0.5))$, if it exists, is non-zero
I agree that $\sum_{k=1}^\infty ({}^{k+1}(0.5)-{}^{k}(0.5))$ might converge, but that's not the same
you can't just use associativity for infinite sums
otherwise you can prove $0=1$ by looking at $\sum_{k=0}^\infty (-1)^k$
13:43
Concerning $^0x$, we can derive $x^0=1$ from the facts that $x^ax^b=x^{a+b}$ and that the inverse operation of multiplication is division. So $x^{a-a}=\frac{x^a}{x^a}=1$. But can you do this with $^ax$? What is $^ax^bx$? It's not equal to $^{a+b}x$, for sure.
@MatheinBoulomenos that won't converge either. Maybe you meant to double the ks?
What about $(^ax)^{(^bx)}$? I don't believe that is equal to $^{a+b}x$ either.
Because you get $(x^{x^{x^\cdots}})^(x^{x^{x^\cdots}})=x^{(x^{x^{x^\cdots}})(x^{x^{x^\cdots}})}$
@Rithaniel I don't think you can keep that in terms of tetration
Indeed, I'm curious if there is any kind of analogue, though. Such as perhaps $(^ax)^{(^bx})$ equaling $^{ab}x$ or something, which seems to be clearly untrue, but might be closer that $^{a+b}x$
13:59
I need to figure out the latex syntax for this
but some rough paper sketches show (^a x)(^b x) to be (^b x)^((^|a-b| x) + 1)
(assuming a > b)
That makes sense. I was seeing a +1 pop out, too.
I have a hunch ^0 x will be 1 (so that with that formula, if a=b, you get (^a x)²)
Hmmm, alright. That would need to be the case if your equality holds in any case.
for quick testing, here's a js func: const tet = (a, x) => a === 1 ? x : x ** tet(a-1, x)
unfortunately js can't handle tet(5,2) because it's too big
oh, but new bigints are a thing: {const tet = (a, x) => a === 1n ? x : x ** tet(a-1n, x); tet(5n, 2n)}
that creates a 38.5KB number
Yeah, that's a pretty big int.
14:10
unfortunately, my equality doesn't seem to hold
Hmmm, darn. Probably need to sit down and work with this in the abstract. Just assume any sized tower of $x$ powers.
So, what values for $a$ and $b$ were you using that gave you the equality to begin with?
5 and 3
it's a pretty simple pattern, I'm just very tired today
Same. I'm on the bus to the university at the moment, and I keep dozing off.
Probably not the best time to be doing math. :P
but a simpler rule to figure out is (^a x)^(^b x)
that should be (^a+b x)
unfortunately, I'm not quite sure how to handle negative tetration
oh, the definition of tetration on the wikipedia page starts with ^0 n == 1
I'm not so sure, because $^1x=x$ and $^ax=x^{^{a-1}x}$. So $(^ax)^{(^bx)}=x^{(^{a-1}x)(^bx)}$. So, if $(^ax)^{(^bx)}=^{a+b}x$ then $(^{a-1}x)(^bx)=^{a+b-1}x$ which I don't think is true
14:24
oh right, that's also failing
I don't see how
by expanding some examples, you're just adding together the numbers of xs in the chain
oh, wait
I think it's a problem with my code
I'd have thought (^a x)^(^b x) == (^b x)^(^a x), right?
what am I missing?
oh, I guess it's an evaluation thing
you are evaluating the chains individually
so that definitely doesn't work
Yeah, tetration, by my understanding, evaluates the towers "from the top down."
@MatheinBoulomenos in the case of alternating sequences, i think that if $\lim_{k \to \infty} (^kx)$ converges to some value, then $\sum_{k=1}^{\infty} (-1)^k(^kx)$ should also converge to some value
(not sure though)
So $3^{3^3}=3^{27}\neq 27^3=(3^3)^3$
but that's what my common sense says looking at this:
$\sum_{k=1}^{\infty} (-1)^k(^k(0.5))=0.207+0.042+0.008+0.001...$
@MatheinBoulomenos, now I understand that a word "bijection"= "surjection"+"injection".
14:58
5 hours ago, by Leaky Nun
I don't think it exists
do you also understand what "no" means
@MatheinBoulomenos, I look for an injective map and not bijective map. the booklet is wrote in another Language. so and I translatete the word by mistake.can you help me now to find you want a cochain map that is injective but homology map that is not injective: a cochain map ψ:A∙→B∙ that exists : ψi:Ai→Bi is an injective map (for i>=0) but ψ∗:Hk(A∙)→Hk(B∙) is not a injective map (for k>=0).
@Sila I already gave you an example
oh wait, that was for surjective
@LeakyNun, of course I know what the meaning of "no". I translated the word bijective by mistake. I wanted injective notbijective.
consider the following cochain complex: $A^0=0$, $A^1=\Bbb Z$ and $A^i=0$ for $i>1$. $B^0=\Bbb Z$, $B^1=\Bbb Z$ and $B^i=0$ for $i>1$. All maps involved are either zero or the identitiy on $\Bbb Z$. Then $\psi:A^\bullet \to B^\bullet$ is injective in all degrees, but $\psi^1:H^1(A^\bullet)=\Bbb Z \to H^1(B^\bullet)=0$ can't be injective
15:08
@MatheinBoulomenos, same question above was to find a surjective map , but the question now is to find a injective map.
@MatheinBoulomenos, thank you, I will try to understand it .
@Sila I feel you. In older literature, you sometimes find "isomorphism into" for an injective homomorphism and "isomorphism onto" for a bijective one. That stumped me and one prof of mine once
@MatheinBoulomenos nice
@LeakyNun That's sounds pretty rude and mean.
2
sorry.
@MatheinBoulomenos, yes, this what is happend, thank you very much!
16:04
En Route to London!
16:15
@ÍgjøgnumMeg what's in london
@Ryan I have a meeting with the body awarding my scholarship
@ÍgjøgnumMeg I thought that was Hberg
@Ryan it's the German academic exchange service, the university itself isn't providing the scholarship
(and they have an office in London that they use to meet with British applicants)
I'm gonna be there for like 5 hours and then have to travel back down to Plymouth again, and London is like 9001 degrees atm (gates of hell opening to welcome Boris Johnson into the office of Prime Minister)
16:28
@ÍgjøgnumMeg I'm glad that all turned out well regarding your application
@Mathein thanks to your help!
@Mathein I think I'm only going to do 2 lectures and a seminar this semester
so I can get back into the swing of things
ANT, Modular Forms, and Quadratic Forms
since none of those clash (I'd really like to have done Diff Top but you mentioned that they clash once a week, I thought about missing one lecture on alternating weeks and just catching up but idk if I'm allowed/prepare to do that) hahaha
@ÍgjøgnumMeg Congrats again man!
Thanks :)
16:45
@ÍgjøgnumMeg oh, about the seminar thing
usually if you want to do a seminar, you go to a "Vorbesprechung" at the end of the semester before
and there you get a topic for your talk
oh really
it's possible that the "Vorbesprechung" already happened
16:47
but you can just email the prof and ask if there are any talks left
usually that works out well unless the seminar is really popular
Fair enough
Is it usually obligatory to attend all lectures?
no
there are no attendance lists or anything
but it's usually really stressful to not attend a lecture
but as long as you do your homework, it's fine really
Okay, (if you have an opinion) what would you think about me missing one lecture for each of mod. forms and diff top on alternating weeks and then catching up? I don't have any other obligations except to study maths for 2 years (since the scholarship pays for everything)
so really I'm missing 1 lecture every 2 weeks
I'd say modular forms are much more important than diff top :P
Lol I'd agree
16:51
no seriously, stuff like that works fine if you put an effort to catch up with the missed lectures
but I feel like my analysis/topology is painfully lacking
if you ask kindly, you can ask your classmates for their notes
I might have to do so lol, otherwise I'm only taking 2 lectures this semester (if I can't attend the seminar) since I don't really have the analysis or geometry background for anything else
and you'll find someone who lets you photograph his notes
@ÍgjøgnumMeg that's one way to lose a scholarship
16:53
also the lecturer for mod forms, Kasten, has excellent TeXed notes on his homepage for every lecture he does
@Mathein nice :)
@towc if you do badly in the exams, sure
where I went, seminar/lecture attendance was somewhat monitored
seminar attendance is important
and if you didn't attend, it would reflect badly on other university stuff
lecture attendance not so much here
16:54
if you're a visa student, your visa is retracted, for example
@MatheinBoulomenos apparently $\lim_{n \to \infty} \sum_{k=1}^{n} (-1)^k(^k(0.5))=0.25884587507436946592396446747751818317$ for even $n$
I'm not
so
and $\lim_{n \to \infty} \sum_{k=1}^{n} (-1)^k(^k(0.5))=-0.38233986943061651856223601463730548340$ for odd $n$
then they don't allow you to get your exams and homework remarked
and other things
I took intro abstract algebra and almost never went to the lecture
16:55
if you're on a scholarship, that's another thing you can lose
so the infinite sum oscillates between to convergent values
@Mathein I spent a large majority of my undergrad degree in the pub and still got a first class hahaha
@Mathphile that makes sense
@Mathphile how did you get thsi?
@towc wrote some code in PARI
16:56
so it doesn't converge if you consider all $n$, just as I said
@MatheinBoulomenos I don't think that was in question
@ÍgjøgnumMeg well, hope things go well for you
also if i understand correct, if $a_{\infty}$ converges to some finite value, then $\sum_k^{\infty} (-1)^k.a_n$ will always oscillate between two values right?
@Mathphile nope, it just doesn't have a numerical answer
@towc I'm on my way to a meeting with the scholarship commission so I'll make sure to ask them about my plans, if they decide it's a bad idea then I'll not do it, otherwise it should be fine :P
@Mathphile consider the series a=1,1,1,...
16:59
@towc but $\lim_{n \to \infty} \sum_{k=1}^{n} (-1)^k(^k(0.5))$ is an example of this phenomenon

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