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?
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
It 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...
@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)?
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?
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...
@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 $0$
@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 $0$
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)?
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 $\sum_{k=0}^\infty (-1)^k$
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.
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 $^{a+b}x$
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
@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
@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).
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
@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
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)
@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
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)
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
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?
@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