@Sawarnik are we doing the angle bisector ? (or are you still mad at me ?) @Sabyasachi you have to get the number 2048 .. just press the arrow keys .. you'll understand
With a function I know the notation $f(A)$ for the image set of $A$ under $f$. What is the standard notation for the image of sequence ? Can I write $s(A)$ for example ? Or $s_A$ ? Or just $\{s_n : n \in A \}$ ?
@robjohn I was thinking that both polynomials are domiated by their largest term. Which is smaller than their absolute values. The absolute values are $R$, because of the contour and hence it behaves like $\mathcal{O}(1/R)$
@DanielFischer How does that differ from @eXtremiity's definition of "The set of all sequences"? It seems to imply to me the set of all finite sequences (for example $\{1,2,3,4,5,6,7,8,9,\dots,11,12,13,14,15,\dots\}$
@eXtremiity You needn't necessarily map it to $[0,1]$... Just show through Cantor's diagonalization argument. I did this in an assignment to show that the set of all binary strings is uncountable, if you allow leading zeroes
I think the $0$ is the only element that makes it uncountable, so long as those 0s can be leading. If you allow zeros but disallow leading zeros, then you should be able to map onto $\mathbb{N}$ quite easy
Removing any other element doesn't remove this problem
If you have at minimum two symbols, one of which is $0$, you cannot enumerate every sequence of digits formed by them.
I think I could demonstrate a bijection from $\{0,1\}^{*}\rightarrow\mathbb{N}$ providing that leading zeros were disallowed. But anything more than those two characters causes issues
@DanielFischer Really basic analysis. Exam questions include things like "$U$ is open iff $x_n\to x\in U$ implies $x_n$ is eventually in $U$" or "if $S$ is bounded then ${\rm diam}\, S={\rm diam}\,\overline S$ and things of the sort.
@robjohn too expensive for me... I'm using a 4 year old Dell, and the company charges a lot for a replacement. Non dell-branded batteries have a tendency to refuse to charge thanks to DRM dell puts in their laptops
@robjohn I found a funny story here mathworld.wolfram.com/HadjicostassFormula.html. They say " It was conjectured by Hadjicostas (2004) and almost immediately proved by Chapman (2004)."
@robjohn thousand and thousands of students can prove that (easily). Perhaps many people didn't know of Hadjicostas formula that he conjectured. Well, I did it immediately as I saw it. There was nothing hard with that.
it turns out (via character theory) that they're a semidirect product, $G=K\rtimes H$. we call $K$ the Frobenius kernel and $H$ the Frobenius complement.
it is a big theorem that $K$ is nilpotent (by John Thompson)
@PedroTamaroff a representation of a group is a homomorphism $\rho:G\rightarrow \operatorname{GL}(V)$, where $V$ is a vector space. generally you're working in finite dimensional spaces, so $\rho(g)$ is a matrix for each $g\in G$. the function $\operatorname{tr}(\rho)$ is the "character" of $\rho$.
I stumbled upon Dirichlet characters, though. Is that a particular case of this, or just the name coincides? I know one can more generally study characters of abelian groups, not only $C_n$.
Hello mods in chat, this "answer" is actually a PSQ in portuguese. Should I translate it first so you can confirm it or should I vote to close it right away?
@AlexanderGruber no, it basically asks you to find $a$ and $b$ given some linearly dependent vectors, totally unrelated. It even says (HINT: remember that $\cos(60^\circ)=1/2$)
Well, my princeton review GRE Mathematics Practice Book just came today..I am so excited to enroll in MIT (hopefully!).
Can we use our calculators on the real test?
Thanks.
Well, my princeton review GRE Mathematics Practice Book just came today..I am so excited to enroll in MIT (hopefully!).
Can we use our calculators on the real test?
Thanks.
