"What is the difference between method and device? A method is a device which you used twice." What is the difference between an analysis and a method? An analysis is a method which you used twice.
@BalarkaSen I don't know how to compute that :P 59 is prime, 109 is prime is reminds me a little of fermat's little theorem that deals with some mod(p) but I can't connect them
Also, is there a faster way to check if x is prime other than eratosthesis sieve?
would need to get hired to do so, @skull ... there are lots of young'uns who need adjunct jobs more than I do ... The college education world is in sad shape in our country.
So, @DanielF, I have come to the realization that it seriously takes some Lebesgue integration/graduate analysis to completely prove the Central Limit Theorem. And half my class can't do high-school level probability/calculus. :(
@TedShifrin I am trying to look at the inverse limit of $S^1$s with the pullback morphisms being $x \mapsto x^{p^i}$ and at the same time at the inverse limit of the graphs $Cay(\Bbb Z/p^i)$s with similar pullback morphisms. The two are quasiisometrically the same if they exists. I believe it has some connection with $\mathbf{Z}_p$ but can't really say anything about it.
The latter is a graph with $|\Bbb Z_p|$ elts so it has a strong chance of being $Cay(\mathbf{Z}_p)$.
Hey guys, very very quick question. I am just too tired to think about it. Reading up on measure theory, I found a sequence $f_n = 1_{[n, n+1]}$ They say this converges point wise to the 0 function
@Mike interesting : replace Z/p^i s by corresponding riemann surfaces of w^p^i = z over C, and take the inverse limit of the inverse system of those with the pullback morphism being covering maps. that also converges to something similar. with the profinite topology, it is in fact homeomorphic to the cantor set and thus to the p-adics.
Yeah that's what I was thinking also. So just in mathematical terms. Lets fix $x, \epsilon$. We want to find a $N$ such that $n > N$ implies $ | f_n(x) - f(x) | < \epsilon$. But why would be a priori set $f(x)= 0$?
I mean, I realize it may be a stupid question and I am just forgetting my first year calc.
From the generating function for the harmonic numbers, we immediately have that
$$\sum_{n=1}^{\infty} \frac{H_n}{n}z^n=\frac{1}{2}\log^2(1-z)+\operatorname{Li_2}(z)$$
and after multiplying both sides by $-2$ and setting $z=i$, we get
$$\sum_{n=1}^{\infty} (-1)^{n+1}\frac{H_{2n}}{n}=-2\left(\f...
@TedShifrin do you mind just quickly going over this thread's answer (there is only one, 3 lines). I don't understand the last line. how can $f_n$ (from above) take on values both 0 and 1?
Why does Egorov's theorem not hold in the case of infinite measure? It turns out that, for example, $f_n = \chi_{[n,n+1]}x$ does not converge nearly uniformly, that is, it does not converge on E such that for a set F m(E\F) < $\epsilon$. Is this simply true because it takes on the value 1 for e...
Oh nevermind, its essentially saying that no matter what closed set you pick, at some point your sequence will get "out" of this closed subset and take on values 1, 0 again. And at that point, you still have pointwise convergence.. but I don't see how that says you can't have uniform converge
@masfenix: Draw pictures!! Uniform convergence means that when you draw an $\epsilon$-fence around the graph of $f$, all the graphs of the $f_n$, for $n>N$, have to be inside.
Yes, I actually did that but I am confused about Egorov's theorem. It says that there exists a closed set inside a measureable set E (m(E) < inf) where the sequence converges uniformly.
okay, but this is the counter example to egorov's thorem. It shows why the set E needs to have finite measure. So we let E = R in our counter example. so m(E) = inf. and we have this sequence $f_n$ that converges pointwise, but never uniformly.
but I am not sure how its trying everything together.. what does this particular sequence have to do with measure of the space? and why are we looking at the complement (in the math.se thread answer)
From the generating function for the harmonic numbers, we immediately have that
$$\sum_{n=1}^{\infty} \frac{H_n}{n}z^n=\frac{1}{2}\log^2(1-z)+\operatorname{Li_2}(z)$$
and after multiplying both sides by $-2$ and setting $z=i$, we get
$$\sum_{n=1}^{\infty} (-1)^{n+1}\frac{H_{2n}}{n}=-2\left(\f...
Why does Egorov's theorem not hold in the case of infinite measure? It turns out that, for example, $f_n = \chi_{[n,n+1]}x$ does not converge nearly uniformly, that is, it does not converge on E such that for a set F m(E\F) < $\epsilon$. Is this simply true because it takes on the value 1 for e...
