@AsafKaragila Actually, to be politically correct, it's a produce of Germany. Anyway: I can't sent you chocolate because that would mean you'd have to reveal your home address on the internet. I don't think that's a good idea.
I'll help you through the course for free. If you still want to learn forcing and set theory after the course is over I'd let you send me a box of candy in return.
@JonasTeuwen though I was on Benelux meeting. There was cool food and loads of free beer. Although the main attraction was trying to determine where the border between Belgium and the Netherlands
@JonasTeuwen No secret, I've actually gone into detail her about it... I write and maintain a program that helps teach sentential logic, give and grade homework and tests, etc
This is the question: What's the asymptotic of $$ \frac{1}{n^{\alpha+1}} \sum_{r=1}^n r^\alpha, $$ where $\alpha > 0$ is an arbitrary real. This question (http://math.stackexchange.com/questions/63986/) does the job for integer $\alpha$.
@Srivatsan The connection is that the giant summation also happens to be the Riemann sum of the function $f(x) = x^\alpha$ over $[0,1]$. Of course, for $\alpha < 1$, the function is continuous, but not Lipschitz or differentiable at $0$.
Haha! Me and my advisor tried to prove the boundedness of an operator for a while because the original author wrote to me that it was "easy". Now I have mailed our (failing) argument and he says "it is a little bit trickier than that" and gets some weird functional calculus out of his sleeve.
@Victor I hashed this over with someone not long ago. to guarantee convergence in arclength, not only do the curves need to converge pointwise, but their derivatives need to converge. Have you looked at the formula for arclength?
@JonasTeuwen So, I can think of two avenues. Stick to defining $f(T)$ by the power series, but restrict our class of $T$ to some nice operators, so that $f(T)$ is nice.
The other is to consider better ways to define $f(T)$ itself.
You pointed to a link where a square of perimeter 4 was being transformed to a circle of diameter 1 and trying to prove that the circumference of the circle was 4, right?
@Victor when the sizes are small, like when dealing with $\mathrm{d}r$ and $\mathrm{d}\theta$, the shapes are essentially rectangles. Their areas are very close to the product of the length and width. As the sizes get smaller the approximations get even better.
I have a question about the first solution here: math.stackexchange.com/questions/89925/… . The post shows the sequence is increasing on (0,z), but I fail to see what that means it converges to z. Why couldn't it converge to some number less than z?
@Potato Not saying that the post is correct, but. If it converges at all, there is only one possibility for the limit since the limit $L$ satisfies $L = \sqrt{1-L}$.
@Victor 1/20 of a degree will change the area by a given percentage, if the area is small, the difference will be small, if the area is big, the difference will be big.
and $1-\cos(\mathrm{d}\theta)$ is about $\theta^2/2$ so as the size of the side gets smaller (about $r\mathrm{d}\theta$) the deviation of the arc is about $r\mathrm{d}\theta^2/2$
Introduction to the answer:
I began writing, hoping to have a relatively short answer. However as I originally expected this is not a simple proof at all. I wrote a short guide to the construction of symmetric extensions and permutation models - the two canonical ways of proving independence fro...
and $1-\cos(\mathrm{d}\theta)$ is about $\theta^2/2$ so as the size of the side gets smaller (about $r\mathrm{d}\theta$) the deviation of the arc is about $r\mathrm{d}\theta^2/2$
@AsafKaragila Can you point me to a tag that has something more than the excerpt? I have been trying to find an example of what is considered a good Wiki, but none of the tags I have tried have more than an excerpt.
@AsafKaragila I have to go. I will do something better later, but it seems that those tags that at least have an excerpt, only have an excerpt. So I will leave it as is. I thought that at least adding a link to the Wikipedia page would be helpful.
