@DanielFischer If you don't mind, I have a question related to what I asked earlier. Would $\cot \pi z$ also be uniformly bounded on the circle $|z|=N+1/2$ or even say the circle $|z|=N+3/4$?
@RandomVariable It would. We don't have that the arcs of the circles near the real axis are translates by a period then, as the parts of the sides of the rectangles were, so we need to consider not only a piece of a single boundary anymore.
But, we could look at a small rectangle that contains the parts of the circles with $\lvert\operatorname{Im} z\rvert \leqslant 1$ for all circles when translated.
@DanielFischer What about along $|z|=N+ \epsilon$ where $\epsilon$ is some really small number. I guess what I'm asking is would $\cot \pi z$ still be uniformly bounded if the contour got very close to the poles on the real axis?
@robjohn I also created the version $$\sum_{p=1}^{\infty} \left(\left( \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{\Gamma(k)\Gamma(n^p+1)}{\Gamma(k+n^p+1)}\left(\frac{1}{n^p+1} + \frac{1}{ n^p +2 } + \cdots + \frac{1}{ n^p + k } \right)\right)-1\right)= \frac{3}{4}$$
@DanielFischer Sometimes my connection is more like 3 Mbps. I can't remember what my modem speed was on my old 386. It was probably 9.6k or 14.4k.
@ZaidAlyafeai I knew that answer to your question involved that contiguous relation. I was trying to prove it before I posted anything. But O.L. beat me to it.
At least I proved that $$E\left(2 \sqrt{-4+3 \sqrt{2}} \right)=\frac{\sqrt{2}}{1+\sqrt{2}}\left[\frac{\Gamma^2\left( \frac{1}{4}\right)}{8\sqrt{\pi}}+\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{\pi}} \right]$$
hmmm, I wonder if I can do this one with high school tools $$\lim_{ n \to \infty} n \int_0^1 \sqrt{\frac{\displaystyle \sum_{i=1}^{k} x^{2n-i}}{\displaystyle 1+\sum_{i=1}^{k}x^{2n+i}}} \ dx=\operatorname{arcsinh}(\sqrt{k})$$
@DanielFischer sorry for pestering you with this, but do you know where I can find the proof that the derivative of a real function is Baire Class one? I couldn't find it via Google
@G.T.R I don't know where you can find a citation, but $$f_n(x) = n\cdot \left(f(x+\tfrac{1}{n}) - f(x)\right)$$ makes it immediate when the function is globally defined. You need some modification when it's defined on an open interval with finite upper bound, and for functions defined on compact intervals it's a little less easy.
Well I left to France when I was 17 to complete what is called a "class prepa" which was a great introduction to theoretical math. And then I applied to american universities, and decided to double major in math and physics
@DanielFischer Thm: let $O$ br open in $R^n$ and let $f:O\to R$ be of class $C^r$. Let $M$ be set of points $x$ for which $f(x)=0$ and let $N$ be the set of points for which $f(x)$\geq 0$. Suppose $M$ is non-empty and $Df(x)$ has rank 1 at each point of $M$. Then $N$ is an n-Manifold in $R^n$ and $boundary(N)=M$
So take $f(x,y,z) = x^2 + y^3 + z^3 - 2xyz - c$ (should that maybe have been $x^3$, by the way?) and look where $Df = 0$. For all $c$ where that doesn't happen, you have a $2$-manifold.
@user52932 No. The zeros of $Df$ are independent of $c$ - I should maybe have left $c$ out in $f$. So let $g(x,y,z) = x^2+y^3+z^3-2xyz$. For each zero $(x_0,y_0,z_0)$ of $Dg$, you get a value $c = g(x_0,y_0,z_0)$. These values are where you (probably) don't have a submanifold.
The qs is this: We have a space $K(R)$ of all compact sets of $R$. We have $S$ as the set of closed intervals in $R$. I want to check whether $S$ is open, closed or neither in $K(R)$
under the hausdorff metric
this is the first i am seeing this metric so it makes 0 sense to me intuitively
The definition that I have is this: For $A,B\in K(X)$, d(A,B) is the smallest $\epsilon$ such that for every point $a$ in $A$ there exists a point $b$ in B such that $d(a,b)\leq \epsilon$. And for every point $b\in B$ there exists a point $a$ in A such that $d(a,b)\leq \epsilon$.
Eisenbud and Harris seem to be saying the dimension of the Zariski tangent space of a scheme at a point is atleast the dimension of the local ring at the same point, even in the non-Noetherian case. Sounds like bs to me.
