@PeterTamaroff I am going to work on understanding the relationship between $k[X_1,\dots,X_n]/I$ and $V(I)\subseteq \bf A^n$. I should probably assume $k$ is algebraically closed and start off with prime ideals $I$.
Suppose $S$ is a set. Consider $S\times S$. Subsets of $S\times S$ can be understood as relations on $S$. Each such relation has a transitive closure. Is there a topology on $S\times S$ in which the topological closure of a subset $R$ of $S\times S$ is exactly the transitive closure of $R$ when $R$ is considered as a relation on $S$?
Hmm, seems not. You need $(A\cup B)^C = A^C \cup B^C$, with $X^C$ denoting the transitive closure.
Define $$ s_0=0\quad\text{and}\quad s_n(x_1,x_2,x_3,\dots,x_n)=\sqrt{x_1+s_{n-1}(x_2,x_3,\dots,x_n)}\tag{1} $$ Then $$ s_n\left(x_1/a,x_2/a^2,x_3/a^4,\dots,x_n/a^{2^{n-1}}\right)=\frac1{\sqrt{a}}s_n(x_1,x_2,x_3,\dots,x_n)\tag{2} $$ and $$ s_n(1,x_2,x_3,\dots,x_n)\ge\sqrt{s_{n-1}(x_2,x_3,\dots,x_n)}\tag{3} $$ **Theorem:** For $x_k\gt0$ and $n\ge1$, $$ \sup\frac{\log(x_1x_2x_3\dots x_n)}{\log(s_n(x_1,x_2,x_3,\dots,x_n))}=2^{n+1}-2\tag{4} $$
**Proof:** For $n=1$, $$ \frac{\log(x_1)}{\log(\sqrt{x_1})}=2\tag{5} $$ Suppose $(4)$ is true for $n-1$, then $$ \begin{align} &\sup\frac{\log(x_1x_2x_3\dots x_n)}{\log(s_n(x_1,x_2,x_3,\dots,x_n))}\\ &=\sup\frac{\log(x_2x_3\dots x_n)+\log(x_1)}{\log(s_n(1,x_2/x_1^2,x_3/x_1^4,\dots,x_n/x_1^{2^{n-1}}))+\frac12\log(x_1)}\tag{6a}\\ &=\sup\frac{\log(x_2x_3\dots x_n)+(2^n-1)\log(x_1)}{\log(s_n(1,x_2,x_3,\dots,x_n))+\frac12\log(x_1)}\tag{6b}\\ &\le\sup\frac{\log(x_2x_3\dots x_n)+(2^n-1)\log(x_1)}{\frac12\log(s_{n-1}(x_2,x_3,\dots,x_n))+\frac12\log(x_1)}\tag{6c}\\[6pt]
$(6a)$ follows by $(2)$. $(6b)$ substitutes $x_k\mapsto x_kx_1^{2^{k-1}}$. $(6c)$ follows by $(3)$. $(6d)$ follows by the induction hypothesis.
Let $x_k=a^{2^k}$, then $$ \begin{align} \sup\frac{\log(x_1x_2x_3\dots x_n)}{\log(s_n(x_1,x_2,x_3,\dots,x_n))} &\ge\sup\frac{(2^{n+1}-2)\log(a)}{\log(s_n(1,1,1,\dots,1))+\log(a)}\tag{7a}\\ &\ge\sup\frac{(2^{n+1}-2)\log(a)}{\log(\phi)+\log(a)}\tag{7b}\\[6pt] &=2^{n+1}-2\tag{7c} \end{align} $$ $(7a)$ follows by $(2)$. $(7b)$ follows by $(3)$. $(7c)$ follows by sending $a\to\infty$.
@Timtech while it is true that $\sqrt{x}^2=|x|$, it is not true that if $y^2=x$, then $y=\sqrt{x}$. On the positive reals, $\sqrt{x}$ is always positive.
@robjohn I thought to use coth(x) expressed in terms of partial fractions to prove $$\sum_{n=1}^\infty\frac{1}{m^2+n^2}=\frac{\pi}{m}\frac{1}{e^{2\pi m}-1}+\frac{\pi}{2m}-\frac{1}{2m^2}$$
@robjohn Ethan's questions made me think of questions like $\displaystyle \sum_{m=1}^\infty \sum_{n=1}^\infty \frac1{m^2+n^2}$ or $\displaystyle \sum_{m=1}^\infty \sum_{n=1}^\infty \sum_{p=1}^\infty \frac1{m^2+n^2+p^2}$
yep, exactly Actually I came across this line on wiki:
In one dimension, a saddle point is a point which is both a stationary point and a point of inflection. Since it is a point of inflection, it is not a local extremum.
But isn't inflection point comes in picture when we talk about the concavity of a graph
@Alyosha Yeah, I left to take a shower, and in the shower, I thought, "I used csc, and it should have been cot"
@FrankScience sorry, that might be correct if taken in the Principal Value sense. It does not converge absolutely, and since a lattice has no natural order of vertices, it is not clear how to sum series that are note absolutely convergent. However, if you sum all the terms in increasing squares about the origin, you can get a conditionally convergent sum.
The one problem with that sum, is that it would not be differentiable since $$ f'(z)=-\sum_{\lambda\in\Lambda}\frac1{(z-\lambda)^2} $$ and that does not converge anywhere.
@robjohn hehe, nice that part with the shower. :-) I remember a cute situation at an interview when I was listening to those people there, but at the same time I was thinking of a new solution for one of my problems. Th interesting part is that at the end of the interview I received an offer! Conclusion: thinking of math problems in various situations brings much luck! :-)
@FrankScience I am not sure what you are asking in the first place. You cite a function; did you have some property you wish to prove? then maybe we could involve the Poisson Summation Formula
Often functions defined as sums on lattices can be used in proofs using Poisson, since Poisson says that the sum of $f$ and $\hat{f}$ are equal over the orthonormal lattice
Okay, so can you apply that on the function and write the equation? (Needn't to prove. Only heuristic is okay. I don't want a result of that. I only want to know the possibility).
@FrankScience I am not sure what you mean "apply that on the function". What I said is that these functions are useful in proofs involving the PSF. Did you have a proof in mind?
Hi all: so there is an answer here that was self-deleted by the poster, apparently because they thought maybe it wasn't appropriate. (They reposted it as a comment.) However, given that the question is a proof-verification type question, I think the answer is fine for all intents and purposes. Please consider voting for undeletion, and then upvoting it when it becomes available. Thanks!
My feeling (it's just a feeling) is that I can related to each other the series from the family $\displaystyle S(n,m)=\sum_{n=1}^{\infty}\frac{1}{n^m (e^{2\pi n}-1)}$ by a recurrence relation.
I'm a bit disappointed. For some months, every time I try to look up for a number using ISC, the result is the same: "Your search resulted in no match". :-(
Hey guys, I was just wondering if someone can suggest to me a calculus problem book that focuses primarily on practical/applied problems rather than problem like "Find the derivative/integral of this" etc
