By Cesaro-Stolz theorem, we have that $\displaystyle\lim_{n\to\infty}\frac{1}{n^2 \log(n)}\sum_{k=1}^{n} k\log(k)=\frac{1}{2}$ and then we see
that for $n$ large, we have that $\displaystyle\frac{1}{n^2}\sum_{k=1}^{n} k\log(k)\approx\frac{\log(n)}{2}\Rightarrow \frac{-4}{n^2}\sum_{k=1}^{n} k\log(...
@Alizter I think you may not have seen the OP's comments. All of the content that I added came directly from his own comments, and almost entirely in his own words.
@Alizter I understand what you're saying, but in this case, there was a lot of back-and-forth clarification in the comments, and the OP added a significant amount of important information. Essentially, the question was answered mainly in his comments, not in the answer itself.
@MikeMiller about the $p^2$ groups, I got $\Bbb Z/p^2\Bbb Z$ and $\Bbb Z/p\Bbb Z\times\Bbb Z/p\Bbb Z$ (dunno how to denote the cartesian product of groups, can't remember how it's notated). No idea if there are more or how to prove that there aren't.
@Alizter It seems like it's difficult to get this across in the edit queue. How would you suggest I approach this situation (and other similar situations I might come across in the future)?
First, you need to show that such groups are abelian; then you need to invoke some form of the classification theorem of finitely abelian groups. You might be able to prove the appropriate special case more easily than the general one, I dunno.
@Alizter Fair enough. That said, I've already taken the time to write this edit so that it reads well and better reflects the OPs entire answer. Would it make sense for me to resubmit it, or should I just leave it alone at this point?
@Alizter Can you suggest an edit comment that would make the situation clearer for reviewers? My previous comment was "Incorporated important information from OP's comments into his answer itself"
@sjrosen I can take the edit as my own by improving it but you will not get reputation. I will accept it. However try not to be wordy with the summary. It is a summary.
@UserX So let $G$ and $H$ both be cyclic of order $k$. Each has a generator; say $g$ is a generator of $G$ and $h$ is a generator of $H$. Any homomorphisms come to mind?
But this has two ordered elements for each element(my terminology sucks). Say we got a generator (g,g). No matter the (a,b) we want to generate , we can't with our generator. So it's not cyclic right?
Okay, I guess I'm left with uniqueness of these 2 groups
Why are order p^2 groups isomorphic to these though? Any hints?
I was just wondering if the following is correct to tackle the Goldbach Conjecture?
Let $G(x)$ be the Goldbach function and $p\in\mathbb{P}$ such that
$$\int_{2}^{x}\left[x\left(\cfrac{p+p}{p}\right)\right]dx=G(x)$$
Using the integral definition of the arctangent function, we may write $$\arctan{\left(\sin{x}\right)}=\int_{0}^{1}\mathrm{d}y\,\frac{\sin{x}}{1+y^2\sin^2{x}},$$
thus, transforming the integral into a double integral. Changing the order of integration, we find:
$$\begin{align}
\mathcal{I}
&=\in...
a nice question for that I have a simple question:
Can we nicely finish it in one line using elementary tools?
Solving Tartary integrals starts to become doing random arithmetic after a while. Especially if there is no good motivation for why you need to solve it.
Hilbert had a student who one day presented him with a paper purporting to prove the Riemann Hypothesis. Hilbert studied the paper carefully and was really impressed by depth of the argument; but unfortunately he found an error in it which even he could not eliminate. The following year the student died. Hilbert asked the grieving parents if he might be permitted to make a funeral oration.
While the student's relatives and friends were weeping beside the grave in the rain, Hilbert came forward. He began by saying what a tragedy it was that such a gifted young man had died before he had had an opportunity to show what he could accomplish. But, he continued, in spite of the fact that this young man's proof of the Riemann Hypothesis contained an error, it was still possible that some day a proof of the famous problem would be obtained along the lines which the deceased had indicated.
"In fact," he continued with enthusiasm, standing there in the rain by the dead student's grave, "let us consider a function of a complex variable...."
I bet that could be Balarka Sen too instead of hilbert
Hello, I just have a phrasing question about perceptrons if anyone could help
I am trying to state simply what a perceptron cannot classify, i know it can classify linearly seperable classes. It does so with a (i want to say) linear hyperplane. Further it cannot generate high order hyperplanes. Is that correct to say?
@BumSkeeter "Arguably the simplest non-trivial example of such a decision function is a line in the plane which separates a training set X \subset \mathbb{R}^2 into two pieces, one for each class label. This is precisely what the perceptron model does, and it generalizes to separating hyperplanes in \mathbb{R}^n."