@DanielFischer That what I said above? We find the mid and then when we are for example at the first part of the array we look at one of the intervals $[0,mid-1]$ or $[mid+1,i]$ ? Or am I wrong?
@MikeMiller Don't know. It seems the OP doesn't see how to use the polarisation identity. On another note, I think your answer could use a couple of "for all".
@DanielFischer I meant the middle for each part... once for the part with the positive numbers and once for the part with the negative numbers... :/ How else could we do this?
@evinda Okay. It's not good to start in the middle of each part. Start at the beginning or end in each part (doesn't really matter which). You need to keep two indices (for each part), moving only in one direction.
@MikeMiller Yes, unfortunately we still don't know what.
"The special case of 0 does not have a unique representation in scientific notation" should read: "The special case of 0 does not have a unique representation in scientific notation" Am I correct @DanielFischer?
@skullpatrol If you take that description as a definition, then no - unless you say $\lfloor \log_{10} 0\rfloor = -\infty$ and write it as e.g. $1\times 0$ - but see "The special case of 0 does not have a unique representation in scientific notation, i.e., ..."
Is there any sort of classification of (say finite) groups with the property that every subgroup is normal?
Of course, any abelian group has this property, but the quaternions show commutativity isn't necessary. More generally, the group
$\rlap{////////////////////////////////////////////////}\...
@evinda Something vaguely like that. Not exactly like that but similar. You need to check whether A[position] == 0. And you have the wrong condition for incrementing j, think about that some more. It's probably better to first deal with only positive numbers.
@BalarkaSen For a compact topological space $X$, there is a covering map $X \to \Sigma_n$ iff $X \cong \Sigma_{1-k+kn}$ for some positive integer $k$. The problem with doing it in general is that there's no classification of hyperbolic RSes.
This must have been known to the ancients, but I am having some trouble finding the references: what can be said (especially geometrically) about the normal closure of an element in a surface group? Especially an elliptic element (in an orbifold group)...
Every covering space of $\Sigma_n$ $n>1$ is a hyperbolic RS. The point is that we can classify which ones actually cover $\Sigma_n$ when they're compact; not so easy in general.
Using series representation you specified above, I got that your integral gets reduced to
$$\sum _{n=1}^{\infty } \frac{(n+1) n^2 (\log (n)-2 \log (n+1)+\log (n+2))+n-1}{2 n}$$
Can you take it from here?
the universal cover of a RS is either the sphere, the complex plane, or the hyperbolic plane, and that corresponds to the three regimes I outlined above
Though @Daniel we're paying close attention to the metric/complex structure in this whole process. Maybe there's a more reasonable classification if we forget those?
Oh, nevermind, no.
Plane minus cantor set is hyperbolic, and if that's in there, like hell we're going to classify them.
Well, @BalarkaSen, you just asked whether every finitely generated Fuchscian group is hyperbolic. You're the one who knows the definition of hyperbolic, so get crackin'.
so take a disc and quotient it by an action of $\mathbb{Z}/n\mathbb{Z}$. As an orbifold, this space is different to the disc, even though they're homeomorphic
@DanielFischer I am confused now... I tried this: `i=0; position=0; while (i<m and B[i]<0){ i++; } position=i; if (y>0){ i=position; while (A[position]==0){ i=position+1; } j=i+1; while (j<m and A[i]<y*A[j]){ j++; } `
Daniel Fischer 8:17 @skullpatrol If you take that description as a definition, then no - unless you say $\lfloor \log_{10} 0\rfloor = -\infty$ and write it as e.g. $1\times 0$ - but see "The special case of 0 does not have a unique representation in scientific notation, i.e., ..." @robjohn
@ccorn I want to describe an algorithm that given an unsorted array $B$ that stores $m$ integers, and any integer number $y$, determines if there are two elements of the array of which the quotient is equal to $y$. The time complexity of the algorithm should be $O(m \log m)$.
@Venus Expand the terms of the series, and then multiply it by 2 to makes things easier. Then try to group the terms of the series in a clever way like $$\left(n^2 \log (n)-n^2 \log (n+1)+n \log (n)-n \log (n+1)\right)+\left(n^2 (-\log (n+1))+n^2 \log (n+2)+n \log (n+2)-n \log (n+1)\right)-\frac{1}{n}+1$$
I've got a question about the integral you answered a few minutes ago, @Venus. How did you guarantee that $\left| \cos \theta \right| = \cos \theta$ when factoring it out of the denominator in line $(2)$?
Let us evaluate the general form of the integral
\begin{align}
\int\frac{\mathrm dx}{(x^2+a^2)\sqrt{x^2-b^2}}&=\int\frac{a\sec^2t}{(a^2\tan^2t+a^2)\sqrt{a^2\tan^2t-b^2}}\mathrm dt\tag1\\[7pt]
&=\frac{1}{a}\int\frac{\cos t}{\sqrt{a^2\sin^2t-b^2\cos^2t}}\mathrm dt\tag2\\[7pt]
&=\frac{1}{a}\int\frac...
Let me know if this seems legit, @Venus. Is it because we're considering $x \in \left( -\infty, +\infty \right)$ (as the integral is indefinite) and with the substitution $x=\tan \theta$, we can deduce that $\tan \theta \in \left( -\infty, +\infty \right)$ and just take the interval $\theta \in \left( \frac{-\pi}{2}, \frac{\pi}{2} \right)$ for convenience and over this interval, $| \cos \theta | = \cos \theta$?
@evinda: Leaving some special cases aside: sort the array by absolute value. Copy the sorted array, with elements multiplied by $y$. Pass through both arrays like the comm utility, looking for a match.
@ChrisOkyen You've described a very hard problem that's actually still being researched! The number of such decompositions of $n$ is the partition function $p(n)$. It's a very mysterious object!
The Wikipedia page I linked might have some stuff you'll find interesting.
@DanielFischer But don't we have to incerement also i? Because like that we just look at the first non-zero element with all the other elements... But we don't look for example at the second with the third one... Or am I wrong?