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May 11

Sum of reciprocals of integers as a fraction

What is the expression of the sum of reciprocals of the first n integers written as a fraction with a common denominator? The same question for the fraction expression of the sum of reciprocals of the first p primes? How does the distribution of primes intervenes in the difference between the two fraction(s) expressions?--82.137.13.173 (talk) 00:02, 11 May 2017 (UTC)[reply]

In the first case you could use n! as the denominator and sum the appropriate numerators. Then there should be a lot of cancellation to get it to lowest terms, but the terms are going to be pretty large, unless n is pretty small. For the second one, you can use Primorial (the product of primes) equivalently. Bubba73 ^{You talkin' to me?} 02:38, 12 May 2017 (UTC)[reply]

This and this relate to your question but don't answer it. Bubba73 ^{You talkin' to me?} 02:47, 12 May 2017 (UTC)[reply]

Iterating on maximum of uniform distributions

Set $c=0$ . Iterate $c:=\operatorname {max} (U(c,d),U(c,d))$ , where $U(a,b)$ is the uniform distribution between $a$ and $b$ . What is the distribution of $c$ after $k$ iterations? For 1 iteration it's easy to see that $p(c=x)=2x/d$ . I'm not sure how to go further. 70.190.164.57 (talk) 02:33, 11 May 2017 (UTC)[reply]

Can you clarify "

c:=\operatorname {max} (U(c,d),U(c,d))

, where

U(a,b)

is the uniform distribution between

a

and

b

" ? (1) This states the max of two identical things, so I wonder if there is a typo. (2) What do you mean by the max over distributions? Loraof (talk) 16:08, 11 May 2017 (UTC)[reply]

And even if the distributions weren't equal, what is the max of two distributions? In the function-max sense, the result isn't a distribution. Tigraan^{Click here to contact me} 16:41, 11 May 2017 (UTC)[reply]

def f(k,d):
    c = 0
    for i in range(k):
        c = max(random.uniform(c,d), random.uniform(c,d))
    return c

Given k and d, what is the resulting distribution? 70.190.164.57 (talk) 18:08, 11 May 2017 (UTC)[reply]

Okay, so you make two independent draws from the same uniform distribution, and the larger of those draws becomes the lower bound on the next two uniform distributions to be drawn from--is that correct? Loraof (talk) 19:58, 11 May 2017 (UTC)[reply]

Do you mean the discrete uniform distribution or the continuous uniform distribution? Loraof (talk) 20:03, 11 May 2017 (UTC)[reply]

Continuous. 70.190.164.57 (talk) 20:26, 11 May 2017 (UTC)[reply]

It looks like a hard problem to get a closed form solution. For the first iteration, though, I get a different answer from yours. The cumulative distribution function for $c_{1}$ is F(x) = Pr(1st draw ≤ x AND 2nd draw ≤ x), and since the draws are independent and from the same distribution, this = [Pr(any draw ≤ x)]² = [(x–0)/(d–0)]² = x²/d². So the probability density function for the first iteration of c is the derivative of that: f(x) = 2x/d². Loraof (talk) 17:59, 12 May 2017 (UTC)[reply]

1°30′

Is 1°30′ equal to 1.5°? I'm 90% sure they are, but just want to double check. ECS LIVA Z (talk) 02:49, 11 May 2017 (UTC)[reply]

Yes. Bubba73 ^{You talkin' to me?} 02:59, 11 May 2017 (UTC)[reply]

Yes. See Minute and second of arc. --76.71.6.254 (talk) 03:53, 11 May 2017 (UTC)[reply]

Thanks, guys.

BTW, how do you pronounce 1°30′? "One degree thirty arc minutes"? Or "One degree and thirty arc minutes"? ECS LIVA Z (talk) 05:37, 11 May 2017 (UTC)[reply]

You don't say "arc", as it's implied when you're clearly talking about angles. As with other mixed units ("five pounds (and) six ounces"), it's possible to use "and", but I would say it's usually not used. So probably "one degree thirty minutes", but with "and" is okay. --76.71.6.254 (talk) 07:39, 11 May 2017 (UTC)[reply]

Thanks, that's completely new information to me. Maybe it's notable enough to be included in the Minute and second of arc article? We'll need a RS for it though.ECS LIVA Z (talk) 05:18, 12 May 2017 (UTC)[reply]

Is there a Gauss–Bonnet theorem for odd-dimensional manifolds?

The generalized Gauss–Bonnet theorem connects the Euler characteristic of even-dimensional manifolds with their curvature. What about odd numbers of dimensions? Take the 3-torus, for example. I know it can be flat (identify opposite faces of a cube). Can it be realized as a homogeneous manifold of positive curvature? Negative? What sort of curvature is the right one to look at, if any? --Trovatore (talk) 03:58, 11 May 2017 (UTC)[reply]

The only global topological invariants obtained from the curvature are those given by the Chern forms, and so always have even degree. There are, however, many different comparison theorems that allow specific estimates on the curvature to be converted into strong statements about the topology. The Bonnet–Myers theorem states that if the Ricci curvature of a complete Riemannian manifold is bounded below by a positive number, then the universal cover is compact. Hence your 3-torus cannot carry a positive curvature metric. An even more famous result is the so-called sphere theorem, which says that if the sectional curvature of a complete simply connected manifold is 1/4-pinched, then the manifold is diffeomorphic to the sphere (this is a relatively recent result due to Schoen and Brendle as I have stated it). There are many similar such theorems, and a good introductory book is "Comparison theorems in Riemannian geometry" by Cheeger and Ebin. Sławomir Biały (talk) 12:50, 12 May 2017 (UTC)[reply]

Also, Cheeger and Ebin prove that a manifold of nonpositive curvature has a solvable fundamental group if and only if it is flat. So the 3-torus cannot have a metric of negative curvature either. Sławomir Biały (talk) 22:17, 13 May 2017 (UTC)[reply]

Ah, very cool; that's what I was looking for. I had vaguely heard (I may not state this perfectly) that someone had proposed that the universe could be a compact manifold even if the curvature is negative, giving the example of a dodecahedron where you identify opposite faces. I'm not sure whether you twist them or reflect them to make them match up. But anyway I was wondering why they didn't go for the simpler 3-torus. This seems to explain it. Off the top of my head I'm not sure what the fundamental group of either would be, but I can easily believe that it's solvable for the 3-torus and not for the dodecahedron thingie. --Trovatore (talk) 22:25, 13 May 2017 (UTC)[reply]

The Poincare dodecahedral space has vanishing

H_{1}

, so its fundamental group is equal to its own derived group. In particular, it is not solvable! Sławomir Biały (talk) 20:35, 15 May 2017 (UTC)[reply]