Wikipedia:Reference desk/Archives/Mathematics/2008 August 12

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August 12

Manifold

In Indra's Pearls, it mentions a way to construct the torus and the two-holed torus from a punctured plane, by taking the quotient of the plane with respect to groups of Mobius transformations. It looked like this could be extended, to give at least the orientable 2-manifolds. Does something like this also work for higher-dimensional manifolds? Black Carrot (talk) 03:24, 12 August 2008 (UTC)

I don't know this stuff, but according to our articles, the generalization of the uniformization theorem to 3-manifolds is the geometrization conjecture (now proved by Perelman). Naturally, the 3d version is enormously more complicated. Algebraist 10:44, 12 August 2008 (UTC)

Convergence

In a proof of pi's irrationality, I saw the statement '${\frac {x^{k}}{k!}}$ tends to zero as k tends to inifinity for any x, as the series $e^{x}$ is convergent.'

First of all, if you have x=100 and k=100 then you have one hundred hundreds on the top all multiplied together and 100 integers on the bottom all multiplied together, none greater than 100 and only one equal to 100. So the top is bigger than the bottom, ie greater than 1, and i would have thought that this would be the case for greater values of x and k, so I don't see why that tends to zero.

Secondly, I don't see why $e^{x}$ is convergent. Could someone explain these two statements to me? 92.4.189.176 (talk) 22:55, 12 August 2008 (UTC)

Firstly:Yes, if you increase x and k together then ${\frac {x^{k}}{k!}}$ increases. However, the statement is that if you hold x constant and increase k, then ${\frac {x^{k}}{k!}}$ tends to zero. This is pretty obvious: if x is 100, and k is 100000000000, then the fraction is already pretty small. For your second question, the simplest way is to use the ratio test. Algebraist 23:03, 12 August 2008 (UTC)

(e/c) For each fixed x, ${\frac {x^{k}}{k!}}$ tends to zero. Thus, in your example, while it is true that (for x=100), this quotient is bigger than 1 if k=100, consider taking k=1000 instead, and you will get a number very close to 0. In fancy terms, ${\frac {x^{k}}{k!}}$ tends to zero pointwise but not uniformly. The power series

$\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}}$

converges by (for instance) the ratio test, since the ratio of two consecutive terms is $x/(k+1)$ which (for fixed x) tends to zero as $k\to \infty$. siℓℓy rabbit (talk) 23:11, 12 August 2008 (UTC)

When it says "the series e^x", it means the Taylor series of e^x (and should say that, really), which is the sum of terms like the one you give. That the terms tend to zero is a necessary condition for a series to converge, so since the series converges, we know that sequence tends to zero. As for why the series converges, I'm not really sure... I would say it converges because of the behaviour of that sequence (you need a little more than in just tending to zero, though), so it ends up being a circular argument. I'm not sure if there's a way of proving the series converges without using the fact that factorials dominate exponentials... --Tango (talk) 23:28, 12 August 2008 (UTC)

The ratio test works. The proof of this test relies only on the convergence of the geometric series. siℓℓy rabbit (talk) 23:29, 12 August 2008 (UTC)

Yeah, so I realised and edit conflicted with you trying to correct myself! Thanks. --Tango (talk) 23:32, 12 August 2008 (UTC)