Wikipedia:Reference desk/Archives/Mathematics/2014 February 28

= February 28 =

Conservative field in high dimension
Given a bounded, continuously differentiable vector field on a bounded simply connected domain in $$\mathbb{R}^n$$ (where n>3); How can I prove that this field is conservative? Differentiating the field gives the Hessian matrix of the potential (if it exists). Clearly it is a necessary condition that the Hessian is symmetric, and for n ≤ 3 a symmetric Hessian is equivalent to zero curl, so in that case it is also a sufficient condition. However, curl is not defined for n>3, and I can't find any citeable reference that will tell me if a symmetric derivative matrix implies a conservative field. Any help will be appreciated. PeR (talk) 09:07, 28 February 2014 (UTC)
 * This can be formulated in the language of differential forms. Instead of your vector field $$F=(F_1,\dots,F_n)$$, you look at the 1-form $$v_F=F_1 dx_1 + F_2 dx_2 +\dots+F_n dx_n$$. By Poincaré's lemma, if F is defined on all of $$\mathbb{R}^n$$, $$v_F$$ is exact if and only if it is closed. Closed means $$dv_F=0$$, which is (look at exterior derivative) another way of writing $$\partial_{x_i} F_j = \partial_{x_j} F_i$$. Exact means there is a 0-form (which is the same as a function) U such that $$v_F=dU$$, which is the same as $$F=\nabla U$$.
 * So a "symmetric derivative matrix" implies that the field is conservative, provided the domain on which it is defined is contractible (although simply connected is enough).
 * I hope that was not too confusing. There are elementary treatments of the fact (and they are not difficult), but all citable references I have on my desk at the moment are in German. For example, H. Heuser, Lehrbuch der Analysis 2, 8th edition, Stuttgart: Teubner 1993, Satz 182.2. Basically, the proof is the same as for two or three dimensions. —Kusma (t·c) 09:53, 28 February 2014 (UTC)
 * Great! Thank you so much for the explanation. PeR (talk) 08:09, 1 March 2014 (UTC)
 * Think you can find most of it in Introduction to smooth manifolds by John M. Lee YohanN7 (talk) 03:52, 3 March 2014 (UTC)

Sum of Reciprocal (prime) powers.
Let D be the set of integers > 1. Let E be the set of Prime Numbers (2,3,5,7,11...).


 * W = (sum over s in D, sum over t in D, (1/(s^t))
 * X = (sum over s in D, sum over t in E, (1/(s^t))
 * Y = (sum over s in E, sum over t in D, (1/(s^t))
 * Z = (sum over s in E, sum over t in E, (1/(s^t))

Any ideas on how to prove any of these being finite or infinite. Note that W>Y>Z and W>X>Z.Naraht (talk) 19:14, 28 February 2014 (UTC)


 * Using the sum of a geometric series,

W=\sum_{n\ge 2} \sum_{k\ge 2} \frac{1}{n^k}=\sum_{n\ge 2} \frac1{n^2(1-\frac1n)}, $$ and that sum converges. So all of your sums converge. —Kusma (t·c) 20:06, 28 February 2014 (UTC)
 * But isn't $$\sum_{n\ge 2} \frac1{n^2(1-\frac1n)}$$ =1?


 * — Preceding unsigned comment added by Naraht (talk • contribs) 23:21, 1 March 2014‎


 * Indeed it is, since $$ \frac1{n^2(1-\frac1n)}=\frac{1}{n(n-1)}=\frac1{n-1}-\frac1n$$. So $$W=1$$. —Kusma (t·c) 06:44, 2 March 2014 (UTC)
 * Any information on the values for X, Y, or Z or which is greater, X or Y?Naraht (talk) 04:48, 3 March 2014 (UTC)
 * I don't know anything about these values. For Y, the closest thing I can find is the "integral" section in Prime zeta function. —Kusma (t·c) 19:54, 4 March 2014 (UTC)
 * Thank you.Naraht (talk) 20:04, 4 March 2014 (UTC)