Wikipedia:Reference desk/Archives/Mathematics/2014 August 19

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

August 19

Thermal resistance from a wire through a disk

Please see attached diagrams in http://i62.tinypic.com/2rdg6yg.jpg

The flow of heat through a solid is governed by the solid's thermal resistivity, R, a property of the solid. For a simple bar shape, the flow of heat from one side to the opposite side as shown in Fig 1 is determined by the particular shape's thermal resistance, R, determined from R and the length, width, and height by the simple formula shown.

For a perfectly heat conducting wire or rod passing through a thermally resistive disk, and in perfect contact with the disk, the thermal resistance from wire to disk perimeter is also given by a standard formula: R = R Ln(Rd/Rw)/(2 Pi T) - See fig 2 in the attached file.

I need to find the thermal resistance from a wire to the disk perimeter edge where a pear-shaped hole is cut into the disk, so that the wire makes contact (assumed to be perfect contact) only over a limited angle, as show in Fig 3. The area of the hole can be assumed large compared to the cross section of the wire but small compared with the disk area. Assume heat only flows through solids, and not across any gap, and that there is negligible radiation.

How can I get or derive a formula, either approximate or exact? 124.178.49.97 (talk) 10:13, 19 August 2014 (UTC)

While the original problem has cylindrical symmetry which makes for a simple analytic solution (see for instance Thermal conduction#Cylindrical shells), with the cutout getting an analytic solution would be more challenging. For more complex geometries like this, it may be best to numerically approximate the solution to the heat equation. Energy2d is a free program that may be sufficient for your needs, but there are many full featured commercial programs, too. --Mark viking (talk) 21:37, 19 August 2014 (UTC)

Well yes, the solution for a plain disk is in almost any textbook on heat flow - I gave it in the drawing I attached. Yes, a numeric solution can be used. But a formula would be MUCH more convenient, and it need not be an exact solution. Within 20% or so would be quite good enough. 124.178.49.97 (talk) 01:30, 20 August 2014 (UTC)

Components Exceed The Average Squared

The following problem has come up in something I'm working on and I can't seem to find a good way to solve it. For reals a1,...an so 0 <= ai <= 1, when is ak >= Average(ai) ** 2 for all k? A specific answer would be wonderful, though, even if the set could simply be characterized, I would be much obliged. Ideally, I need to find a continuous transform of the unit n-cube into the solution that moves around the points as little as possible (various other constraints as well, but I'm not worrying about that at the moment). If the above is simply solved, I'd be even more interested in the general problem: given ai and linear combinations f1,...fm of them, when is ai >= f1 * f2 *...fm for all i? Thank you for all help - this problem is outside of what I'm good at, so any help is quite greatly appreciated:-)Phoenixia1177 (talk) 19:00, 19 August 2014 (UTC)

If I understand you correctly, any constant sequence will do: $a_{1},\dots ,a_{n}=p$ gives average of $p$ and $p\geq p^{2}$. Am I missing something...? --CiaPan (talk) 20:15, 19 August 2014 (UTC)

Phoenixia1177 wants a classification of all such points, not just an example. Is it obvious that the set is even convex?--80.109.80.78 (talk) 20:45, 19 August 2014 (UTC)

Simplification of bincoef series?

Keeping in mind that

$\sum _{j=m}^{\infty }{\frac {1}{\binom {j}{k}}}={\frac {k}{(k-1){\binom {m-1}{k-1}}}}$ and $\sum _{j=m}^{M-1}{\frac {1}{k{\binom {j}{k}}}}={\frac {1}{(k-1){\binom {m-1}{k-1}}}}-{\frac {1}{(k-1){\binom {M-1}{k-1}}}}$,

is there a similar simplification for

$\sum _{k=1}^{M}{\frac {1}{\binom {2k}{2}}}={\frac {1}{\binom {2}{2}}}+{\frac {1}{\binom {4}{2}}}+{\frac {1}{\binom {6}{2}}}+{\frac {1}{\binom {8}{2}}}+{\frac {1}{\binom {10}{2}}}+\dots +{\frac {1}{\binom {2M}{2}}}\,\!$?

I do know that $\sum _{k=1}^{\infty }{\frac {1}{\binom {2k}{2}}}=\ln(4)\,\!$.

 ~Kaimbridge~ (talk) 21:01, 19 August 2014 (UTC)

Given that the last series converges to an irrational it's unlikely there would be a closed form for the partial sums. Note that it's a variation on 1/1-1/2+1/3-...=ln 2. You could perhaps get an asymptotic formula for the partial sums using the Euler–Maclaurin formula. --RDBury (talk) 04:14, 20 August 2014 (UTC)

The formula is $~H_{_{M-{\frac {1}{2}}}}-H_{_{M}}+\ln 4,$ where the values of harmonic numbers of fractional argument are computed with the help of this formula. — 79.113.222.166 (talk) 09:51, 20 August 2014 (UTC)

While my version of Mathematica seems to find this summation easy (and verifies your answer, 79.113.222.166), I can't see how to prove it myself. Could anyone enlighten me? 129.234.186.11 (talk) 14:42, 20 August 2014 (UTC)

You should be able to prove it by induction using the properties $H_{\alpha }=H_{\alpha -1}+{\frac {1}{\alpha }}$ and $H_{\frac {1}{2}}=2-2\ln {2}$ found in this section. Egnau (talk) 16:28, 20 August 2014 (UTC)