Wikipedia:Reference desk/Archives/Mathematics/2012 May 5

= May 5 =

Uniform dual polyhedra
If all the Catalan solids have constant dihedral angles, do all the duals of the uniform polyhedra (the nonconvex ones) also have constant dihedral angles? (I suspect the answer is yes.) Double sharp (talk) 03:12, 4 May 2012 (UTC)
 * Yeah, the canonical dual having constant dihedral angles follows from the original polyhedron having regular polygons as faces. It shouldn't depend on convexity of the polyhedron, or even convexity of the faces. Rckrone (talk) 04:58, 5 May 2012 (UTC)

Summation
Could someone please explain to me why $$\sum_{i=0}^{m-r-1} {m \choose m-i} = \sum_{j=r+1}^m {m \choose j}$$. I can see we're using the 2 j=i+r+1 but I don't understand why the binomial coefficient ends up the way it does. Thanks. 131.111.184.11 (talk) 12:51, 5 May 2012 (UTC)


 * The binomial coefficients are reversed in the order of summation with the substitution you have used. Rather consider the substitution $j = m − i$, and you will have to swap the limits because generally $m ≥ r + 1$.  — Quondum☏ 13:23, 5 May 2012 (UTC)


 * double counting? -- Taku (talk) 10:33, 6 May 2012 (UTC)

Normalise a DE
What does it mean to "normalise" a DE? I am told to normalise $$y'' - xy = 0$$ by the substitution $$t = \frac{2}{3}x^{3/2}$$. My working out is $$\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = \frac{dy}{dt}x^\frac{1}{2}$$. $$\frac{d^2y}{dx^2} = \frac{dt}{dx}\frac{d}{dt}\frac{dy}{dx} = \frac{dt}{dx}\frac{d}{dt}(\frac{dy}{dt}\frac{dt}{dx}) = x^\frac{1}{2}\frac{d}{dt}(\frac{dy}{dt}x^\frac{1}{2}) = x^\frac{1}{2}(\frac{d^2y}{dt^2}x^\frac{1}{2}+\frac{dy}{dt}\frac{1}{2}x^{-\frac{1}{2}}\frac{dx}{dt}) = x^\frac{1}{2}(\frac{d^2y}{dt^2}x^\frac{1}{2}+\frac{1}{2}x^{-1}\frac{dy}{dt}) = x\frac{d^2y}{dt^2}+\frac{1}{2}x^{-\frac{1}{2}}\frac{dy}{dt}$$. So the DE is $$x\frac{d^2y}{dt^2}+(-x^\frac{3}{2}+\frac{1}{2}x^{-\frac{1}{2}})\frac{dy}{dt} = 0$$. $$t = \frac{2}{3}x^\frac{3}{2} \Rightarrow x = \left(\frac{3}{2}t\right)^\frac{2}{3}$$.

So the DE is $$\left(\frac{3}{2}t\right)^\frac{2}{3} \frac{d^2y}{dt^2} + (-\frac{3}{2}t + \frac{1}{2}\left(\frac{3}{2}t\right)^\frac{2}{3})\frac{dy}{dt}$$. Doesn't look very normal to me. 150.203.114.37 (talk) 20:11, 5 May 2012 (UTC)
 * Did you mean $$y'' - xy' = 0$$? Otherwise your substitution looks off. 129.234.53.19 (talk) 21:41, 5 May 2012 (UTC)
 * Yes, that is what I meant. 150.203.114.37 (talk) 22:12, 5 May 2012 (UTC)
 * Except it actually is $$y'' - xy = 0$$. WHOOPS. I'll correct it and come back if I'm still stuck. 150.203.114.37 (talk) 22:14, 5 May 2012 (UTC)
 * OK, so the DE is $$\frac{d^2y}{dt^2} + \frac{1}{3t}\frac{dy}{dy} - y = 0$$ I think. Still curious to know what "normalise" means. 150.203.114.37 (talk) 01:11, 6 May 2012 (UTC)