On Mon, Nov 29, 2010 at 5:24 PM, Bill Gosper <billgosper@gmail.com> wrote:
Veit>Bill, you can work out many of these moments of inertia mentally if you
use the "parallel axis theorem": http://en.wikipedia.org/wiki/Parallel_axis_theorem
Veit
Veit, the center of oscillation seems to be the ratio of moment of inertia to some other moment. The math isn't the problem--it's my physics. --Bill
Well, the math CAN be a problem-- Proof that pi = 149/60 :) Out[312]= 2 1 w 4 4 EllipticF[- ArcCos[1 - --], --] 2 2 2 w --------------------------------- w
In[313]:= Limit[%, w -> 0]
Out[313]= 149/60
In[314]:= $MaxExtraPrecision = \[Infinity]; N[%% /. w -> 10^-69, 69]
Out[314]= 3.14159265358979323846264338327950288419716939937510582097494459230782 + 0.*10^-69 I
And luckily for me, the Doc search for "pendulum" omitted EllipticK, whose first two applications give the period formula with Sin[alpha/2] instead of Sin[alpha/2]^2.
And 7.0 doesn't seem to know that
3/4 3/2 2 (-1) Pi EllipticK[2] == -------------------- . 1 3 Gamma[-(-)] Gamma[-] 4 4 --rwg
Thought for today (simplfied by Julian) sum(sech(((%pi * n)/(2 * sqrt(3)))),n, - inf,inf) = ((3 * 2^(1/6) * G^3((1/3)))/(2 * 3^(1/4) * (sqrt(3) + 1)^(5/2) * sqrt(5 * sqrt(3) - 4 * sqrt(2) - 3) * (%pi)^2)) inf ==== \ %pi n
sech(---------) =
/ 2 sqrt(3) ==== n = - inf 1/6 3 1 3 2 gamma (-) 3 ------------------------------------------------------------ 1/4 5/2 2 2 3 (sqrt(3) + 1) sqrt(5 sqrt(3) - 4 sqrt(2) - 3) %pi Empirical conjecture: (1/2)*Integrate[(r^2*(Pi*r*Cos[\[Theta]]* Hypergeometric2F1[1/4, 3/4, 2, (4*r^2*Cos[\[Theta]]^2)/(1 + r^2)^2] + 4*(1 + r^2)*HypergeometricPFQ[{-(1/4), 1/4, 1}, {1/2, 3/2}, (4*r^2*Cos[\[Theta]]^2)/(1 + r^2)^2]))/ Sqrt[1 + r^2], {r, 0, 1}, {\[Theta], 0, 2*Pi}] == (8*Pi)/5