Kirill Z. answered • 10/31/13

Physics, math tutor with great knowledge and teaching skills

^{6}(t)*(1+sin

^{2}(t)+sin

^{4}(t))];

^{-½}=∑

_{0}

^{∞}(2k-1)!!/(2k)!!*x

^{k}, summed over k. This series converges for x<1.

^{6}(t)(1+sin

^{2}(t)+sin

^{4}(t)) needs to be less than 1. This puts restrictions on the a parameter. [f(t)]

^{k}is integrable, however, integrals will be very ugly looking. For example,

{------- - --------------- + ------------- - ------------- +

256 12 256 1024

7 Sin[8 x] Sin[10 x]

------------ - ----------- }

2048 5120

Alilouche A.

int (1/sqrt[1 - x^2 + a (2(bx)^6 - (bx)^12)]) dx

I'm sorry for this error.

11/01/13

Andre W.

^{6}-(bx)

^{12}). You should also mention that a and b are very small positive numbers. The equilibrium point now is at x

_{0}=1/b. Since x=1/r, b is the (stable) equilibrium distance between the interacting atoms/molecules, usually a few Angstroms. With U(1/b)=a, U'(1/b)=0, and U''(1/b)=-72ab, you get the quadratic approximation U(x)≈a(1-36 b(x-1/b)²) and can find the small frequency oscillations about the equilibrium.

11/01/13

Andre W.

10/31/13