Do either (1) or (2):
(1) On each element ds of a closed plane curve there is a force 1/R ds, where R is the radius of curvature. The force is towards the center of curvature at each point. Show that the curve is in equilibrium.
(2) Prove that x + 2/3 x3 + 2·4/3·5 x5 + ... + 2·4· ... .2n/(3·5. ... .2n+1) x2n+1 + ... = (1 - x2)-1/2 sin-1x.
Solution
(1)
(2) Let f(x) = x + 2/3 x3 + 8/15 x5 + ... + 1/2 n! n!/2n+1! (2x)2n+1 + ... .
Hence x f(x) = x2 + 2/3 x4 + 8/15 x6 + ... + 1/4 n! n!/2n+1! (2x)2n+2 + ... .
Also f '(x) = 1 + 2x2 + 8/3 x4 + ... + n! n!/2n! (2x)2n + ... .
and (1 - x2) f '(x) = 1 + x2 + 2/3 x4 + ... + n-1! n-1!/2n! n/2 (2x)2n + ...
= 1 + x2 + 2/3 x4 + ... + 1/4 n-1!n-1!/2n-1! (2x)2n + ... .
Hence the derivative of √(1 - x2) f(x) is 1/√(1 - x2) (- x f(x) + (1 - x2) f '(x) ) = 1/√(1 - x2). So √(1 - x2) f(x) = const + sin-1(x). Putting x = 0, we find that the constant is 0.
© John Scholes
jscholes@kalva.demon.co.uk
11 Mar 2002