*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 x^{3} + 2·4/3·5 x^{5} + ... + 2·4· ... .2n/(3·5. ... .2n+1) x^{2n+1} + ... = (1 - x^{2})^{-1/2} sin^{-1}x.

**Solution**

(1)

(2) Let f(x) = x + 2/3 x^{3} + 8/15 x^{5} + ... + 1/2 n! n!/2n+1! (2x)^{2n+1} + ... .

Hence x f(x) = x^{2} + 2/3 x^{4} + 8/15 x^{6} + ... + 1/4 n! n!/2n+1! (2x)^{2n+2} + ... .

Also f '(x) = 1 + 2x^{2} + 8/3 x^{4} + ... + n! n!/2n! (2x)^{2n} + ... .

and (1 - x^{2}) f '(x) = 1 + x^{2} + 2/3 x^{4} + ... + n-1! n-1!/2n! n/2 (2x)^{2n} + ...

= 1 + x^{2} + 2/3 x^{4} + ... + 1/4 n-1!n-1!/2n-1! (2x)^{2n} + ... .

Hence the derivative of √(1 - x^{2}) f(x) is 1/√(1 - x^{2}) (- x f(x) + (1 - x^{2}) f '(x) ) = 1/√(1 - x^{2}). So √(1 - x^{2}) 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