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        riddles >> medium >> Functions satisfying  2 {f(x)}^2 - f(2x) = 1 
        
(Message started by: Michael_Dagg on Mar 2^nd, 2008, 2:01pm)

Title: Functions satisfying 2 {f(x)}^2 - f(2x) = 1
Post by Michael_Dagg on Mar 2^nd, 2008, 2:01pm

Determine the complex-valued functions f(x) which have power series
expansions that converge near 0 and which satisfy 2 {f(x)}^2 - f(2x) = 1
inside the circle of convergence.

Title: Re: Functions satisfying 2 {f(x)}^2 - f(2x)
Post by pex on Mar 4^th, 2008, 2:56pm

Amazing! If I haven't messed up too badly, [hideb]either
f(x) = -1/2 independent of x,
f(x) = cos(kx) for some k, or
f(x) = cosh(kx) for some k.
The latter two solutions include the constant function f(x) = 1 as the special case k = 0.[/hideb]