math problem 2

IG

for k is greater than .25, any function that is (1): symmetric to the y-axis, ie f(-x)=f(x), and (2): f(x) is continuous for all x between [0,k], and (3): f(0)=f(k) will be continuous for all x. The resulting function is cyclical, with interval(?) k, k^2+k, ((k^2+k)^2)+k,... www.ddhw.com We are given that f(A)=f(A^2+k)=f((A^2+k)^2+k)=... what if A=(A^2+k)? this implies A=(1+(1-4k)^.5)/2 or (1-(1-4k)^.5)/2, so if k is greater than .25, A will never equal to A^2+k. However if k is less than or equal to .25, and (case 1): for all A less than (1+(1-4k)^.5)/2, the sequence, A, A^2+k,((A^2+k)^2+k,... will converge to (1-(1-4k)^.5)/2 since... (I think somehow we prove it if can prove it's monotonic decr/incr and (1-(1-4k)^.5)/2 is the root of A=A^2+k ) This implies that f(x) must be a constant for all x between [0, (1+(1-4k))^.5)/2 since they will all equal to f((1-(1-4k)^.5)/2)www.ddhw.com (case 2): for all A greater than or equal to (1+(1-4k)^.5)/2, if we trace it backward ie f(A)=f((A-k)^.5)=f(((A-k)^.5-k)^.5)=...., the sequence A,(A-k)^.5, ((A-k)^.5-k)^.5,... will converge to (1+(1-4k)^.5)/2, (again, I think somehow we prove it.) This implies that f(x) must be a constant for all x greater than or equal to (1+(1-4k)^.5)/2 since they will all equal to f((1+(1-4k)^.5)/2) from (1) and (2). therefore when k is less than or equal to .25, if f(x) is a continuous function,f(x) = constant for all x.www.ddhw.com