53. (a) For s = 1.8 and x0 = 0.5, we obtain x1 = 0.9, x2 = 0.18, and x3 = 0.324.

(b) Because x0 = 0.5, the larger the s value, the smaller will be x1. From (a) we see x1 can be so small that the next two iterations are insufficient to climb back to the fixed point. So start with perturbing s to 1.6, yielding x1 = 0.8. From the results of (a), we try to make x2 and x3 increase as much as possible, hoping to get close to the fixed point. So, now perturb s to 2, obtaining x2 = 0.4 and x3 = 0.8. Now we find x3 has overshot the fixed point. If we perturbed s back to 1.6 for the computation of x3, we get x3 = .64, very close to the fixed point.

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