53. (a) For s = 1.8 and x_{0} = 0.5, we obtain x_{1} = 0.9,
x_{2} = 0.18, and x_{3} = 0.324.

(b) Because x_{0} = 0.5, the larger the s value, the smaller will be
x_{1}. From (a) we see x_{1} 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 x_{1} = 0.8. From the results of (a),
we try to make x_{2} and x_{3} increase as much as possible,
hoping to get close to the fixed point. So, now perturb s to 2, obtaining
x_{2} = 0.4 and x_{3} = 0.8. Now we find x_{3} has overshot
the fixed point. If we perturbed s back to 1.6 for the computation of x_{3},
we get x_{3} = .64, very close to the fixed point.

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