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Fixed Point Iterations

1. The equation x - 3 ln x = 2 has exactly two solutions A and B, with 0 < A < B.
(You do not have to show this.)

(a) Show that A is in [0.5,0.7], and B is in [8.3,8.5].

(b) Consider the following fixed-point iteration for finding a solution of the given
equation: xn+1 = e(1/3(Xn-2)):
Show that if X0 = B+ E, where E is a positive real number, then X1 > X0.
Deduce that this Fixed point iteration does not converge to ¯Before all X0 > B.
(Hint: You may need to use the facts that B is a solution of x-3 ln x = 2,
and that B> 3.)

(c) Consider the following fixed point iteration for finding a solution of the given
Xn+1 = 2 + 3 ln Xn:
(i) Show that this fixed point iteration converges to B for all X0 is in [6.5,9.5].
(ii) Starting with X0 = 8.4, use this fixed point iteration to find B correct to
four significant figures.

Solution Preview

(a) Let f(x) = x - 3lnx - 2, then the zeros of f(x) are the solutions of the
equation x - 3lnx = 2. We have
f(0.5)=0.5794>0, f(0.7)=-0.2300<0, f(8.3)=-0.0488<0, f(8.5)=0.0798>0
By the mean value theorem, f has a zero between 0.5 and 0.7 and another zero
between 8.3 and 8.5. Since 0<A<B, then A is in [0.5,0.7], B is in [8.3,8.5]
(b) The iteration is X(n+1)=exp((1/3)(X(n)-2))
Let g(x)=exp((1/3)(x-2))-x, then g'(x)=(1/3)exp((1/3)(x-2))-1
When x>=8.3, ...

Solution Summary

Fixed Point Iterations are investigated. The response received a rating of "5/5" from the student who originally posted the question.