Question

Problem 1. (The golden mean] In this problem you will find the exact value of the number 7, often called the golden mean or the golden ratio (sometimes this terminology is used for 7-1). The golden mean is defined by the following expression: 7= 1+- 1+ - 1 1+ 1+... (a) Consider the iteration Xn+1 = f(xn), where x1 = 1, and 1 f(x) = 1+2 1 1+: Recall the following result. Theorem. (i) If the function g : [a, b] → R is continuous and g(x) € [a, b] for all x € [a, b], then g has at least one fixed point in [a, b]. (ii) If, in addition, g'(x) exists on (a,b) and a positive constant k < 1 exists with g'(x)| <k for all x € (a,b), then there is exactly one fixed point in [a, b]. Find an interval [a, b] that is mapped into itself by f and use this theorem to prove the existence of a fixed point of f in [a, b]. (b) Use the above theorem to prove the uniqueness of the fixed point of f. (C) Is the fixed point of f stable or unstable? Hint: Recall Lecture 4. (d) Sketch (by hand) a cobweb plot for the function f to illustrate how the iterates of f behave. (e) Find the exact value of the fixed point of f. (f) How is your result in part (e) related to the value of the golden mean? Explain briefly.

Answer 1

Solution :

(a)

Observe that if x is in [1/2,1], then f(x) = 1/(1+x) is in [1/2,2/3] which is a subset of [1/2,1]. Further, f is continuous on [1/2,1].

Hence, it suffices to consider the interval [1/2,1] as [a,b].

By part (i) of the above theorem, it follows that f has a fixed point in [1/2,1] .

(b)

Observe that f'(x) = - 1/(1+x)² on [1/2,1]. Hence f' exists on (1/2,1). Further, |f'(x)| = 1/(1+x)² 4/9 on (1/2,1).

Hence, it suffices to take k = 4/9 < 1 .

Thus, by part (ii) of the above theorem (which by the way, is called the Banach Fixed Point Theorem), it follows that f has a unique fixed point in [1/2,1].

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