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Section 8.4 Examples and questions

These are additional examples for reviewing the topic we have covered. When reading each example, try to find your own solution before clicking “Answer”. There are also questions for reviewing the concepts of this section.

The equation \(x^5 - 2x = 1\) has a solution in the interval \((1, 2)\text{,}\) as one can see from the Intermediate Value Theorem. Rewrite the equation as a fixed-point problem so that this root is an attracting fixed point.

Solution

The form \((x^5 -1)/ 2 = x\) does not work here because the function \(g(x) = (x^5 -1)/ 2 \) has \(|g'(x)| = 5x^4/2\) which is greater than 1 on the interval \((1, 2)\text{.}\) But \((2x+1)^{1/5} = x\) works because the function \(g(x) = (2x+1)^{1/5} \) has \(|g'(x)| = 2|2x+1|^{-4/5}/5\) which is less than \(2/5\text{.}\)

The function \(g(x) = \tan x\) has infinitely many fixed points, because the graph of tangent intersects the line \(y=x\) infinitely many times. Show that one of the fixed points is neutral and the rest are repelling.

Solution

The neutral point is at 0, because \(\tan 0 = 0\) and \(g'(0)=1\text{.}\) How to find \(g'\) at other fixed points, if we do not know them? Recall that \(g'(x) = \sec^2 x = \tan^2 x + 1 = g(x)^2 + 1\text{.}\) Therefore, if \(x^*\) is a fixed point, then \(g'(x^*) = (x^*)^2 + 1\text{.}\) This expression is strictly greater than one when \(x^* \ne 0\text{.}\) So, all other fixed points are repelling.

Both functions \(\sin x\) and \(\tan x\) have 0 as a neutral fixed point. For one of these functions, the iteration method converges to 0, although very slowly. For another it does not. Why? (Hint: consider the plots of these functions.)

Recall that a sequence \(x_n\) converges to its limit \(x^*\) at a linear rate, the graph of \(\log|x-x_n|\) as a function of \(n\) looks like a line. What will it look like when the rate of convergence is quadratic?