# Fixed Point Iteration $x = g(x)$ method for $y_1 = e ^{-x}$ and $y_2= \cos x$

Find the x and y coordinates of the intersection points by means of the $x = g(x)$ method. ( I believe they are referring to the Fixed Point Iteration method)

I would greatly appreciate it if anyone could help me get from the question part to the answer part.

I think it would start something like this:

And then one would have to somehow get x alone??

• The curves intersect and a countable number of points (not two). Is there some additional restriction. – copper.hat Jan 27 '15 at 18:10

For Fixed Point Iteration, we want to get the form:

$$x = g(x)$$

This can produce several results where some converge and some do not converge. Do you know the rules of when you get convergence or divergence? The first step is to figure out how to get the given problem into this desired form.

Case 1:

$$\cos x = e^{-x} \implies \cos^{-1}( \cos x) = \cos^{-1}(e^{-x}) \implies x = \cos^{-1}(e^{-x})$$

Case 2:

$$e^{-x} = \cos x \implies -x \ln(e) = \ln(\cos x) \implies x = -\ln(\cos x)$$

This gives two test cases to see if Fixed Point Iteration converges, thus the two iteration results are:

$$x_{n+1} = \cos^{-1}(e^{-x_n})~~ \mbox{or} ~~ x_{n+1} = -\ln(\cos x_n)$$

Perhaps they are giving you a range to test it, for example, see the plot using Wolfram Alpha, but I will leave those results to you. If there is not some restriction of the range, there are many intersection points, but you should graph and test the two iteration formulas, plus numerically solve the original to compare results.

• Thank you so much, Amzoti!! – FranB Jan 27 '15 at 18:10