two fixed point problem

Question

The function $f$ given by $f(x)=x^{x - sin(x)}$ has a fixed point at 1 and another fixed point near 2.

If the I want to estimate fixed pt near 2 using a graph of the function, then:

This is what the function looks like the one. But my estimate is actually 1.0000 if x0=0.5 and by using the fixed point iteration code on Maple: FixedPointIteration(fixedpointiterator = f(x), x = .5, tolerance = 10^(-4), output = sequence, stoppingcriterion = function_value, maxiterations = 30).

So how can I actually estimate the fixed pt near 2 by using a graph of the function? The problem is that if I use x0=2 then Maple stops working.

Answer 1

I am reading that you want to identify values where $f(x)=x$ by visual inspection of the graph. I made this graph for you. You can zoom in to this graph and manually change (play with) the numbers until they line up where you want, and to the visual precision desired, then zoom back out for the look I have shown here. With this method you can effectively get some pretty good approximations using only visual inspection of the graph.

In Maple, you will want to read documentation on aspect ratios. That may help.

Answer 2

The standard error analysis (e.g. http://www-solar.mcs.st-and.ac.uk/~alan/MT2003/Numerical/node11.html) indicates that fixed point iteration will not find the root $r \approx 1.93$, because the derivative $f'(r) \approx 2.73$ satisfies $\left| f'(r) \right| > 1$.

Other root-finding methods (e.g. bisection method, Newton's method, secant method) would likely work. The only caveat is that you should choose good initial values.

Answer 3

As said in other comments and answers, the good candidate for this kind of problem is Newton iterative scheme. By inspection, you noticed that the second fixed point is close to $2$. Then let us make it our guess : $x_{old} = 2$ and define $g(x)=x^{[x - sin(x)]- x}$

Now, apply Newton scheme which write
$x_{new} = x_{old} - \frac{f(x_{old})}{f'(x_{old})}$
Now the iterates are $1.94237$, $1.93469$, $1.93456$. At this point $g(x)=-0.00000555709$. You could continue iterating until you reach the required or desired level of accuracy.