Determine an interval $[a,b]$ on which the fixed-point ITERATION will converge.

$x = g(x) = (2 - e^x + x^2)/3$

I've determined that $g'(x) = (2x -e^x)/3$, but I don't know how to determine the interval without the guess-and-check method (which is not what I'm supposed to do). The question provided a hint saying "You may need some calculus concepts to work this out." However, I haven't the slightest as to which concepts it's referring to. I have the Fixed Point Convergence Theorem sitting next to me, but it is not offering much guidance.


  • $\begingroup$ Maybe this is just me, but the statement "...on which the fixed-point interval will converge" isn't clear to me. Are you iterating this function and look for an iterative fixed point? Same for "Fixed Point Convergence Theorem." Is this a calculus version of Banach? Or what exactly has your teacher called like this? If yes, you probably are supposed to check where the derivative is bounded by $1$ (then use it as a Lipshitz constant). But I might get this all wrong. $\endgroup$ – gnometorule Jan 27 '13 at 4:47

Hint: Find the interval on which $\vert g^\prime(x)\vert<1$.


From the Fixed Point Theorem, we suppose that $g'$ exists on $(a, b)$ with

$$|g'(x)| \le k \lt 1$$

for all $x \in (a, b)$

Can you choose a $k \lt 1$ and $[a, b]$ based on the derivative you found?

For example, $k = \frac{1}{2}$ and $[a, b] = [-1, 2]$.

Check if these conditions satisfy the FPT.

Do you see that you have flexibility in these choices as long as the FPT conditions are met?

The theorem states, that if you meet those conditions, any of the choices you made will converge to the fixed point.

Now, can you perform the algorithm and find the fixed point with the above information?

You could also plot using WA to see this.

Lastly, play around with first and second derivatives (what do those tell you and how can those help you), various values of $a, b, k$ and see you can you use all sorts of information to guide you.


  • $\begingroup$ $\large \uparrow \;+ 1$ $\endgroup$ – Namaste May 6 '13 at 0:32
  • $\begingroup$ No one the responded heard back, so never sure if theyconverged. $\endgroup$ – Amzoti May 6 '13 at 0:41

Looking at a plot of $g$ you see that $$g(0)={1\over 3}>0\ ,\qquad g(1)=(3-e)/3<1\ .$$ Therefore the equation $g(x)=x$ has a solution in $\ ]0,1[\ $, by the intermediate value theorem. We now try to prove that this solution can be obtained applying the fixed point theorem to $g$ and the interval $[0,1]$.

As $$g'(x)={2x-e^x\over 3}\leq{2x-(1+x)\over3}\leq0\qquad(0\leq x\leq 1)\qquad(*)$$ it follows that $g$ is monotonically decreasing on $[0,1]$ from $g(0)={1\over3}<1$ to $g(1)>0$. Therefore $g$ maps the interval $[0,1]$ into itself.

It remains to prove that $g'(x)\geq -q$ on $[0,1]$ for some $q<1$. But from $(*)$ we can conclude that $$g'(x)\geq -q:=-{e\over3}\qquad(0\leq x\leq 1)\ ,$$ and as $e<3$ we are done.


A related problem. Use the mean value theorem as

$$ |x_{n+1}-x_{n}|=|g(x_n)-g(x_{n-1})|=|g'(\eta)||x_n-x_{n-1}|, $$

where $x_{n+1}=g(x_n)$. From the above, to have a contraction you need to impose the condition $|g'(x)|<1$. To solve for $x$ and find the interval, you can have a solution in terms of the Lambert W function.


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