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Okay, here's the question:

Suppose we want to find a solution to $\frac{1} {2} e^{x} - x = 0 $ on the interval [0,1]. Show how to restate this problem as a fixed point problem.

This is the first part of the question, but I think I'll know how to do the second part once I know this. I think I get what is being asked.. We need to find a suitable g(x) that converges. But I can't find a dumbed down version in my notes of the process for this but I know you use MVT or the Contraction Mapping Theorem. It's an exam type question that comes up commonly so if someone could simplify this method or even just show me by example that would be great! :) Thanks in advance!

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  • $\begingroup$ Aren’t you just looking for a fixed point of the function $f(x)=\frac12e^x$ on the interval $[0,1]$? After all, $x$ is a fixed point of $f$ iff $f(x)-x=0$. $\endgroup$ – Brian M. Scott Dec 1 '12 at 0:07
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You are asked to rewrite the equation in the form $g(x)=x$. There is an obvious $g(x)$, which is the one you are expected to write down.

There are also many less obvious ways to rewrite the equation in the form $g(x)=x$. for example, we can rewrite the equation as $e^x=2x$, and then as $x=\ln(2x)$. That is $g(x)=x$ where $g9x)=\ln(2x)$.

Depending on how you choose to rewrite the equation in the form $g(x)=x$, the fixed point iteration may be more efficient, less efficient, or may not work at all. In general, it is good if $|g'(x)|$ is much less than $1$ near the root that we are trying to approximate. .

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