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Determine all fixed points of the discrete dynamical system $$x_{n + 1} = e^{x_{n}} - 2 .$$

Determine if they are asymptotically stable.

How can I find the fixed points?

Thank you in advance.

Angela
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  • Check the definitions of "fixed point" and "asymptotically stable" in your notes. Voting to close. – Did Sep 03 '16 at 08:57

2 Answers2

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Hint: Draw the graphs of $e^x$ and $2+x$. Show (using e.g. the IVT and convexity of exp) that there are precisely two intersections and study the derivative of $e^x$ at each intersection point (I don't think there is any closed form expression for these points).

H. H. Rugh
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  • I still do not understand how to do it. May you please give me some more details? – Angela Aug 31 '16 at 08:37
  • Begin by studiying the function $f(x)=e^x - x- 2$ and try to show it vanishes (ivt, convexity...). – anonymus Aug 31 '16 at 11:37
  • Some hints (also reasonable given the graph): The function $f(x)=e^x-x-2$ is strictly convex (why?), $f(0)<0$ and $f$ tends to $+\infty$ when $x$ goes to plus or minus infinity. From this you should deduce (IVT) that there is a positive and a negative zero of $f$ (whence fixed point of your iteration). From convexity that these fixed points are the unique ones. The stability is determined by whether or not the derivative of $e^x-2$ is greater or smaller than one in absolute value which is not so complicated again. – H. H. Rugh Aug 31 '16 at 13:38
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You have an iteration of the form $$ x_{n+1} = f(x_n) $$ with $$ f(x) = e^x - 2 $$ A fixed point $x_n$ would fullfil $$ f(x_n) = x_n \quad (*) $$ A graphical solution is to draw the graphs of $f$ and $\DeclareMathOperator{id}{id}\id$, where $\id(x) = x$ is the identity function.

graphs

Where those graphs intersect, the condition $(*)$ holds, that intersection point is a fixed point.

mvw
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