$x_1>x_0$ for $x_{n+1} = e^{\frac13(x_n-2)}$

Question

I have this fixed point iteration for finding a solution of the given equation: $x_{n+1} = e^{\frac13(x_n-2)}$.

I have to show that if $x_0=\beta+\epsilon$, where $\epsilon$ is a positive real number, then $x_1>x_0$.

And also to deduce that this fixed point iteration does not converge to $\beta$ for all $x_0>\beta$.

(Hint: You may need to use the facts that $\beta$ is a solution of $x-3lnx=2$, and that $\beta>3$)

OK. My approach is to start from the equation $x-3lnx=2$ and use $\beta$ as a solution. And then I add $\epsilon$ to one side of the equation to arrive at the inequality $x_0<x_1+\epsilon$, which is not what I exactly want.

Any suggestions?

Thanks

score 2 · Accepted Answer · answered Apr 28 '13 at 11:36

2

Consider the function $u:x\mapsto\exp(\frac13(x-2))-x$ then $u(\beta)=0$ and $u'(x)=\frac13\exp(\frac13(x-2))-1$ hence $u'$ is increasing. If $u'(\beta)\gt0$ then $u'(x)\gt0$ for every $x\gt\beta$ hence $u(x)\gt u(\beta)=0$ for every $x\gt\beta$. In particular, if $x_0\gt\beta$ then $x_1=x_0+u(x_0)\gt x_0$.

To show that $u'(\beta)\gt0$, note that $u'(x)\gt0$ is equivalent to $\exp(\frac13(x-2))\gt3$, that is, $x\gt2+3\log3$. But $\beta=2+3\ln\beta$ and one knows that $\beta\gt3$, hence $\beta\gt2+3\ln3$, QED.

answered Apr 28 '13 at 11:36

Did

279,727

Thanks. And the deduction follows from $u(x)$ being increasing. – Bran Apr 28 '13 at 12:45
I approved it as accepted. Obviously, I did not have enough score to upvote. Now I could. – Bran Apr 28 '13 at 13:53
Thanks for the explanation. I had forgotten the reputation limit to be able to upvote. – Did Apr 28 '13 at 13:58

$x_1>x_0$ for $x_{n+1} = e^{\frac13(x_n-2)}$

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