Find interval for contraction map

[Unredeemed]

Homework Statement

Find an interval [a, b] for which the Contraction Mapping
Theorem guarantees convergence to the positive fixed point or verify that there is no
such interval.

Homework Equations

$x = g(x) = \frac{14}{13} - \frac{x^{3}}{13}$

The Attempt at a Solution

I know the solution is slightly greater than 2. So, I assumed the upper bound on my interval would be 3 and g(3)=-1. So, I picked my lower bound to be -1 so that the function mapped from [-1,3] to [-1,3]. But, I'm having trouble showing that:

$abs(g'(x)) \leq γ$ for some 0 < γ < 1.

Any help would be massively appreciated.

 

[voko]

The latter inequality is what you should use to find the interval. Find g'(x), and see in what interval the inequality is satisfied.

 

[Unredeemed]

I've found that $abs(g'(x)) < \frac{\sqrt{39}}{3}$

But, this value is less than $\sqrt[3]{14}$ which is obviously the root. So is there no interval?

 

[voko]

Where does $\frac {\sqrt {39} } {9}$ even come from?

Regardless, this is NOT what you want. You need to let $|g'(x)| < 1$ and find the interval satisfying that.

 

[Unredeemed]

Where does $\frac {\sqrt {39} } {9}$ even come from?

Regardless, this is NOT what you want. You need to let $|g'(x)| < 1$ and find the interval satisfying that.

But if we want $|g'(x)| < γ$
And $|g'(x)| = \frac{-3x^{2}}{13}$
So $|\frac{-3x^{2}}{13}| < γ$
And then $|x^{2}| < \frac{13γ}{3}$
Implies $|x| < \sqrt{\frac{13γ}{3}}$
Then $|x| < \frac{\sqrt{39}}{3}$ since $γ < 1$

No?

 

[voko]

You got $|x| < r$, which is what you want. Compare that to #3.

 

[Unredeemed]

Sorry, I don't understand what you mean?

 

[voko]

In #5, you got the interval for x where g'(x) satisfies the contraction map criteria. In #3, you got some bogus inequality.

 

[Unredeemed]

So is my interval $[-\frac{\sqrt{39}}{3},\frac{\sqrt{39}}{3}]$ ?

Because that doesn't contain a root of the equation?

Sorry, I'm very confused.

 

[voko]

$\sqrt[3]{14}$ which is obviously the root

What makes you think so?

 

[Unredeemed]

Well $\frac{14}{13} - \frac{(\sqrt[3]{14})^{3}}{13}=\frac{14}{13}-\frac{14}{13} = 0$

So that's the root?

 

[voko]

That's the zero on the RHS. You still have a non-zero LHS.