[SOLVED] Contraction Mapping Principle

**Paperwings** · July 31st 2008, 04:25 PM

Problem:
1.Show that none of the following mappings $f:X \rightarrow X$ have a fixed point and explain why the contraction mapping principle is not contradicted.
a. $(0,1) \subseteq R$ and $f(x)=\frac{x}{2}$
b. X = R and $f(x)=x+1$ for x in X.
c. ${(x,y) \in R^2 | x^2+y^2=1}$ and $f(x,y)=(-y,x)$ for (x,y) in X

2. Define the function $f:[1, \infty) \rightarrow R$ by
$f(x)=1+\sqrt{x}$ for $x \geq 1$
Show that this function has exactly one fixed point.
================================================== ======================
Attempt:
Problem 1:
By a definition a point x in X is a fixed point for the mapping $f:X \rightarrow X$ provided that $f(x) = x$
For (a) and (b) is it sufficient to show that $x \neq \frac{x}{2}$ and $x \neq x+1$ ?
For (c), the only fixed point would be $x= \frac{\sqrt{2}}{2},y= \frac{\sqrt{2}}{2}$ , but $f \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right) \neq f \left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$

Problem 2:
If this function has a fixed point, then $f(x)=x= 1 +\sqrt{x}$

$\implies x - \sqrt{x} = 1 \implies -x^{\frac{1}{2}} + x - 1 =0$
I should slap myself, because I can't figure a way to solve for x. First glance, I thought of using the quadratic formula, but I can't do that.

Thank you for your time.

**CaptainBlack** · July 31st 2008, 10:34 PM

Originally Posted by Paperwings

2. Define the function $f:[1, \infty) \rightarrow R$ by
$f(x)=1+\sqrt{x}$ for $x \geq 1$

Show that this function has exactly one fixed point.

If this has a fixed point it is a solution of:

$1+\sqrt{x}=x$

or:

$\sqrt{x}=x-1$ ,

now square:

$x^2-3x-1=0$

Which can be solved using the quadratic formula.

However we can show that this has exactly one real root $>1$ using Descartes rule of signs after the change of variables $y=x-1$ , which gives:

$y^2-1-1=0$

which by Descartes rule of signs has exactly one positive root.

RoNL

**Paperwings** · August 1st 2008, 07:01 AM

Originally Posted by CaptainBlack

If this has a fixed point it is a solution of:

$1+\sqrt{x}=x$

or:

$\sqrt{x}=x-1$ ,

now square:

$x^2-x-1=0$

Hello Captain Black,
I am unsure how you went from this step to the "now square step:"

Since
$\sqrt{x}=x-1$ , then by squaring both sides

I get

$\left( \sqrt{x}\right) ^2 = \left( x-1 \right) ^2$

$\implies x = x^2 - 2x + 1 \implies x^2-3x+1=0$

By the quadratic formula, then $x= \frac{3 \pm \sqrt{5}}{2}$ , which are two real roots.

But since the function is defined as $x \geq 1$ , then the only fixed point is $\frac{3+\sqrt{5}}{2}$

**CaptainBlack** · August 1st 2008, 07:47 AM

Originally Posted by Paperwings

Hello Captain Black,
I am unsure how you went from this step to the "now square step:"

Since
$\sqrt{x}=x-1$ , then by squaring both sides

I get

$\left( \sqrt{x}\right) ^2 = \left( x-1 \right) ^2$

$\implies x = x^2 - 2x + 1 \implies x^2-3x+1=0$

By the quadratic formula, then $x= \frac{3 \pm \sqrt{5}}{2}$ , which are two real roots.

But since the function is defined as $x \geq 1$ , then the only fixed point is $\frac{3+\sqrt{5}}{2}$

Typing trouble, I'm pretty sure at some point that is what I had, but the post was lost, and I had to retype. I guess the error crept in at that point. The Decartes rule of signs argument still appliers though.

RonL

**Paperwings** · August 1st 2008, 07:45 PM

Ah, ok. Thank you.

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