# Thread: [SOLVED] Contraction Mapping Principle

1. ## [SOLVED] Contraction Mapping Principle

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

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

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

$\displaystyle \implies x - \sqrt{x} = 1 \implies -x^{\frac{1}{2}} + x - 1 =0$
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.

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

Show that this function has exactly one fixed point.
If this has a fixed point it is a solution of:

$\displaystyle 1+\sqrt{x}=x$

or:

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

now square:

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

Which can be solved using the quadratic formula.

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

$\displaystyle y^2-1-1=0$

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

RoNL

3. Originally Posted by CaptainBlack
If this has a fixed point it is a solution of:

$\displaystyle 1+\sqrt{x}=x$

or:

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

now square:

$\displaystyle x^2-x-1=0$
Hello Captain Black,
I am unsure how you went from this step to the "now square step:"

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

I get

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

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

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

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

4. Originally Posted by Paperwings
Hello Captain Black,
I am unsure how you went from this step to the "now square step:"

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

I get

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

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

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

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

5. Ah, ok. Thank you.