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Analysis: fixed point, contraction mapping

  • #1
Let [tex]p,q : \mathbb{C} \to \mathbb{C}[/tex] be
defined by
[tex]
\begin{align*}
p(z) =& z^7 + z^3 - 9z - i, \\
q(z) =& \frac{z^7 + z^3 - i}{9}
\end{align*}
[/tex]

1. Prove that [tex]p[/tex] has a zero at [tex]z_0[/tex] if and only if [tex]z_0[/tex] is a
fixed point for [tex]q[/tex].

If [tex]z_0[/tex] is a fixed point for [tex]q[/tex] then
[tex]
\begin{align*}
q(z_0) = \frac{z_0^7 + z_0^3 - i}{9} =& z_0 \\
z_0^7 + z_0^3 - i =& 9 z_0 \\
z_0^7 + z_0^3 - 9 z_0 - i =& p(z_0) = 0
\end{align*}
[/tex]
Hence [tex]z_0[/tex] is a zero for [tex]p[/tex].

If [tex]p[/tex] has a zero at [tex]z_0[/tex] then
[tex]
\begin{align*}
p(z_0) = z_0^7 + z_0^3 - 9 z_0 - i =& 0 \\
z_0^7 + z_0^3 - i =& 9 z_0 \\
\frac{z_0^7 + z_0^3 - i}{9} =& q(z_0) = z_0
\end{align*}
[/tex]
Hence [tex]z_0[/tex] is a fixed point for [tex]q[/tex].

2. Hence or otherwise show that [tex]p[/tex] has exactly one zero in
the closed unit disk [tex]D = \{ z \in \mathbb{C} : |z| \leq 1 \}[/tex].

How can I solve this?

3. Where are the other zeros?

Does this follow part 2?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
236
0
Show that q is a contraction map from D into D, i.e. that [tex]|q(w)-q(z)|<\alpha|w-z|[/tex] with [tex]\alpha<1[/tex]. Then use a theorem about contraction maps.
 
  • #3
Show that q is a contraction map from D into D, i.e. that [tex]|q(w)-q(z)|<\alpha|w-z|[/tex] with [tex]\alpha<1[/tex]. Then use a theorem about contraction maps.
What metric should I use for complex number space? The question did not mention any metric function at all.
 
  • #4
In [tex]D[/tex] the derivative [tex]q'[/tex] is not less than 1.

[tex]
\begin{align*}
|q'(z)| =
|\frac{7z^6 + 3z^2}{9}| \leq \frac{1}{9}
(|7z^6| + |3z^2|) \leq \frac{1}{9} (7 + 3) = \frac{10}{9}.
\end{align*}
[/tex]

Is [tex]q : D \to D[/tex] is a contraction mapping in D?
 

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