Fixed Points Examples 1
 Table of Contents

# Fixed Points Examples 1

Recall from the Fixed Points page that if we can rewritten $f(x) = 0$ as $x = g(x)$, then if $\alpha$ is such that $\alpha = g(\alpha)$, then $\alpha$ is called a fixed point of $g$ and consequently will be a root of $f$.

We will now look at some examples regarding fixed points.

## Example 1

Find three different ways of writing $f(x) = x^3 + 2x^2 - x + 4 = 0$ in the fixed point iteration form $x = g_1(x)$, $x = g_2(x)$, and $x = g_3(x)$ where $g_1$, $g_2$, and $g_3$ are obtained by isolating $x$, $x^2$, and $x^3$ respectively.

We isolate $x$ in the equation above to immediately get $f(x) = 0$ into the form $x = g_1(x)$.

(1)
\begin{align} \quad x = x^3 + 2x^2 + 4 \end{align}

Therefore we have that $g_1(x) = x^3 + 2x^2 + 4$. We now isolate the $x^2$ in the equation above and square root both sides to obtain that:

(2)
\begin{align} \quad 2x^2 = -x^3 + x - 4 \\ \quad x^2 = \frac{-x^3 + x - 4}{2} \\ \quad x = \left ( \frac{-x^3 + x - 4}{2} \right )^{1/2} \end{align}

Therefore we have that $g_2(x) = \left ( \frac{-x^3 + x - 4}{2} \right )^{1/2}$. Lastly, we isolate the $x^3$ in the equation above and then cube root both sides to obtain that:

(3)
\begin{align} \quad x^3 = -2x^2 + x - 4 \\ \quad x = \left ( -2x^2 + x - 4 \right )^{1/3} \end{align}

Therefore we have that $g_3(x) = \left ( -2x^2 + x - 1 \right )^{1/3}$.

## Example 2

Find three different ways of writing $f(x) = x^4 +2x^2 - x - 1 = 0$ in the fixed point iteration form $x = g_1(x)$, $x = g_2(x)$, and $x = g_3(x)$ where $g_1$, $g_2$, and $g_3$ are obtained by isolating $x$, $x^2$, and $x^4$ respectively.

We isolate $x$ in the equation above to immediately get $f(x) = 0$ into the form $x = g_1(x)$.

(4)
\begin{align} \quad x = x^4 + 2x^2 - 1 \end{align}

Therefore we have that $g_1(x) = x^4 + 2x^2 - 3$. We now isolate the $x^2$ in the equation above and square root both sides to obtain that:

(5)
\begin{align} \quad -2x^2 = x^4 - x - 1 \\ \quad x^2 = \frac{x^4 - x - 1}{-2} \\ \quad x = \left ( \frac{x^4 - x - 1}{-2} \right )^{1/2} \end{align}

Therefore we have that $g_2(x) = \left ( \frac{x^4 - x - 1}{-2} \right )^{1/2}$. Lastly, we isolate the $x^4$ in the equation above and then fourth root both sides of the equation to obtain that:

(6)
\begin{align} \quad x^4 = -2x^2 +x + 1 \\ \quad x = \left ( -2x^2 + x + 1 \right )^{1/4} \end{align}

Therefore we have that $g_3(x) = \left ( -2x^2 + x + 1 \right )^{1/4}$.

## Example 3

Show that $g(x) = 2^{-x}$ has a unique fixed point on the interval $\left [ \frac{1}{3} , 1 \right ]$.

We first note that $g(x)$ is a strictly decreasing function on the interval $\left [ \frac{1}{3}, 1 \right ]$. Since $g$ is continuous on $\left [ \frac{1}{3}, 1 \right ]$, we have by the Extreme Value Theorem that $g$ attains a maximum of $g \left ( \frac{1}{3} \right ) \approx 0.7937...$ and a minimum of $g(1) = \frac{1}{2}$. Note that thus for all $x \in \left [ \frac{1}{3} , 1 \right ]$ we have that $g(x) \in \left [ \frac{1}{3} , 1 \right ]$. Thus there exists a fixed point $\alpha$ in this interval.

Now we need to guarantee the uniqueness of $\alpha$. We note that $g$ is differentiable on $\left ( \frac{1}{3}, 1 \right )$ and that:

(7)
\begin{align} \quad g'(x)= - \ln (2) 2^{-x} \end{align}

$g'$ is a strictly increasing continuous function and obtains a maximum and minimum value on $\left [ \frac{1}{3}, 1 \right ]$. We therefore we have that the minimum value on this interval is $g' \left ( \frac{1}{3} \right ) \approx -0.5501...$ and the maximum value on this interval is $g'(1) = -0.4901...$. Therefore for all $x \in \left ( \frac{1}{3}, 1 \right )$ we have that:

(8)
\begin{align} \mid g'(x) \mid ≤ 0.5501... \end{align}

Therefore the fixed point $\alpha$ is unique.

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