Applying The Fixed Point Method

Be sure to review the following pages before looking at the example:

• The Fixed Point Method for Approximating Roots

• The Convergence of The Fixed Point Method

Example 1

Show that there exists a root $\alpha \in (1.5, 2)$ for the function $f(x) = x^3 - 2x - 1$, and use the Fixed Point Method to approximate $\alpha$. Use either $x = g_1(x)$ or $x = g_2(x)$ where $g_1(x) = \frac{1}{2} (x^3 - 1)$ and $g_2(x) = (2x + 1)^{1/3}$ (justify your choice for the convergence of the Fixed Point Method). Use $x_0 = 1.75$ as an initial approximation of $\alpha$ until the accuracy of $\epsilon = 10^{-2}$ is achieved and verify the accuracy in this approximation.

To show that a root $\alpha$ exists in $(1.5, 2)$, we first note that $f(1.5) = (1.5)^3 - 2(1.5) - 1 = -0.625 < 0$ and $f(2) = (2)^3 - 2(2) - 1 = 3 > 0$. Therefore $f(1.5) \cdot f(2) < 0$ and thus a root $\alpha$ exists in $(1.5, 2)$.

We now want to choose either $g_1$ or $g_2$ to apply the Fixed Point Method to. Note that since $g_1(x) = \frac{1}{2} (x^3 - 1)$ then the derivative of $g_1(x)$ is $g_1'(x) = \frac{3}{2} x^2$. We note that $g_1'(x) > 0$ on the interval $[1.5, 2]$ and so $\mid g_1'(x) \mid = g_1'(x)$ on this interval. Furthermore, $g_1'$ is continuous and is an increasing function on $[1.5, 2]$ and thus the maximum is achieved at the right endpoint of this interval, that is $\max_{1.5 ≤x ≤ 2} \mid g_1'(x) \mid = \max_{1.5 ≤ x ≤ 2} g_1'(x) = \frac{3}{2} (2)^2 \mid = 6$.

(1)

\begin{align} \quad \lambda = \max_{1.5 ≤ x ≤ 2} \mid g_1'(x) \mid = 6 > 1 \end{align}

As we see, the choice of $x = g_1(x)$ does not satisfy the condition for convergence. Now note that since $g_2(x) = (2x + 1)^{1/3}$ we have that $g_2'(x) = \frac{2}{3} (2x + 1)^{-2/3}$. Once again we see that $g_2'(x) > 0$ on the interval $[1.5, 2]$ and so $\mid g_2'(x) \mid = g_2'(x)$ on this interval. Furthermore, $g_2'$ is continuous and is a decreasing function on $[1.5, 2]$ and thus the maximum is achieved at the left endpoint of this interval, that is $\max_{1.5 ≤ x ≤ 2} \mid g_2'(x) \mid = \max_{1.5 ≤ x ≤ 2} \mid g_2'(x) \mid = \frac{2}{3} (2(1.5) + 1)^{-2/3} = 0.264566… < 1$ and o $x = g_2(x)$ is a good choice for applying the Fixed Point method.

We will now apply the fixed point method to $x = g_2(x) = (2x + 1)^{1/3}$.

The first iteration using the initial approximation $x_0 = 1.75$ is:

(2)

\begin{align} \quad x_1 = g(x_0) \\ \quad x_1 = (2(1.75) + 1)^{1/3} \\ \quad x_1 = 1.650963… \end{align}

The second iteration using the approximation $x_1 = 1.650963$ is:

(3)

\begin{align} \quad x_2 = g(x_1) \\ \quad x_2 = (2(1.650963) + 1)^{1/3} \\ \quad x_2 = 1.626376… \end{align}

The third iteration using the approximation $x_2 = 1.626376$ is:

(4)

\begin{align} \quad x_3 = g(x_2) \\ \quad x_3 = (2(1.626376) + 1)^{1/3} \\ \quad x_3 = 1.620155… \end{align}

Note that $\mid x_3 - x_2 \mid = \mid -0.006221 \mid = 0.006221 < 0.01 = 10^{-2} = \epsilon$. We will now verify that the root of interest $\alpha$ is between $x_3 - \epsilon$ and $x_3 + \epsilon$ to ensure that $x_3$ is within $\epsilon = 10^{-2}$ of $\alpha$. We note that:

(5)

\begin{align} \quad f(x_3 - \epsilon) = f(1.620155 - 0.01) = f(1.610155) = (1.610155)^3 - 2(1.610155) - 1 = -0.045823... < 0 \\ \quad f(x_3 + \epsilon) = f(1.620155 + 0.01) = f(1.630155) = (1.630155)^3 - 2(1.630155) - 1 = 0.071672… > 0 \end{align}

Thus we have verified that the a root $\alpha$ exists in $[x_3 - \epsilon, x_3 + \epsilon] = [1.610155, 1.630155]$ and so we have verified that $x_3$ is within $\epsilon = 10^{-2}$ of $\alpha$.