Chapter 6, Problem 1P

Use simple fixed-point iteration to locate the root of

$f\left(x\right)=\text{sin }\left(\sqrt{x}\right)-x$

Use an initial guess of $x_0=0.5$ and iterate until ${\epsilon }_a\le 0.01\%$. Verify that the process is linearly convergent as described in Box 6.1.

To determine

To calculate: The root of the function $f\left(x\right)=\sin \left(\sqrt{x}\right)-x$ by the use of simple fixed-point iteration with $x_0=0.5$ as the initial condition and iterate until ${\epsilon }_a\le 0.01\%$. Also, verify that the process is linearly convergent.

Answer to Problem 1P

Solution:

The root of the function $f\left(x\right)=\sin \left(\sqrt{x}\right)-x$ is 0.7686.

Explanation of Solution

Given:

The function, $f\left(x\right)=\sin \left(\sqrt{x}\right)-x$.

The initial condition, $x_0=0.5$ and iterate until ${\epsilon }_a\le 0.01\%$.

Formula used:

The simple fixed-point iteration formula for the function $x=g\left(x\right)$,

$x_{i+1}=g\left(x_i\right)$

And, formula for approximate error is,

${\epsilon }_a=\left|\frac{x_{i+1}-x_i}{x_{i+1}}\right|100\%$

Calculation:

Consider the function,

$f\left(x\right)=\sin \left(\sqrt{x}\right)-x$

The function can be formulated as fixed-point iteration as,

$\begin{array}{c}0=\sin \left(\sqrt{x}\right)-x\\ x=\sin \left(\sqrt{x}\right)\\ x_{i+1}=\sin \left(\sqrt{x_i}\right)\end{array}$

Use initial guess of $x_0=0.5$, the first iteration is,

$\begin{array}{c}{x}_{0+1}=\sin \left(\sqrt{x_0}\right)\\ x_1=\sin \left(\sqrt{0.5}\right)\\ =\sin \left(0.7071\right)\\ =0.6496\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{0.6496-0.5}{0.6496}\right|\times 100\%\\ =\left|\frac{0.1496}{0.6496}\right|\times 100\%\\ =\left|0.2303\right|\times 100\%\\ =23.03\%\end{array}$

Use $x_1=0.6496$, the second iteration is,

$\begin{array}{c}{x}_{1+1}=\sin \left(\sqrt{x_1}\right)\\ x_2=\sin \left(\sqrt{0.6496}\right)\\ =\sin \left(0.80598\right)\\ =0.7215\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{0.7215-0.6496}{0.7215}\right|\times 100\%\\ =\left|\frac{0.0719}{0.7215}\right|\times 100\%\\ =\left|0.09965\right|\times 100\%\\ =9.965\%\end{array}$

Use $x_2=0.7215$, the second iteration is,

$\begin{array}{c}{x}_{2+1}=\sin \left(\sqrt{x_2}\right)\\ x_3=\sin \left(\sqrt{0.7215}\right)\\ =\sin \left(0.8494\right)\\ =0.7509\end{array}$

Therefore, the approximate error is,

$\begin{array}{c}{\epsilon }_a=\left|\frac{0.7509-0.7215}{0.7509}\right|\times 100\%\\ =\left|\frac{0.0294}{0.7509}\right|\times 100\%\\ =\left|0.03915\right|\times 100\%\\ =3.915\%\end{array}$

Similarly, all the iteration can be summarized as below,

$i$	$x_i$	${\epsilon }_a=\left|\frac{x_{i+1}-x_i}{x_{i+1}}\right|100\%$
0	0.5
1	0.6496	23.03%
2	0.7215	9.965%
3	0.7509	3.915%
4	0.7621	1.47%
5	0.7662	0.535%
6	0.7678	0.208%
7	0.7683	0.0651%
8	0.76852	0.029%
9	0.7686	0.01%

Since, the approximate error in the ninth iteration is 0.01%. So, stop the iteration.

Hence, the root of the function is 0.7686.

Now, to verify that the process is linearly convergent, the condition to be satisfied is $\left|g^{\prime }\left(x\right)\right|<1$ for $x=0.7686$.

The fixed-point iteration is,

$x_{i+1}=\sin \left(\sqrt{x_i}\right)$

Therefore,

$g\left(x\right)=\sin \left(\sqrt{x}\right)$

Differentiate the above function with respect to x,

$\begin{array}{c}{g}^{\prime }\left(x\right)=\frac{d}{dx}\left[\sin \left(\sqrt{x}\right)\right]\\ =\cos \left(\sqrt{x}\right)\frac{d}{dx}\left(\sqrt{x}\right)\\ =\cos \left(\sqrt{x}\right)\left(-\frac{1}{2\sqrt{x}}\right)\\ =-\frac{\cos \left(\sqrt{x}\right)}{2\sqrt{x}}\end{array}$

Therefore, $\left|g^{\prime }\left(x\right)\right|$ at $x=0.7686$ is,

$\begin{array}{c}\left|g^{\prime }\left(0.7686\right)\right|=\left|-\frac{\cos \left(\sqrt{0.7686}\right)}{2\sqrt{0.7686}}\right|\\ =\left|-\frac{0.6397}{2\times 0.8767}\right|\\ =\left|-0.3648\right|\\ =0.3648\end{array}$

Since, $\left|g^{\prime }\left(0.7686\right)\right|<1$. Hence, it is verified that the process is linearly convergent.