Question

Solve ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}$using the Jacobi iterative method with $\mathbf{x}\left(0\right)\mathbf{=}\mathbf{1}$.

Continue until (6.2.2) is satisfied with $\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$. Check using the quadratic

formula.

Short answer

The approximation solution of the quadratic equation 0.27 .

Step-by-step solution

Write the given data from question.

The specific tolerance value,$\mathrm{\epsilon }=0.01$

The initial condition,localid="1655361066633" $\mathrm{x}\left(0\right)=1$

The system equation,

${\mathrm{x}}^2-4\mathrm{x}+1=0$

Determine the solution of system and quadratic equation.

Rewrite the system equation as,

$$\begin{gathered} {\mathrm{x}}^2-4\mathrm{x}+1=0 \\ 4\mathrm{x}={\mathrm{x}}^2+1 \\ \mathrm{x}=\frac{1}{4}{\mathrm{x}}^2+\frac{1}{4} \end{gathered}$$

The system equation can be expressed in Jacobi method as,

$\mathrm{x}\left(\mathrm{n}-1\right)=\frac{1}{4}{\left[\mathrm{x}\left(\mathrm{n}\right)\right]}^2+\frac{1}{4}$

Substitute 1 for$x\left(0\right)$into above equation.

$$\begin{gathered} \mathrm{x}\left(1\right)=\frac{1}{4}{\left(1\right)}^2+\frac{1}{4} \\ \mathrm{x}\left(1\right)=0.5 \end{gathered}$$

The specific tolerance value at$n=0$

$\mathrm{\epsilon }\left(0\right)=\left|\frac{\mathrm{x}\left(0\right)-\mathrm{x}\left(1\right)}{\mathrm{x}\left(0\right)}\right|$

Substitute 1 for x ( 0 ) and 0.5 for x(1) into above equation.

$$\begin{gathered} \mathrm{\epsilon }\left(0\right)=\left|\frac{1-0.05}{0.5}\right| \\ \mathrm{\epsilon }\left(0\right)=0.01 \\ 0.5>0.01 \end{gathered}$$

The second iteration,

$$\begin{gathered} x\left(1+1\right)=\frac{1}{4}{\left[x\left(1\right)\right]}^2+\frac{1}{4} \\ x\left(2\right)=\frac{1}{4}{\left[x\left(1\right)\right]}^2+\frac{1}{4} \end{gathered}$$

Substitute 0.5 for x(1) into above equation.

$$\begin{gathered} \mathrm{x}\left(2\right)=\frac{1}{4}{\left(0.5\right)}^2+\frac{1}{4} \\ \mathrm{x}\left(2\right)=0.3125 \end{gathered}$$

The specific tolerance value at$n=1$.

$\epsilon \left(1\right)=\left|\frac{x\left(1\right)-x\left(2\right)}{x\left(1\right)}\right|$

Substitute 0.3125 for$\mathrm{x}\left(2\right)$and 0.5 for$\mathrm{x}\left(1\right)$into above equation.

$$\begin{gathered} \epsilon \left(1\right)=\left|\frac{0.5-0.3125}{0.5}\right| \\ \epsilon \left(1\right)=0.375 \\ 0.375>0.01 \end{gathered}$$

Repeat the above process until the required precision. The table of iteration is shown

below.

From the above table, it can be seen that at the$4^{th}$iteration the specific tolerance

value is less than$0.01\left[\epsilon \left(4\right)=0.0028<0.01\right]$. The value of x after$4^{th}$teration is

0.2681.

The solution of the quadratic equation. $$\begin{gathered} x=\frac{-\left(-4\right)\pm \sqrt{{\left(-4\right)}^2-4\left(1\right)\left(1\right)}}{2\times 1} \\ x=\frac{4\pm \sqrt{12}}{2} \\ x=2\pm \sqrt{3} \\ x=3.73,0.27 \end{gathered}$$

Hence from the above the approximation solution of the quadratic equation 0.27.