Question

Find a root of the following equation by using the Gauss-Seidel

method: (use an initial estimate of x = 2 ) f$\left(x\right)\mathbf{=}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{6}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{9}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{=}\mathbf{0}$.

Short answer

The root of the given equation is$f\left(x\right)=4$.

Step-by-step solution

Write the given data

The given equation is $f\left(x\right)=x^3-6x^2+9x-4=0$; and the initial value is $x\left(0\right)=2$.

Rewrite the equation in Gauss-Seidel form

Write the rearranged equation in below given form,

Divide equation $f\left(x\right)$by$\left(-9x\right)$to get,

$$\begin{gathered} x=-\frac{1}{9}{x}^3+\frac{2}{3}{x}^2+\frac{4}{9} \\ g\left(x\right)=-\frac{1}{9}{x}^3+\frac{2}{3}{x}^2+\frac{4}{9}.........\left(1\right) \end{gathered}$$

Further, write the equation $g\left(x\right)$in Gauss-Seidel form,

$$\begin{gathered} \mathbf{x}\left(n+1\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{9}}{\left[x\left(n\right)\right]}^{\mathbf{3}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{3}}{\left[x\left(n\right)\right]}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{4}}{\mathbf{9}} \\ \mathbf{g}\left(x\left(n\right)\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{9}}{\left[x\left(n\right)\right]}^{\mathbf{3}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{3}}{\left[x\left(n\right)\right]}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{4}}{\mathbf{9}}\mathbf{}.........\left(2\right) \end{gathered}$$

First iteration, determine the value of x( 1)

Substitute the value 0 for n in equation (2).

$$\begin{gathered} \mathrm{g}(\mathrm{x}\left(\mathrm{n}\right))=-\frac{1}{9}{\left[\mathrm{x}\left(\mathrm{n}\right)\right]}^3+\frac{2}{3}{\left[\mathrm{x}\left(\mathrm{n}\right)\right]}^2+\frac{4}{9} \\ \mathrm{x}(0+1)=-\frac{1}{9}{\left[\mathrm{x}\left(0\right)\right]}^3+\frac{2}{3}{\left[\mathrm{x}\left(0\right)\right]}^2+\frac{4}{9} \end{gathered}$$

Further, using the initial value$x\left(0\right)=2$in the given date,

$$\begin{gathered} x\left(1\right)=-\frac{1}{9}{\left[2\right]}^3+\frac{2}{3}{\left[2\right]}^2+\frac{4}{9} \\ =-\frac{8}{9}+\frac{8}{4}+\frac{4}{9} \\ =2.22 \end{gathered}$$

Second iteration, determine the value of x( 2 )

Substitute the value 1 for n in equation (2).

$$\begin{gathered} x\left(n+1\right)=g\left(x\left(n\right)\right)=-\frac{1}{9}{\left[x\left(n\right)\right]}^3+\frac{2}{3}{\left[x\left(n\right)\right]}^2+\frac{4}{9} \\ x\left(1+1\right)=-\frac{1}{9}{\left[x\left(1\right)\right]}^3+\frac{2}{3}{\left[x\left(1\right)\right]}^2+\frac{4}{9} \end{gathered}$$

Further, using the value of $x\left(1\right)=2.22$in the above equation

$$\begin{gathered} x\left(2\right)=-\frac{1}{9}{\left[x\left(2\right)\right]}^3+\frac{2}{3}{\left[x\left(2\right)\right]}^2+\frac{4}{9} \\ =-\frac{1}{9}{\left[2.22\right]}^3+\frac{2}{3}{\left[2.22\right]}^2+\frac{4}{9} \\ =\frac{-10.941}{9}+\frac{9.8568}{3}+\frac{4}{9} \\ =2.51 \end{gathered}$$

Third iteration, determine the value of x ( 3 )

Substitute the value 2 for n in equation (2)

$$\begin{gathered} \mathrm{g}(\mathrm{x}\left(\mathrm{n}\right))=-\frac{1}{9}{\left[\mathrm{x}\left(\mathrm{n}\right)\right]}^3+\frac{2}{3}{\left[\mathrm{x}\left(\mathrm{n}\right)\right]}^2+\frac{4}{9} \\ \mathrm{x}(2+1)=-\frac{1}{9}{\left[\mathrm{x}\left(2\right)\right]}^3+\frac{2}{3}{\left[\mathrm{x}\left(2\right)\right]}^2+\frac{4}{9} \end{gathered}$$

Further, using the value of $\mathrm{x}\left(2\right)=2.51$in the above equation

$$\begin{gathered} x\left(3\right)=-\frac{1}{9}{\left[x\left(3\right)\right]}^3+\frac{2}{3}{\left[x\left(3\right)\right]}^2+\frac{4}{9} \\ =-\frac{1}{9}{\left[2.51\right]}^3+\frac{2}{3}{\left[2.51\right]}^2+\frac{4}{9} \\ =\frac{-15.81}{9}+\frac{12.60}{3}+\frac{4}{9} \\ =2.88 \end{gathered}$$

Now iterate for n = 3,4,5 ….

At the 8^thiteration the value of the root for the equation converges at 4.

Thus, root for the function $f\left(x\right)=x^3-6x^2+9x-4is4$