Question

Use Newton-Raphson to find one solution to the polynomial equation $\mathbf{f}\left(x\right)\mathbf{=}\mathbf{y}$where$\mathbf{y}\mathbf{=}\mathbf{7}$and$\mathbf{f}\left(x\right)\mathbf{=}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{15}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{19}\mathbf{x}\mathbf{+}\mathbf{30}$. Start with$\mathbf{x}\left(0\right)\mathbf{=}\mathbf{0}$ and continue until (6.2.2) is satisfied with$\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{001}$.

Short answer

The solution of the polynomial is$0.8042$.

Step-by-step solution

Write the given data from the question.

The polynomial equation,$f\left(x\right)=y$

Where $y=7$

Function, $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4+3\mathrm{x}-15{\mathrm{x}}^2-19\mathrm{x}+30$

Initial condition,$x\left(0\right)=0$

Specific tolerance value,$\epsilon =0.001$

Determine the formula to calculate the solution of the polynomial equation by using Newton-Raphson method.

The generalise form of the Newton-Raphson method is given as follows.

$\mathbf{x}\left(i+1\right)\mathbf{=}\mathbf{x}\left(i\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{J}\left(i\right)}\left[y-f\left(x\left(i\right)\right)\right]$ …… (1)

The equation to calculate the is given as follows.

$\mathbf{J}\left(i\right)\mathbf{=}{\overline{)\frac{\mathbf{df}}{\mathbf{dx}}}}_{\mathbf{x}\mathbf{=}\mathbf{xi}}$ …… (2)

The equation to calculate the specific tolerance value is given as follows.

The tolerance value can be calculated as,

$\left|\frac{x(i+1)-x\left(i\right)}{x\left(i\right)}\right|\mathbf{<}\mathbf{\epsilon }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{for}\mathbf{}\mathbf{all}\mathbf{}\mathit{i}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{N}$ ……. (3)

Calculate the solution of the polynomial by using Newton Raphson method.

Calculate the value of $J\left(i\right)$.

Substitute $x^4+3x^3-15x^2-19x+30$for f into equation (2).

$$\begin{gathered} J\left(i\right)={\overline{)\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^4+3{\mathrm{x}}^3-15{\mathrm{x}}^2-19\mathrm{x}+30\right)}}_{\mathrm{x}-\mathrm{xi}} \\ J\left(i\right)={\left[4x^3+9x^2-30x-19\right]}_{\mathrm{x}-\mathrm{xi}} \end{gathered}$$

Use the general form of Newton Raphson method equation.

Substitute value of $j\left(i\right)$, $y$and $f\left(x\right)$ into equation (1).

localid="1655195094255" $x\left(i+1\right)=X\left(i\right)+\left[\frac{7-{\left(x\left(i\right)\right)}^4-3{\left(x\left(i\right)\right)}^3+15{\left(x\left(i\right)\right)}^2-19x\left(i\right)-30}{4{\left(x\left(i\right)\right)}^3+9{\left(x\left(i\right)\right)}^2-30x\left(i\right)-19}\right]$ ……

(4)

Substitute $0$for $\left(i\right)$into above equation (4).

localid="1655195103516" $$\begin{gathered} x\left(0+1\right)=x\left(0\right)+\left[\frac{7-{\left(x\left(0\right)\right)}^4-3{\left(x\left(0\right)\right)}^3+15{\left(x\left(0\right)\right)}^2-19x\left(0\right)-30}{4{\left(x\left(0\right)\right)}^3+9{\left(x\left(0\right)\right)}^2-30x\left(0\right)-19}\right] \\ x\left(1\right)=x\left(0\right)+\left[\frac{7-{\left(x\left(0\right)\right)}^4-3{\left(x\left(0\right)\right)}^3+15{\left(x\left(0\right)\right)}^2-19xx\left(0\right)-30}{4{\left(x\left(0\right)\right)}^3+9{\left(x\left(0\right)\right)}^2-30x\left(0\right)-19}\right] \end{gathered}$$

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

$$\begin{gathered} x\left(1\right)=0+\left[\frac{7-{\left(0\right)}^4-3{\left(0\right)}^3+15{\left(0\right)}^2-19\left(0\right)-30}{4{\left(0\right)}^3+9{\left(0\right)}^2-30\left(0\right)-19}\right] \\ x\left(1\right)=\frac{7-30}{-19} \\ x\left(1\right)=\frac{-23}{-19} \\ x\left(1\right)=1.2105 \end{gathered}$$

Substitute $1$for $i$into above equation (4).

$$\begin{gathered} x\left(1+1\right)=x\left(1\right)+\left[\frac{7-{\left(x\left(1\right)\right)}^4-3{\left(x\left(1\right)\right)}^3+15{\left(x\left(1\right)\right)}^2-19x\left(1\right)-30}{4x{\left(1\right)}^3+9x{\left(1\right)}^2-30x\left(1\right)-19}\right] \\ x\left(2\right)=x\left(1\right)+\left[\frac{7-{\left(x\left(1\right)\right)}^4-3{\left(x\left(1\right)\right)}^3+15{\left(x\left(1\right)\right)}^2-19x\left(1\right)-30}{4x{\left(1\right)}^3+9x{\left(1\right)}^2-30x\left(1\right)-19}\right] \end{gathered}$$

Substitute $1.2105$for $x\left(1\right)$into above equation.

$$\begin{gathered} x\left(2\right)=1.2105+\left[\frac{7-{\left(1.2105\right)}^3-3\left(1.2105\right)+15{\left(1.2105\right)}^2-19x\left(1.2105\right)-30}{4{\left(1.2105\right)}^3+9{\left(1.2105\right)}^2-30x\left(1.2105\right)-19}\right] \\ x\left(2\right)=1.2105+\frac{7-2.4171-5.3212+21.979+22.99-30}{7.095+13.1877-36.32-19} \\ x\left(2\right)=1.2105+\frac{14.2307}{-35.0373} \\ x\left(2\right)=0.8043 \end{gathered}$$

Calculate the specific tolerance value.

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

Substitute $0.8043$for $x\left(2\right)\mathrm{and}1.2105\mathrm{for}\mathrm{x}\left(1\right)$into above equation.

$$\begin{gathered} \epsilon =\left|\frac{0.80443-1.2105}{1.2105}\right| \\ \epsilon =\left|\frac{0.4062}{1.2105}\right| \\ \epsilon =0.3355 \end{gathered}$$

The specific tolerance value is greater than $0.001$. So, go for the next iteration.

Substitute for $i$into above equation (4).

$$\begin{gathered} x\left(2+1\right)=x\left(\right)+\left[\frac{7-{\left(x\left(2\right)\right)}^4-3{\left(x\left(2\right)\right)}^3+15{\left(x\left(2\right)\right)}^2-19x\left(2\right)-30}{4x{\left(2\right)}^3+9x{\left(2\right)}^2-30x\left(2\right)-19}\right] \\ x\left(2\right)=x\left(1\right)+\left[\frac{7-{\left(x\left(2\right)\right)}^4-3{\left(x\left(2\right)\right)}^3+15{\left(x\left(2\right)\right)}^2-19x\left(2\right)-30}{4x{\left(2\right)}^3+9x{\left(2\right)}^2-30x\left(2\right)-19}\right] \end{gathered}$$

Substitute $0.8043$for $x\left(2\right)$into above equation.

$$\begin{gathered} x\left(3\right)=0.8043+\left[\frac{7-{\left(0.8043\right)}^3-3\left(0.8043\right)+15{\left(0.8043\right)}^2-19x\left(0.8043\right)-30}{4{\left(0.8043\right)}^3+9{\left(0.8043\right)}^2-30x\left(0.8043\right)-19}\right] \\ x\left(3\right)=0.8043+\frac{7-0.4285-1.5609+9.7035+15.2817-30}{2.0812+5.8220-24.129-19} \\ x\left(3\right)=0.8043+\frac{0.0058}{-35.2258} \\ x\left(3\right)=0.8042 \end{gathered}$$

Calculate the specific tolerance value.

$$\begin{gathered} \epsilon =\left|\frac{x\left(2+1\right)-x\left(2\right)}{x\left(2\right)}\right| \\ \epsilon =\left|\frac{x\left(3\right)-x\left(2\right)}{x\left(2\right)}\right| \end{gathered}$$

Substitute $0.8043$for $x\left(2\right)$and $0.8042$for $x\left(3\right)$into above equation.

$$\begin{gathered} \epsilon =\left|\frac{0.8042-0.8043}{0.8043}\right| \\ \epsilon =\left|\frac{0.0001}{0.8043}\right| \\ \epsilon =1.243 \end{gathered}$$

The specific tolerance value is lessthan$0.001$. Therefore, the solution of the polynomial is$0.8042$.