Question

Repeat Problem 6.20 with an initial guess of $\mathbf{x}\left(0\right)\mathbf{=}\mathbf{41}$.

Short answer

The solution of the polynomial is$3.0818$.

Step-by-step solution

Write the given data from the question.

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

Here,$y=7$

Function, $f\left(X\right)=X^4+3X^3-15X^2-19X+30$

Initial condition,$x\left(0\right)=4$

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.

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

The equation to calculate the $J\left(i\right)$is given as follows.

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

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

The tolerance value can be calculated as,

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

localid="1655198617382" $x\left(i+1\right)=X\left(i\right)+\left[\frac{7-{\left(x\left(i\right)\right)}^4-3{\left(x\left(i\right)\right)}^3+15{\left(x\left(i\right)\right)}^2-19x\left(i\right)-30}{4{\left(x\left(i\right)\right)}^3+9{\left(x\left(i\right)\right)}^2-30x\left(i\right)-19}\right]$

(4)

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

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

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

localid="1655199928858" $$\begin{gathered} x\left(1\right)=4+\left[\frac{7-{\left(4\right)}^4-3{\left(4\right)}^3+15{\left(4\right)}^2-19\left(4\right)-30}{4{\left(4\right)}^3+9{\left(4\right)}^2-30\left(4\right)-19}\right] \\ x\left(1\right)=4+\frac{7-256-192+240+76-30}{256+144-120-19} \\ x\left(1\right)=4+\left(\frac{-155}{261}\right) \\ x\left(1\right)=3.4062 \end{gathered}$$

Substitute $3.4062$for $x\left(1\right)$into above equation.

role="math" localid="1655200138359" $$\begin{gathered} x\left(2\right)=3.4062+\left[\frac{7-{\left(3.4062\right)}^4-3\left(4^3\right)+15{\left(3.4062\right)}^2-19\left(3.4062\right)-30}{4{\left(3.4062\right)}^3+9{\left(3.4062\right)}^2-30\left(3.4062\right)-19}\right] \\ x\left(2\right)=3.4062+\frac{7-134.61-118.56+174.03+64.72-30}{158.07+104.42-102.18-19} \\ x\left(2\right)=3.4062+\left(\frac{-37.42}{141.31}\right) \\ x\left(2\right)=3.1414 \end{gathered}$$

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 $3.1414$for $x\left(2\right)$and $3.4062$for $x\left(1\right)$into above equation.

$$\begin{gathered} \epsilon =\left|\frac{3.1414-3.4062}{3.4062}\right| \\ \epsilon =\left|\frac{0.2648}{3.4062}\right| \\ \epsilon =0.0777 \end{gathered}$$

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

Substitute $2$for $i$into above equation (4).

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

Substitute $3.1414$for $x\left(2\right)$into above equation.

$$\begin{gathered} x\left(3\right)=3.1414+\left[\frac{7-{\left(3.1414\right)}^3-3\left(3.1414\right)+15{\left(3.1414\right)}^2-19x\left(3.1414\right)-30}{4{\left(3.1414\right)}^3+9{\left(3.1414\right)}^2-30x\left(3.1414\right)-19}\right] \\ x\left(3\right)=3.1414+\frac{7-97.38-93+148.03+59.68-30}{124+88.82-94.24-19} \\ x\left(3\right)=3.1414+\frac{\left(-5.67\right)}{99.58} \\ x\left(3\right)=3.0845 \end{gathered}$$

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$3.1414$for$x\left(2\right)$and$3.045$for$x\left(3\right)$into above equation.

$$\begin{gathered} \epsilon =\left|\frac{3.0845-3.1414}{3.1414}\right| \\ \epsilon =\left|\frac{0.0569}{3.1414}\right| \\ \epsilon =0.0181 \end{gathered}$$

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

Substitute $3$for $i$into above equation (4).

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

Substitute$3.0845$for$x\left(3\right)$into above equation.

$$\begin{gathered} x\left(4\right)=3.0845+\left[\frac{7-{\left(3.0845\right)}^3-3\left(3.0845\right)+15{\left(3.0845\right)}^2-19x\left(3.0845\right)-30}{4{\left(3.0845\right)}^3+9{\left(3.0845\right)}^2-30x\left(3.0845\right)-19}\right] \\ x\left(4\right)=3.0845+\frac{7-90.53-93+88.04+142.71-30}{117+85.63-92.54-19} \\ x\left(4\right)=3.0845+\frac{\left(-025\right)}{91.47} \\ x\left(4\right)=3.0818 \end{gathered}$$

Calculate the specific tolerance value.

$$\begin{gathered} \epsilon =\left|\frac{x\left(3+1\right)-x\left(3\right)}{x\left(3\right)}\right| \\ \epsilon =\left|\frac{x\left(4\right)-x\left(3\right)}{x\left(3\right)}\right| \end{gathered}$$

Substitute $3.0845$for $x\left(3\right)$and $3.0818$for $x\left(4\right)$into above equation.

$$\begin{gathered} \epsilon =\left|\frac{3.0818-3.0845}{3.0845}\right| \\ \epsilon =\left|\frac{0.0027}{3.845}\right| \\ \epsilon =0.0008 \end{gathered}$$

The specific tolerance value is less than$0.001$. Therefore, the solution of the polynomial is$3.0818$.