Question

Use Newton-Raphson to find a solution to the polynomial equation $\mathbf{f}\left(x\right)\mathbf{=}\mathbf{y}$where$\mathbf{y}\mathbf{=}\mathbf{0}$and$\mathbf{f}\left(X\right)\mathbf{=}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{8}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{40}$. Start with$\mathit{x}\left(0\right)\mathbf{=}\mathbf{1}$and continue until (6.2.2) is satisfied with$\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{001}$.

Short answer

The value of$x=2.126$with tolerance level$\mathrm{\epsilon }=0.001$

Step-by-step solution

Write the given data

A polynomial equation $f\left(x\right)=y$with$y=0$and$f\left(x\right)=x^{3}8x^2+2x-40$.

Initial value$x\left(0\right)=1$with tolerance level$\mathrm{\epsilon }=0.001$

Use Newton-Raphson method

The general formula for Newton – Raphson method is

$\mathbf{x}\mathbf{(}\mathbf{i}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{x}\mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{+}\mathbf{J}{\mathbf{\left(}\mathbf{i}\mathbf{\right)}}^{\mathbf{-}\mathbf{1}}\mathbf{[}\mathbf{y}\mathbf{-}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{\right)}\mathbf{]}$ ….. (1localid="1655369532681" $\mathrm{Here},\mathbf{J}\mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{=}{\overline{)\frac{\mathbf{df}}{\mathbf{dx}}}}_{\mathbf{x}\mathbf{-}\mathbf{x}\mathbf{\left(}\mathbf{i}\mathbf{\right)}}\mathbf{=}{\mathbf{\left[}\begin{array}{cccc}\frac{{\mathbf{af}}_{\mathbf{1}}}{{\mathbf{ax}}_{\mathbf{1}}}& \frac{{\mathbf{af}}_{\mathbf{1}}}{{\mathbf{ax}}_{\mathbf{2}}}& \mathbf{\dots }& \frac{{\mathbf{af}}_{\mathbf{1}}}{{\mathbf{ax}}_{\mathbf{1}}}\\ \frac{{\mathbf{af}}_{\mathbf{2}}}{{\mathbf{ax}}_{\mathbf{1}}}& \frac{{\mathbf{af}}_{\mathbf{2}}}{{\mathbf{ax}}_{\mathbf{2}}}& & \frac{{\mathbf{af}}_{\mathbf{1}}}{{\mathbf{ax}}_{\mathbf{N}}}\\ \mathbf{⋮}& \mathbf{⋮}& \mathbf{⋮}& \mathbf{⋮}\\ \frac{{\mathbf{af}}_{\mathbf{N}}}{{\mathbf{ax}}_{\mathbf{1}}}& \frac{{\mathbf{af}}_{\mathbf{N}}}{{\mathbf{ax}}_{\mathbf{2}}}& \mathbf{\dots }& \frac{{\mathbf{af}}_{\mathbf{N}}}{{\mathbf{ax}}_{\mathbf{N}}}\end{array}\mathbf{\right]}}_{\mathbf{x}\mathbf{-}\mathbf{x}\mathbf{\left(}\mathbf{i}\mathbf{\right)}}\mathrm{is}\mathrm{the}\mathrm{Jacobian}\mathrm{matrix}.$

Rewrite the polynomial f(x)in general form of Newton-Raphson method

Substitute $(X^3+8x^2+2x-40)$for $f\left(x\right)$in equation (1) and 0 for y as given in the data,

$x\left(i+1\right)=x\left(i\right)+\frac{1}{J\left(i\right)}\left\{0-\left[x{\left(i\right)}^3+8x{\left(i\right)}^2+2x\left(i\right)-40\right]\right\}$ …… (2)

Now calculate $J\left(i\right)$

$$\begin{gathered} J\left(i\right)={\overline{)\frac{df}{dx}}}_{x-x\left(i\right)} \\ ={\overline{)\frac{d\left(x^3+8x^2+2x-40\right)}{dx}}}_{x-x\left(i\right)} \\ ={\left[3x^2+16x+2\right]}_{x-x\left(i\right)} \end{gathered}$$

Now substitute the value of $J\left(i\right)$in equation (2).

localid="1655370002213" $$\begin{gathered} x\left(i+1\right)=x\left(i\right)+\frac{1}{\left[3x^2+16x+2\right]}\left\{0-\left[x{\left(i\right)}^3+8x{\left(i\right)}^2+2x\left(i\right)-40\right]\right\} \\ =x\left(i\right)+\frac{-x\left(i\right)-8x-2x{\left(i\right)}^2+40}{3x{\left(i\right)}^2+16x\left(i\right)+2} \end{gathered}$$ …. (3)

Step 4:First iteration, placing initial conditions.

For$i=1$,

Substitute 1 for$x\left(0\right)$in equation (3).

$$\begin{gathered} x\left(i+1\right)=x\left(i\right)+\frac{-x{\left(i\right)}^3-8x{\left(i\right)}^2-2x\left(i\right)+40}{3x{\left(i\right)}^2+16x\left(i\right)+2} \\ x\left(0+1\right)=x\left(0\right)+\frac{-x{\left(0\right)}^3-8x{\left(0\right)}^2-2x\left(0\right)+40}{3x{\left(0\right)}^2+16x\left(0\right)+2} \\ x\left(1\right)=1+\frac{-{\left(1\right)}^3-8{\left(1\right)}^2-2\left(1\right)+40}{3{\left(1\right)}^2+16\left(1\right)+2} \\ x\left(1\right)=1+\frac{-1-8-2+40}{3+16+2} \end{gathered}$$

Further solving,

$$\begin{gathered} x\left(1\right)=1+\frac{-1-8-2+40}{3+16+2} \\ =1+\frac{29}{21} \\ =\frac{50}{21} \\ =2.38 \end{gathered}$$

Step 5:Solve for the second iteration

For $i=2$,

Substitute 1 for$x\left(0\right)$in equation (3).

$$\begin{gathered} x\left(i+1\right)=x\left(i\right)+\frac{-x{\left(i\right)}^3-8x{\left(i\right)}^2-2x\left(i\right)+40}{3x{\left(i\right)}^2+16x\left(i\right)+2} \\ x\left(i+1\right)=x\left(1\right)+\frac{-x{\left(1\right)}^3-8x{\left(1\right)}^2-2x\left(1\right)+40}{3x{\left(1\right)}^2+16x\left(1\right)+2} \\ x\left(2\right)=1+\frac{-{\left(2.38\right)}^3-8{\left(2.38\right)}^2-2\left(2.38\right)+40}{3{\left(2.38\right)}^2+16\left(2.38\right)+2} \\ x\left(2\right)=1+\frac{-13.48-8\left(5.66\right)-2\left(2.38\right)+40}{3\left(5.66\right)+16\left(2.38\right)+2} \end{gathered}$$

Further solving,

$$\begin{gathered} x\left(2\right)=2.38+\frac{-13.48-8\left(5.66\right)-2\left(2.38\right)+40}{3\left(5.66\right)+16\left(2.38\right)+2} \\ =2.38+\frac{-23.55}{57.06} \\ =\frac{112.25}{57.06} \\ =1.96 \end{gathered}$$

Determine the value of the tolerance.

$$\begin{gathered} \left|\frac{x\left(2\right)-x\left(1\right)}{x\left(1\right)}\right|<\epsilon \\ \left|\frac{1.96-2.38}{2.38}\right|<\epsilon \\ \left|\frac{0.42}{2.38}\right|<\epsilon \end{gathered}$$

The above value comes down to 0.17 which is not less than the value of tolerance level that is 0.001. So go to next iteration process, till the value of calculated $\epsilon$becomes less than 0.001.

Using the Newton Raphson method, after 5 iterations, the value of tolerance level becomes less than 0.001, and the final value is achieved to be$x=1.91$.