Question

Repeat 6.18using $\mathbf{x}\left(0\right)\mathbf{=}\mathbf{2}$

Short answer

The value of x from the given polynomial equation is$x=-3.61$ with tolerance level$\epsilon =0.001$.

Step-by-step solution

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,$x\left(0\right)=-2$with tolerance level$\mathrm{\epsilon }=0.001$.

Use Newton-Raphson method

The general formula for Newton – Raphson method is,

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

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

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)

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

Step 4:Determine the first iteration, placing initial conditions

For$i=0$,

$$\begin{gathered} x\left(i+1\right)=x\left(i\right)+\frac{-x{\left(i\right)}^3-8x{\left(i\right)}^2-2x\left(i\right)+40}{3x{\left(i\right)}^2+16x\left(i\right)+2} \\ x\left(0+1\right)=x\left(0\right)+\frac{-x{\left(0\right)}^3-8x{\left(0\right)}^2-2x\left(0\right)+40}{3x{\left(0\right)}^2+16x\left(0\right)+2} \\ x\left(1\right)=2+\frac{-{\left(2\right)}^3-8{\left(2\right)}^2-2\left(2\right)+40}{3{\left(2\right)}^2+16\left(2\right)+2} \\ x\left(1\right)=-2+\frac{8-32-+4+40}{12-32+2} \end{gathered}$$

Further solving,

$$\begin{gathered} x\left(1\right)=-2+\frac{8-32-+4+40}{12-32+2} \\ =-2+\frac{20}{-18} \\ =-\frac{56}{18} \\ =-3.11 \end{gathered}$$

Step 5:Determine the second iteration

For $i=1$,

Substitute -3.11 for$X\left(1\right)$in equation (3),

$$\begin{gathered} x\left(i+1\right)=x\left(i\right)+\frac{-x{\left(i\right)}^3-8x{\left(i\right)}^2-2x\left(i\right)+40}{3x{\left(i\right)}^2+16x\left(i\right)+2} \\ x\left(i+1\right)=x\left(1\right)+\frac{-x{\left(1\right)}^3-8x{\left(1\right)}^2-2x\left(1\right)+40}{3x{\left(1\right)}^2+16x\left(1\right)+2} \\ x\left(2\right)=3.11+\frac{-{\left(3.11\right)}^3-8{\left(3.11\right)}^2-2\left(3.11\right)+40}{3{\left(3.11\right)}^2+16\left(3.11\right)+2} \\ x\left(2\right)=3.11+\frac{30.08-8\left(9.67\right)-2\left(3.11\right)+40}{3\left(5.66\right)+16\left(-3.11\right)+2} \end{gathered}$$

Further solving,

$$\begin{gathered} x\left(2\right)=3.11+\frac{30.08-8\left(9.67\right)-2\left(3.11\right)+40}{3\left(5.66\right)+16\left(-3.11\right)+2} \\ =-3.11+\frac{-1.06}{-30.78} \\ =-\frac{94.66}{30.78} \\ =-3.07 \end{gathered}$$

Calculate tolerance level,

$$\begin{gathered} \left|\frac{x\left(2\right)-x\left(1\right)}{x\left(1\right)}\right|<\epsilon \\ \left|\frac{-3.07+3.11}{-3.11}\right|<\epsilon \\ \left|\frac{0.04}{3.11}\right|<\epsilon \\ 0.012<\epsilon \end{gathered}$$

The above value comes down to 0.012 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 becomes less than 0.001.

Using the Newton Raphson method, after 3 iterations, the value of tolerance level becomes less than 0.001, and the final value is achieved to be$x=-3.61$.