Answer to Question #141046 in Quantitative Methods for Subhasis

Question

Perform iterations of Newton-Raphson method to approximate a root of the
equation f(x) = x⁴ - x³ + x - 1 = 0, until the roots at successive iterations are
closer than 10⁻⁵
. How many iterations do you need for this much accuracy ?

Answer 1

$$f(x)=x^4-x^3+x-1$$

$$f'(x)=4x^3-3x^2+1$$ $$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$$

Initial solution $x_0 =-0.5$

$$\begin{matrix} n & x_n & f(x_n) & x_{n+1} \\ 0 & -0.5 & -1.3125 & -5.75\\ 1 & -5.75 & 1276.48828125 & -4.26333436 \\ 2 & -4.26333436 & 402.49581545 & -3.15575240 \\ 3 & -3.15575240 & 126.44883333 & -2.33776871 \\ 4 & -2.33776871 & 39.30651893& -1.74669885 \\ 5 & -1.74669885 & 11.89074191 & -1.34320270 \\ 6 & -1.34320270 & 3.33530905 & -1.10675944 \\ 7 & -1.10675944 & 0.74934873 & -1.01421870 \\ 8 & -1.01421870 & 0.08714613 & -1.00029532 \\ 9 & -1.00029532 & 0.00177271 &-1.00000013 \\ 10 & -1.00000013 & 0.00000078 & -1.00000000 \\ \end{matrix}$$

$$\Delta =|x_{n+1}-x_n|$$

$$\Delta =|-5.75-(-0.5)|=5.25$$

$$\Delta =|-4.26333436-(-5.75)|=1.48666564$$

$$\Delta =|-3.15575240-(-4.26333436)|=1.10758196$$

$$\Delta =|-2.33776871-(-3.15575240)|=0.81798369$$

$$\Delta =|-1.74669885-(-2.33776871)|=0.59106986$$

$$\Delta =|-1.34320270-(-1.74669885)|=0.40349615$$

$$\Delta =| -1.10675944-(-1.34320270)|=0.23644326$$

$$\Delta =| -1.01421870-(-1.10675944)|=0.09254074$$

$$\Delta =|-1.00029532-( -1.01421870)|=0.01392338$$

$$\Delta =|-1.00000013-(-1.00029532)|=0.00029519$$

$$\Delta =|-1.00000000-(-1.00000013)|=0.00000013$$

The root is $-1.00000$

It needs 10 iterations for this much accuracy.