Answer to Question #141046 in Quantitative Methods for Subhasis

Question #141046
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 ?
1
Expert's answer
2020-11-02T20:17:37-0500
f(x)=x4x3+x1f(x)=x^4-x^3+x-1

f(x)=4x33x2+1f'(x)=4x^3-3x^2+1xn+1=xnf(xn)f(xn)x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}

Initial solution x0=0.5x_0 =-0.5

nxnf(xn)xn+100.51.31255.7515.751276.488281254.2633343624.26333436402.495815453.1557524033.15575240126.448833332.3377687142.3377687139.306518931.7466988551.7466988511.890741911.3432027061.343202703.335309051.1067594471.106759440.749348731.0142187081.014218700.087146131.0002953291.000295320.001772711.00000013101.000000130.000000781.00000000\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}


Δ=xn+1xn\Delta =|x_{n+1}-x_n|

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

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

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

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

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

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

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

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

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


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

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

The root is 1.00000-1.00000

It needs 10 iterations for this much accuracy.



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