Question

Using the starting value 2(1 + i ), solve
x4 − 5x3 + 20x2 − 40x + 60 = 0 by Newton-Raphson method, given that all the roots of the given equation are complex.

Answer 1

$$f(x)=x^4-5x^3+20x^2-40x+60$$

$$f'(x)=4x^3-15x^2+40x-40$$

$$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$$

Initial solution $x_0 =2+2i$

$n=0, x_0=2+2i,$

$f(x_0)=-4,$

$x_1=1.91666667+1,91666667i$

$n=1, x_1=1.91666667+1.91666667i,$

$f(x_1)=-0.23746142-0.13310185i$

$x_2=1.91486064+1.90780412i$

$n=2,x_2=1.91486064+1.90780412i$

$f(x_2)=0.00016433-0.00142961i$

$x_3=1.91490059+1.90777750i$

$n=3,x_3=1.91490059+1.90777750i$

$f(x_3)=0.00000002+0.00000003i$

$x_4=1.91490059+1.90777750i$

$$\varepsilon =|f(x_n)|\cdot 100\%$$

$$\varepsilon =|-4|\cdot 100\%=400\%$$

$$\varepsilon =|-0.23746142-0.13310185i|\cdot 100\%=$$ $$=27.22205511\%$$

$$\varepsilon =|0.00016433-0.00142961i|\cdot 100\%=$$ $$=0.1439023662\%$$

$$\varepsilon =|0.00000002+0.00000003i|\cdot 100\%=$$ $$=0.000003\%$$

The root is $1.91490059+1.90777750i$