Answer to Question #140244 in Quantitative Methods for Usman

Question #140244
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.
1
Expert's answer
2020-10-26T20:09:05-0400
"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"



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