How can Vieta's methods be applied to determine all the roots of the polynomial f(x)=x^3−3x^2−1after the Newton-Raphson method has already obtained the root x=3.1038 with a precision of 10−4?
How can Vieta's methods be applied to determine all the roots of the polynomial f(x)=x^3−3x^2−1after the Newton-Raphson method has already obtained the root x=3.1038 with a precision of 10−4?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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How can Vieta's methods be applied to determine all the roots of the polynomial f(x)=x^3−3x^2−1after the Newton-Raphson method has already obtained the root x=3.1038 with a precision of 10−4?
Transcribed Image Text:To Find: Roots of polynomial to a precision of 10-4
Given f(x) = x²³-3x²-1. ε = 10-4
E
Using Newton-Raphson method,
f(x)
Formula used:
Хан Хи-
1
1
ха
f (xn)
1st iteration: n =o
x₁ = x₁= f (xo)
f'(xo)
f'(x)
= x3-
3rd iteration: n=2
x3 = x₂- f (x₂)
f'(x₂)
f(3)= -1 <0
f (4) = 15>0
Root lies between 3 and 4
3+4
3,5
xo
2nd iteration: n=1
x₂ = x₁ - f (x₁) = 3.1146-
f'(x₁)
: 3.5- f (3.5) 3.5- 5.125
15.75
f (3.5)
4th iteration; n = 3;
f (x3)
f'(x3)
f(x) = x³ 3x²-1
f'(x) = _d (x²³²-3x²-1) = 3x²³²-6x
dx
= 3.10669-
f(3.1746)
f(3-1746)
3
= 3-10381 -
= 3.10669
f(3.10669)
f'(3.10669)
f(3.10381)
f'(3.10381)
= 3.1746
= 3.10381
= 3.1038
114-31 = 13.1038-3.10381 | ≤ 10 = E
+
Root obtained with accuracy & = 10 + after 4
iterations, x ~3.1038.
:: Root of polynomial, ~3.1038.
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