The polynomial f(x)=230x^4 +18x^3 +9x^2 221x9, (E) has two real roots, one in [1
ID: 3167976 • Letter: T
Question
The polynomial
f(x)=230x^4 +18x^3 +9x^2 221x9, (E)
has two real roots, one in [1,0] (p = 0.0406592883) and the other one in [0,1] (p = 0.962396622)
1. Use the tangent lines to the curve of y = f (x) on [1, 1] to explain why the sequence generated by the Newton’s method with p0=0.5 converges to the root in [1,0].
Newton’s method with p0=0.5 generated with code:
This is Newtons Method
Input the function F(x) in terms of x
230*x^4 + 18*x^3 + 9*x^2 -221*x - 9
Input the derivative of F(x) in terms of x
920*x^3 + 54*x^2 + 18*x - 221
Input initial approximation
0.5
Input tolerance
10^-5
Input maximum number of iterations - no decimal point
1000
Newtons Method
I P F(P)
1 -7.05089820e-01 2.0183630e+02
2 -3.23791114e-01 6.5418427e+01
3 -6.46031310e-02 5.3140071e+00
4 -4.06861512e-02 5.9556165e-03
5 -4.06592883e-02 6.5571477e-09
6 -4.06592883e-02 1.7763568e-15
Approximate solution = -4.0659288316e-02
with F(P) = 1.7763568394e-15
Number of iterations = 6
Tolerance = 1.0000000000e-05
Explanation / Answer
clc;
clear all;
%x0 initial guess
f=@(x)230*x^4 + 18*x^3 + 9*x^2 -221*x - 9;%f function
fp=@(x)920*x^3 + 54*x^2 + 18*x - 221;% fp derivative of f
tol=10^(-5);%tol tolerence
%N max iterations
% f = inline(f);
% fp = inline(fp);
x0=0.5;
n=0;
err=0.1;
p(1)=x0;
N=100;
while (abs(err>tol)& (n<=N))
y1=x0-(f(x0)/fp(x0));%Newton method
err=abs((y1-x0)); %erorr
% if abs(erorr<1e-3)
% break
%end
n=1+n;
x0=y1;
p(n+1)=x0;
end
disp('num_iter p_value F(p)')
disp('_______________________________________________________________________________')
for i=1:n
fprintf('%d %20f %20f ',i ,p(i),f(p(i)))
end
p(end)
f(p(end))
n
num_iter p_value F(p)
_______________________________________________________________________________
1 0.500000 -100.625000
2 -0.705090 201.836304
3 -0.323791 65.418427
4 -0.064603 5.314007
5 -0.040686 0.005956
6 -0.040659 0.000000
ans =
-0.0407
ans =
1.7764e-015
n =
6
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