Question

The fourth-degree polynomial

f(x) = 230x4 + 18x3 + 9x2 - 221x - 9

has two real zeros, one in [-1; 0] and the other in [0; 1]. Attempt to approximate these zeros to within

10-2 using the

(a)

Secant method(Use the endpoints of each interval as the initial approximations),

(b)

Newtons method(Use the midpoints of each interval as the initial approximation)

Answer 1

$f(x)=230x^4+18x^3+9x^2-221x-9\\ \left[-1,0\right] , \left[0,1\right]$

a)

$a_0=-1, b_0=0\\ f(a_0)=f(-1)=230-18+9+221-9=433\\ f(b_0)=f(0)=-9\\ p_2=a_0-\frac{b_0-a_0}{f(b_0)-f(a_0)}f(a_0)=\\ =-1-\frac{0+1}{-9-433}\cdot433=-0.0204\\ a_1=-0.0204, b_1=0\\ f(a_1)=f(-0.0204)=\\ =230(-0.0204)^4+18(-0.0204)^3+\\ +9(-0.0204)^2-221(-0.0204)-9=-4.488\\ f(b_1)=f(0)=-9\\ p_3=a_1-\frac{b_1-a_1}{f(b_1)-f(a_1)}f(a_1)=\\ =-0.0204-\frac{0+0.0204}{-9+4.488}\cdot(-4.488)=-0.0408\\ |p_3-p_2|=0.0204\\ a_2=-0.0408, b_2=0\\ f(a_2)=f(-0.0408)=\\ =230(-0.0408)^4+18(-0.0408)^3+\\ +9(-0.0408)^2-221(-0.0408)-9=0.034\\ f(b_2)=f(0)=-9\\ p_4=a_2-\frac{b_2-a_2}{f(b_2)-f(a_2)}f(a_2)=\\ =-0.0408-\frac{0+0.0408}{-9-0.034}\cdot(0.034)=-0.0408\\ |p_4-p_3|=0\\ x=-0.0408$

$a_0=0, b_0=1\\ f(a_0)=f(0)=-9\\ f(b_0)=f(1)=230+18+9-221-9=27\\ p_2=a_0-\frac{b_0-a_0}{f(b_0)-f(a_0)}f(a_0)=\\ =0-\frac{1-0}{27+9}\cdot(-9)=0.25\\ a_1=0.25, b_1=1\\ f(a_1)=f(0.25)=\\ =230(0.25)^4+18(0.25)^3+\\ +9(0.25)^2-221(0.25)-9=-62.508\\ f(b_1)=f(1)=27\\ p_3=a_1-\frac{b_1-a_1}{f(b_1)-f(a_1)}f(a_1)=\\ =0.25-\frac{1-0.25}{27+62.508}\cdot(-62.508)=0.771\\ |p_3-p_2|=0.521\\ a_2=0.771, b_2=1\\ f(a_2)=f(0.771)=\\ =230(0.771)^4+18(0.771)^3+\\ +9(0.771)^2-221(0.771)-9=-84.519\\ f(b_2)=f(1)=27\\ p_4=a_2-\frac{b_2-a_2}{f(b_2)-f(a_2)}f(a_2)=\\ =0.771-\frac{1-0.771}{27+84.519}\cdot(-84.519)=0.945\\ |p_4-p_3|=0.174\\$

$a_3=0.945, b_3=1\\ f(a_3)=f(0.945)=\\ =230(0.945)^4+18(0.945)^3+\\ +9(0.945)^2-221(0.945)-9=-11.194\\ f(b_3)=f(1)=27\\ p_5=a_3-\frac{b_3-a_3}{f(b_3)-f(a_3)}f(a_3)=\\ =0.945-\frac{1-0.945}{27+11.194}\cdot(-11.194)=0.961\\ |p_5-p_4|=0.016\\$

$a_4=0.961, b_4=1\\ f(a_4)=f(0.961)=\\ =230(0.961)^4+18(0.961)^3+\\ +9(0.961)^2-221(0.961)-9=-0.929\\ f(b_4)=f(1)=27\\ p_6=a_4-\frac{b_4-a_4}{f(b_4)-f(a_4)}f(a_4)=\\ =0.961-\frac{1-0.961}{27+0.929}\cdot(-0.929)=0.962\\ |p_6-p_5|=0.001\\ x=0.962$

b)

$f(x)=230x^4+18x^3+9x^2-221x-9\\ f'(x)=920x^3+54x^2+18x-221\\ a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)}\\ a_0=-1\\ f(-1)=433\\ f'(-1)=-920+54-18-221=-1105\\ a_1=a_0-\frac{f(a_0)}{f'(a_0)}=-1-\frac{433}{-1105}=-0.608\\ f(a_1)=f(-0.608)=\\ =230(-0.608)^4+18(-0.608)^3+\\ +9(-0.608)^2-221(-0.608)-9=156,079\\ f'(a_1)=f'(-0.608)=\\ =920(-0.608)^3+54(-0.608)^2+\\ +18(-0.608)-221=-418.757\\ a_2=a_1-\frac{f(a_1)}{f'(a_1)}=-0.608-\frac{156.079}{-418.575}=-0.235\\ |a_2-a_1|=0.373$

$f(a_2)=f(-0.235)=\\ =230(-0.235)^4+18(-0.235)^3+\\ +9(-0.235)^2-221(-0.235)-9=43.9\\ f'(a_2)=f'(-0.235)=\\ =920(-0.235)^3+54(-0.235)^2+\\ +18(-0.235)-221=-234.187\\ a_3=a_2-\frac{f(a_2)}{f'(a_2)}=-0.235-\frac{43.9}{-234.187}=-0.0448\\ |a_3-a_2|=0.19$

$f(a_3)=f(-0.0448)=\\ =230(-0.0448)^4+18(-0.0448)^3+\\ +9(-0.0448)^2-221(-0.0448)-9=0.918\\ f'(a_3)=f'(-0.0448)=\\ =920(-0.0448)^3+54(-0.0448)^2+\\ +18(-0.0448)-221=-221.78\\ a_4=a_3-\frac{f(a_3)}{f'(a_3)}=-0.0448-\frac{0.918}{-221.78}=-0.0401\\ |a_4-a_3|=0.0047\\ x=-0.0401$

$a_0=1\\ f(a_0)=f(1)=\\ =230+18+9-221-9=27\\ f'(a_0)=f'(1)=\\ =920+54+18-221=771\\ a_1=a_0-\frac{f(a_0)}{f'(a_0)}=1-\frac{27}{771}=0.965\\$

$f(a_1)=f(0.965)=\\ =230(0.965)^4+18(0.965)^3+\\ +9(0.965)^2-221(0.965)-9=1.743\\ f'(a_1)=f'(0.965)=\\ =920(0.965)^3+54(0.965)^2+\\ +18(0.965)-221=673.398\\ a_2=a_1-\frac{f(a_1)}{f'(a_1)}=0.965-\frac{1.743}{673.398}=0.962\\ |a_2-a_1|=0.003\\ x=0.962$