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Problem: muller method 2x^3-2x-5 [ Calculator, Method and examples ]

Solution:
Your problem -> muller method 2x^3-2x-5

Here 2x^3-2x-5=0

Let f(x) = 2x^3-2x-5

Here
 x f(x) 0 1 2 -5 -5 7

x_0 = 1

x_1 = 2

x_2 = 1.5

1^(st) iteration :

f(x_0)=f(1)=2*1^3-2*1-5=-5

f(x_1)=f(2)=2*(2)^3-2*(2)-5=7

f(x_2)=f(1.5)=2*(1.5)^3-2*(1.5)-5=-1.25

h_1=x_1-x_0=2-1=1

h_2=x_2-x_1=1.5-2=-0.5

delta_1=(f(x_1)-f(x_0))/h_1=(7--5)/1=12

delta_2=(f(x_2)-f(x_1))/h_2=(-1.25-7)/-0.5=16.5

a=(delta_2-delta_1)/(h_2+h_1)=(16.5-12)/(-0.5+1)=9

b=a xx h_2 + d_2=9xx-0.5+16.5=12

c=f(x_2)=-1.25

x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))

x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))

=1.5+(-2 xx -1.25)/(12 + sqrt(12^2 - 4xx 9 xx -1.25))

=1.5+(2.5)/(12 + sqrt(189))

=1.5+(2.5)/(12 + 13.7477)

=1.5971

Relative percent error
varepsilon_(a^1)=|(x_3-x_2)/x_3| xx 100%=|(1.5971-1.5)/1.5971| xx 100%=6.0795%

Now,
x_0=x_1=2

x_1=x_2=1.5

x_2=x_3=1.5971

2^(nd) iteration :

f(x_0)=f(2)=2*(2)^3-2*(2)-5=7

f(x_1)=f(1.5)=2*(1.5)^3-2*(1.5)-5=-1.25

f(x_2)=f(1.5971)=2*(1.5971)^3-2*(1.5971)-5=-0.0467

h_1=x_1-x_0=1.5-2=-0.5

h_2=x_2-x_1=1.5971-1.5=0.0971

delta_1=(f(x_1)-f(x_0))/h_1=(-1.25-7)/-0.5=16.5

delta_2=(f(x_2)-f(x_1))/h_2=(-0.0467--1.25)/0.0971=12.3927

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