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 Problem: muller method 2x^3-2x-5, a=0, b=1, c=2 [ Calculator, Method and examples ]Solution:Your problem -> muller method 2x^3-2x-5, a=0, b=1, c=2Here 2x^3-2x-5=0Let f(x) = 2x^3-2x-5x_0 = 0x_1 = 1x_2 = 21^(st) iteration :f(x_0)=f(0)=2*(0)^3-2*(0)-5=-5f(x_1)=f(1)=2*1^3-2*1-5=-5f(x_2)=f(2)=2*(2)^3-2*(2)-5=7h_1=x_1-x_0=1-0=1h_2=x_2-x_1=2-1=1delta_1=(f(x_1)-f(x_0))/h_1=(-5--5)/1=0delta_2=(f(x_2)-f(x_1))/h_2=(7--5)/1=12a=(delta_2-delta_1)/(h_2+h_1)=(12-0)/(1+1)=6b=a xx h_2 + d_2=6xx1+12=18c=f(x_2)=7x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))=2+(-2 xx 7)/(18 + sqrt(18^2 - 4xx 6 xx 7))=2+(-14)/(18 + sqrt(156))=2+(-14)/(18 + 12.49)=1.5408Relative percent errorvarepsilon_(a^1)=|(x_3-x_2)/x_3| xx 100%=|(1.5408-2)/1.5408| xx 100%=29.7999%Now,x_0=x_1=1x_1=x_2=2x_2=x_3=1.54082^(nd) iteration :f(x_0)=f(1)=2*1^3-2*1-5=-5f(x_1)=f(2)=2*(2)^3-2*(2)-5=7f(x_2)=f(1.5408)=2*(1.5408)^3-2*(1.5408)-5=-0.7653h_1=x_1-x_0=2-1=1h_2=x_2-x_1=1.5408-2=-0.4592delta_1=(f(x_1)-f(x_0))/h_1=(7--5)/1=12delta_2=(f(x_2)-f(x_1))/h_2=(-0.7653-7)/-0.4592=16.9117a=(delta_2-delta_1)/(h_2+h_1)=(16.9117-12)/(-0.4592+1)=9.0817b=a xx h_2 + d_2=9.0817xx-0.4592+16.9117=12.7417c=f(x_2)=-0.7653x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))=1.5408+(-2 xx -0.7653)/(12.7417 + sqrt(12.7417^2 - 4xx 9.0817 xx -0.7653))=1.5408+(1.5306)/(12.7417 + sqrt(190.15))=1.5408+(1.5306)/(12.7417 + 13.7895)=1.5985Relative percent errorvarepsilon_(a^2)=|(x_3-x_2)/x_3| xx 100%=|(1.5985-1.5408)/1.5985| xx 100%=3.6089%Now,

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