Find a root of an equation `f(x)=2x^3-2x-5` using Muller method
Solution:
Here `2x^3-2x-5=0`
Let `f(x) = 2x^3-2x-5`
Here
`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 + 13.74773)`
`=1.5971`
Relative percent error
`varepsilon_(a^1)=|(x_3-x_2)/x_3| xx 100%=|(1.5971-1.5)/1.5971| xx 100%=6.07953%`
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=-0.04672`
`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=(-0.04672--1.25)/0.0971=12.39272`
`a=(delta_2-delta_1)/(h_2+h_1)=(12.39272-16.5)/(0.0971+-0.5)=10.19419`
`b=a xx h_2 + d_2=10.19419xx0.0971+12.39272=13.38253`
`c=f(x_2)=-0.04672`
`x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))`
`x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))`
`=1.5971+(-2 xx -0.04672)/(13.38253 + sqrt(13.38253^2 - 4xx 10.19419 xx -0.04672))`
`=1.5971+(0.09343)/(13.38253 + sqrt(180.99718))`
`=1.5971+(0.09343)/(13.38253 + 13.45352)`
`=1.60058`
Relative percent error
`varepsilon_(a^2)=|(x_3-x_2)/x_3| xx 100%=|(1.60058-1.5971)/1.60058| xx 100%=0.21753%`
Now,
`x_0=x_1=1.5`
`x_1=x_2=1.5971`
`x_2=x_3=1.60058`
`3^(rd)` iteration :
`f(x_0)=f(1.5)=2(1.5)^(3)-2(1.5)-5=-1.25`
`f(x_1)=f(1.5971)=2(1.5971)^(3)-2(1.5971)-5=-0.04672`
`f(x_2)=f(1.60058)=2(1.60058)^(3)-2(1.60058)-5=-0.00028`
`h_1=x_1-x_0=1.5971-1.5=0.0971`
`h_2=x_2-x_1=1.60058-1.5971=0.00348`
`delta_1=(f(x_1)-f(x_0))/h_1=(-0.04672--1.25)/0.0971=12.39272`
`delta_2=(f(x_2)-f(x_1))/h_2=(-0.00028--0.04672)/0.00348=13.33768`
`a=(delta_2-delta_1)/(h_2+h_1)=(13.33768-12.39272)/(0.00348+0.0971)=9.39535`
`b=a xx h_2 + d_2=9.39535xx0.00348+13.33768=13.37039`
`c=f(x_2)=-0.00028`
`x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))`
`x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))`
`=1.60058+(-2 xx -0.00028)/(13.37039 + sqrt(13.37039^2 - 4xx 9.39535 xx -0.00028))`
`=1.60058+(0.00056)/(13.37039 + sqrt(178.7779))`
`=1.60058+(0.00056)/(13.37039 + 13.37079)`
`=1.6006`
Relative percent error
`varepsilon_(a^3)=|(x_3-x_2)/x_3| xx 100%=|(1.6006-1.60058)/1.6006| xx 100%=0.00131%`
Approximate root of the equation `2x^3-2x-5=0` using Muller method is `1.6006`
| `n` | `x_0` | `x_1` | `x_2` | `f(x_0)` | `f(x_1)` | `f(x_2)` | `a` | `b` | `c` | `x_3` | `varepsilon_(a^n` |
| 1 | 1 | 2 | 1.5 | -5 | 7 | -1.25 | 9 | 12 | -1.25 | 1.5971 | 6.07953 |
| 2 | 2 | 1.5 | 1.5971 | 7 | -1.25 | -0.04672 | 10.19419 | 13.38253 | -0.04672 | 1.60058 | 0.21753 |
| 3 | 1.5 | 1.5971 | 1.60058 | -1.25 | -0.04672 | -0.00028 | 9.39535 | 13.37039 | -0.00028 | 1.6006 | 0.00131 |
This material is intended as a summary. Use your textbook for detail explanation.
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