Find a root of an equation `f(x)=2x^3-2x-5` using Steffensen's method
Solution:
Here `2x^3-2x-5=0`
Let `f(x) = 2x^3-2x-5`
Here
Here `f(1) = -5 < 0 and f(2) = 7 > 0`
`:.` Root lies between `1` and `2`
`x_0 = (1 + 2)/2 = 1.5`
`x_0 = 1.5`
`1^(st)` iteration :
`f(x_0)=f(1.5)=2*1.5^3-2*1.5-5=-1.25`
`f(x_0+f(x_0))=f(1.5+-1.25)=-5.4688`
`x_1=x_0-(f(x_0)^2)/(f(x_0+f(x_0))-f(x_0))`
`x_1=1.5-((-1.25)^2)/(-5.4688--1.25)`
`x_1=1.8704`
`2^(nd)` iteration :
`f(x_1)=f(1.8704)=2*1.8704^3-2*1.8704-5=4.3454`
`f(x_1+f(x_1))=f(1.8704+4.3454)=462.8797`
`x_2=x_1-(f(x_1)^2)/(f(x_1+f(x_1))-f(x_1))`
`x_2=1.8704-((4.3454)^2)/(462.8797-4.3454)`
`x_2=1.8292`
`3^(rd)` iteration :
`f(x_2)=f(1.8292)=2*1.8292^3-2*1.8292-5=3.5823`
`f(x_2+f(x_2))=f(1.8292+3.5823)=301.1224`
`x_3=x_2-(f(x_2)^2)/(f(x_2+f(x_2))-f(x_2))`
`x_3=1.8292-((3.5823)^2)/(301.1224-3.5823)`
`x_3=1.7861`
`4^(th)` iteration :
`f(x_3)=f(1.7861)=2*1.7861^3-2*1.7861-5=2.823`
`f(x_3+f(x_3))=f(1.7861+2.823)=181.6021`
`x_4=x_3-(f(x_3)^2)/(f(x_3+f(x_3))-f(x_3))`
`x_4=1.7861-((2.823)^2)/(181.6021-2.823)`
`x_4=1.7415`
`5^(th)` iteration :
`f(x_4)=f(1.7415)=2*1.7415^3-2*1.7415-5=2.0801`
`f(x_4+f(x_4))=f(1.7415+2.0801)=98.978`
`x_5=x_4-(f(x_4)^2)/(f(x_4+f(x_4))-f(x_4))`
`x_5=1.7415-((2.0801)^2)/(98.978-2.0801)`
`x_5=1.6968`
`6^(th)` iteration :
`f(x_5)=f(1.6968)=2*1.6968^3-2*1.6968-5=1.3775`
`f(x_5+f(x_5))=f(1.6968+1.3775)=46.9661`
`x_6=x_5-(f(x_5)^2)/(f(x_5+f(x_5))-f(x_5))`
`x_6=1.6968-((1.3775)^2)/(46.9661-1.3775)`
`x_6=1.6552`
`7^(th)` iteration :
`f(x_6)=f(1.6552)=2*1.6552^3-2*1.6552-5=0.7592`
`f(x_6+f(x_6))=f(1.6552+0.7592)=18.32`
`x_7=x_6-(f(x_6)^2)/(f(x_6+f(x_6))-f(x_6))`
`x_7=1.6552-((0.7592)^2)/(18.32-0.7592)`
`x_7=1.6224`
`8^(th)` iteration :
`f(x_7)=f(1.6224)=2*1.6224^3-2*1.6224-5=0.2959`
`f(x_7+f(x_7))=f(1.6224+0.2959)=5.282`
`x_8=x_7-(f(x_7)^2)/(f(x_7+f(x_7))-f(x_7))`
`x_8=1.6224-((0.2959)^2)/(5.282-0.2959)`
`x_8=1.6048`
`9^(th)` iteration :
`f(x_8)=f(1.6048)=2*1.6048^3-2*1.6048-5=0.0567`
`f(x_8+f(x_8))=f(1.6048+0.0567)=0.8503`
`x_9=x_8-(f(x_8)^2)/(f(x_8+f(x_8))-f(x_8))`
`x_9=1.6048-((0.0567)^2)/(0.8503-0.0567)`
`x_9=1.6008`
`10^(th)` iteration :
`f(x_9)=f(1.6008)=2*1.6008^3-2*1.6008-5=0.0024`
`f(x_9+f(x_9))=f(1.6008+0.0024)=0.0344`
`x_10=x_9-(f(x_9)^2)/(f(x_9+f(x_9))-f(x_9))`
`x_10=1.6008-((0.0024)^2)/(0.0344-0.0024)`
`x_10=1.6006`
`11^(th)` iteration :
`f(x_10)=f(1.6006)=2*1.6006^3-2*1.6006-5=0`
`f(x_10+f(x_10))=f(1.6006+0)=0.0001`
`x_11=x_10-(f(x_10)^2)/(f(x_10+f(x_10))-f(x_10))`
`x_11=1.6006-((0)^2)/(0.0001-0)`
`x_11=1.6006`
Approximate root of the equation `2x^3-2x-5=0` using Steffensen's method is `1.6006` (After 11 iterations)
| `n` | `x_0` | `f(x_0)` | `f(x_0+f(x_0))` | `x_1` | Update |
| 1 | 1.5 | -1.25 | -5.4688 | 1.8704 | `x_0 = x_1` |
| 2 | 1.8704 | 4.3454 | 462.8797 | 1.8292 | `x_0 = x_1` |
| 3 | 1.8292 | 3.5823 | 301.1224 | 1.7861 | `x_0 = x_1` |
| 4 | 1.7861 | 2.823 | 181.6021 | 1.7415 | `x_0 = x_1` |
| 5 | 1.7415 | 2.0801 | 98.978 | 1.6968 | `x_0 = x_1` |
| 6 | 1.6968 | 1.3775 | 46.9661 | 1.6552 | `x_0 = x_1` |
| 7 | 1.6552 | 0.7592 | 18.32 | 1.6224 | `x_0 = x_1` |
| 8 | 1.6224 | 0.2959 | 5.282 | 1.6048 | `x_0 = x_1` |
| 9 | 1.6048 | 0.0567 | 0.8503 | 1.6008 | `x_0 = x_1` |
| 10 | 1.6008 | 0.0024 | 0.0344 | 1.6006 | `x_0 = x_1` |
| 11 | 1.6006 | 0 | 0.0001 | 1.6006 | `x_0 = x_1` |
This material is intended as a summary. Use your textbook for detail explanation.
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