Find `root(3)(48)` using Secant method
Solution:
Let `x=48^(1/3)`
`:.x^3=48`
`:.x^3-48=0`
i.e. `f(x)=x^3-48`
Here
| `x` | 0 | 1 | 2 | 3 | 4 |
|---|
| `f(x)` | -48 | -47 | -40 | -21 | 16 |
|---|
`1^(st)` iteration :
`x_0 = 3` and `x_1 = 4`
`f(x_0) = f(3) = -21` and `f(x_1) = f(4) = 16`
`:. x_2 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`
`x_2 = 3 - (-21) * (4 - 3)/(16 - (-21))`
`x_2 = 3.5676`
`:. f(x_2)=f(3.5676)=3.5676^3-48=-2.5936`
`2^(nd)` iteration :
`x_1 = 4` and `x_2 = 3.5676`
`f(x_1) = f(4) = 16` and `f(x_2) = f(3.5676) = -2.5936`
`:. x_3 = x_1 - f(x_1) * (x_2 - x_1)/(f(x_2) - f(x_1))`
`x_3 = 4 - 16 * (3.5676 - 4)/(-2.5936 - 16)`
`x_3 = 3.6279`
`:. f(x_3)=f(3.6279)=3.6279^3-48=-0.2513`
`3^(rd)` iteration :
`x_2 = 3.5676` and `x_3 = 3.6279`
`f(x_2) = f(3.5676) = -2.5936` and `f(x_3) = f(3.6279) = -0.2513`
`:. x_4 = x_2 - f(x_2) * (x_3 - x_2)/(f(x_3) - f(x_2))`
`x_4 = 3.5676 - (-2.5936) * (3.6279 - 3.5676)/(-0.2513 - (-2.5936))`
`x_4 = 3.6344`
`:. f(x_4)=f(3.6344)=3.6344^3-48=0.0047`
`4^(th)` iteration :
`x_3 = 3.6279` and `x_4 = 3.6344`
`f(x_3) = f(3.6279) = -0.2513` and `f(x_4) = f(3.6344) = 0.0047`
`:. x_5 = x_3 - f(x_3) * (x_4 - x_3)/(f(x_4) - f(x_3))`
`x_5 = 3.6279 - (-0.2513) * (3.6344 - 3.6279)/(0.0047 - (-0.2513))`
`x_5 = 3.6342`
`:. f(x_5)=f(3.6342)=3.6342^3-48=0`
Approximate root of the equation `x^3-48=0` using Secant method is `3.6342` (After 4 iterations)
| `n` | `x_0` | `f(x_0)` | `x_1` | `f(x_1)` | `x_2` | `f(x_2)` | Update |
| 1 | 3 | -21 | 4 | 16 | 3.5676 | -2.5936 | `x_0 = x_1`
`x_1 = x_2` |
| 2 | 4 | 16 | 3.5676 | -2.5936 | 3.6279 | -0.2513 | `x_0 = x_1`
`x_1 = x_2` |
| 3 | 3.5676 | -2.5936 | 3.6279 | -0.2513 | 3.6344 | 0.0047 | `x_0 = x_1`
`x_1 = x_2` |
| 4 | 3.6279 | -0.2513 | 3.6344 | 0.0047 | 3.6342 | 0 | `x_0 = x_1`
`x_1 = x_2` |
This material is intended as a summary. Use your textbook for detail explanation.
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