Find `sqrt(12)` using False Position method (regula falsi method)
Solution:
Let `x=sqrt(12)`
`:.x^2=12`
`:.x^2-12=0`
i.e. `f(x)=x^2-12`
Here
| `x` | 0 | 1 | 2 | 3 | 4 |
|---|
| `f(x)` | -12 | -11 | -8 | -3 | 4 |
|---|
`1^(st)` iteration :
Here `f(3) = -3 < 0` and `f(4) = 4 > 0`
`:.` Now, Root lies between `x_0 = 3` and `x_1 = 4`
`x_2 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`
`x_2=3 - (-3) * (4 - 3)/(4 - (-3))`
`x_2=3.4286`
`f(x_2)=f(3.4286)=3.4286^2-12=-0.2449 < 0`
`2^(nd)` iteration :
Here `f(3.4286) = -0.2449 < 0` and `f(4) = 4 > 0`
`:.` Now, Root lies between `x_0 = 3.4286` and `x_1 = 4`
`x_3 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`
`x_3=3.4286 - (-0.2449) * (4 - 3.4286)/(4 - (-0.2449))`
`x_3=3.4615`
`f(x_3)=f(3.4615)=3.4615^2-12=-0.0178 < 0`
`3^(rd)` iteration :
Here `f(3.4615) = -0.0178 < 0` and `f(4) = 4 > 0`
`:.` Now, Root lies between `x_0 = 3.4615` and `x_1 = 4`
`x_4 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`
`x_4=3.4615 - (-0.0178) * (4 - 3.4615)/(4 - (-0.0178))`
`x_4=3.4639`
`f(x_4)=f(3.4639)=3.4639^2-12=-0.0013 < 0`
`4^(th)` iteration :
Here `f(3.4639) = -0.0013 < 0` and `f(4) = 4 > 0`
`:.` Now, Root lies between `x_0 = 3.4639` and `x_1 = 4`
`x_5 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`
`x_5=3.4639 - (-0.0013) * (4 - 3.4639)/(4 - (-0.0013))`
`x_5=3.4641`
`f(x_5)=f(3.4641)=3.4641^2-12=0 < 0`
Approximate root of the equation `x^2-12=0` using False Position method is `3.4641` (After 4 iterations)
| `n` | `x_0` | `f(x_0)` | `x_1` | `f(x_1)` | `x_2` | `f(x_2)` | Update |
| 1 | 3 | -3 | 4 | 4 | 3.4286 | -0.2449 | `x_0 = x_2` |
| 2 | 3.4286 | -0.2449 | 4 | 4 | 3.4615 | -0.0178 | `x_0 = x_2` |
| 3 | 3.4615 | -0.0178 | 4 | 4 | 3.4639 | -0.0013 | `x_0 = x_2` |
| 4 | 3.4639 | -0.0013 | 4 | 4 | 3.4641 | 0 | `x_0 = x_2` |
This material is intended as a summary. Use your textbook for detail explanation.
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