Consider a function
f(x) = x³ – 2x² + 3x – 5 = 0
Find the value of x.
The derivative is
f^{\prime}(x) = 3x² – 4x + 3
Let the initial value of x = 3, then f(x) = 13 and f^{\prime}(x) = 18. For k = 1, from Equation 12.4,
0 = f(x_{0}) + f^{\prime}(x_{0})(x_{1} – x_{0})
x_{1} = x_{0} – \frac{f(x_{0})}{f^{\prime}(x_{0})} (12.4)
x_{1} = 3 – (13/18) = 2.278. Table 12.1 is compiled to k = 4 and gives x = 1.843. As a verification, substituting this value into the original equation, the identity is approximately satisfied.
TABLE 12.1 Iterative Solution of a Function of One Variable (Example 12.1) |
|||
k | x_{k} | f(x_{k}) | f^{\prime}(x_{k}) |
0 | 3 | 13 | 18 |
1 | 2.278 | 3.277 | 9.456 |
2 | 1.931 | 0.536 | 6.462 |
3 | 1.848 | 0.025 | 5.853 |
4 | 1.844 | ≈ 0 |