04. Solution of Non Linear Equations (Regula-Falsi Method)
Dated: 15-08-2025
Assume a function1 is defined over \([x_n, x_{n - 1}]\) such that \(f(x_n) \times f(x_{n - 1}) < 0\) implying that the function1 has opposite signs on both extremes of the interval.2 Now draw a chord connecting \(f(x_n)\) and \(f(x_{n - 1})\). It will intersect the \(x\) axis at some point which will be our initial approximation.
Equation for this chord is
\[
\frac{y - f(x_n)}{f(x_{n-1}) - f(x_n)} = \frac{x - x_n}{x_{n-1} - x_n}
\]
Setting \(y = 0\), we get
\[
x_{n + 1} = x = x_n - \frac{x_n - x_{n-1}}{f(x_n) - f(x_{n-1})} f(x_n)
\]
This gives us the first approximation to the root3
Example
Use the Regula-Falsi method to compute a real4 root3 of the equation \(x^3 – 9x + 1 = 0\) defined over \([2, 4]\).
Solution
\[x_1 = 2 \quad \text{and} \quad x_2 = 4\]
\[
f(2) = 2^3 - 9(2) + 1 = 8 - 18 + 1 = -9
\]
\[
f(4) = 4^3 - 9(4) + 1 = 64 - 36 + 1 = 29
\]
\[
x_3 = x_2 - \frac{x_2 - x_1}{f(x_2) - f(x_1)}f(x_2) = 4 - \frac{4-2}{29-(-9)}(29) = 4 - \frac{2(29)}{38}
\]
\[
= 4 - \frac{58}{38} = 4 - 1.5263 = 2.4736
\]
\[f (x_3) = 2.4736^3 - 9(2.4736) + 1 = 15.13520-22.2624 + 1 = -6.12644\]
\[\because f(x_2) \times f(x_3) < 0\]
\[
x_4 = x_3 - \frac{x_3 - x_2}{f(x_3) - f(x_2)} f(x_3) = 2.4736 - \frac{2.4736-4}{f(2.4736)-29}(-6.12644)
\]
\[
= 2.4736 - \frac{-1.5264}{-6.12644-29}(-6.12644) = 2.4736-(0.04345)(-6.12644)
\]
\[
= 2.4736 + 0.26619 = 2.73989
\]
\[f (x_4) = 2.73989 - 9(2.73989) + 1=20.5683-24.65901+1 =- 3. 090707\]
\[\because f(x_2) \times f(x_4) < 0\]
\[
x_5 = x_4 - \frac{x_4 - x_2}{f(x_4) - f(x_2)} f(x_4) = 2.73989 - \frac{2.73989 - 4}{f(2.73989) - 29}(-3.090707) = 2.86125
\]
\[
= 2.73989 - \frac{-1.26011}{-3.090707 - 29}(-3.090707)
\]
\[
= 2.73989 + 0.039267(3.090707) = 2.73989 + 0.121363 = 2.86125
\]
\[
f(x_5) = 2.86125 - 9(2.86125) + 1=23.42434-25.75125+1= -1.326868
\]
The initial interval \((x_1, x_2)\) in which the root3 of the equation lies should be sufficiently small.
References
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