Question

Using Xo = 0 find an approximation to one of the zeros of x^3 − 4x +1 = 0 by using Birge Vieta Method. Perform two iterations

Answer 1

Answer on Question #65700 – Math – Algorithms | Quantitative Methods

Question

Using $X_0 = 0$ find an approximation to one of the zeros of $x^3 - 4x + 1 = 0$ by using Birge Vieta Method. Perform two iterations

Solution

$$f(x) = x^3 - 4x + 1, \quad f'(x) = 3x^2 - 4, \quad x_0 = 0, \quad f(0) = 1, \quad f'(0) = -4$$ $$f(x) = xQ(x) + R(x) = x(x^2 - 4) + 1, \quad Q(x) = x^2 - 4, \quad R(x) = 1$$$$\frac{f(x)}{x - x_0} = \frac{x^3 - 4x + 1}{x} = x^2 - 4 + \frac{1}{x}, \quad x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 0 - \frac{1}{-4} = 0.25.$$$$| 1 \quad 0.25 + 0.25*1 = 0.5 \quad -3.9375 + 0.25*0.5 = -3.8125 = f'(0.25)$$ $$f(x) = x^3 - 4x + 1, \quad f'(x) = 3x^2 - 4, \quad x_1 = 0.25, \quad f(0.25) = 0.015625,$$ $$f'(0.25) = -3.8125$$ $$f(x) = xQ_1(x) + R_1(x) = (x - 0.25)(x^2 + 0.25x - 3.9375) + 0.015625,$$ $$Q_1(x) = x^2 + 0.25x - 3.9375, \quad R_1(x) = 0.015625$$ $$\frac{f(x)}{x - x_1} = \frac{x^3 - 4x + 1}{x - 0.25} = x^2 + \frac{x}{4} - \frac{63}{16} + \frac{\frac{1}{64}}{x - 0.25},$$ $$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.25 - \frac{\frac{1}{64}}{-3.8125} = 0.2541.$$

Answer: 0.25; 0.2541.

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