Question

Using Regula Falsi method approximates the root of the following equation upto four decimal places.

f(x)=x=e power -x

Answer 1

Answer on Question #42259, Math, Linear Algebra

Problem. Using Regula Falsi method approximates the root of the following equation upto four decimal places.

$$x = e ^ {- x}.$$

Solution. Let $f(x) = x - e^{-x}$. The function $f(x)$ has the root in the interval (0,1) by Intermediate-Value Theorem, as $f(0) = -1 < 0$ and $f(1) = 1 - \frac{1}{e} > 0$. Let $a_1 = 0$ and $b_1 = 1$. By Regula Falsi method

$$c _ {1} = b _ {1} - \frac {f (b _ {1}) (b _ {1} - a _ {1})}{f (b _ {1}) - f (a _ {1})} = 0.612699837.$$ $$f (c _ {1}) = 0.070813948 > 0, \text{ so } a _ {2} = a _ {1} \text{ and } b _ {2} = c _ {1}.$$ $$c _ {2} = b _ {2} - \frac {f (b _ {2}) (b _ {2} - a _ {2})}{f (b _ {2}) - f (a _ {2})} = 0.572181412.$$ $$f (c _ {2}) = 0.007888273 > 0, \text{ so } a _ {3} = a _ {2} \text{ and } b _ {3} = c _ {2}.$$ $$c _ {3} = b _ {3} - \frac {f (b _ {3}) (b _ {3} - a _ {3})}{f (b _ {3}) - f (a _ {3})} = 0.567703214.$$ $$f (c _ {3}) = 0.000877392 > 0, \text{ so } a _ {4} = a _ {3} \text{ and } b _ {4} = c _ {3}.$$ $$c _ {4} = b _ {4} - \frac {f (b _ {4}) (b _ {4} - a _ {4})}{f (b _ {4}) - f (a _ {4})} = 0.567205553$$ $$f (c _ {4}) = 0.000097572 > 0, \text{ so } a _ {5} = a _ {4} \text{ and } b _ {5} = c _ {4}.$$ $$c _ {5} = b _ {5} - \frac {f (b _ {5}) (b _ {5} - a _ {5})}{f (b _ {5}) - f (a _ {5})} = 0.567150214.$$ $$f (c _ {5}) = 0.000010850 > 0, \text{ so } a _ {6} = a _ {5} \text{ and } b _ {6} = c _ {5}.$$ $$c _ {6} = b _ {5} - \frac {f (b _ {6}) (b _ {6} - a _ {6})}{f (b _ {6}) - f (a _ {6})} = 0.56714406.$$ $$\left| c _ {6} - c _ {5} \right| < 0.0001, \text{ so the root } x^{*} \approx 0.5671.$$

Answer. $x^{*} \approx 0.5671$.

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