Question

Numerical Analysis

hr2 h 2 f(x) = a. x3. e-(0.1)x -- +4. x. In(x) – 1500 = 0 VX + 2 We want to find the root of the above equation. (In order to ease the reading, points are used between variables. Only the number above “e” is equal to “zero point one”.) b) If "a=1.5” and “b=0.8" at the above function, find the root between Xa=50 and Xu=70 using method of false position “Regula-Falsi” until Ea of approximation satisfies the tolerance as Ex<&s=0.1.

Answer 1

ANSWER :

Given data : a=1.5 , b=0.8 , Xa=50, Xu = 70

$F(x)=ax^{^{3}}e^{-0.1x}-\frac{bx^{2}}{2+\sqrt{x}}+4xln(x)-1500=0$putting the value in equation

$F(x)=1.5x^{^{3}}e^{-0.1x}-\frac{0.8x^{2}}{2+\sqrt{x}}+4xln(x)-1500$

Regula- Falsi method

$x_{a}=50$ ,  

, = 70

$x_{k+1}=x_{k}-\frac{(x_{k}-x_{k-1})}{f(x_{k})-f(x_{k-1})}f(x_{k})$

$x_{1}=x_{u}-\frac{(x_{u}-x_{a})}{f(x_{u})-f(x_{a})}f(x_{a})$

  $=70-\frac{(20)(-219.3955313)}{(-219.3955313-315.2884542)}$

  $=61.94411669$

$(x_{a},x_{1})$

$x_{1}=60.0576905$

Now

$(x_{a},x_{2})$

$x_{3}=59.7188014$

$(x_{a},x_{3})$

$x_{4}=59.6619262$

$\left | x_{4}-x_{3} \right |<0.1$

$\approx 59.6619262$