Answer to Question #110544 in Quantitative Methods for Anju Jayachandran

Question

Find the interval of unit length that contains the smallest positive root of the
equation f(x) = x
3 −5x
2 +1 = 0. Starting with this interval, find an interval of
length 0.05 or less that contains the root, by Bisection method.

Answer 1

$f(x)=x^3-5x^2+1$

We distinguish the roots by Sturm's theorem:

$f_1(x)=f'(x)=3x^2-10x$

$f:f_1\\ x^3-5x^2+1\quad|3x^2-10x\\ 3x^3-15x^2+3\quad x-5\\ 3x^3-10x^2\\ -5x^2+3\\ -15x^2+9\\ -15x^2+50x\\ -50x+9$

$f_2(x)=50x-9$

$f_1:f_2\\ 3x^2-10x\quad\quad|50x-9\\ 150x^2-500x\quad3x-\frac{473}{50}\\ 150x^2-27x\\ -473x\\ -473x+\frac{9\cdot473}{50}\\ -A, A>0$

$f_3(x)=A, A>0$

$\begin{matrix} & f&f_1&f_2&f_3 &\\ 0 & +&0&-&+&2\\ 1&-&-&+&+&1\\ 2&-&-&+&+&1\\ 3&-&-&+&+&1 \end{matrix}$

 The smallest positive root of $f(x):x\in(0,1)$ .

By Bisection method:

$x\in(0,1)\\ f(0)=1>0\\ f(1)=-3<0\\ x_1=\frac{0+1}{2}=0.5\\ f(x_1)=f(0.5)=-0.125$  ,which is negative.  

Hence, the root lies in between 0 and 0.5

$x_2=\frac{0+0.5}{2}=0.25\\ f(x_2)=f(0.25)=0.703$ , which is positive

Hence, the root lies in between 0.25 and 0.5

$x_3=\frac{0.25+0.5}{2}=0.375\\ f(x_3)=f(0.375)=0.349$ ,which is positive

Hence, the root lies in between 0.375 and 0.5

$x_4=\frac{0.375+0.5}{2}=0.4375\\ f(x_4)=f(0.4375)=0.1267$ ,which is positive

Hence, the root lies in between 0.4375 and 0.5

$x_5=\frac{0.4375+0.5}{2}=0.46875\\ f(x_5)=f(0.46875)=0.004$

0.5-0.46875=0.03125<0.05

$x\in(0.46875,0.5)$