Secant method and false position method exercise

Question

We have $f(x)=x^2-6$. I have to find $p_3$ if $p_0 = 3$ and $p_1 = 2$ by using

a) Secant method

b) False position method

So for the first one I have

$p_2=p_0- \dfrac {f(p_0)(p_1-p_0)}{f(p_1)-f(p_0)}$= $3- \dfrac {3(2-3)}{(-2-3)}$=3-0.6=2.4

$p_3=p_1- \dfrac {f(p_1)(p_2-p_1)}{f(p_2)-f(p_1)}$=2.45454

Now how do I find an approximate root using false position method?

Answer 1

Your Secant Method calculations are correct.

Using Algorithm $1.19$ for the Method of False Position, we have (note, the algorithm uses the point numbers over for efficiency):

• $p_0 = 3, p_1 = 2$
• $f(p_0) = 3, f(p_1) = -2 \implies ~\mbox{root in} ~(2, 3)$
• $p_2 = p_1 - f(p_1)\dfrac{(p_1-p_0)}{f(p_1)-f(p_0)} = 3 - f(3)\dfrac{(2-3)}{f(2)-f(3)} = 2.4$
• $f(2.4) = -0.24 \implies ~\mbox{root in} ~(2.4, 3)$
• $p_3 = 3 - f(3)\dfrac{(3-2.4)}{f(3)-f(2.4)} = 2.4444444444444446$

Note, if you want to practice the method and calculate more points, here they are in exact form:

$$\left\{3,2,\frac{12}{5},\frac{22}{9},\frac{120}{49},\frac{218}{89},\frac{1188}{485},\frac{2158}{881}\right\}$$