Question

Find n for which the nth iteration by the Bisection Method guarantees to approximate the root of f(x) = 2x^2 − 3x − 2 on [−2, 1] with accuracy within 10^−8 .

Answer 1

Answer:

n = 29 iterations would be enough to obtain a root of $f(x)=2x^2-3x-2$ that is at most $10^{-8}$ away from the correct solution.

Step-by-step explanation:

You can use this formula which relates the number of iterations, n, required by the bisection method to converge to within an absolute error tolerance of ε starting from the initial interval (a, b).

$n\geq \frac{log(\frac{b-a}{\epsilon} )}{log(2)}$

We know

a = -2, b = 1 and ε = $10^{-8}$ so

$n\geq \frac{log(\frac{1+2}{10^{-8}} )}{log(2)}\\n \geq 29$

Thus, n = 29 iterations would be enough to obtain a root of $f(x)=2x^2-3x-2$ that is at most $10^{-8}$ away from the correct solution.

You can prove this result by doing the computation as follows:

From the information given we know:

• $f(x)=2x^2-3x-2$
• $\epsilon = 10^{-8}$

This is the algorithm for the Bisection method:

1. Find two numbers a and b at which f has different signs.
2. Define $c=\frac{a+b}{2}$
3. If $b-c\leq \epsilon$ then accept c as the root and stop
4. If $f(a)f(c)\leq 0$ then set c as the new b. Otherwise, set c as the new a. Return to step 1.

We know that $f(-2)=2(-2)^2-3(-2)-2=12$ and $f(1)=2(1)^2-3(1)-2=-3$ so we take $a=-2$ and $b=1$ then $c=\frac{-2+1}{2} =-0.5$

Because $1-(-0.5)\geq 10^{-8}$ we set $c=-0.5$ as the new b.

The bisection algorithm is detailed in the following table.

After the 29 steps we have that $6\cdot 10^{-9}\leq 10^{-8}$ hence the required root approximation is c = -0.50