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

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.

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 (log((b-a)/(\epsilon) ))/(log(2))`

We know

a = -2, b = 1 and ε =
`10^(-8)` so

`n\geq (log((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=(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=(-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