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enot [183]
3 years ago
5

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 .

Mathematics
1 answer:
Lady_Fox [76]3 years ago
8 0

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.

<u>You can prove this result by doing the computation as follows:</u>

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 <em>a</em> and <em>b</em> at which <em>f</em> 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 <em>c </em>as the new<em> b</em>. Otherwise, set <em>c </em>as the new <em>a</em>. 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 <em>b.</em>

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

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