The Bisection Method is an iterative numerical technique used to find roots of a function $f(x)$. The stopping criterion often involves ensuring that the relative approximate error $\varepsilon_a$ falls below a certain tolerance level. A key formula associated with the Bisection Method is:

$n \geq \frac{\log \left( \frac{b - a}{\varepsilon} \right)}{\log(2)}$

Where:

• $n$ is the number of iterations,
• $b - a$ is the initial interval length (given as 2 in this case),
• $\varepsilon$ is the desired tolerance (in this case, $\varepsilon_a \leq 10^{-6}\%$).

Step 1: Convert tolerance to decimal form

The relative approximate error tolerance $\varepsilon_a$ is $10^{-6}\%$, so we need to convert it to a decimal: $\varepsilon = \frac{10^{-6}}{100} = 10^{-8}$

Step 2: Use the Bisection Method formula

Substitute the given values $b - a = 2$ and $\varepsilon = 10^{-8}$ into the formula: $n \geq \frac{\log \left( \frac{2}{10^{-8}} \right)}{\log(2)}$

Step 3: Solve for $n$

Now we calculate the value of $n$.

$n \geq \frac{\log(2 \times 10^8)}{\log(2)}$ $n \geq \frac{\log(2) + 8\log(10)}{\log(2)}$ $n \geq \frac{0.3010 + 8 \times 1}{0.3010}$ $n \geq \frac{8.3010}{0.3010} \approx 27.58$

Since the number of iterations must be an integer, we round up to the next whole number: $n \geq 28$

Final Answer:

It will take at least 28 iterations to ensure that the relative approximate error $\varepsilon_a \leq 10^{-6}\%$.

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Would you like further details or clarifications on any of the steps?

Here are 5 related questions that expand on the topic:

1. How does the Bisection Method ensure convergence to a root?
2. What happens if the initial interval does not bracket a root in the Bisection Method?
3. How does the choice of the tolerance $\varepsilon$ affect the accuracy and speed of convergence?
4. Can the Bisection Method fail? Under what conditions might this happen?
5. How does the Bisection Method compare to other root-finding methods like Newton's method?

Tip: The Bisection Method is guaranteed to converge to a root if the function changes sign over the initial interval $[a, b]$.