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The Intermediate Value Theorem can be used to approximate a root. The following is an example of binary search in computer science.

Suppose you want to approximate [tex]\sqrt{7}[/tex]. You know that it is between 2 and 3. If you consider the function [tex]f(x) = x^2 - 7[/tex], then note that [tex]f(2) \ \textless \ 0[/tex] and [tex]f(3) \ \textgreater \ 0[/tex]. Therefore, by the Intermediate Value Theorem, there is a value [tex]2 \leq c \leq 3[/tex] such that [tex]f(c) = 0[/tex].

Next, choose the midpoint of these two values, 2.5, which is guaranteed to be within 0.5 of the actual root. [tex]f(2.5)[/tex] will either be less than 0 or greater than 0. You can use the Intermediate Value Theorem again, replacing 2.5 with the previous endpoint that has the same sign as 2.5. Continuing this process gives a sequence of approximations [tex]x_n[/tex] with [tex]x_1 = 2.5[/tex].

How many iterations must you do in order to be within 0.0009765625 of the root?



Answer :

GinnyAnswer
To approximate [tex]\(\sqrt{7}\)[/tex] using the binary search method, we need to iteratively narrow down the interval [tex]\([2, 3]\)[/tex] until the length of the interval is within the specified tolerance of 0.0009765625.

Given our function [tex]\( f(x) = x^2 - 7 \)[/tex]:
1. Define the initial interval [tex]\([2, 3]\)[/tex].
2. Calculate the midpoint of the interval.
3. Evaluate [tex]\(f\)[/tex] at the midpoint.
4. Determine which subinterval contains the root based on the sign of [tex]\(f\)[/tex].
5. Repeat steps 2-4 until the interval width is less than or equal to the tolerance.

We start with the interval [tex]\([2, 3]\)[/tex]:

1. Calculate the midpoint: [tex]\( \text{midpoint} = \frac{2 + 3}{2} = 2.5 \)[/tex]
2. [tex]\( f(2.5) = 2.5^2 - 7 = 6.25 - 7 = -0.75 \)[/tex]
3. Since [tex]\( f(2) < 0 \)[/tex] and [tex]\( f(2.5) < 0 \)[/tex], the root must be in [tex]\([2.5, 3]\)[/tex]. However, [tex]\( f(2.5) * f(3) < 0 \)[/tex], so the root must be in [tex]\([2.5, 3]\)[/tex].

Now update the interval to [tex]\([2.5, 3]\)[/tex]:
1. Calculate the new midpoint: [tex]\( \text{midpoint} = \frac{2.5 + 3}{2} = 2.75 \)[/tex]
2. [tex]\( f(2.75) = 2.75^2 - 7 = 7.5625 - 7 = 0.5625 \)[/tex]
3. Since [tex]\( f(2.5) < 0 \)[/tex] and [tex]\( f(2.75) > 0 \)[/tex], the root must be in [tex]\([2.5, 2.75]\)[/tex].

Following this process, you continue until the length of the interval is less than 0.0009765625:

Each iteration halves the interval size, so we can find the number of iterations [tex]\[iterations\][/tex] required to satisfy the interval tolerance criterion [tex]\( \frac{3 - 2}{2^n} \leq 0.0009765625 \)[/tex].

This gives [tex]\( 0.0009765625 \leq \frac{1}{2^n} \)[/tex].

Taking the logarithm base 2:
[tex]\[ \log_2 (0.0009765625) \leq -n \][/tex]

Calculating the precise number for [tex]\( \log_2(0.0009765625) \)[/tex]:

[tex]\[ 0.0009765625 = 2^{-10} \][/tex]

So, [tex]\( -n = -10 \)[/tex] leading to:
[tex]\[ n = 10 \][/tex]

Thus, it takes approximately 10 iterations to reach the desired tolerance, but in this case, since the function was evaluated, it took 9 iterations to be within the tolerance. Therefore, the answer is 9 iterations.

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