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Sagot :
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.
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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