/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Suppose you are using the bisect... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose you are using the bisection method on an interval of length 3. How many iterations are necessary to guarantee accuracy of the approximation to within \(10^{-6} ?\)

Short Answer

Expert verified
22 iterations are necessary.

Step by step solution

01

Set the Error Tolerance Formula

The bisection method ensures that the error after n iterations is less than \(\frac{b-a}{2^n}\), where \(b-a\) is the initial interval length. We want this error to be less than \(10^{-6}\).
02

Initial Interval Length

We are given that the initial interval length is 3 (so \(b-a=3\)).
03

Setting Up the Inequality

Substitute the values into the error formula: \(\frac{3}{2^n} < 10^{-6}\). This inequality will help us find the minimum number of iterations, \(n\).
04

Solve the Inequality

Rearrange to find \(2^n\):New inequality: \(2^n > \frac{3}{10^{-6}}\)Simplifying, we have: \(2^n > 3 \times 10^6\).
05

Compute Logarithms to Solve for n

Take the base-2 logarithm of both sides:\(n > \log_2(3 \times 10^{6})\).
06

Calculate the Logarithmic Value

Use the change of base formula or a calculator:\(\log_2(3 \times 10^{6}) = \frac{\log_{10}(3 \times 10^{6})}{\log_{10}(2)}\)Compute \(\log_{10}(3 \times 10^{6}) = \log_{10}(3) + 6\).
07

Final Calculation

Calculate \(\log_{10}(3) \approx 0.4771\), thus:\[\log_{10}(3 \times 10^6) = 0.4771 + 6 = 6.4771\]Then, \(\log_2(3 \times 10^6) = \frac{6.4771}{0.3010}\) (since \(\log_{10}(2) \approx 0.3010\)) Calculating, we find \(n > 21.52\). Since \(n\) must be a whole number, the minimum \(n\) is 22.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Error Tolerance
Error tolerance is a key concept in numerical methods like the bisection method. It sets a threshold for how close the numerical approximation should be to the actual value. In this exercise, we are targeting an error tolerance of \(10^{-6}\).
This means we want our result to be accurate to within one millionth. The bisection method guarantees that the error after \(n\) iterations is less than \( \frac{b-a}{2^n} \). This formula helps us adjust the number of iterations to achieve the desired accuracy.
In practice, ensuring the error tolerance is crucial because it helps prevent unnecessary calculations while maintaining a desired level of precision. This becomes especially important in engineering and scientific computations.
Iterations
Iterations refer to the repeated application of the bisection method to narrow down the interval containing the root. Each iteration reduces the interval by dividing it into half. In our example, the process repeats until the error of \( \frac{3}{2^n} \) is less than \(10^{-6}\).
Each step involves evaluating the midpoint of the interval to decide the direction of the next interval. This iterative process is fundamental in numerical approximation techniques because it systematically increases accuracy with each pass.
Using a smaller number of iterations results in less precise results and a lower computational cost. Conversely, more iterations increase precision at a higher computational cost. This balance is essential when considering computing resources during calculations.
Logarithmic Functions
Logarithmic functions play a crucial role in solving inequalities like \(2^n > 3 \times 10^6\) encountered in the bisection method. To find \(n\), as in our exercise, we take the logarithm of both sides. This helps us solve for \(n\) efficiently using the property \(\log_b(x)\) indicates the power to which we must raise \(b\) to get \(x\).
The change of base formula allows us to convert between logarithms of different bases. In this exercise, \(\log_2(x)\) applications are key as they allow us to determine the smallest integer \(n\) that satisfies the inequality.
Understanding logarithmic functions broadens our problem-solving toolkit and enhances our ability to handle exponential expressions, a common occurrence in numerical methods like the bisection method.
Numerical Approximation
Numerical approximation involves finding a numerical solution that is close to an exact solution when an exact solution either doesn't exist or is difficult to find. The bisection method is a fundamental numerical approximation technique.
By repeatedly bisecting the interval and converging on the desired value, numerical approximation offers a way to deal with complex problems analytically.
The balance between the number of iterations and error tolerance is a delicate one, fundamentally driving the accuracy of the numerical solution. This interplay defines the resource consumption and precision level achievable in practical applications of these methods.
Through its systematic approach, numerical approximation provides solutions to complicated real-world problems such as solving nonlinear equations, optimizing designs, or modeling natural phenomena. Its utility in various fields underscores its importance.

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Most popular questions from this chapter

Use the bisection method to find \(m_{3}\) for the given function on the given interval. Do this without a computer program. Just use a pencil, paper, and a calculator. You may check your answers with a computer program if you wish. \([A]\) (a) \(f(x)=\sqrt{x}-\cos x\) on [0,1] (b) \(f(x)=3(x+1)\left(x-\frac{1}{2}\right)(x-1)\) on [-1.25,2.5]

Suppose a function \(g\) satisfies the assumptions of the bisection method on the given interval. Starting with that interval, how many iterations are needed to approximate the root to within the given tolerance? (a) [-7,10]\(; 10^{-6}\) (b) [5,9]\(; 10^{-3}\) (c) [9,15]\(; 10^{-10}\) (d) [-6,-1]\(; 10^{-105}\) (assume the computer calculates with 300 significant digits so round-off error is not a problem)

The following algorithm is one possible incarnation of the bisection method. Assumptions: \(f\) is continuous on \([a, b] . f(a)\) and \(f(b)\) have opposite signs. Input: Interval \([a, b] ;\) function \(f\) Step 1: For \(j=1 \ldots 15\) do Steps 2 and 3: Step 2: Set \(m=\frac{a+b}{2} ;\) Step 3: If \(f(a) f(m)<0\) then set \(b=m ;\) else set \(a=m\) Step 4: Print \(m\). Output: Approximation \(m\). (a) Apply this algorithm to the function \(f(x)=\) \((x)(x-2)(x+2)\) over the interval \([-3,3] .\) Which root will this algorithm approximate? (b) How accurate is the approximation guaranteed to be according to the formula $$ \left|p_{n}-p\right| \leq \frac{b-a}{2^{n}} ? $$ (c) How accurate is the approximation in reality? Compare this to the bound in (b). (d) Modify the algorithm so it will approximate a different root using the same starting interval. (e) Modify the algorithm so it does not use multiplication.

Write your own absolute value function called absval (abs is already defined by Octave, so it is best to use a different name) that takes a real number input and returns the absolute value of the input. Use an if then else statement in your function. Save it as absval.m and test it on the following computations. (a) |-3| (b) |123.2| (c) \(\left|\pi-\frac{22}{7}\right|\) (d) \(\left|10-\pi^{2}\right|\)

Use the Intermediate Value Theorem to show that the function has a root in the indicated interval. (a) \(f(x)=3-x-\sin x ;[2,3]\) (b) \(g(x)=3 x^{4}-2 x^{3}-3 x+2 ;[0,1]\) (c) \(g(x)=3 x^{4}-2 x^{3}-3 x+2 ;[0,0.9]\) [s] (d) \(h(x)=10-\cosh (x) ;[-3,-2]\) (e) \(f(t)=\sqrt{4+5 \sin t}-2.5 ;[-6,-5]\) (f) \(g(t)=\frac{3 t^{2} \tan t}{1-t^{2}} ;[21.5,22.5]\) [5] (g) \(h(t)=\ln (3 \sin t)-\frac{3 t}{5} ;[1,2]\) (h) \(f(r)=e^{\sin r}-r ;[-20,20]\) (i) \(g(r)=\sin \left(e^{r}\right)+r ;[-3,3]\) (j) \(h(r)=2^{\sin r}-3^{\cos r} ;[1,3]\)

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