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Chapter 2: Problem 5

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]

Short Answer

Expert verified
(a) Root is approximately 0.684. (b) Root is approximately -0.200.

Step by step solution

01

Understand the Bisection Method

The bisection method is a numerical method for finding roots of a continuous function. For a function \(f(x)\) on the interval \([a, b]\), the method requires \(f(a)\) and \(f(b)\) to have opposite signs. This indicates there is at least one root in \([a, b]\). We repeatedly halve the interval until it becomes sufficiently small, thereby approximating the root.
02

Apply Bisection Method to Part (a)

**Function:** \(f(x) = \sqrt{x} - \cos x\) on \([0, 1]\).1. Evaluate \(f(0) = \sqrt{0} - \cos(0) = 0 - 1 = -1\) and \(f(1) = \sqrt{1} - \cos(1) = 1 - 0.5403 \approx 0.4597\).2. Since \(f(0) < 0\) and \(f(1) > 0\), there is a root in \([0, 1]\).**Iterations:**- **Midpoint 1:** \(c = \frac{0 + 1}{2} = 0.5\) - \(f(0.5) = \sqrt{0.5} - \cos(0.5) \approx 0.7071 - 0.8776 = -0.1705\) - Since \(f(0.5) < 0\), new interval is \([0.5, 1]\).- **Midpoint 2:** \(c = \frac{0.5 + 1}{2} = 0.75\) - \(f(0.75) = \sqrt{0.75} - \cos(0.75) \approx 0.8660 - 0.7317 = 0.1343\) - Since \(f(0.75) > 0\), new interval is \([0.5, 0.75]\).- **Midpoint 3:** \(c = \frac{0.5 + 0.75}{2} = 0.625\) - \(f(0.625) = \sqrt{0.625} - \cos(0.625) \approx 0.7906 - 0.8101 = -0.0195\) - Since \(f(0.625) < 0\), new interval is \([0.625, 0.75]\).Further iterations are performed until the desired precision is reached (approx. \(m_3 = 0.684\)).
03

Apply Bisection Method to Part (b)

**Function:** \(f(x) = 3(x+1)\left(x-\frac{1}{2}\right)(x-1)\) on \([-1.25, 2.5]\).1. Evaluate \(f(-1.25) = 3(-0.25)(-1.75)(-2.25) = -2.953125\), \(f(2.5) = 3(3.5)(2)(1.5) = 31.5\).2. Since \(f(-1.25) < 0\) and \(f(2.5) > 0\), a root exists in this interval.**Iterations:**- **Midpoint 1:** \(c = \frac{-1.25 + 2.5}{2} = 0.625\) - \(f(0.625) = 3(1.625)(0.125)(-0.375) = -0.228515625\) - Since \(f(0.625) < 0\), new interval is \([0.625, 2.5]\).- **Midpoint 2:** \(c = \frac{0.625 + 2.5}{2} = 1.5625\) - \(f(1.5625) = 3(2.5625)(1.0625)(0.5625) \approx 5.117919922\) - Since \(f(1.5625) > 0\), new interval is \([0.625, 1.5625]\).- **Midpoint 3:** \(c = \frac{0.625 + 1.5625}{2} = 1.09375\) - \(f(1.09375) = 3(2.09375)(0.59375)(0.09375) \approx 0.34942627\) - Since \(f(1.09375) > 0\), new interval is \([0.625, 1.09375]\).Continue iterating to approximate \(m_3 \approx -0.200\) in desired precision.

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Key Concepts

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

Root Finding Algorithms
Root finding algorithms are techniques used to identify solutions, or 'roots,' of equations where the function value becomes zero. Among these, the bisection method is particularly popular due to its simplicity and robustness. Here's a closer look at why root finding is so essential:
  • Roots represent the values for which a function is equal to zero, which can be crucial in many fields like physics, engineering, and economics.
  • Finding these roots helps in solving equations that can't be simplified algebraically.
  • Understanding root finding methods enhances deeper mathematical comprehension and problem-solving skills.
The bisection method, as utilized in our exercise, allows for finding roots by methodically bisecting intervals and checking the signs of the function at these points. It mandates that the function is continuous and changes sign over the interval, ensuring a root by the Intermediate Value Theorem.
Numerical Methods in Mathematics
Numerical methods form the cornerstone of approximating solutions to mathematical problems when exact answers are hard to come by. Especially with functions that are complex or transcendental, numerical methods offer practical solutions. In the realm of numerical methods, the bisection method shines as an easy-to-understand and highly reliable technique.
  • It works well for continuous functions with known sign changes, providing a systematic approach to hone in on a root.
  • The algorithm repeatedly bisects an interval and applies the concept of interval halving, which can efficiently narrow down the location of the root.
  • Its major strength lies in the guarantee of convergence, meaning it will always lead to a solution within the precision limits set by the user.
While detailed and rigorous, the bisection method is often slower than other techniques like Newton's method, yet its rate of convergence is steady and reliable, making it a stalwart choice in teaching and initial problem analysis.
Calculus Problem Solving
Calculus provides the theoretical foundation behind the methods used in solving root-finding problems. It ensures that we confidently rely on approaches like the bisection method for real-life applications. Here's how calculus interconnects:
  • The notion of continuity is central, leading us to understand the Intermediate Value Theorem, which guarantees a root where a function changes signs.
  • Derivative insights, though not directly used in the bisection method, bolster our understanding of function behavior and can expedite analysis in more complex methods.
  • Using calculus, we can better perceive asymptotic behavior, local extrema, and concavity, which are often important when selecting or verifying interval boundaries for root finding.
In practical problem solving, calculus equips us with a wide array of tools not only for finding roots but also for analyzing and predicting the behavior of complex functions, which can significantly impact computation efficiency and accuracy.

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Most popular questions from this chapter

Use fixed point iteration with \(p_{0}=-1\) to approximate a root of \(g(x)=x^{3}-3 x+3\) accurate to the nearest \(10^{-4}\)

Suppose you are using the secant method with \(x_{0}=1\) and \(x_{1}=1.1\) to find a root of \(f(x)\). (a) Find \(x_{2}\) given that \(f(1)=0.3\) and \(f(1.1)=0.23\). (b) Create a sketch (graph) that illustrates the calculation. HINT: \(x_{2}\) will be located where the line through \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(x_{1}, f\left(x_{1}\right)\right)\) crosses the \(x\) axis.

The errors in three consecutive iterations of M眉ller's method are shown in the table. Use this information to estimate the order of convergence. $$ \begin{array}{|c|c|} \hline n & \left|x_{n}-x\right| \\ \hline \hline 12 & 1.53627(10)^{-349} \\ \hline 13 & 1.67365(10)^{-642} \\ \hline 14 & 1.83922(10)^{-1181} \\ \hline \end{array} $$

Use Aitken's delta squared method to find \(p=\lim _{n \rightarrow \infty} p_{n}\) accurate to 3 decimal places. $$ \begin{array}{c} p_{n}=\\{-2,-1.85271,-1.74274,-1.66045 \\ -1.59884,-1.55266,-1.51804 \\ -1.49208,-1.47261, \ldots\\} \end{array} $$

Use Steffensen's method to find the root of \(g(x)=\) \(x^{4}-2 x^{3}-4 x^{2}+4 x+4\) in [2,3] accurate to five siginificant digits.
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