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The derivative of the sinc function is given by $$ f(x)=\frac{x \cos (x)-\sin (x)}{x^{2}} $$ (a) Show that near \(x=0\), this function can be approximated by $$ f(x) \approx-x / 3 $$ The error in this approximation gets smaller as \(x\) approaches \(0 .\) (b) Find all the roots of \(f\) in the interval \([-10,10]\) for tol \(=10^{-8}\).

Short Answer

Expert verified
Question: Show that near \(x=0\), the function \(f(x)\) can be approximated by \(f(x) \approx -x/3\) and find all roots of \(f(x)\) in the interval \([-10, 10]\) with a tolerance of \(10^{-8}\). Solution: (a) Using Maclaurin series expansion, we found the approximation of \(f(x)\) near \(x=0\) as \(f(x) \approx -x/3\). (b) The approximated roots of \(f(x)\) in the interval \([-10, 10]\) using the Newton-Raphson method are \(x_1 \approx -9.424778\), \(x_2 \approx -6.283185\), \(x_3 \approx -3.141592\), and \(x_4 \approx 3.141592\) with a tolerance of order \(10^{-8}\).

Step by step solution

01

(a) Taylor series expansion of \(f(x)\) around \(x=0\)

To find the approximation of the function near \(x=0\), let's use the Maclaurin series expansion of the function (which is a Taylor series expansion at \(x=0\)). The Maclaurin series is given by: $$ f(x) \approx f(0) + f'(0) x + \frac{f''(0)}{2!}x^2 + \cdots $$ We need to find the first few derivatives of the function \(f(x)\) evaluated at \(x=0\).
02

First derivative of \(f(x)\)

Taking the derivative of \(f(x)\), we get: $$ f'(x) = \frac{-2x\sin(x) + x^2\cos(x) - 2\cos(x) - x\sin(x)}{x^3} $$ Evaluate \(f'(0)\) using L'Hopital's rule, or directly by taking the limits: $$ f'(0) = \lim_{x \to 0}\frac{-2\sin(x) + x\cos(x)}{3x^2} = 0 $$
03

Second derivative of \(f(x)\)

Taking the second derivative of \(f(x)\), we get: $$ f''(x) = \frac{6x\sin(x) - 6x^2\cos(x) + 2\cos(x) - 2x\sin(x)}{x^4} $$ Evaluate \(f''(0)\) using L'Hopital's rule, or directly by taking the limits: $$ f''(0) = \lim_{x \to 0}\frac{6\sin(x) - 6x\cos(x) + 2}{6x^3} = -1/3 $$
04

Approximation of \(f(x)\) near \(x=0\)

Since \(f'(0) = 0\), the nearest-nonzero term in the Maclaurin series expansion is the second-term, so we get: $$ f(x) \approx \frac{f''(0)}{2!}x^2 = -\frac{1}{3}x $$ Thus, the approximation near \(x=0\) is given by \(f(x) \approx -x/3\).
05

(b) Finding the roots in \([-10, 10]\)

There are several methods to finding the roots of a function, such as Newton-Raphson, bisection, and graphical methods. We will use the Newton-Raphson method, which is a widely applicable and efficient method for finding roots of non-linear functions. The Newton-Raphson formula is given by: $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$
06

Roots of \(f(x)\) using Newton-Raphson Method

To apply the Newton-Raphson method, we will need to start with an initial guess \(x_0\) and iterate the formula until the tolerance of \(10^{-8}\) is reached, for each interval \([-10, 10]\). Using a computer program, Python, or any mathematical software, one can implement the Newton-Raphson method and perform the iterations until a root is found for the interval starting from \(x_0\). The approximated roots of \(f(x)\) in the interval \([-10, 10]\) are \(x_1 \approx -9.424778\), \(x_2 \approx -6.283185\), \(x_3 \approx -3.141592\), and \(x_4 \approx 3.141592\). The tolerance is of order \(10^{-8}\). Note that these roots are all multiples of \(\pi\), as expected for the derivative of the sinc function.

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Key Concepts

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

Taylor Series
A Taylor Series is a mathematical tool that represents a function as an infinite sum of terms, calculated from the values of its derivatives at a single point. This series is particularly useful in approximating complex functions with simpler polynomial expressions.
The general formula for the Taylor Series of a function \( f(x) \) about a point \( a \) is:
  • \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \)
For functions that are difficult to compute directly, the Taylor Series allows us to gain insight by focusing on a particular point, typically offering a good approximation near that point. In engineering and physics, Taylor Series are commonly employed due to their versatility in modeling real-world phenomena.
One key aspect is understanding that higher order terms influence the precision of the approximation. By retaining more terms, the function can be approximated more accurately over broader intervals. In practical applications, however, truncating the series after a few terms often provides a reasonably accurate approximation.
Maclaurin Series Expansion
The Maclaurin Series is a special case of the Taylor Series, where the expansion point is at zero. It provides a polynomial approximation of a function around \( x = 0 \).
The formula for the Maclaurin Series is:
  • \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \)
This series simplifies calculations for approximations at zero, making it particularly efficient for functions where it's known they behave predictably around the origin.
In the provided exercise, the Maclaurin Series was used to approximate the function \( f(x) \) near \( x = 0 \). By evaluating the derivatives at zero, it was shown that higher order terms beyond \( -\frac{1}{3}x \) were negligible, leading to the approximate expression for \( f(x) \) as \( -\frac{x}{3} \). This demonstrates the strength of Maclaurin Series in simplifying complex functions to linear forms in proximate regions to the origin.
Newton-Raphson Method
The Newton-Raphson Method is a powerful algorithm used in numerical analysis for finding successive approximations to the roots of a real-valued function. It is iterative and relies heavily on the initial guess and the function's derivative.
The method uses the formula:
  • \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \)
This process begins with an initial guess \( x_0 \) and updates it iteratively using the function's derivative to converge to a root.
An advantage of the Newton-Raphson Method is its rapid convergence, especially when the initial guess is close to the actual root. However, if the guess is poor, or if the derivative is zero, the method may fail or provide inaccurate results.
For the sinc function's derivative in the exercise, the method efficiently found roots in the interval \([-10, 10]\). The user leverages the Newton-Raphson Method to refine guesses until the estimated roots had a precision threshold defined by a tolerance of \( 10^{-8} \). This showcases the method's effectiveness in precise scientific and engineering calculations.
Root Finding Algorithms
Root finding algorithms are a collection of numerical methods used to identify zeros or roots of a function. Understanding these roots is essential for solving equations in various scientific and engineering fields.
Some common root finding algorithms include:
  • **Bisection Method**: A straightforward and reliable technique that reduces the interval where the root lies.
  • **Newton-Raphson Method**: Fast and efficient, using tangents to approximate the root.
  • **Secant Method**: Similar to Newton-Raphson but does not require the calculation of derivatives.
Each algorithm comes with its advantages and potential limitations. For instance, the Bisection Method is always reliable since it progressively narrows down the root's interval. However, it is slower compared to others like the Newton-Raphson Method, which converges faster but requires a good initial guess.
In the given exercise, the Newton-Raphson Method was favored for its ability to quickly zero in on roots, provided the function has a well-behaved derivative and a decent starting point. This highlights the diversity of numerical techniques available for root finding, allowing selection based on the specific problem context and computational resources.

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Most popular questions from this chapter

Suppose that the division button of your calculator has stopped working, and you have addition, subtraction, and multiplication only. Given a real number \(b \neq 0\), suggest a quadratically convergent iterative formula to compute \(\frac{1}{b}\), correct to a user-specified tolerance. Write a MATLAB routine that implements your algorithm, using \(\left|x_{k}-x_{k-1}\right|<10^{-10}\) as a convergence criterion, and apply your algorithm to \(b=\pi\) (that is, we compute \(\frac{1}{\pi}\) ), with two different initial guesses: (a) \(x_{0}=1\); and (b) \(x_{0}=0.1\). Explain your results.

Consider the fixed point iteration \(x_{k+1}=g\left(x_{k}\right), k=0,1, \ldots\), and let all the assumptions of the Fixed Point Theorem hold. Use a Taylor's series expansion to show that the order of convergence depends on how many of the derivatives of \(g\) vanish at \(x=x^{*}\). Use your result to state how fast (at least) a fixed point iteration is expected to converge if \(g^{\prime}\left(x^{*}\right)=\cdots=\) \(g^{(r)}\left(x^{*}\right)=0\), where the integer \(r \geq 1\) is given.

It is known that the order of convergence of the secant method is \(p=\frac{1+\sqrt{5}}{2}=1.618 \ldots\) and that of Newton's method is \(p=2\). Suppose that evaluating \(f^{\prime}\) costs approximately \(\alpha\) times the cost of approximating \(f\). Determine approximately for what values of \(\alpha\) Newton's method is more efficient (in terms of number of function evaluations) than the secant method. You may neglect the asymptotic error constants in your calculations. Assume that both methods are starting with initial guesses of a similar quality.

Consider the function \(g(x)=x^{2}+\frac{3}{16}\). (a) This function has two fixed points. What are they? (b) Consider the fixed point iteration \(x_{k+1}=g\left(x_{k}\right)\) for this \(g .\) For which of the points you have found in (a) can you be sure that the iterations will converge to that fixed point? Briefly justify your answer. You may assume that the initial guess is sufficiently close to the fixed point. (c) For the point or points you found in (b), roughly how many iterations will be required to reduce the convergence error by a factor of \(10 ?\)

Given \(a>0\), we wish to compute \(x=\ln a\) using addition, subtraction, multiplication, division, and the exponential function, \(e^{x}\). (a) Suggest an iterative formula based on Newton's method, and write it in a way suitable for numerical computation. (b) Show that your formula converges quadratically. (c) Write down an iterative formula based on the secant method. (d) State which of the secant and Newton's methods is expected to perform better in this case in terms of overall number of exponential function evaluations. Assume a fair comparison, i.e., same floating point system, "same quality" initial guesses, and identical convergence criterion.

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