Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The sine of an angle (specified in radians) can be computed by making use of the approximation \(\sin x \approx x\) if \(x\) is sufficiently small, and the trigonometric identity \(\sin x=3 \sin \frac{x}{3}-4 \sin ^{3} \frac{x}{3}\) to reduce the size of the argument of \(\sin\). (For purposes of this exercise an angle is considered "sufficiently small" if its magnitude is not greater than \(0.1\) radians.) These ideas are incorporated in the following procedures: (define (cube \(x\) ) \((* x x x)\) ) (define (p \(x\) ) \((-(* 3 x)(* 4(\) cube \(x)))\) ) (define (sine angle) (if (not (> (abs angle) 0.1)) angle (p (sine (/ angle 3.0))))) a. How many times is the procedure \(\mathrm{p}\) applied when (sine 12.15) is evaluated? b. What is the order of growth in space and number of steps (as a function of \(a\) ) used by the process generated by the sine procedure when (sine a) is evaluated?

Short Answer

Expert verified
a. 5 times; b. Order of growth is \( \Theta(\log a) \).

Step by step solution

01

Understanding the Sine Procedure

The provided procedure calculates the sine of an angle using a recursive approach. It continually reduces the angle by dividing it by 3 until it's 'sufficiently small,' i.e., its absolute value is no greater than 0.1. At every recursive step, it uses the function \( p \) to approximate the sine value based on smaller angles.
02

Evaluating Sine(12.15) Using Recursion

To find out how many times \( p \) is applied when calculating \( \text{sine}(12.15) \), determine how many times 12.15 must be divided by 3 before it is less than 0.1. This involves calculating how many steps are needed so that \( \frac{12.15}{3^n} \leq 0.1 \). Solving \( 12.15/3^n = 0.1 \) gives \( n = \log_3(12.15/0.1) \), which must be rounded up to the nearest whole number.
03

Calculating the Number of Applications of p

Calculate \( \log_3(12.15/0.1) \):\[ \log_3(12.15/0.1) = \log_3(121.5) \]\[ \log_3(121.5) \approx 4.81 \]Rounding 4.81 up results in \( n = 5 \). Thus, \( p \) is applied 5 times.
04

Analyzing the Order of Growth in Space and Steps

Each recursive call of the \( \text{sine} \) procedure reduces the angle size by dividing by 3. This results in a logarithmic growth in the number of calls needed relative to the size of the starting angle. As for space, each call must be stored until all recursive operations are complete, also implying a logarithmic growth pattern for the stack size due to the recursive nature of the procedure.
05

Conclusion on Order of Growth

The order of growth in space and number of steps (as a function of \( a \)) is \( \Theta(\log a) \) due to the logarithmic division of the angle in each recursive call.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Algorithm Analysis
When dealing with recursive functions, algorithm analysis is crucial to understanding the efficiency of the process. In our example with the \( \text{sine} \) procedure, we use recursion to break down the problem until it becomes manageable, specifically until the angle is sufficiently small.
One key aspect of algorithm analysis is determining how the input size impacts the number of operations needed, and thereby figuring out the function's efficiency. In the case of the sine approximation, each recursive call reduces the angle by a factor of three, leading to several operations. This pattern exemplifies linear recursive processes that can be analyzed to deduce both the number of recursive steps and how much computational space is required.
An effective analysis reveals that the sine function calculations follow a logarithmic growth, meaning the number of steps scales with the logarithm of the initial angle divided by the threshold value, which is 0.1 in this case. Consequently, a proper understanding of the recursion pattern and algorithm analysis tells us that the function is efficient for larger angles, as it quickly reduces the computation necessary to achieve the desired sine value.
Trigonometric Functions
Trigonometric functions are a staple in mathematics and computer science for handling cyclical phenomena. Here, the sine function is central to the task. The challenge rises from the fact that trigonometric functions have traditionally complex computations.
In computing the sine of an angle, approximations come into play. For very small angles (\(\leq 0.1\) radians), the sine of an angle can be approximated simply as the angle itself. Beyond this, we use the identity: \( \sin x=3 \sin \left( \frac{x}{3} \right)-4 \sin^3 \left( \frac{x}{3} \right) \), which helps reduce the angle in chunks, enabling easier computation. Understanding these identities and approximations significantly simplifies programming tasks involving trigonometric functions.
By iterating and employing these identities, the sine function transforms challenging trigonometric calculations into more straightforward, manageable computations. These methods are not only insightful for computer science but also illustrate how mathematical properties simplify real-world problem solving.
Order of Growth
Order of growth is a concept in algorithm analysis that describes how the resource consumption of an algorithm (such as time or space) increases as the input size becomes larger. It is generally expressed in terms of big O notation, a mathematical notation that describes the limiting behavior of a function when the argument tends towards a particular value or infinity.
In the sine computation example, the order of growth in terms of space and the number of iterations is logarithmic, represented as \(\Theta(\log a)\). This is because the algorithm divides the angle by three repeatedly until the angle is small enough. Logarithmic growth, especially when compared to linear or exponential growth, indicates good efficiency as it minimizes both the number of steps and memory usage.
Specifically, for larger angles, a logarithmic order of growth ensures that resources are conserved. This means that if you double the starting angle, the increase in required steps grows at a much slower rate than the increase in the input size, which makes this method efficient for computing sine for large angles. Knowing about the order of growth can help design algorithms that perform well even as the data size scales.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(f\) and \(g\) be two one-argument functions. The composition \(f\) after \(g\) is defined to be the function \(x \mapsto f(g(x))\). Define a procedure compose that implements composition. For example, if inc is a procedure that adds 1 to its argument, ((compose square inc) 6) 49

Design a procedure that evolves an iterative exponentiation process that uses successive squaring and uses a logarithmic number of steps, as does \(f\) ast- expt. (Hint: Using the observation that \(\left(b^{n / 2}\right)^{2}=\left(b^{2}\right)^{n / 2}\), keep, along with the exponent \(n\) and the base \(b\), an additional state variable \(a\), and define the state transformation in such a way that the product \(a b^{n}\) is unchanged from state to state. At the beginning of the process \(a\) is taken to be 1 , and the answer is given by the value of \(a\) at the end of the process. In general, the technique of defining an invariant quantity that remains unchanged from state to state is a powerful way to think about the design of iterative algorithms.)

Alyssa P. Hacker complains that we went to a lot of extra work in writing expmod. After all, she says, since we already know how to compute exponentials, we could have simply written (define (expmod base exp \(\mathrm{m}\) ) (remainder (fast-expt base exp) \(\mathrm{m}\) )) Is she correct? Would this procedure serve as well for our fast prime tester? Explain.

Newton's method for cube roots is based on the fact that if \(y\) is an approximation to the cube root of \(x\), then a better approximation is given by the value \(\frac{x / y^{2}+2 y}{3}\) Use this formula to implement a cube-root procedure analogous to the squareroot procedure. (In section \(1.3 .4\) we will see how to implement Newton's method in general as an abstraction of these square-root and cube- root procedures.)

Translate the following expression into prefix form $$ \frac{5+4+\left(2-\left(3-\left(6+\frac{4}{5}\right)\right)\right)}{3(6-2)(2-7)} $$

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free