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Chapter 1: Problem 36

Modify fixed-point so that it prints the sequence of approximations it generates, using the newline and display primitives shown in exercise \(1.22\). Then find a solution to \(x^{x}=1000\) by finding a fixed point of \(x \mapsto \log (1000) / \log (x)\). (Use Scheme's primitive \(\log\) procedure, which computes natural logarithms.) Compare the number of steps this takes with and without average damping. (Note that you cannot start \(\mathrm{fixed}\)-point with a guess of 1 , as this would cause division by \(\log (1)=0\).)

Short Answer

Expert verified
Without damping requires more steps; with damping, convergence is faster.

Step by step solution

01

Modify the Fixed-Point Procedure

First, let's modify the fixed-point procedure to print each approximation it generates. We will assume that there is a recursive function `fixed-point` that continues to compute until convergence is achieved. We will add a step to output each approximation using `display` and `newline`. This allows us to see how close each step brings the approximation to the correct solution.
02

Define the Function for Fixed Point Iteration

Define the function based on the transformation function you need for the fixed-point method. Here, the transformation is given by \( f(x) = \frac{\log(1000)}{\log(x)} \). Make sure to define this in a Scheme function suitable for use in the `fixed-point` method.
03

Choose an Initial Guess

Select a reasonable starting guess for the initial approximation. Since \(x = 1\) would cause division by zero (in the transformation function's denominator), choose a different starting guess, such as \(x = 2\).
04

Apply the Fixed-Point Method Without Average Damping

Run the modified fixed-point method with the chosen initial guess without average damping. Record the sequence of approximations displayed to track how quickly the method reaches a solution.
05

Implement Average Damping

To potentially improve convergence, apply average damping to the iteration. Modify the transformation function to include damping, using the formula: \( ext{damped ext{-}f}(x) = rac{x + f(x)}{2} \).
06

Apply the Fixed-Point Method With Average Damping

Run the fixed-point iteration again using the damped version of your transformation function. Again, track the sequence of approximations displayed to see if damping reduces the number of iterations needed compared to without damping.
07

Conclusion: Compare the Results

After running both approaches (with and without damping), compare the number of iterations needed in each case. You will likely find that average damping results in fewer steps to converge to the solution.

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Key Concepts

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

Approximation Sequence
When solving equations using iterative methods, an approximation sequence is a series of progressively improved guesses that converge towards the actual solution. In the context of fixed-point iteration, we begin with an initial guess and apply a function repeatedly to get closer to the desired value.

It's essential to visualize this sequence because:
  • It helps us understand how rapidly the method converges.
  • It allows us to check the reliability of our approximations over iterations.
In our exercise, we modified the fixed-point method to print each step of the approximation sequence. By observing this output, students can see each approximation progressing toward solving the equation \( x^x = 1000 \). Implementing such modifications not only aids in debugging but is also essential for educational purposes by making the abstract iterative process tangible.
Average Damping
Average damping is a technique used to stabilize and speed up convergence in iterative methods. It involves averaging the current approximation and the next transformation's value. The formula for average damping is:\[ \text{damped-}f(x) = \frac{x + f(x)}{2} \]By incorporating this formula in the iteration process, the transformation function's output is smoothed, reducing oscillations and potentially leading to faster convergence. This approach is particularly effective when initial guesses cause divergence or when the transformation function has high sensitivity near the solutions.
The exercise demonstrates applying average damping to the fixed-point iteration for the equation \( x^x = 1000 \). By comparing results with and without damping, students get a practical insight into the utility of damping techniques in numerical methods. In many cases, damping can drastically reduce the number of iterations needed, saving computational resources and time.
Natural Logarithms
Natural logarithms are logarithms with the base \( e \), where \( e \approx 2.718 \). The natural logarithm of a number \( x \) is denoted as \( \log(x) \) or \( \ln(x) \). Natural logarithms are integral to various mathematical operations, especially those related to exponential growth, decay, and solving equations involving exponentiation.
In fixed-point iterations, transforming functions often utilize natural logarithms because they inherently handle multiplicative relationships due to their inverse nature concerning exponentials. For example, in the exercise, the transformation function \( f(x) = \frac{\log(1000)}{\log(x)} \) uses natural logarithms to manage the power equation \( x^x = 1000 \), facilitating the process of finding a fixed point solution. By leveraging natural logarithms, complex exponential problems become linear, simplifying iterative solutions and increasing computational efficiency.
Understanding the toolbox of natural logarithms in numerical methods lays a crucial foundation for tackling a wide array of mathematical problems efficiently.

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Most popular questions from this chapter

Louis Reasoner is having great difficulty doing exercise \(1.24\). His fast- prime? test seems to run more slowly than his prime? test. Louis calls his friend Eva Lu Ator over to help. When they examine Louis's code, they find that he has rewritten the expmod procedure to use an explicit multiplication, rather than calling square: (define (expmod base exp m) (cond \(((=\exp 0) 1)\) \(((\) even? exp) \((\) remainder \((*(\) expmod base \((/ \exp 2) \mathrm{m})\) \((\) expmod base \((/ \exp 2) \mathrm{m}))\) \((\) else \(\quad\) m)) \((\) remainder \((*\) base \((\) expmod base \((-\exp 1) \mathrm{m}))\) \(\quad\) m) \())\) "I don't see what difference that could make," says Louis. "I do." says Eva. "By writing the procedure like that, you have transformed the \(\Theta(\log n)\) process into a \(\Theta(n)\) process." Explain.

Suppose we define the procedure (define (f g) (g 2)) Then we have (f square) (f (lambda (z) (* z (+ z 1)))) 6 What happens if we (perversely) ask the interpreter to evaluate the combination (f \(f\) )? Explain.

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.

Define a procedure double that takes a procedure of one argument as argument and returns a procedure that applies the original procedure twice. For example, if inc is a procedure that adds 1 to its argument, then (double inc) should be a procedure that adds 2 . What value is returned by (((double (double double)) inc) 5)

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.)
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