We introduced compound procedures functions in section 1.1.4 as a mechanism for abstracting patterns of numerical operations so as to make them independent of the particular numbers involved. With higher-order procedures, functions, such as the integral procedure function of section 1.3.1, we began to see a more powerful kind of abstraction: procedures functions used to express general methods of computation, independent of the particular functions involved. In this section we discuss two more elaborate examples—general methods for finding zeros and fixed points of functions—and show how these methods can be expressed directly as procedures. functions.

Finding roots of equations by the half-interval method

The half-interval method is a simple but powerful technique for finding roots of an equation $f(x)=0$, where $f$ is a continuous function. The idea is that, if we are given points $a$ and $b$ such that $f(a) < 0 < f(b)$, then $f$ must have at least one zero between $a$ and $b$. To locate a zero, let $x$ be the average of $a$ and $b$ and compute $f(x)$. If $f(x) > 0$, then $f$ must have a zero between $a$ and $x$. If $f(x) < 0$, then $f$ must have a zero between $x$ and $b$. Continuing in this way, we can identify smaller and smaller intervals on which $f$ must have a zero. When we reach a point where the interval is small enough, the process stops. Since the interval of uncertainty is reduced by half at each step of the process, the maximal number of steps required grows as $\Theta(\log( L/T))$, where $L$ is the length of the original interval and $T$ is the error tolerance (that is, the size of the interval we will consider small enough). Here is a procedure that implements this strategy: function that implements this strategy:

Original	JavaScript
(define (search f neg-point pos-point) (let ((midpoint (average neg-point pos-point))) (if (close-enough? neg-point pos-point) midpoint (let ((test-value (f midpoint))) (cond ((positive? test-value) (search f neg-point midpoint)) ((negative? test-value) (search f midpoint pos-point)) (else midpoint))))))	function search(f, neg_point, pos_point) { const midpoint = average(neg_point, pos_point); if (close_enough(neg_point, pos_point)) { return midpoint; } else { const test_value = f(midpoint); return positive(test_value) ? search(f, neg_point, midpoint) : negative(test_value) ? search(f, midpoint, pos_point) : midpoint; } }

We assume that we are initially given the function $f$ together with points at which its values are negative and positive. We first compute the midpoint of the two given points. Next we check to see if the given interval is small enough, and if so we simply return the midpoint as our answer. Otherwise, we compute as a test value the value of $f$ at the midpoint. If the test value is positive, then we continue the process with a new interval running from the original negative point to the midpoint. If the test value is negative, we continue with the interval from the midpoint to the positive point. Finally, there is the possibility that the test value is 0, in which case the midpoint is itself the root we are searching for. To test whether the endpoints are close enough we can use a procedure function similar to the one used in section 1.1.7 for computing square roots:[1]

Original	JavaScript
(define (close-enough? x y) (< (abs (- x y)) 0.001))	function close_enough(x, y) { return abs(x - y) < 0.001; }

Search The function search is awkward to use directly, because we can accidentally give it points at which $f$'s values do not have the required sign, in which case we get a wrong answer. Instead we will use search via the following procedure, function, which checks to see which of the endpoints has a negative function value and which has a positive value, and calls the search procedure function accordingly. If the function has the same sign on the two given points, the half-interval method cannot be used, in which case the procedure function signals an error.[2]

Original	JavaScript
(define (half-interval-method f a b) (let ((a-value (f a)) (b-value (f b))) (cond ((and (negative? a-value) (positive? b-value)) (search f a b)) ((and (negative? b-value) (positive? a-value)) (search f b a)) (else (error "Values are not of opposite sign" a b)))))	function half_interval_method(f, a, b) { const a_value = f(a); const b_value = f(b); return negative(a_value) && positive(b_value) ? search(f, a, b) : negative(b_value) && positive(a_value) ? search(f, b, a) : error("values are not of opposite sign"); }

The following example uses the

Original		JavaScript

half-interval method to approximate $\pi$ as the root between 2 and 4 of $\sin\, x = 0$:

Original	JavaScript
(half-interval-method sin 2.0 4.0)	half_interval_method(math_sin, 2, 4); 3.14111328125

Here is another example, using the half-interval method to search for a root of the equation $x^3 - 2x - 3 = 0$ between 1 and 2:

Original	JavaScript
(half-interval-method (lambda (x) (- (* x x x) (* 2 x) 3)) 1.0 2.0)	half_interval_method(x => x * x * x - 2 * x - 3, 1, 2); 1.89306640625

Finding fixed points of functions

A number $x$ is called a fixed point of a function $f$ if $x$ satisfies the equation $f(x)=x$. For some functions $f$ we can locate a fixed point by beginning with an initial guess and applying $f$ repeatedly, \[ \begin{array}{l} f(x), \ f(f(x)), \ f(f(f(x))), \ \ldots \end{array} \] until the value does not change very much. Using this idea, we can devise a procedure function fixed-point fixed_point that takes as inputs a function and an initial guess and produces an approximation to a fixed point of the function. We apply the function repeatedly until we find two successive values whose difference is less than some prescribed tolerance:

Original	JavaScript
(define tolerance 0.00001) (define (fixed-point f first-guess) (define (close-enough? v1 v2) (< (abs (- v1 v2)) tolerance)) (define (try guess) (let ((next (f guess))) (if (close-enough? guess next) next (try next)))) (try first-guess))	const tolerance = 0.00001; function fixed_point(f, first_guess) { function close_enough(x, y) { return abs(x - y) < tolerance; } function try_with(guess) { const next = f(guess); return close_enough(guess, next) ? next : try_with(next); } return try_with(first_guess); }

For example, we can use this method to approximate the fixed point of the cosine function, starting with 1 as an initial approximation:[3]

Original	JavaScript
(fixed-point cos 1.0) 0.7390822985224023	fixed_point(math_cos, 1); 0.7390822985224023

Similarly, we can find a solution to the equation $y=\sin y + \cos y$:

Original	JavaScript
(fixed-point (lambda (y) (+ (sin y) (cos y))) 1.0) 1.2587315962971173	fixed_point(y => math_sin(y) + math_cos(y), 1); 1.2587315962971173

The fixed-point process is reminiscent of the process we used for finding square roots in section 1.1.7. Both are based on the idea of repeatedly improving a guess until the result satisfies some criterion. In fact, we can readily formulate the square-root computation as a fixed-point search. Computing the square root of some number $x$ requires finding a $y$ such that $y^2 = x$. Putting this equation into the equivalent form $y = x/y$, we recognize that we are looking for a fixed point of the function[4] $y \mapsto x/y$, and we can therefore try to compute square roots as

Original	JavaScript
(define (sqrt x) (fixed-point (lambda (y) (/ x y)) 1.0))	function sqrt(x) { return fixed_point(y => x / y, 1); }

Unfortunately, this fixed-point search does not converge. Consider an initial guess $y_1$. The next guess is $y_2 = x/y_1$ and the next guess is $y_3 = x/y_2 = x/(x/y_1) = y_1$. This results in an infinite loop in which the two guesses $y_1$ and $y_2$ repeat over and over, oscillating about the answer.

One way to control such oscillations is to prevent the guesses from changing so much. Since the answer is always between our guess $y$ and $x/y$, we can make a new guess that is not as far from $y$ as $x/y$ by averaging $y$ with $x/y$, so that the next guess after $y$ is $\frac{1}{2}(y+x/y)$ instead of $x/y$. The process of making such a sequence of guesses is simply the process of looking for a fixed point of $y \mapsto \frac{1}{2}(y+x/y)$:

Original	JavaScript
(define (sqrt x) (fixed-point (lambda (y) (average y (/ x y))) 1.0))	function sqrt(x) { return fixed_point(y => average(y, x / y), 1); }

(Note that $y=\frac{1}{2}(y+x/y)$ is a simple transformation of the equation $y=x/y$; to derive it, add $y$ to both sides of the equation and divide by 2.)

With this modification, the square-root procedure function works. In fact, if we unravel the definitions, we can see that the sequence of approximations to the square root generated here is precisely the same as the one generated by our original square-root procedure function of section 1.1.7. This approach of averaging successive approximations to a solution, a technique we call average damping, often aids the convergence of fixed-point searches.

Exercise 1.35 Show that the golden ratio $\phi$ (section 1.2.2) is a fixed point of the transformation $x \mapsto 1 + 1/x$, and use this fact to compute $\phi$ by means of the fixed-point procedure. fixed_point function.

Solution

The fixed point of the function is \[ 1 + 1 / x = x \] Solving for x, we get \[ x^2 = x + 1 \] \[ x^2 - x - 1 = 0 \] Using the quadratic equation to solve for $x$, we find that one of the roots of this equation is the golden ratio $(1+\sqrt{5})/2$.

Original	JavaScript
	fixed_point(x => 1 + (1 / x), 1);

Exercise 1.36 Modify fixed-point fixed_point so that it prints the sequence of approximations it generates, using the primitive function display 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.) (Use the primitive function math_log, which computes natural logarithms.) Compare the number of steps this takes with and without average damping. (Note that you cannot start fixed-point fixed_point with a guess of 1, as this would cause division by $\log(1)=0$.)

Solution

We modify the function fixed_point as follows:

Original	JavaScript
(define tolerance 0.00001) (define (fixed-point f first-guess) (define (close-enough? v1 v2) (< (abs (- v1 v2)) tolerance)) (define (try guess) (let ((next (f guess))) (if (close-enough? guess next) next (try next)))) (try first-guess))	const tolerance = 0.00001; function fixed_point(f, first_guess) { function close_enough(x, y) { return abs(x - y) < tolerance; } function try_with(guess) { display(guess); const next = f(guess); return close_enough(guess, next) ? next : try_with(next); } return try_with(first_guess); }

Here is a version with average dampening built-in:

Original	JavaScript
(define tolerance 0.00001) (define (fixed-point f first-guess) (define (close-enough? v1 v2) (< (abs (- v1 v2)) tolerance)) (define (try guess) (let ((next (f guess))) (if (close-enough? guess next) next (try next)))) (try first-guess))	function fixed_point_with_average_dampening(f, first_guess) { function close_enough(x, y) { return abs(x - y) < tolerance; } function try_with(guess) { display(guess); const next = (guess + f(guess)) / 2; return close_enough(guess, next) ? next : try_with(next); } return try_with(first_guess); }

Exercise 1.37 An infinite continued fraction is an expression of the form \[ \begin{array}{lll} f & = & {\dfrac{N_1}{D_1+ \dfrac{N_2}{D_2+ \dfrac{N_3}{D_3+\cdots }}}} \end{array} \] As an example, one can show that the infinite continued fraction expansion with the $N_i$ and the $D_i$ all equal to 1 produces $1/\phi$, where $\phi$ is the golden ratio (described in section 1.2.2). One way to approximate an infinite continued fraction is to truncate the expansion after a given number of terms. Such a truncation—a so-called $k$-term finite continued fraction—has the form \[ {\dfrac{N_1}{D_1 + \dfrac{N_2}{\ddots + \dfrac{N_K}{D_K}}}} \]

1. Suppose that n and d are procedures functions of one argument (the term index $i$) that return the $N_i$ and $D_i$ of the terms of the continued fraction. Define a procedure Declare a function cont-frac cont_frac such that evaluating (cont-frac n d k) cont_frac(n, d, k) computes the value of the $k$-term finite continued fraction. Check your procedure function by approximating $1/\phi$ using

   Original	JavaScript
   (cont-frac (lambda (i) 1.0) (lambda (i) 1.0) k)	cont_frac(i => 1, i => 1, k);

   for successive values of k. How large must you make k in order to get an approximation that is accurate to 4 decimal places?
2. If your cont-frac cont_frac procedure function generates a recursive process, write one that generates an iterative process. If it generates an iterative process, write one that generates a recursive process.

Solution

//recursive process function cont_frac(n, d, k) { function fraction(i) { return i > k ? 0 : n(i) / (d(i) + fraction(i + 1)); } return fraction(1); } //iterative process function cont_frac(n, d, k) { function fraction(i, current) { return i === 0 ? current : fraction(i - 1, n(i) / (d(i) + current)); } return fraction(k, 0); }

Exercise 1.38 In 1737, the Swiss mathematician Leonhard Euler published a memoir De Fractionibus Continuis, which included a continued fraction expansion for $e-2$, where $e$ is the base of the natural logarithms. In this fraction, the $N_i$ are all 1, and the $D_i$ are successively 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, …. Write a program that uses your cont-frac procedure cont_frac function from exercise 1.37 to approximate $e$, based on Euler's expansion.

Solution

2 + cont_frac(i => 1, i => (i + 1) % 3 < 1 ? 2 * (i + 1) / 3 : 1, 20);

Exercise 1.39 A continued fraction representation of the tangent function was published in 1770 by the German mathematician J.H. Lambert: \[ \begin{array}{lll} \tan x & = & {\dfrac{x}{1- \dfrac{x^2}{3- \dfrac{x^2}{5- \dfrac{x^2}{ \ddots }}}}} \end{array} \] where $x$ is in radians. Define a procedure (tan-cf x k) Declare a function tan_cf(x, k) that computes an approximation to the tangent function based on Lambert's formula. K specifies the number of terms to compute, as in exercise 1.37. As in exercise 1.37, k specifies the number of terms to compute.

Solution

function tan_cf(x, k) { return cont_frac(i => i === 1 ? x : - x * x, i => 2 * i - 1, k); }