We introduced compound procedures 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, such as
the `integral`

procedure of
section 1.3.1, we began to see a more
powerful kind of abstraction: procedures 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.

### 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
number of steps required grows as Θ(`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:

```
(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))))))
```

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.

```
(define (close-enough? x y)
(< (abs (- x y)) 0.001))
```

`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, which checks to see which of the endpoints has a negative function value and
which has a positive value, and calls the `search`

procedure
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 signals an error.[56]

```
(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)))))
```

The following example uses the half-interval method to approximate π
as the root between 2 and 4 of `sin`

`x` = 0:

`(half-interval-method sin 2.0 4.0)`

3.14111328125

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

`(half-interval-method (lambda (x) (- (* x x x) (* 2 x) 3)) 1.0 2.0)`

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,

until the value does not change very much. Using this idea, we can
devise a procedure `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:

```
(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))
```

For example, we can use this method to approximate the fixed point of the cosine function, starting with 1 as an initial approximation:[57]

`(fixed-point cos 1.0)`

.7390822985224023

Similarly, we can find a solution to the equation
`y` = `sin`

`y` + `cos`

`y`:

`(fixed-point (lambda (y) (+ (sin y) (cos y))) 1.0)`

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[58] `y` → `x`/`y`, and we can therefore try to compute square roots as

```
(define (sqrt x)
(fixed-point (lambda (y) (/ x y))
1.0))
```

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 (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` → (1/2)(`y` + `x`/`y`):

```
(define (sqrt x)
(fixed-point (lambda (y) (average y (/ x y)))
1.0))
```

(Note that `y` = (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 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 of section 1.1.7. This approach of averaging successive approximations to a solution, a technique we that we call average damping, often aids the convergence of fixed-point searches.

`error`

, which takes as arguments a number of items that are printed as error messages. [back]`cos`

button until you obtain the fixed point. [back]`lambda`

. `y`→

`x`/

`y`means

`(lambda(y) (/ x y))`

, that is, the function whose value at `y`is

`x`/

`y`. [back]

Exercises

## Exercise 1.35

Show that the golden ratio φ (section 1.2.2)
is a fixed point of the transformation `x` → 1 + 1/`x`, and use
this fact to compute φ by means of the `fixed-point`

procedure.

## Exercise 1.36

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` → `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 `fixed-point`

with a guess of 1, as this would cause division by `log`

(1) = 0.)

## Exercise 1.37

An infinite continued fraction is an expression of the form

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/φ, where φ 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 formSuppose that

`n`

and`d`

are procedures 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`cont-frac`

such that evaluating`(cont-frac n d k)`

computes the value of the`k`-term finite continued fraction. Check your procedure by approximating 1/φ using`(cont-frac (lambda (i) 1.0) (lambda (i) 1.0) 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?If your

`cont-frac`

procedure generates a recursive process, write one that generates an iterative process. If it generates an iterative process, write one that generates a recursive process.

## 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 from
exercise 1.37 to approximate `e`, based on Euler’s expansion.

## Exercise 1.39

A continued fraction representation of the tangent function was published in 1770 by the German mathematician J.H. Lambert:

where `x` is in radians. Define a procedure `(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.

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