1.1.7 Example: Square Roots by Newton's Method
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Procedures, as introduced above, are much like ordinary mathematical functions. They specify a value that is determined by one or more parameters. But there is an important difference between mathematical functions and computer procedures. Procedures must be effective.
As a case in point, consider the problem of computing square roots. We can define the square-root function as
This describes a perfectly legitimate mathematical function. We could use it to recognize whether one number is the square root of another, or to derive facts about square roots in general. On the other hand, the definition does not describe a procedure. Indeed, it tells us almost nothing about how to actually find the square root of a given number. It will not help matters to rephrase this definition in pseudo-Lisp:
(define (sqrt x)
(the y (and (>= y 0)
(= (square y) x))))
This only begs the question.
The contrast between function and procedure is a reflection of the general distinction between describing properties of things and describing how to do things, or, as it is sometimes referred to, the distinction between declarative knowledge and imperative knowledge. In mathematics we are usually concerned with declarative (what is) descriptions, whereas in computer science we are usually concerned with imperative (how to) descriptions.[20]
How does one compute square roots? The most common way is to use Newton’s method of successive approximations, which says that whenever we have a guess y for the value of the square root of a number x, we can perform a simple manipulation to get a better guess (one closer to the actual square root) by averaging y with x/y.[21] For example, we can compute the square root of 2 as follows. Suppose our initial guess is 1:
Guess | Quotient | Average |
---|---|---|
1 | (2/1) = 2 | ((2 + 1)/2) = 1.5 |
1.5 | (2/1.5) = 1.3333 | ((1.3333 + 1.5)/2) = 1.4167 |
1.4167 | (2/1.4167) = 1.4118 | ((1.4167 + 1.4118)/2) = 1.4142 |
1.4142 | … | … |
Continuing this process, we obtain better and better approximations to the square root.
Now let’s formalize the process in terms of procedures. We start with a value for the radicand (the number whose square root we are trying to compute) and a value for the guess. If the guess is good enough for our purposes, we are done; if not, we must repeat the process with an improved guess. We write this basic strategy as a procedure:
(define (sqrt-iter guess x)
(if (good-enough? guess x)
guess
(sqrt-iter (improve guess x)
x)))
A guess is improved by averaging it with the quotient of the radicand and the old guess:
(define (improve guess x)
(average guess (/ x guess)))
where
(define (average x y)
(/ (+ x y) 2))
We also have to say what we mean by “good enough.” The following will do for illustration, but it is not really a very good test. (See exercise 1.7.) The idea is to improve the answer until it is close enough so that its square differs from the radicand by less than a predetermined tolerance (here 0.001): [22]
(define (good-enough? guess x)
(< (abs (- (square guess) x)) 0.001))
Finally, we need a way to get started. For instance, we can always guess that the square root of any number is 1: [23]
(define (sqrt x)
(sqrt-iter 1.0 x))
If we type these definitions to the interpreter, we can use sqrt just as we can use any procedure:
(sqrt 9)
3.00009155413138
(sqrt (+ 100 37))
11.704699917758145
(sqrt (+ (sqrt 2) (sqrt 3)))
1.7739279023207892
(square (sqrt 1000))
1000.000369924366
The sqrt
program also illustrates that the simple procedural language we have introduced so far is sufficient for writing any purely numerical program that one could write in, say, C or Pascal. This might seem surprising, since we have not included in our language any iterative (looping) constructs that direct the computer to do something over and over again. Sqrt-iter
, on the other hand, demonstrates how iteration can be accomplished using no special construct other than the ordinary ability to call a procedure.[24]
Exercises
Comments
supermanhelp.com
ALGORITHM Newton
REAL :: Input, X, NewX, Tolerance
INTEGER :: Count
READ(*,*) Input, Tolerance
Count = 0 ! count starts with 0
X = Input ! X starts with the input value
DO ! for each iteration
Count = Count + 1 ! increase the iteration count
NewX = 0.5*(ans + Input/X) ! compute a new approximation
IF (ABS(X - NewX) < Tolerance) EXIT ! if they are very close, exit
X = NewX ! otherwise, keep the new one
END DO
WRITE(*,*) ‘After ‘, Count, ’ iterations:’
WRITE(*,*) ’ The estimated square root is ‘, NewX
WRITE(*,*) ’ The square root from SQRT() is ‘, SQRT(Input)
WRITE(*,*) ’ Absolute error = ‘, ABS(SQRT(Input) - NewX)
END PROGRAM SquareRoot
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