The Resource Magazine For Apple, Atari, and Commodore ...

34 COMPUTE! September/October. 198O. Issue 6
may drain from your rosy cheeks when you realize
that this equation has two answers, and the method
of successive substitutions will not work to find the
other
answer.
How do we know that the equation has two
answers? If you write a short program to print ihc
value of ex -4x for some values of x, you obtain the
results in Table 4. Note that the function ex - 4x is
positive at x = 0.2 while it is negative at x = 0.4.
That means that somewhere between 0.2 and 0.4 the
function ex - 4x went through zero, and at that point
(he equation was satisfied. That is the answer we
found above, namely x = 0.3574029. Note also that
at x =2 the function is negative, while at x = 2.2 the
function is positive, indicating that another answer is
to be found between 2.0 and 2.2. Try to find this
answer with successive substitutions.
Table 4. The value
of cx -4x versus x.
X
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
EXP(X) - 4'X
1.00
0.42
-(1.11
-0.58
-0.97
-1.28
-1.4")
-1.54
-1.44
-1.15
-0.61
0.22
1.42
The Method Of Interval Halving
The failure of the method of successive substitutions
to converge to an answer in certain situations is
reason enough to look for another method. The
method of interval halving is particularly attractive
because you are (almost) guaranteed that it will find
an answer if you know that the answer lies between
two numbers. Refer again to our problem of finding
the solution to the equation ex -4x = 0 and Table 4.
Table 4 indicates that one answer is between 0.2 and
0.4 and another answer is between 2.0 and 2.2,
because the function cx -4x changes sign on these internals.
The first step in the interval halving method is to put
the equation to be solved in the form
f(x) = 0 (7)
and to find two values of x (call (hem xl and xr, L
and R for left and right) such that (he function f(x) is
positive for one of these values of x and it is negative
for the other.
Suppose we deal with our example, ex - 4x = 0.
It already is in the form f(x) = 0. Furthermore, let
us concentrate for the moment on the solution that
we could not find with the method of successive
substitutions. That solution we know to be between x
= 2.0 and x = 2.2. Thus, xt = 2.0 and xR = 2.2
The second step is to try a value of x half-way
between x^, and x]^. This is where the name "inter
val halving" originates. This value of x, call it xjyj
(M for middle) is given by the simple expression,
*M = (*L + *R)/2 (8)
Now comes the tricky part. Suppose f"| is the value
of the function when xj^ is substituted (plugged in)
into f(x), and suppose ij^j is the value of (he function
when XjYj is plugged into the function f(x). If the
product (1"l . f\j) is positive, then xjyj lies to the led
of the answer just like xj^. We know this because the
product can only be positive if jFl and f j^ have
the same sign. In this case, we replace xj with a new
value, namely x^. On the other hand, if the product
Cl ■ '\l) 's negative, (hen f^ and l\/[ have opposite
signs, and xjyj is to the right of the answer. In that
case we replace xr with xjyf, giving a new value for
xr. In either case, we have bracketed the answer in
an interval half as wide as the interval we started
with.
Repeating this process allows us to bracket the
answer in as small an interval as we wish, with the
answer as accurate as we wish. Each time we
calculate a new xj^ we must test the sign of the pro
duct (Fl . fjyj) to see if xjyj is to the right or left of
the answer. If the answer was originally known to be
in an interval of width w, where w is the difference
between our first xr and xj^, then after n iterations
or repetitions of the interval halving process, the er
ror in the answer is w/2n. Thus, after 10 iterations
the error in the answer is about 1/1000 of the
original uncertainty in the answer.
A program to solve the equation ex - 4x =0
with the method of interval halving is given in
Listing 4. With little modification, this program can
be used lor other equations as well. The variables in
(he program in Listing 4 are closely related to our
previous discussion, so no further explanation will be
given. I expect that most people can understand
BASIC" about as well as algebra. Table* 5 shows the
answers we obtain for both of the solutions to this
equation. A total of 20 iterations are done in the pro
gram, giving an error of less than 0.00000096 if the
distance between the original xj^ and xj^ were less
than one. The roundoff error in many machines will
exceed
this.
Listing 4. Interval halving used to solve ex - 4x. = 0.
10 INPUT XL, XR
20 FOR 1 = 1 TO 20
30 XM = (XI. + XR)/2
40 FL = F.XP(XL) -4'XL
50 FM = KXF(XM) = 4'X.\I
60 IF FL*FM XM
100 PRINT XM
110 NEXT 1
Are there any practical applications of these techni
ques for ordinary citizens? The answer is yes. Sup-