An important reason for introducing the concept of a continuous function was to make precise the idea of a function whose graph is connected—that is, it lacks any breaks. The following result, which can be proved by using theε-δ definition of limit, expresses this property in mathematical language.
T H E O R E M 7 . 1 0 . 1 ( T H E I N T E R M E D I A T E V A L U E T H E O R E M ) Letf be a function continuous in the closed interval [a, b].
(i) Iff (a)andf (b)have different signs, then there is at least onecin(a, b) such thatf (c)=0.
(ii) Iff (a) = f (b), then for everyintermediate valuey in the open interval betweenf (a)andf (b)there is at least onecin(a, b)such thatf (c)=y.
The conclusion in part (ii) follows from applying part (i) to the functiong(x)=f (x)−y.
You should draw a figure to help convince yourself that a function for which there is no such cmust have at least one discontinuity.
Theorem 7.10.1 is important in assuring the existence of solutions to some equations that cannot be solved explicitly.
E X A M P L E 1 Prove that the following equation has at least one solutioncbetween 0 and 1:
x6+3x2−2x−1=0
Solution: The polynomial
f (x)=x6+3x2−2x−1
is continuous for allx—in particular forx in [0,1]. Moreover,f (0)= −1 andf (1)=1.
According to Theorem 7.10.1, there exists at least one numbercin(0,1)such thatf (c)=0.
Sometimes we need to prove that certain equations have unique solutions. Consider the following example.
E X A M P L E 2 Prove that the equation
2x−5e−x(1+x2)=0 has a unique solution, which lies in the interval(0,2).
Solution: Defineg(x)=2x−5e−x(1+x2). Theng(0)= −5 andg(2)=4−25/e2. In factg(2) >0 becausee >5/2. According to the intermediate value theorem, therefore, the continuous functiongmust have at least one zero in(0,2). Moreover, note that
g(x)=2+5e−x(1+x2)−10xe−x =2+5e−x(1−2x+x2)=2+5e−x(x−1)2 But theng(x) >0 for allx, sogis strictly increasing. It follows thatgcan have only one zero.
Newton’s Method
The intermediate value theorem can often be used to show that an equationf (x)=0 has a solution in a given interval, but it gives no additional information about the location of the zero. In this subsection we shall explain a method which usually leads to a good approximate solution in an efficient way. The method was first suggested by Newton and it has an easy geometric explanation.
Consider the graph of the functiony=f (x)shown in Fig. 1. It has a zero atx=a, but this zero is not known. To find it, start with an initial estimate,x0, ofa. (It is an advantage to start withx0not too far froma, if possible.) In order to improve the estimate, construct the tangent line to the graph at the point(x0, f (x0)), and find the pointx1at which the tangent crosses thex-axis, as shown in Fig. 1.
x0 x1
a
y ⫽ f(x) y
x
Figure 1 Illustration of Newton’s method
S E C T I O N 7 . 1 0 / I N T E R M E D I A T E V A L U E T H E O R E M . N E W T O N ’ S M E T H O D 247 Usuallyx1is a significantly better estimate ofathanx0. After having foundx1, repeat the procedure by constructing the tangent line to the curve at the point(x1, f (x1)). Denote by x2 the point where this new tangent line crosses thex-axis. Repeating this procedure, we obtain a sequence of points which usually converges very quickly toa.
It is easy to find formulas for x1, x2, . . .. The slope of the tangent at x0 is f(x0).
According to the point–slope formula, the equation for the tangent line through the point (x0, f (x0))with slopef(x0)is given by
y−f (x0)=f(x0)(x−x0)
At the point where this tangent line crosses thex-axis, we havey =0 andx =x1. Hence
−f (x0)=f(x0)(x1−x0). Solving this equation forx1, we get x1=x0− f (x0)
f(x0) Similarly, givenx1, the formula forx2is
x2=x1− f (x1) f(x1)
In general, one has the following formula for the(n+1)th approximationxn+1, expressed in terms of thenth approximationxn:
N E W T O N ’ S M E T H O D
As long asf(xn)=0, Newton’s method generates the sequence of points given by the formula
xn+1=xn− f (xn)
f(xn), n=0,1, . . . Usually, the sequence{xn}converges quickly to a zero off.
(1)
E X A M P L E 3 Find an approximate value for the zero of
f (x)=x6+3x2−2x−1
in the interval [0,1], using Newton’s method once (see Example 1).
Solution: Choosex0 =1. Thenf (x0)=f (1)=1. Becausef(x)=6x5+6x−2, we havef(1)=10. Hence, equation (1) forn=0 yields
x1 =1− f (1)
f(1) =1− 1 10 = 9
10 =0.9
E X A M P L E 4 Use Newton’s method twice to find an approximate value for 15√ 2.
Solution: We need an equation of the formf (x)=0 which hasx = 15√
2 =21/15as a root. The equationx15 =2 has this root, so we letf (x)=x15−2. Choosex0=1. Then f (x0)=f (1)= −1, and becausef(x)=15x14, we havef(1)=15. Thus, forn=0, (1) gives
x1 =1− f (1)
f(1) =1−−1 15 = 16
15 ≈1.0667 Moreover,
x2 =x1− f (x1) f(x1)= 16
15− f (16/15) f(16/15)= 16
15−(16/15)15−2
15(16/15)14 ≈1.04729412 This is actually correct to 8 decimal places.3
NOTE 1 In most cases Newton’s method is very efficient, but it can happen that the sequence {xn}defined in (1) does not converge. Figure 2 shows an example wherex1is a much worse approximation toa thanx0 was. Of course, the formula (1) breaks down iff(xn) = 0.
Usually, Newton’s method fails only if the absolute value off(xn)becomes too small, for somen.
x0
x1 a
y
x f
Figure 2
How Fast Does Newton’s Method Converge (If it Does)?
T H E O R E M 7 . 1 0 . 2 ( C O N V E R G E N C E O F N E W T O N ’ S M E T H O D )
Suppose in (1) thatf is twice differentiable withf (x∗)=0 andf(x∗)=0. Suppose too there exist bothK >0 andδ >0 withKδ <1 such that
|f (x)f(x)|
f(x)2 ≤K|x−x∗|
for allxin the open intervalI=(x∗−δ, x∗+δ). Then, provided that the sequence{xn} in (1) starts at anx0inI, it will converge tox∗, with an errorxn−x∗that satisfies
|xn−x∗| ≤(δK)2n/K.
3 A frequently used rule of thumb says that, to obtain an approximation that is correct tondecimal places, use Newton’s method until it gives the samendecimal places twice in a row.
S E C T I O N 7 . 1 1 / I N F I N I T E S E Q U E N C E S 249
P R O B L E M S F O R S E C T I O N 7 . 1 0
1. Show that each of the following equations has at least one root in the given interval.
(a) x7−5x5+x3−1=0 in (−1,1) (b) x3+3x−8=0 in (1,3) (c)
x2+1=3x in (0,1) (d) ex−1=2x in (0,1) 2. Explain why anybody taller than 1 metre was once exactly 1 metre tall.
3. Find a better approximation to√3
17≈2.5 by using Newton’s method once.
⊂SM⊃4. The equationx4+3x3−3x2−8x+3=0 has an integer root. Find it. The three additional roots are close to−1.9, 0.4, and 1.5. Find better approximations by using Newton’s method once for each root that is not an integer.
5. The equation (2x)x = 15 has a solution which is approximately an integer. Find a better approximation by using Newton’s method once.
6. In Fig. 1,f (x0) >0 andf(x0) <0. Moreover,x1is to the right ofx0. Verify that this agrees with the formula (1) forn=0. Check the other combinations of signs forf (x0)andf(x0)to see both geometrically and analytically on which side ofx0the pointx1lies.