MATH 4/5470: SOLUTIONS TO SOME HOMEWORK
QUESTIONS ON NEWTON’S METHOD
WINFRIED JUST, OHIO UNIVERSITY
Recall that if f : R → R is a differentiable function, then we can define a function
F : R ∪ {∞} → R ∪ {∞} by requiring
f (x)
if f 0 (x) 6= 0,
f 0 (x)
F (x) = x if f (x) = 0,
F (x) = x −
(1)
F (x) = ∞ in all other cases.
Newton’s method for finding x∗ with f (x∗ ) = 0 is based on the recursion xn+1 =
∗
F (xn ), and the sequence {xn }∞
n=0 is supposed to converge to x whenever x0 is
∗
0
sufficiently near x , and neither f (x), f (x) take the value 0 for x that are sufficiently
close to but different from x∗ . In our terminology this would mean that x∗ is a
locally asymptotically stable equilibrium of the dynamical system (R, F ). In this
exercise you will be asked to prove, under some mild additional assumptions, that
this is actually true.
Let us assume that f 00 exists and is a continuous function on R, and and f 0 (x)
does not take the value 0 on an interval (x∗ − ε, x∗ + ε), where f (x∗ ) = 0. Then f 0
does not change its sign on this interval.
Let x0 ∈ (x∗ − ε, x∗ + ε) and consider the sequence {xn }∞
n=0 that is defined by
(2)
xn+1 = F (xn ) = xn −
f (xn )
.
f 0 (xn )
Proposition 1. Under the assumptions made above, the sequence {xn }∞
n=0 is guaranteed to be a Cauchy sequence that stays within [x∗ − ε, x∗ + ε] as long as ε is
sufficiently small.
Proof: We have already shown that for some ξ in the interval between xn and
xn+1 ,
f 00 (ξ)
(xn+1 − xn )2 .
2
Let ε0 > 0 be such that f 0 (x∗ ) 6= 0 on [x∗ − ε0 , x∗ + ε0 ]. Since f 00 (x), f 0 (x)
were
00 assumed
continuous, by the Extreme Value Theorem there exists M such that
f (x)f 0 (y) ≤ M for all x, y ∈ [x∗ − ε0 , x∗ + ε0 ]. Let ε be such that 0 < 2ε <
2
(3)
f (xn+1 ) =
ε0
min{0.5, ε0 , M
}. Notice that xn+1 − xn+2 =
then (3) implies
(4)
f (xn+1 )
f 0 (xn+1 ) .
If xn , xn+1 ∈ [x∗ − ε, x∗ + ε]
00
f (ξ)f 0 (xn+1 ) (xn+1 − xn )2 < ε|xn+1 − xn |.
|xn+1 − xn+2 | = 2
1
2
WINFRIED JUST, OHIO UNIVERSITY
It follows by induction and our choice of ε that
|xn+1 − xn | < εn (x1 − x0 )
(5)
|xn+k − xn | <
k
X
and
εk (xn+1 − xn ) < 2ε(xn+1 − xn ).
i=1
Thus the sequence {xn }∞
n=0 will be a Cauchy sequence and will stay in the
interval [x∗ − ε, x∗ + ε] as long as x0 is sufficiently close to x∗ so that x1 is in this
interval. But the latter can be easily assured by continuity of F at x∗ . Now consider the system (R, F ) for f (x) = x2 + c, where c > 0 is fixed. As
we argued in class, the map F is continuous and strictly increasing on (0, ∞) and
maps this interval onto (−∞, ∞). Moreover, there exists a sequence 0 < a0 < a1 <
· · · < an < . . . such that F (a0 ) = 0 and F (an+1 ) = an for all n. This implies that
F maps (0, a0 ) onto (−∞, 0) and maps F (an , an+1 ) onto (an−1 , an ) for all n > 0.
Proposition 2. Show that for all n > 0 there exists an x ∈ (an−1 , an ) such that x
is a periodic point of period n + 1.
Proof: Consider the map Gn+1 (x) = F n+1 (x) − x. Consider x ∈ (an−1 , an ). Then
x = F 0 (x) ∈ (an−1 , an ),
F (x) ∈ (an−2 , an−1 ), . . .
(6)
F k (x) ∈ (an−k−1 , an−k ), . . .
F n−1 (x) ∈ (a0 , a1 ),
F n (x) < 0.
Since F n (an ) = 0, by continuity of F , as long as x is sufficiently close to but
slightly less than an , we will have F n+1 (x) > 0, and, moreover,
lim F n+1 (x) = ∞.
x→a−
n
Since x stays bounded from above as it approaches an from the left, the latter
implies also that
(7)
lim Gn+1 (x) = ∞.
x→a−
n
Since F maps (−∞, 0) onto R, there exists some b < 0 such that F (b) < an−1 .
By induction one can show that F n maps (an−1 , an ) onto (−∞, 0); hence there
exists c ∈ (an−1 , an ) with F n (c) = b, that is, F n+1 (c) = F (b) < an−1 . For such c
we have
(8)
Gn+1 (c) < 0.
By continuity of Gn+1 on (an−1 , an ), the Intermediate Value Theorem implies
that there exists x ∈ (c, an ) with Gn+1 (x) = 0, that is, F n+1 (x) = x. Thus the
orbit of x consists of the points x, F (x), . . . , F n (x), which must be pairwise distinct
in view of (6) and the fact that the intervals (−∞, 0), (0, a0 ), . . . , (an−1 , an ) are
pairwise disjoint.