The Computable Differential Equation Lecture ... - Bruce E. Shapiro

CHAPTER 3. APPROXIMATE SOLUTIONS 61
theorem for f(t) = 1 − √ 1 − t. Construct the sequence 3 of Polynomials
on [0, 1].
P 0 (t) = 0 (3.150)
P n+1 (t) = 1 (
Pn (t) 2 + t ) 2
(3.151)
We observe first that 0 ≤ P (t) ≤ 1. This may be proven by induction. First, ⇐=
P 1 = t/2 ≤ 1 on [0, 1]. Next, assume that P n < 1. Then P n+1 = 1 2 (P 2 n + t) ≤
1
2 (12 + 1) = 1.
Furthermore, each P n is monotonically increasing. Again, prove by induction.
Certainly P 1 = t/2 is monotonically increasing with slope 1/2. Assume P ′ n > 0.
Then P ′ n+1 = 1 2 (2P nP ′ n + 1) > 0 because the P n are all positive. Hence the function
is monotonically increasing.
Next we observe that the sequence of polynomials is itself increasing. To prove
this, observe that
P n+2 (t) − P n+1 (t) = 1 (
P
2
2 n+1 (t) + t − P n 2 (t) − t ) (3.152)
= 1 (
P
2
2 n+1 (t) − P n 2 (t) ) (3.153)
= 1 2 (P n+1(t) + P n (t)) (P n+1 (t) − P n (t)) (3.154)
Since t/2 = P 1 (t) ≥ P 0 (x) = 0 for all t ∈ [0, 1], by applying this recursively we see
that the sequence is monotonically increasing: P n+1 (t) ≥ P n t on [0, 1]
.
Since the sequence P 0 , P 1 , . . . is bounded and monotonically increasing it converges
to some limit P (t). But
lim P 1
n+1(t) = lim
n→∞ n→∞ 2 (P n(t) 2 + t) (3.155)
P (t) = 1 2 (P 2 (t) + t) (3.156)
0 = P 2 (t) − 2P (t) + t (3.157)
P (t) = 1 ± √ 1 − t (3.158)
We need to take the negative square root because of the restriction P (t) ≤ 1. Hence
lim P n(t) = 1 − √ 1 − t (3.159)
n→∞
3 Use of the fixed point iteration x 0 = 0, x n+1 = 1 2 (u + x2 n) to find √ z, where z = 1 − u and
x = 1 − √ z, was discovered in ancient Babylonia.
c○2007, B.E.Shapiro
Last revised: May 23, 2007
Math 582B, Spring 2007
California State University Northridge