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

62 CHAPTER 3. APPROXIMATE SOLUTIONS
Finally, since the P n (t) are a monotonic sequence with P n+1 (t) ≥ P n (t), we know
that P m (t) ≥ P n (t) whenever m > n. Therefore
and hence
P m (t) − P m (1) ≤ P n (t) − P n (1) (3.160)
P m (t) − P n (t) ≤ P m (1) − P n (1) (3.161)
lim (P m(t) − P n (t)) ≤ lim (P m(1) − P n (1)) (3.162)
m→∞ m→∞
P (t) − P n (t) ≤ 1 − P n (1) (3.163)
1 − √ 1 − t − P n (t) ≤ 1 − P n (1) (3.164)
=⇒
Pick any ɛ > 0. Then, since P n (1) → P (1) as n → ∞, there exists some n such that
1 − P n (1) < ɛ, in which case
1 − √ 1 − t − P n (t) < ɛ (3.165)
i.e.,
‖f(t) − P n (t)‖ < ɛ (3.166)
This proves the Weierstrass Approximation Theorem for the special case f(t) =
1 − √ 1 − t.
Next, suppose that f(t) = |t − c| for some number c ∈ [0, 1]. Let u = |t − c| and
pick any ɛ > 0. Since u, ɛ ≥ 0, we know that
√
Now define a new variable v,
u 2 + ɛ2 4 < u2 + uɛ + ɛ2 4 = ( ɛ
2 + u ) 2
(3.167)
√
u 2 + ɛ2 4 < ɛ 2 + u (3.168)
u 2 + ɛ2 4 − u < ɛ 2
(3.169)
v = 1 − u 2 − ɛ 2 /4 (3.170)
By the previous case there is some polynomial Q(v) that approximates 1 − √ 1 − v
arbitrarily closely:
∣
∣1 − √ 1 − v − Q(v) ∣ ∣ < ɛ/2 (3.171)
Define a new polynomial P (v) = 1 − Q(v); then we have
∣ √ 1 − v − P (v) ∣ < ɛ/2 (3.172)
√
∣ ∣∣∣∣ u
∣
2 + ɛ2 4 − P (v) < ɛ/2 (3.173)
Math 582B, Spring 2007
California State University Northridge
c○2007, B.E.Shapiro
Last revised: May 23, 2007