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

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CHAPTER 3. APPROXIMATE SOLUTIONS 63<br />

But since P (v) is a polynomial in v, and v is a polynomial in u, then P is also a<br />

polynomial in u. But u is a polynomial in t, hence P is a polynomial in t, which we<br />

write as follows:<br />

∣ ∣∣∣∣<br />

√<br />

u 2 + ɛ2 4 − P (t) ∣ ∣∣∣∣<br />

< ɛ 2<br />

(3.174)<br />

Now since<br />

u 2 + ɛ2 4 < u2 + ɛ2 4 + ɛ|u| = (<br />

|u| + ɛ 2) 2<br />

(3.175)<br />

√<br />

u 2 + ɛ2 4 < |u| + ɛ 2<br />

√<br />

u 2 + ɛ2 4 − |u| < ɛ 2<br />

(3.176)<br />

(3.177)<br />

Adding equations 3.174 and 3.177, we get<br />

√<br />

∣ ∣∣∣∣<br />

√<br />

u<br />

∣<br />

2 + ɛ2 4 − P (t) + u 2 + ɛ2 4<br />

− |u| < ɛ (3.178)<br />

Since u 2 + ɛ 2 /4 > u 2 , we have<br />

√<br />

∣ ∣∣∣∣ √<br />

∣ ∣∣∣∣ u<br />

∣<br />

2 + ɛ2 4 − P (t) +<br />

u<br />

∣<br />

2 + ɛ2 4 − |u| < ɛ (3.179)<br />

√<br />

∣ ∣∣∣∣ u<br />

∣<br />

2 + ɛ2 4 − P (t) +<br />

√u<br />

∣ |u| − 2 + ɛ2 4 ∣ < ɛ (3.180)<br />

Using the triangle inequality on the last expression gives<br />

|P (t) − u| < ɛ (3.181)<br />

which proves the Weierstrass Approximation <strong>The</strong>orem for f(t) = u = |t − c|.<br />

Now consider any piecewise linear function<br />

N∑<br />

f(t) = b + a k |t − c k | (3.182)<br />

By the previous case there exist N + 1 polynomials Q 0 , Q 1 , . . . , Q N such that<br />

k=0<br />

|a k |t − c m | − Q m (t)| < ɛ N<br />

(3.183)<br />

By the triangle inequality,<br />

where<br />

c○2007, B.E.<strong>Shapiro</strong><br />

Last revised: May 23, 2007<br />

|f(t) − P (t)| < ɛ (3.184)<br />

P (t) =<br />

N∑<br />

Q k (t) (3.185)<br />

k=1<br />

Math 582B, Spring 2007<br />

California State University Northridge

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