1-4-10 mat 4223 weierstrass approximation theorem ch 11

Real Analysis II Chapter 11 Dr. Manuel Berriozábal Weierstrass Approximation Theorem Theorem (Weierstrass Approximation Theorem) Let f(x) be a continuous function defined on a closed interval [a,b]. Then ∀ε> 0, ∃ a

polynomial P(x) such that ∀x∈[a,b] |f(x)-P(x)| < ε Proof

LEMMA 1. ∀x∈ [a,b], n

k=0∑ c k

n xk(1-x)n-k = 1 where c k

n = n!k!(n-k)!

.

Proof By Newton’s binomial formula, ∀u, v∈R and n∈N

(u+v)n = n

k=0∑ c k

n unvn-k

Let u=x and v=1-x

Then 1=1n=[x+(1-x)]n = (u+v)n =n

k=0∑ c k

n unvn-k = n

k=0∑ c k

n xk(1-x)n-k

#

LEMMA 2. ∀x∈R and n∈N0{0}. ∑=

n

k 0Cn

k (k-n x)2 xk (1-x)n-k ≤ 4m

Proof: Case 1: x=1

Then ∑=

n

k 0Cn

k (k-nx)2 xk(1-x)n-k

= ∑=

n

vk

Cnk (k-n)2 1k (1-1)n-k

= Cnn |0|2 .1.1 = 0 ≤

4n

Case 2: x≠1. By use of Newton’s binomial formula, we note that ∀z∈R

(1+z)n = (z+1)n = ∑=

n

k 0Cn

k zk 1n-k = ∑=

n

k 0Cn

k zk, we agree O0=1.

A. Thus ∑=

n

k 0Cn

kzk = (1+z)n identically

Differentiating both sides with respect to z, we obtain

∑=

n

k 1

kCnkzk-1 = n(1+z)n-1

Real Analysis II Chapter 11 Dr. Manuel Berriozábal Weierstrass Approximation Theorem Multiplying both sides by z, we obtain

B. ∑=

n

k 1kCn

kzk = nz(1+z)n-1

Or

∑=

n

k 0

kCnkzk = 0 + ∑

=

n

k 1

kCnkzk = ∑

=

n

k 1

kCnkzk = nz(1+z)n-1

Differentiating both sides of the identity and multiplying the result by z,

C. We obtain ∑=

n

k 0kz Cn

kzk = nz(1+nz)(1+z)n-2

Let z = x

x−1

in A, B, C and multiply each by (1-x)n

We thus obtain

A’: ∑=

n

k 0

Cnkxk(1-x)n-k = 1

B’: ∑=

n

k 0k Cn

kxk(1-x)n-k = nx

C’: ∑=

n

k 0kz k Cn

kxk(1-x)n-k = nx (1-x+nx)

Multiply A’ by n2x2, B’ by -2nx, we obtain

A’’ ∑=

n

k 0

Cnkxk(1-x)n-k . n2x2 = n2x2

B’’ ∑=

n

k 0 k Cn

kxk(1-x)n-k(-2nx) = -2n2x2

C’’ (same as C’) ∑=

n

k 0k2 k Cn

kxk(1-x)n-k = nx (1-x+nx)

Real Analysis II Chapter 11 Dr. Manuel Berriozábal Weierstrass Approximation Theorem Adding the corresponding sides,

∑=

n

k 0 Cn

kxk(1-x)n-k[u2-2nkx+k2] = nx(1-x)

Or

∑=

n

k 0

Cnkxk(1-x)n-k(k-nx)2 = nx(1-x)

We claim x(1-x) ≤41 or 4x(1-x)

2≤ 1

Now 4x(1-x) = (-4x2+4x-1)+1 = -(2x-1)2+1≤0+1=1

Thus ∑=

n

k 0 Cn

k(k-nx)2xk(1-x)n-k = ∑=

n

k 0 Cn

kxk(1-x)n-k(k-nx)2≤4n

Real Analysis II Chapter 11 Dr. Manuel Berriozábal Weierstrass Approximation Theorem Definition: Let f(x) be a real valued function defined on [0,1].

Let Bn(x) = ∑=

n

k 0f(

nk )Cn

kxk(1-x)n-k

Bn(x) is the Bernstein polynomial of degree n for f(x). Lemma 2 (Bernstein’s Theorem) If f: [0,1] R is continuous, then Bn(x) ⇒ f(x)

Proof: Let M = max |f(x)| on [0,1].

Take ε>0. We wish to show ∃ p∈n such that ∀ n>p and x∈[0, 1], |B(x) – F(x)| < ε.

Since f is continuous on [0,1] then f is uniformly continuous on [0,1]. Thus ∃ δ > 0 such that ∀ x’,x’’∈ [0,1], where |x’-x”|<δ, |f(x”)-f(x’)| <.ε/2.

Pick p∈N such that p > 2

Mεδ

. We claim ∀ n>p, |Bn(x)-f(x)| < ε

By lemma 1, ∀ x∈[0,1] and ∀ n>p f(x) = ∑=

n

k 0f(x) C k

n xk(1-x)n-k

Take and fix x∈[0,1]

D. Thus, |Bn(x)-f(x)| = |∑=

n

k 0

f( kn

) Ckn x

k(1-x)n-k - ∑=

n

k 0

f(x) Ckn x

k(1-x)n-k|

= |∑=

n

k 0[f( k

n) – f(x)] f(x) Ck

nxk(1-x)n-k|

≤ ∑=

n

k 0|f( k

n) – f(x)| C k

n xk(1-x)n-k

Let G = { k∈{0,1,…n} | | kn

-x| < δ } and H = {k∈{0,1,…,n} | | kn

-x| ≥ δ}.

Then H∪G = {0,1,…,n} and H∩G = ∅

Real Analysis II Chapter 11 Dr. Manuel Berriozábal Weierstrass Approximation Theorem Theorem (Weierstrass). Let f:[a,b] R be continuous Then ∀ ε>0 ∃ P(x) such that ∀ x∈[a,b] |f(x)-P(x)| < ε

Proof: Case 1: [a,b] = [0,1]. This is Bernstein’s Theorem. Case 2: [a,b] ≠ [0,1] Let h(z) = a+z(b-a). Then h(0) = a and h(1) = b. Also h is strictly increasing and continuous on [a,b]. Thus f(h(z)) = (f . h)(z) is continuous on [0,1]. Take ε > 0. By Bernstein’s Theorem ∃ Q(z) such that ∀ z∈[0,1] ⏐f(h(z))-Q(z)|<ε.

z∈[0,1]

Now x ∈ [a,b] if and only if x-ab-a

1. Thus h(z) = h ( x-ab-a

) = a + x-ab-a

(b-a) = x

Thus Q(z) = Q ( x-ab-a

)

Let P(x) = Q( x-ab-a

). Hence ∀ x∈ [a,b], |f(x) – P(x)| < ε

x∈ [a,b]

#

1-4-10 mat 4223 weierstrass approximation theorem ch 11

Documents