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The Weierstrass Approximation Theorem - UUMath Flipbook PDF
The Weierstrass Approximation Theorem−1− There should be a technical proof in every math course. I encourage you to stud
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The Weierstrass Approximation Theorem−1− There should be a technical proof in every math course. I encourage you to study this proof and to make sure you understand every detail. Once you do you will have greatly increased your technical abilities. You will also know why the name of Bernstein is in the Bernstein-B´ezier form of a polynomial. Let me know if you have any questions. Polynomials are widely used. Reasons for their utility include the fact that they are easily evaluated, differentiated, integrated, and patched together. Another major reason is the fact that they are dense in C[a, b]. This is the contents of the celebrated Theorem (Weierstrass Approximation Theorem) (1885). Let f ∈ C[a, b]. Given any ǫ > 0 there exists a polynomial pn of sufficiently high degree n for which |f (x) − pn (x)| ≤ ǫ
for all x ∈ [a, b].
(1)
Many proofs of the Weierstrass Theorem are known. In these notes we present one due to Bernstein, because it is constructive (it actually tells what that polynomial is), and it gives an introduction to an extremely useful form of a polynomial, the Bernstein-B´ezier form of a polynomial (which we will discuss in depth later in this class). For details of the proof see, e.g., the classic book by Phil Davis, Interpolation and Approximation, Dover, 1975. Before starting, notice the following: • f only needs to be continuous. We saw an error expression for an interpolant of f that involved high order derivatives of f . The Weierstrass Theorem is much more general than that. • It is essential that the domain [a, b] be closed and bounded. Easy examples (exercise!) show that the assumptions of the Theorem are necessary. • While we do not require that f be differentiable, it can be shown (see Davis) that if f is differentiable then the Bernstein approximation given below approximates the function and its derivatives simultaneously. Derivatives of the approximation converge to the corresponding derivatives of f as n goes to infinity. Without loss of generality, let [a, b] = [0, 1].
(2)
(For any other interval we’d use a linear change of variables which preserves polynomials and their degrees. Easy exercise!) The n-th Bernstein Polynomial for f is defined to be n X k n k Bn (f, x) = f x (1 − x)n−k n k
(3)
k=0
where, as usual
n! n = k k!(n − k)!
and n! =
n
1 1 × 2 × 3...× n
Bn is called the Bernstein Operator. The binomial coefficient
n k
if k = 0 otherwise
(4)
is pronounced n choose k.
It’s easy to check that the monomials xk (1 − x)n−k are non-negative in [0, 1] and assume their maximum at x = nk . Figure 1 shows graphs of the Benrstein Basis functions bk = nk xk (1 − x)n−k for n = 10 and k = 0, . . . , 10. The following statement was proved by Bernstein. The Weierstrass Theorem is a corollary of this result. −1−
TEX processed by Peter Alfeld, 801-581-6842, [email protected], November 8, 2013 1
1
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
x
Figure 1. The Bernstein Basis functions, n = 10. Theorem (Bernstein). If f ∈ C[0, 1] then lim Bn (f, x) = f (x),
(5)
n−→∞
uniformly for x ∈ [0, 1]. The phrase uniformly in this statement means that in order to make |f (x) − Bn (f, x)| less than ǫ one can choose n > N (ǫ) where N (ǫ) is independent of x. To begin with, we rewrite Bn (f, x) =
n X
n t △t f (0) x t t=0
(6)
where △0 f (x) = f (x) △t+1 f (x) = △t f
1 − △t f (x) x+ n
△ is called a forward difference operator. We obtain, for example, 2
t = 0, 1, 2, . . .
(7)
△0 f (0) = f (0) 1 − f (0) △1 f (0) = f n 1 − △1 f (0) △2 f (0) = △1 f n 2 1 1 =f −f − f − f (0) n n n 1 2 − 2f + f (0) =f n n
(8)
In general (exercise, use induction!) t X k t △ f (0) = f (−1)t−k . n k t
(9)
k=0
The proof of (6) is elementary, but a little technical. We write
Bn (f, x) =
n X k n k f x (1 − x)n−k n k
k=0
=
n X
k=0
=
by (3) n−k n k X n−k k (−1)n−k−j xn−k−j x f j n k j=0
(10)
by the Binomial Theorem k n n−k f (−1)n−k−j xn−j n k j j=0
n n−k X X
k=0
reorganizing We now change summation indices, by replacing n − j with t. Thus as j runs from 0 to n − k, t runs from k to n. We obtain n n X X n n−k k f (−1)t−k xt Bn (f, x) = n k n−t k=0 t=k . n t X X k n n−k t t−k = x f (−1) n k n−t t=0
(11)
k=0
since, for any numbers atk n X n X
atk =
n X t X
atk .
(12)
t=0 k=0
k=0 t=k
Using n t n n−k n! = , = (n − t)!(t − k)!k! t k k n−t 3
(13)
we get n X
t X k n n−k f (−1)t−k n k n − t t=0 k=0 X n t X t k t n (−1)t−k x = f n k t t=0 k=0 n X n △t f (0) = xt t t=0 n X n = △t f (0)xt t t=0
Bn (f, x) =
xt
(14)
as required. Note that if f is a polynomial of degree m then since △t f (0) = 0 for t > m (exercise). Moreover, Bn (f, x) is also a polynomial of degree m as long as n > m. This might lead one to suppose that the Bernstein operator Bn reproduces polynomials. This is not true in general. However, we obtain: n 0 x = 1, 0 n 0 1 n 1 Bn (x, x) = 0 x + x = x, 0 n 1 Bn (1, x) = 1
and
2 1 n 1 n 2 4 n 0 1 −0 − 2 +0 x + x = x2 + (x − x2 ) Bn (x , x) = 0 × x + n2 n2 n n 1 2 0 2
(15) (16)
(17)
So Bernstein Polynomials reproduce linear functions exactly, but not higher degree polynomials. We want to show that for n sufficiently large |Bn (f, x) − f (x)| < ǫ. To get a handle on this we use the fact that n n X X n k n k n−k f (x) = f (x) × 1 = f (x) × x (1 − x) = f (x) x (1 − x)n−k (18) k k k=0
k=0
and break the sum in f (x) − Bn (f, x) into two pieces: X n k k f (x) − Bn (f, x) = x (1 − x)n−k + f (x) − f n k | nk −x|≥δ X k n k + f (x) − f x (1 − x)n−k . n k | nk −x|