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The Weierstrass Approximation Theorem
Transcript of The Weierstrass Approximation Theorem
MATH 4540: Analysis Two
The Weierstrass Approximation Theorem
James K. Peterson
Department of Biological Sciences and Department of Mathematical SciencesClemson University
February 26, 2018
MATH 4540: Analysis Two
Outline
1 The Wierstrass Approximation Theorem
2 MatLab Implementation
3 Compositions of Riemann Integrable Functions
MATH 4540: Analysis Two
The Wierstrass Approximation Theorem
The next result is indispensable in modern analysis. Fundamentally, itstates that a continuous real-valued function defined on a compact setcan be uniformly approximated by a smooth function. This is usedthroughout analysis to prove results about various functions.
We can often verify a property of a continuous function, f , by proving ananalogous property of a smooth function that is uniformly close to f .
We will only prove the result for a closed finite interval in <. The general
result for a compact subset of a more general set called a Topological
Space is a modification of this proof which is actually not that more
difficult, but that is another story.
MATH 4540: Analysis Two
The Wierstrass Approximation Theorem
Theorem
Let f be a continuous real-valued function defined on [0, 1]. Forany ε > 0, there is a polynomial, p, such that |f (t)− p(t)| < εfor all t ∈ [0, 1], that is || p − f ||∞< ε
Proof
We first derive some equalities. We will denote the interval [0, 1]by I . By the binomial theorem, for any x ∈ I , we have
n∑k=0
(n
k
)xk(1− x)n−k = (x + 1− x)n = 1. (α)
MATH 4540: Analysis Two
The Wierstrass Approximation Theorem
Proof
Differentiating both sides of Equation α, we get
0 =n∑
k=0
(n
k
)(kxk−1(1− x)n−k − xk(n − k)(1− x)n−k−1
)
=n∑
k=0
(n
k
)xk−1(1− x)n−k−1
(k(1− x) − x(n − k)
)
=n∑
k=0
(n
k
)xk−1(1− x)n−k−1
(k − nx
)
Now, multiply through by x(1− x), to find
0 =n∑
k=0
(n
k
)xk(1− x)n−k(k − nx).
Differentiating again, we obtain
MATH 4540: Analysis Two
The Wierstrass Approximation Theorem
Proof
0 =n∑
k=0
(n
k
)d
dx
(xk(1− x)n−k(k − nx)
).
This leads to a series of simplifications. It is pretty messy and many textsdo not show the details, but we think it is instructive.
0 =n∑
k=0
(n
k
)[−nxk(1− x)n−k
+(k − nx)(
(k − n)xk(1− x)n−k−1 + kxk−1(1− x)n−k)]
=n∑
k=0
(n
k
)[−nxk(1− x)n−k
+(k − nx)(1− x)n−k−1xk−1(
(k − n)x + k(1− x))]
MATH 4540: Analysis Two
The Wierstrass Approximation Theorem
Proofor
0 =n∑
k=0
(n
k
)(− nxk(1− x)n−k + (k − nx)2(1− x)n−k−1xk−1
)= −n
n∑k=0
(n
k
)xk(1− x)n−k +
n∑k=0
(n
k
)(k − nx)2xk−1(1− x)n−k−1
Thus, since the first sum is 1, we have
n =n∑
k=0
(n
k
)(k − nx)2xk−1(1− x)n−k−1
and multiplying through by x(1− x), we have
nx(1− x) =n∑
k=0
(n
k
)(k − nx)2xk(1− x)n−k
MATH 4540: Analysis Two
The Wierstrass Approximation Theorem
Proofor
x(1− x)
n=
n∑k=0
(n
k
)(k − nx
n
)2xk(1− x)n−k
This last equality then leads to the
n∑k=0
(n
k
)(x − k
n
)2xk(1− x)n−k =
x(1− x)
n(β)
We now define the nth order Bernstein Polynomial associated with f by
Bn(x) =n∑
k=0
(n
k
)xk(1− x)n−k f
(kn
).
Note f (x)− Bn(x) =∑n
k=0
(nk
)xk(1− x)n−k
[f (x)− f
(kn
)].
MATH 4540: Analysis Two
The Wierstrass Approximation Theorem
Proof
Also note that f (0)− Bn(0) = f (1)− Bn(1) = 0, so f and Bn match atthe endpoints. It follows that
| f (x)− Bn(x) | ≤n∑
k=0
(n
k
)xk(1− x)n−k
∣∣∣f (x)− f(kn
)∣∣∣. (γ)
Now, f is uniformly continuous on I since it is continuous. So, givenε > 0, there is a δ > 0 such that |x − k
n | < δ ⇒ |f (x)− f ( kn )| < ε
2 .Consider x to be fixed in [0, 1]. The sum in Equation γ has only n + 1terms, so we can split this sum up as follows. Let {K1,K2} be a partitionof the index set {0, 1, ..., n} such that k ∈ K1 ⇒ |x − k
n | < δ and
k ∈ K2 ⇒ |x − kn | ≥ δ.
MATH 4540: Analysis Two
The Wierstrass Approximation Theorem
Proof
Then
| f (x)− Bn(x) | ≤∑k∈K1
(n
k
)xk(1− x)n−k
∣∣∣f (x)− f(kn
)∣∣∣+∑k∈K2
(n
k
)xk(1− x)n−k
∣∣∣f (x)− f(kn
)∣∣∣.which implies
|f (x)− Bn(x)| ≤ ε
2
∑k∈K1
(n
k
)xk(1− x)n−k
+∑k∈K2
(n
k
)xk(1− x)n−k
∣∣∣f (x)− f(kn
)∣∣∣=
ε
2+∑k∈K2
(n
k
)xk(1− x)n−k
∣∣∣f (x)− f(kn
)∣∣∣.
MATH 4540: Analysis Two
The Wierstrass Approximation Theorem
Proof
Now, f is bounded on I , so there is a real number M > 0 such that|f (x)| ≤ M for all x ∈ I . Hence∑
k∈K2
(n
k
)xk(1− x)n−k
∣∣∣f (x)− f(kn
)∣∣∣ ≤ 2M∑k∈K2
(n
k
)xk(1− x)n−k .
Since k ∈ K2 ⇒ |x − kn | ≥ δ, using Equation β, we have
δ2∑k∈K2
(n
k
)xk(1−x)n−k ≤
∑k∈K2
(n
k
)(x − k
n
)2xk(1−x)n−k ≤ x(1− x)
n.
This implies that ∑k∈K2
(n
k
)xk(1− x)n−k ≤ x(1− x)
δ2n.
MATH 4540: Analysis Two
The Wierstrass Approximation Theorem
Proof
and so combining inequalities
2M∑k∈K2
(n
k
)xk(1− x)n−k ≤ 2Mx(1− x)
δ2n
We conclude then that∑k∈K2
(n
k
)xk(1− x)n−k
∣∣∣f (x)− f(kn
)∣∣∣ ≤ 2Mx(1− x)
δ2n.
Now, the maximum value of x(1− x) on I is 14 , so
∑k∈K2
(n
k
)xk(1− x)n−k
∣∣∣f (x)− f(kn
)∣∣∣ ≤ M
2δ2n.
MATH 4540: Analysis Two
The Wierstrass Approximation Theorem
Proof
Finally, choose n so that n > Mδ2ε . Then M
nδ2 < ε implies M2nδ2 <
ε2 . So,
Equation γ becomes
| f (x)− Bn(x) |≤ ε
2+ε
2= ε.
Note that the polynomial Bn does not depend on x ∈ I , since n onlydepends on M, δ, and ε, all of which, in turn, are independent of x ∈ I .So, Bn is the desired polynomial, as it is uniformly within ε of f .
Comment
A change of variable translates this result to any closed interval [a, b].
MATH 4540: Analysis Two
MatLab Implementation
Let’s write some code to implement Bernstein polynomials In Octave/MatLab. We use the function Bernstein (f,a,b,n) like this:
X = l i n s p a c e (0 ,10 ,401) ;f = @( x ) e .ˆ( − . 3∗ x ) . ∗ cos (2∗ x + 0 . 3 ) ;B3 = Be r n s t e i n ( f , 0 , 1 0 , 3 ) ;p l o t (X, f (X) ,X, B3(X) ) ;
5 B10 = Be r n s t e i n ( f , 0 , 1 0 , 10 ) ;p l o t (X, f (X) ,X, B3(X) ,X, B10 (X) ) ;B25 = Be r n s t e i n ( f , 0 , 1 0 , 25 ) ;p l o t (X, f (X) ,X, B3(X) ,X, B10 (X) ,X, B25 (X) ) ;B45 = Be r n s t e i n ( f , 0 , 1 0 , 45 ) ;
10 p l o t (X, f (X) ,X, B3(X) ,X, B10 (X) ,X, B25 (X) ,X, B45 (X) ) ;B150 = Be r n s t e i n ( f , 0 , 10 , 150 ) ;p l o t (X, f (X) ,X, B3(X) ,X, B10 (X) ,X, B25 (X) ,X, B45 (X) ,X
, B150 (X) ) ;l egend ( ’f ’ , ’B3 ’ , ’B10 ’ , ’B25 ’ , ’B45 ’ , ’B140 ’ ) ;x l a b e l ( ’x ’ ) ; y l a b e l ( ’y ’ ) ;
15 t i t l e ( ’f(x) = e^{ -3 x} cos (2 x + 0.3) on [0 ,10] and
Bernstein Polynomials ’ ) ;
MATH 4540: Analysis Two
MatLab Implementation
f u n c t i o n p = Be r n s t e i n ( f , a , b , n )% compute Be r n s t e i n po l y nom i a l app rox ima t i on% of o r d e r n on the i n t e r v a l [ a , b ] to f% f i s the f u n c t i o n
5 % n i s the o r d e r o f the Be r n s t e i n po l y nom i a lp = @( x ) 0 ;f o r i = 1 : n+1
k = i −1;% conve r t the i n t e r v a l [ a , b ] to [ 0 , 1 ] he re
10 y = @( x ) ( x−a ) /(b−a ) ;% conve r t the p o i n t s we e v a l u a t e f at to be
i n [ a , b ]% i n s t e a d o f [ 0 , 1 ]z = a + k ∗ ( b−a ) /n ;q = @( x ) nchoosek (n , k ) ∗ ( y ( x ) . ˆ k ) .∗((1 − y ( x ) )
. ˆ ( n−k ) ) ∗ f ( z ) ;15 p = @( x ) ( p ( x ) + q ( x ) ) ;
endend
MATH 4540: Analysis Two
MatLab Implementation
We deliberately tried to approximate an oscillating function with theBernstein polynomials and we expected this to be difficult because wechose the interval [0, 10] to work on.
The Bernstein polynomials require us to calculate the binomial
coefficients. When n is large. such as in our example with n = 150,
running this code generates errors about the possible loss of precision in
the use of nchoosek which is the binomial coefficient function. You can
see how we did with the approximations in the next figure.
MATH 4540: Analysis Two
MatLab Implementation
MATH 4540: Analysis Two
MatLab Implementation
Example
If∫ ba f (s)snds = 0 for all n with f continuous, then f = 0 on [a, b].
Solution
Let Bn(f ) be the Bernstein polynomial of order n associated with
f . From the assumption, we see∫ ba f (s)Bn(f )(s)ds = 0 for all n
also. Now consider∣∣∣∣∫ b
af 2(s)ds −
∫ b
af (s)Bn(f )(s)ds
∣∣∣∣ =
∣∣∣∣∫ b
af (s)(f (s)− Bn(f )(s))ds
∣∣∣∣≤
∫ b
a||f − Bn(f )||∞ |f (s)|ds
Then given ε > 0, there is a N so that n > N implies||f − Bn(f )||∞ < ε/((||f ||∞ + 1)(b − a)).Then we have n > N implies
MATH 4540: Analysis Two
MatLab Implementation
Solution
∣∣∣∣∫ b
a
f 2(s)ds −∫ b
a
f (s)Bn(f )(s)ds
∣∣∣∣ < ε
(||f ||∞ + 1)(b − a)
∫ b
a
|f (s)|ds
<ε
(||f ||∞ + 1)(b − a)||f ||∞ (b − a) < ε
This tells us∫ b
af (s)Bn(f )(s)ds →
∫ b
af 2(s)ds. Since∫ b
af (s)Bn(f )(s)ds = 0 for all n, we see
∫ b
af 2(s)ds = 0. It then follows
that f = 0 on [a, b] using this argument.
Assume f 2 is not zero at some point c in [a, b]. We can assume c is aninterior point as the argument at an endpoint is similar. Since f 2(c) > 0and f is continuous, there is r with (r − c , r + c) ⊂ [a, b] andf 2(x) > f 2(c)/2 on that interval. Hence
MATH 4540: Analysis Two
MatLab Implementation
Solution
0 =
∫ b
a
f 2(s)ds =
∫ c−r
a
f 2ds +
∫ c+r
c−rf 2(s)ds +
∫ b
c+r
f 2(s)ds
> (2r)f 2(c)/2 > 0
This is a contradiction and so f 2 = 0 on [a, b] implying f = 0 on [a, b].
MATH 4540: Analysis Two
Compositions of Riemann Integrable Functions
We already know that continuous functions and monotone functions are
classes of functions which are Riemann Integrable on the interval [a, b].
Hence, since f (x) =√x is continuous on [0,M] for any positive M, we
know f is Riemann integrable on this interval. What about the
composition√g where g is just known to be non negative and Riemann
integrable on [a, b]? If g were continuous, since compositions of
continuous functions are also continuous, we would have immediately
that√g is Riemann Integrable. However, it is not so easy to handle the
case where we only know g is Riemann integrable.
MATH 4540: Analysis Two
Compositions of Riemann Integrable Functions
Let’s try this approach. Using the Weierstrass Approximation Theorem,we know given a finite interval [c , d ], there is a sequence of polynomials{pn(x)} which converge in || · ||∞ to
√x on [c , d ]. Of course, the
polynomials in this sequence will change if we change the interval [c , d ],but you get the idea.
To apply this here, note that since g is Riemann Integrable on [a, b], gmust be bounded. Since we assume g is non negative, we know thatthere is a positive number M so that g(x) is in [0,M] for all x in [a, b].Thus, there is a sequence of polynomials {pn} which converge in || · ||∞to√· on [0,M].
Next, we know a polynomial in g is also Riemann integrable on [a, b](f 2 = f · f so it is integrable and so on). Hence, pn(g) is Riemannintegrable on [a, b]. Then given ε > 0, we know there is a positive N sothat
| pn(u)−√u | < ε, if n > N and u ∈ [0,M].
MATH 4540: Analysis Two
Compositions of Riemann Integrable Functions
Thus, in particular, since g(x) ∈ [0,M], we have
| pn(g(x))−√g(x) | < ε, if n > N and x ∈ [a, b].
We have therefore proved that pn ◦ g converges || · ||∞ to√g on [0,M].
What we need next is a new theorem: If fn converges to f in || · ||∞ on[a, b] with all fn Riemann integrable, then the limit function f is also
Riemann integrable and∫ b
afn(s)ds →
∫ b
af (s)ds.
We can indeed prove such a theorem. Hence, we know√g is Riemann
integrable when g is Riemann Integrable. Note the kind of arguments weuse here. The ideas of functions converging in the sup norm and thequestions what properties are retained in the limit are of greatimportance.
Note this kind of proof makes us think that compositions of a continuous
function ( like√· ) with a Riemann integrable function will give us new
Riemann integrable functions as the argument we used above would
translate easily to the new context. We will explore this more later.
MATH 4540: Analysis Two
Compositions of Riemann Integrable Functions
In general, the composition of Riemann Integrable functions is notRiemann integrable. Here is the standard counterexample. Define f on[0, 1] by
f (y) =
{1 if y = 00 if 0 < y ≤ 1
and g on [0, 1] by
g(x) =
1 if x = 01/q if x = p/q, (p, q) = 1, x ∈ (0, 1] and x is rational0 if x ∈ (0, 1] and x is irrational
Let’s show g is RI on [0,1]. First define g to be 0 at x = 1. If we show
this modification of g is RI, g will be too. Let q be a prime number
bigger than 2. Then form the uniform partition
πq = {0, 1/q, . . . , (q − 1)/q, 1}. On each subinterval of this partition, we
see mj = 0. Inside the subinterval [(j − 1)/q, j/q], the maximum value of
g is 1/q. Hence, we have Mj = 1/q.
MATH 4540: Analysis Two
Compositions of Riemann Integrable Functions
This gives
U(g ,πq)− L(g ,πq) =∑πq
(1/q)∆xi = 1/q
Given ε > 0, there is an N so that 1/N < ε. Thus if q0 is a prime withq0 > N, we have U(g ,πq0)− L(g ,πq0) < ε.
So if π is any refinement of πq0 , we have U(g ,π)− L(g ,π) < ε also,Thus g satisfies the Riemann Criterion and so g is RI on [0, 1]. It is also
easy to see∫ 1
0g(s)ds = 0. Now f ◦ g becomes
f (g(x)) =
f (1) if x = 0f (1/q) if x = p/q, (p, q) = 1, x ∈ (0, 1] and x rationalf (0) if 0 < x < 1 and x irrational
=
1 if x = 00 if if x rational ∈ (0, 1)1 if if x irrational ∈ (0, 1)
The function f ◦ g above is not Riemann integrable as U(f ◦ g) = 1 and
L(f ◦ g) = 0. Thus, we have found two Riemann integrable functions
whose composition is not Riemann integrable!
MATH 4540: Analysis Two
Compositions of Riemann Integrable Functions
Homework 18
18.1 For f (x) = sin2(3x), graph the Bernstein polynomials B5(f ), B10(f )and B15(f ) along with f simultaneously on the interval [−2, 4] onthe same graph in MatLab. This is a word doc report so write yourcode, document it and print out the graph as part of your report.
18.2 For f (x) = sin(x) +√x , graph the Bernstein polynomials B5(f ),
B10(f ) and B15(f ) along with f simultaneously on the interval [0, 6]on the same graph in MatLab. This is a word doc report so writeyour code, document it and print out the graph as part of yourreport.
MATH 4540: Analysis Two
Compositions of Riemann Integrable Functions
Homework 18
18.3 Fix x 6= 0 in <. Let the sequence (an) be defined by an = cos(3nx).Prove this sequence does not always have a limit but it does have atleast one subsequential limit. Of course, because of periodicity, it isenough to look at points x ∈ [0, 2π]. For example, there is a limit ifx = π/2 and other special points. Let’s look at x = 1 carefully.Prove this sequence does not have a limit but it does have at leastone subsequential limit.
Hint
The Bolzano - Weierstrass Theorem tells us there is a subsequence(cos(3nk ) which converges to a number α in [−1, 1].
There are identities which show us cos(3θ) = 4 cos3(θ)− 4 cos(θ).Hence, we know
cos(3 (3nk )) −→ 4α3 − 4α
MATH 4540: Analysis Two
Compositions of Riemann Integrable Functions
Homework 18
18.3 Continued
Hint
If α 6= 0, use the above fact to show the subsequences cos(3nk+1)and cos(3nk ) converge to different values. This implies the limit cannot exist.
If α = 0, the argument is harder. This tells us sin(3nk ) −→ 1. Nowshow sin(3θ) = sin(θ) (4 cos2(θ)− 1) and also show that the sin(3θ)identity implies sin(3n) can not have a limit.
Finally, since sin(θ) = ±√
1− cos2(θ), if we assume the limit ofcos(3n) exists and equals β, then by looking at all the quadrantchoices that the value of β implies, we find the limit of sin(3n) mustexist. This is a contradiction and so the limit of cos(3n) does notexist.
MATH 4540: Analysis Two
Compositions of Riemann Integrable Functions
Homework 18
18.3 Continued
Hint
This sequence is hard to analyze as using the 2π periodicty of cos we canrecast the sequence using remainders.
3nk = Lnk 2π + rnk , rnk ∈ (0, 1)
cos(3nk ) = cos(Lnk 2π + rnk ) = cos(2πrnk )
We have no idea how the remainder terms rnk are distributed in (0, 1)and that is why this analyis is very nontrivial.