Weierstrass Approximation Theorem. Bernstein's Polynomials.

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Weierstrass
Approximation Theorem.
Bernstein's Polynomials.

There are many proofs of a basic theorem of analysis - theorem of Weierstrass - which says that

Theorem 1 (Weierstrass). Let f:[0,1]-> R be a continuous function defined on the closed interval [0,1] and let

be a positive number. There exists a polynomial W:R->R such that | W(x) - f(x) |

for any x from the interval [0,1]. In other words: for every continuous function f:[0,1]->R there exists a sequence

of polynomials uniformly approaching f on the [0,1].

There are generalizations of this theorem and non-constructive proofs. However polynomials

can be defined explicitly Rozmiar: 4501 bajtów

This is a polynomial of n-th degree. It is called n-th Bernstein polynomial for the function f.

Theorem 2. Let f:[0,1]->R be a continuous function and let e>0 be a positive number. There exists a number N such that for n>N

for x from [0,1].

Usually proofs of this theorem use probability methods. That is why 'Literka' decided to present a non-probabilistic proof. Another reason is that the following proof uses, 'Literka' thinks, new ideas (see Lemma 1).

Let us define, for

,

These functions play an important role in probability theory. They define beta functions. Functions

are increasing on the interval

and decreasing on the interval

Lemma 1. Let

or

. Then

Proof. Let us assume that

It will be done, if we prove that the function

is negative for

A simple computation of the derivative of H gives us Rozmiar: 5829 bajtów

because x*(1-x)

. This ends the proof, since F(b)=0.
The case

can be proved in a similar way.

Lemma 2. Let

. There exists N such that for n>N Rozmiar: 4526 bajtów

for

Notice that if the summation of the statement of Lemma 2 extends over all

then the result of this summation is 1.

Proof. Let A be a set of all

such that

Let B be a part of A of elements less than p, let C be a part of A of elements bigger than p.
Define

Notice that

where

For k belonging to B and

we have

so that Lemma 1 can be applied:

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if

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Finally

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A similar inequality can be derived with a set C instead of B. Adding both inequalities we receive the assertion of Lemma 2.

Proof of Theorem 2. Let

and take a positive integer n. Since f(x) is uniformly continuous on [0,1], there is a

such that if

then

Define an integer p to be

Then

only if n is large enough so that Lemma 2 is satisfied. This ends the proof of Theorem 2.

8637

See the list and descriptions of mathematical pages from Mathematical Countryside.

See other pages of 'Literka' from Mathematical Countryside:
Monotonic subsequences.
A remarkable monotonic property of the gamma function .
Weight centers of simple geometrical figures.
An elementary problem can be unsolvable.
Roots of cubic equation. Cardano's formula.
Rudin's Theorem of Complex Analysis.
Exact values of trigonometric functions of angles (n*pi)/17.
Equalities for values of trigonometric functions of angles (n*pi)/17.
Factorization of a polynomial, which defines values of sine function (angles n*pi/17).
Polynomials with roots cos(2*k*pi/n).
Factorization of polynomials with roots cos(2*k*pi/n), where n is Fermat number.
Values of trigonometric functions of angles (n*pi)/257. Part I.
Values of trigonometric functions of angles (n*pi)/257. Part II, Part III, Part IV, Part V, Part VI, Part VII.
Values of trigonometric functions of angles (n*pi)/65537. Part I.
Values of trigonometric functions of angles (n*pi)/65537. Part II, Part III, Part IV, Part V, Part VI, Part VII, Part VIII, Part IX, Part X, Part XI, Part XII, Part XIII, Part XIV.
Construction of a regular pentagon.
Construction of a regular heptadecagon.
Construction of a regular polygon with 257 sides.

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