Weierstrass Approximation Theorem

Weierstrass Approximation Theorem

Among the great mathematical ideas, Karl Weierstrass, in 1885, gave the original version of the Weierstrass Approximation Theorem. The idea of this theorem was facilitated by Marshall H. Stone in 1948. He also gave the theorem in a generalized form in 1937. Later, the theorem was referred to as the Stone–Weierstrass theorem. 

According to Stone, the Weierstrass approximation theorem can be directed in two directions:

1. Rather than the real interval [a, b], considering an arbitrary compact Hausdorff space, X would give vital results.

2. Instead of the algebra of polynomial functions, a variety of other families of continuous functions should suffice.

What is the Weierstrass Approximation Theorem?

Weierstrass approximation theorem states that every continuous real-valued function on a bounded closed interval [a, b] can be uniformly approximated by a polynomial function with arbitrarily high preciseness. The theorem was given because polynomials are the simplest functions evaluated by computers. In general, you can use the theorem as a solution to various Weierstrass approximation theorem questions.

Therefore, continuous function can be taken as 

f = f(x), x ∈[a, b]. 

The polynomial should be represented by P(x) for number ε > 0.

Therefore,

maxx∈|a,b|| f (x) − P(x)| < ε

Proof of Weierstrass Approximation Theorem

Here, we present two ways to prove the Weierstrass Approximation Theorem to help you solve the Weierstrass approximation theorem HOTS questions. 

Consider these proofs when verifying the theorem:

Proof 1: 

It was given by Weierstrass and can be used to prove comparable properties of a smooth function. 

Take f = f(x) as a result.

Step 1) The function in the result should equal zero for x ∈[a − 1, b + 1]. 

Step 2) Let u = u(t, x). So, 

ut = uxx, t > 0, x ∈ ℝ

Initial values can be taken as u(0, x) = f(x).

u(t,x) =    1   ∫+∞ f(y)e−(x−y)2/(4t) dy.ー

√4πtZ  −∞         

Step 3)

 lim max |u(t, x) − f(x)| = 0

t→0+x∈[a, b]

Step 4) Now,  t > 0, the function u = u(t, x) 

Step 5) For step 5, use Step 3 to find a t0 > 0 we obtain,

max |u(t0, x) − f(x)| < ε/2

x∈[a,b]                               

Now, approximate u(t0, x) by the corresponding Taylor polynomial.

Proof 2: 

Bernstein’s proof of the Weierstrass Approximation Theorem.

You can start by supposing that f ∈ C([a, b], ℝ). 

Also, the sequence of polynomials pn(x) converges uniformly to f(x) on [a, b].

Now as we know all the functions, let’s consider first f ∈ C ([0, 1], ℝ).

As it was given that ε > 0. Along with this, δ > 0. 

So, we can write 

|x − y| ≤ δ =⇒ |f(x) − f(y)|≤ ε ∀x, y ∈ [0, 1].

Let M:= ∥f∥∞. Fix ξ ∈ [0, 1]. Then, if |x−ξ|≤ δ, then |f(x)−f(ξ)| ≤ ε∕2

It can be also be written as, f |x − ξ| ≥ δ, then

|f(x) − f(ξ)| ≤ 2M ≤ 2M（x − ξ）/δ + ε∕2

When we combine the two equations, we notice that 

|f(x) − f(ξ)| ≤ 2M ( x − ξ) 2 +ε∕2    ∀x ∈ [0, 1]

δ

Note that, n

Bn(x, f − f(ξ)) =∑(f − f(ξ)) (k/n) (n/k)xk(1−x)n−k = Bn(x, f)−f(ξ)Bn(x, 1).

 k=0

And,

 n

Bn(x,1) =∑(n/k)xk(1−x)n−k = (x + (1 − x))n = 1

k=0

≤ Bn x, 2M  (x,2M(x – ξ )2  +ε∕2) 

 Now, we can write the next step as:

Bn(x,(x − ξ)2) = x2 + 1 (x − x2) − 2ξx + ξ2

 (Bn(ξ, f) − f(ξ)| ≤ ε∕2 + 2M 1   (ξ − ξ2)

When [0, 1], the maximum of z − z2 would be 1/4

Hence, |(Bn(ξ, f) − f(ξ)| ≤ ε∕2 + M

 2δ2n

Now, if N ≥ M

2δ2 ε

For n ≥ N, we can write 

∥(Bn(ξ, f) − f(ξ)∥∞ ≤  ε.

Hence we proved the theorem for continuous functions on [0, 1]. 

If the function is φ : [0, 1] → [a, b]

From the proof of the functions [0, 1], the case for an arbitrary interval [a, b] follows.

Conclusion 

The Weierstrass approximation theorem is significant in studying the algebra of continuous functions. Further, with an understanding of the theorem, you can easily solve Weierstrass approximation theorem HOTS questions. While solving the Weierstrass approximation theorem questions, make sure you compute only polynomials. There are various ways to prove the theorem that we have covered in the above article.