What are they doing here? (polynomials-functions)

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Discussion Overview

The discussion centers around the Stone–Weierstrass approximation theorem, particularly focusing on the proof and transformations involved in approximating continuous functions on the interval [0,1] using polynomial functions. Participants express confusion regarding specific steps in the proof and the implications of certain transformations.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the necessity of the condition that the function must be zero at the endpoints of the interval [0,1].
  • Another participant seeks clarification on the definition of the function f and the hypotheses of the theorem.
  • A participant explains that the theorem aims to show that any continuous function can be approximated by a polynomial, but expresses confusion about the transformations used in the proof.
  • One participant introduces a decomposition of the function into two parts, suggesting that any continuous function can be represented as a sum of two functions, where one is zero at the endpoints, facilitating the proof.
  • A later reply indicates that the participant has gained clarity on the topic after the explanations provided.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and confusion regarding the proof and transformations involved in the theorem. There is no consensus on the clarity of the proof or the necessity of certain conditions.

Contextual Notes

Some participants highlight missing assumptions and the need for clearer definitions regarding the function f and the hypotheses of the theorem, indicating potential gaps in understanding the proof.

nalkapo
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I am currently studying Stone–Weierstrass approximation theorem, and some parts really confuse me.
basically, the Weierstrass approximation theorem states that every continuous function defined on an interval [a,b] can be uniformly approximated as closely as desired by a polynomial function.

In the proof,
at first they say on [a,b]=[0,1]
f(0)=f(1)=0. Why do they want to say that at the end points function must be zero?

second, if the theorem is proved for this case, consider
g(x)= f(x)- f(0) - x[f(1)-f(0)] (0<= x <= 1)
Here g(0)=g(1)=0

Can anyone explain the meaning and purpose of these transformation?

My guess is that they want to define function f and then find a g(x) which looks like polynomials. Is that right? and also, is there any relationship between these transformation and uniformly convergence?
 
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nalkapo said:
I am currently studying Stone–Weierstrass approximation theorem, and some parts really confuse me.
basically, the Weierstrass approximation theorem states that every continuous function defined on an interval [a,b] can be uniformly approximated as closely as desired by a polynomial function.

In the proof,
at first they say on [a,b]=[0,1]
f(0)=f(1)=0. Why do they want to say that at the end points function must be zero?

I don't understand what you are saying. What is "f"? What are the hypotheses here?

second, if the theorem is proved for this case, consider
g(x)= f(x)- f(0) - x[f(1)-f(0)] (0<= x <= 1)
Here g(0)=g(1)=0

Can anyone explain the meaning and purpose of these transformation?

My guess is that they want to define function f and then find a g(x) which looks like polynomials. Is that right? and also, is there any relationship between these transformation and uniformly convergence?
 
HallsofIvy said:
I don't understand what you are saying. What is "f"? What are the hypotheses here?

f is any continuous function f(x) on an interval [a,b] . in the proof they chose [0,1] .
they want to prove that for any continuous function, there is a polynomial that can be approximated to this function as much as desired.
i couldn't understand the change between those function.
 
Any function f(x) on a closed interval [a,b] can be represented as the sum of two functions. h(x)=f(a) + [(x-a)/(b-a)]f(b) and g(x)=f(x)-h(x). g(a)=g(b)=0. The proof is then done in two steps: first prove it for any continuous function g(x), which is 0 at the two ends, then use the decomposition to prove it for any continuous function f(x) by adding h(x) to the polynomial approximations for g(x).
 
thanks mathman,
now i think i got the idea...
 

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