S u p x i n [ a , b ] | P n ( x ) − f ( x ) | < ϵ

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Discussion Overview

The discussion revolves around the mathematical concept of approximating a continuous function \( f \) on the interval \([a, b]\) using polynomial functions \( P_n(x) \). Participants explore the implications of the condition \( \sup_{x \in [a,b]} |P_n(x) - f(x)| < \epsilon \) and its relation to proving that \( f(x) = 0 \) on the interval.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the meaning of the condition \( \sup_{x \in [a,b]} |P_n(x) - f(x)| < \epsilon \) and its implications for the inequalities \( f(x) - \epsilon < P_n(x) < f(x) + \epsilon \).
  • Another participant seeks clarification on how to express their question more clearly for others to understand.
  • A participant states that if \( f \) is in \( C[a,b] \) and \( n = 0, 1, 2, \ldots \), they have the integral condition \( \int_a^b x^n f(x) \, dx = 0 \) and wonders if they can use the supremum condition to show \( f(x) = 0 \) on \([a,b]\).
  • One participant expresses skepticism about directly using the supremum condition and references the Weierstrass approximation theorem, suggesting that proving the statement may be difficult from first principles.
  • Another participant admits to not knowing the Weierstrass approximation theorem and asks if it is possible to show \( f(x) = 0 \) without it, requesting assistance.
  • A later reply reiterates the lack of knowledge about the Weierstrass approximation theorem and suggests looking it up for a proof, while also noting the need to establish that \( f \) is bounded on \([a,b]\).

Areas of Agreement / Disagreement

Participants express uncertainty regarding the use of the supremum condition to prove \( f(x) = 0 \). There is no consensus on whether the Weierstrass approximation theorem is necessary or if an alternative approach exists.

Contextual Notes

Participants acknowledge the need for additional assumptions, such as the boundedness of \( f \) on \([a,b]\), which may affect the validity of their arguments.

feerrr
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why sup x in [a,b] |Pn(x) - f(x) | < ϵ , Pn(x)=a0+a1x+...+anx^n
why f(x)-ϵ<Pn(x)<f(x)+ϵ
 
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Are you able to express that in a way that let's other people understand what you are asking?
 
f in C[a,b] and n=0,1,2,...
we have : ∫ x^n*f(x)dx=0 (integral from a to b )
im trying to show that f(x)=0 in [a,b]
can i use : sup x in [a,b] |Pn(x) - f(x) | < ϵ ( Pn(x)=a0+a1x+...+anx^n ) ? and why ?
 
I don't follow that you can go straight to that.

Do you know the Weierstrass approximation theorem?

I suspect this is hard to prove from first principles.
 
no i don't know the Weierstrass approximation theorem .
can i show that f(x)=0 without using the Weierstrass approximation theorem ? i need some help
 
feerrr said:
no i don't know the Weierstrass approximation theorem .
can i show that f(x)=0 without using the Weierstrass approximation theorem ? i need some help
What I suggest is to look up the theorem on line and find a proof using it. It's still not easy.

I can't immediately see another way.

Also, you need to know or prove that ##f## is bounded on ##[a,b]##.
 

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