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

  1. Mar 28, 2017 #1
    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
     
  2. jcsd
  3. Mar 28, 2017 #2

    PeroK

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    Are you able to express that in a way that lets other people understand what you are asking?
     
  4. Mar 28, 2017 #3
    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 ?
     
  5. Mar 28, 2017 #4

    PeroK

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    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.
     
  6. Mar 28, 2017 #5
    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
     
  7. Mar 28, 2017 #6

    PeroK

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    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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