Why does sup|Pn(x) - f(x)| < ε imply uniform approximation?

  • Context: High School 
  • Thread starter Thread starter feerrr
  • Start date Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
5 replies · 2K views
feerrr
Messages
3
Reaction score
0
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
 

Discussion

Physics news on Phys.org
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]##.