why sup x in [a,b] |Pn(x) - f(x) | < ϵ , Pn(x)=a0+a1x+....+anx^n
Are you able to express that in a way that lets 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
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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