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Does every continuous function has a power series expansion on a closed interval 
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#1
Oct3109, 03:37 AM

P: 679

By Weierstrass approximation theorem, it seems to be obvious that every continuous function has a power expansion on a closed interval, but I'm not 100% sure about this. Is this genuinely true or there're some counterexamples?



#2
Oct3109, 07:36 AM

P: 607

This is incorrect. The sum of a power series is an analytic function. So, for example, the function x on the interval [1,1] is not the sum of a power series, since it is not even differentiable at x=0.



#3
Oct3109, 10:01 AM

P: 679

But by Weierstrass you can always construct a sequence of polynomials uniformly converging to it, isn't it? I thought differentiability is required only if you want to do something like Taylor series expansion.



#4
Oct3109, 11:21 AM

P: 679

Does every continuous function has a power series expansion on a closed interval
Em, now I see what you mean. As long as a function can be represented by a power series, it can be differentiated as many times as you want, so x can not be represented by power series.
But by Weierstrass theorem you can always construct a sequence of polynomials uniformly converging to it, so in what sense is this different with saying that it can be written as a power series? I'm very confused 


#5
Oct3109, 07:43 PM

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PF Gold
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Also there exist infinitely smooth functions that aren't represented by their Taylor series. Consider [tex]f(x) = e^{ \frac 1 {x^2}},\ x \neq 0[/tex] and f(0) = 0. This has derivatives of all orders equal to 0 when x = 0 so its Taylor series sums to 0, not to f(x) except at 0. Yet on any finite interval it can be uniformly approximated by polynomials. 


#6
Nov109, 07:36 AM

P: 679




#7
Nov109, 06:57 PM

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PF Gold
P: 7,663

[tex]f(x) = \sum_{n=0}^\infty a_nx^n[/tex] with a positive radius of convergence then f must be as differentiable as the right side and, moreover, differentiating both sides shows the only choice for the coefficients are the Taylor coefficients. So Taylor series is the only game in town for a power series. 


#8
Nov109, 10:45 PM

P: 679

And, why in the Bernstein case we don't need the differentiability condition? Please bear with if the question is too naive; I'm not majoring in math so I'm really not good at these analysis stuff. Thanks. 


#9
Nov109, 11:53 PM

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PF Gold
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The question is: given a sequence of Bernstein polynomials, do they actually converge to a power series? In general the coefficients will change from one polynomial to the next



#10
Nov209, 09:05 AM

P: 679

But for those can be written in power series, the coefficients must converge to what we collect from Bernstein polynomials, is that right? 


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