Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Define $B$ to be the set of functions $f:[0,1]\rightarrow \mathbb{R}$ for which there exists a dense set $C\subset [0,1]$ of computables numbers and an algorithm $F$ such that for any $x_0\in C,$ in the input $(f,x_0),$ $F$ return $(a_0,a_1,a_2,\ldots),$ where $a_n$ is the $n$th Taylor coefficient of $f$ around $x_0$ and for any $n,$ $a_n$ is a computable number.

Is it true that for any metric $d$ (non trivial) on the set of real analytic functions on $[0,1]$, any $f$ in this set and any $\epsilon>0$ there exists a $g\in B$ with $d(f,g)<\epsilon?$

share|improve this question
    
Since you changed the nature of your question, from "does there exists a topology such that [...] is not true" to "is it true that for any distance [...] is true", I must remove my first answer. –  Loïc Teyssier Feb 8 '13 at 14:08
1  
The question is as unclear as ever. First, it is totally unclear what kind of formulas are allowed. What is worse is that there are only countably many computable numbers, but continuum many points of $[0,1]$. It follows easily that regardless of what $F$ is, the Taylor series for an analytic function can have computable coefficients around every $x_0\in[0,1]$ only if $f$ is constant. –  Emil Jeřábek Feb 8 '13 at 15:33
    
I edited it again. Many thanks because of your comments. –  Umberto Feb 8 '13 at 16:33
1  
How is the algorithm supposed to get $f$ as an input? –  Emil Jeřábek Feb 8 '13 at 16:58
    
This is another question I had (see How are enconded the real analytic functions). –  Umberto Feb 8 '13 at 17:53

1 Answer 1

Let me just present the following example which I think relates somehow to the OP question, although it does not answer (anymore) the (now modified) question.

Considering the subset $C$ of power series $f(x):=\sum_{n=0}^\infty{a_{n}x^n}$ where the sequence $(a_n)_n$ is bounded (thus the series is convergent). Then $C$ can be equipped with the normed topology given by $\|f\|:=\sup_{n}|a_n|$. In that case the subspace consisting in all real polynomials is not dense in $C$, since e.g. $f(x):=\frac{1}{1-x}$ is not in the adherence of the polynomials for this topology.

share|improve this answer
    
but for $f\in C,$ $f(1)$ is not well defined. I do not want in $A$ to be maps with singularities in $[0,1].$ –  Umberto Feb 8 '13 at 13:25
    
Ok, consider $\frac{1}{1+x^2}$ if it suits you better, that is not the point. The point is that there do exist "natural" topologies for which "intuitive" approximation schemes do not hold. I believe this was your question, right? –  Loïc Teyssier Feb 8 '13 at 13:28
    
But $1/(1+x^2)\in B,$ because we can find an explicit formula for the Taylor coefficients. –  Umberto Feb 8 '13 at 13:36
    
I really don't understand what you mean by a metric that "consider all the coordinates". Anyway, my example is clear enough I think, and it does not "consider all the coordinates" since $x$ does not appear in the definition of the norm (was that what you meant??). That being said, if you don't want singularities in the closure of your interval, then apply a scaling $x\mapsto{x/2}$ to all elements of $C$, this does not change the argument I presented above. And, again, your set $B$ is not defined correctly! –  Loïc Teyssier Feb 8 '13 at 13:41
    
It does not answer my question because the set B is strictly larger than the set of polynomials. –  Umberto Feb 11 '13 at 20:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.