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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 $d(f,g)<\epsilon?$

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  • $\begingroup$ 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. $\endgroup$
    – Loïc Teyssier
    Commented Feb 8, 2013 at 14:08
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    $\begingroup$ 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. $\endgroup$
    – Emil Jeřábek
    Commented Feb 8, 2013 at 15:33
  • $\begingroup$ I edited it again. Many thanks because of your comments. $\endgroup$
    – Umberto
    Commented Feb 8, 2013 at 16:33
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    $\begingroup$ How is the algorithm supposed to get $f$ as an input? $\endgroup$
    – Emil Jeřábek
    Commented Feb 8, 2013 at 16:58
  • $\begingroup$ This is another question I had (see How are enconded the real analytic functions). $\endgroup$
    – Umberto
    Commented Feb 8, 2013 at 17:53

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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.

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  • $\begingroup$ but for $f\in C,$ $f(1)$ is not well defined. I do not want in $A$ to be maps with singularities in $[0,1].$ $\endgroup$
    – Umberto
    Commented Feb 8, 2013 at 13:25
  • $\begingroup$ 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? $\endgroup$
    – Loïc Teyssier
    Commented Feb 8, 2013 at 13:28
  • $\begingroup$ But $1/(1+x^2)\in B,$ because we can find an explicit formula for the Taylor coefficients. $\endgroup$
    – Umberto
    Commented Feb 8, 2013 at 13:36
  • $\begingroup$ 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! $\endgroup$
    – Loïc Teyssier
    Commented Feb 8, 2013 at 13:41
  • $\begingroup$ It does not answer my question because the set B is strictly larger than the set of polynomials. $\endgroup$
    – Umberto
    Commented Feb 11, 2013 at 20:33

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