Applying the [Müntz–Szász theorem](https://en.wikipedia.org/wiki/M%C3%BCntz%E2%80%93Sz%C3%A1sz_theorem) on $[0,1]$ repeatedly, we can represent
$$
x= \sum\_{n\geq 2} c\_n x^n
$$ 
as a uniformly convergent series (**edit:** only over some subsequence, see edits below) on $[0,1]$ of higher powers $x^n$ for $n\geq 2$. What can one say about the coefficients? Is there an explicit choice of $c\_n$?


**Edit:** Comments below suggest that this is not possible. What is wrong with the following argument? Take $\epsilon>0$ and approximate $x$ by a finite combination of higher powers and a constant uniformly with an error $\epsilon/2.$ Plugging $x=0$ we see that the constant is smaller than $\epsilon/2$ so dropping it we get an approximation up to $\epsilon$ by a finite sum $\sum\_{n=2}^{N\_1} c\_n x^n.$ Next, consider $x-\sum\_{n=2}^{N\_1} c\_n x^n$ and approximate it by a linear combination $\sum\_{n=N\_1+1}^{N\_2} c\_n x^n$ up to an error $\epsilon/2.$ This gives us an $\epsilon/2$-approximation $\sum\_{n=1}^{N\_2}c\_n x^n.$ Continue this construction repeatedly.


**Edit II:** Theorem holds for $a=0$ if we include constants, but Robert Israel's comment below contained the main point: the series only converges over some subsequence $(N\_k)\_{k\geq 1}$ as in the above construction. Let me rephrase the question accordingly:


Is there anything interesting one can say about $c\_n$? Can one choose the subsequence and $c\_n$ in a way that $(c\_n)\in\ell^p$ for some $p$, or uniformly bounded?



Representing $x$ as a linear combination of higher powers $x^n$
It can't be in $\ell^p$ or bounded, in fact you can't have $|c\_n| = O(t^{-n})$ for any $t > a$ where the subsequence converges on $[a,1]$. This is because if $|c\_n| = O(t^{-n})$, $\sum\_n c\_n z^n$ is analytic in $|z|<t$, and by uniqueness...


