This question seems simple but I can't manage to disprove it. Let $N\in \mathbb{N}$. We know that by its analyticity that this precise linear combination of monomials
$
\sum\_{n=0}^K \frac1{n!} x^n
$
converges uniformly to $g$ on $[0,1]$ and that for $K$ large enough
$$
\max\_{x \in [0,1]}\, \|\sum\_{n=0}^K \frac1{n!} x^n - \exp(x)\|<N^{-1}.
$$
However, $K$ becomes unbounded as a function of $N$.


My question: Does there exist a single $K\in \mathbb{N}$ for which, for any $N\in \mathbb{N}$ always exists $\alpha\_1,\dots,\alpha\_K\in [0,1]$ and $\beta\_1,\dots,\beta\_K\in \mathbb{R}$ such that
$$
\max\_{x \in [0,1]}\, \|\sum\_{n=0}^K \beta\_n x^{\alpha\_n} - \exp(x)\|<N^{-1}.
$$
That is, can we bound the number of these "monomials" required to approximate the function $\exp$.



Approximation of analytic function by a fixed number of monomials
The general description of all continuous functions which are uniform limits of fewnomials seems to be given by the following


**Theorem.** Fix $0<a<b$. The following two properties of a continuous function $f\in C[a,b]$ are equivalent:


(i) $f$ is log-quasipolynomial, that is, $f(x)=P(x,\log x)$ for a certain quasi-polynomial $P(u,v)=\sum\_{i=1}^N c\_i u^{\kappa\_i} v^{m\_i}$, where $\kappa\_i$'s are real and $m\_i$'s non-negative integers;


(ii) $f$ is a uniform limit of a sequence of fewnomials (that is, of functions of the form $f\_n(x)=\sum\_{k=1}^K \beta\_k^{(n)} x^{\alpha\_k^{(n)}}$ with the same $K$).


**Proof.** $(i)\Rightarrow (ii)$ follows from $\log x=\lim n(x^{1/n}-1)$ uniformly on $[a,b]$.


$(i)\Rightarrow (ii)$ Induction in $K$. Base $K=0$, then $f\_n(x)\equiv 0$ and $f\equiv 0$ too.


Induction step from $K-1$ to $K$. Passing to a subsequence, we may suppose that $\alpha\_K^{(n)}$ has a limit $\alpha\_K$ which may be equal to $+\infty$, $-\infty$ or be a real number.


If, say, $\alpha\_K=\infty$, we may fix $c\in (a/b,1)$ and consider the functions $g\_n(x)=f\_n(cx)-c^{\alpha\_K^{(n)}}f\_n(x)$ (it has at most $K-1$ monomials), they converge to $f(cx)$ uniformly on $[a/c,b]$. Thus by induction hypothesis $f(cx)$ is log-quasipolynomial on $[a/c,b]$. In other words, $f$ is log-quasipolynomial on $[a,cb]$. Such log-quasipolynomials for distinct $c$ must be consistent, so $f$ is log-quasipolynomial on $[a,b]$. Analogously if $\alpha\_K=-\infty$.


So assume that $\alpha\_K$ is a real number. Consider the functions $\tilde{f}\_n(x)=x^{-\alpha\_K^{(n)}}f\_n(x)=\sum\_{k=1}^K \beta\_k^{(n)} x^{\tilde{\alpha}\_k^{(n)}}$, where $\tilde{\alpha}\_k^{(n)}=\alpha\_k^{(n)}-\alpha\_K^{(n)}$. They uniformly on $[a,b]$ converge to $\tilde{f}(x):=x^{-\alpha\_K}f(x)$. If $\tilde{f}(x)$ is log-quasipolynomial, then so is $f$. Changing notations back (omit tilde), we now on suppose that $\alpha\_K^{(n)}=0$.


Again, we fix $c\in (a/b,1)$ and consider the functions $g\_n(x)=f\_n(cx)-f\_n(x)$. They converge to $f(cx)-f(x)$ uniformly on $[a/c,b]$. By induction hypothesis $f(cx)-f(x)$ is a log-quasipolynomial on $[a/c,b]$. Any log-quasipolynomial may be represented as $H(cx)-H(x)$ for another log-quasipolynomial $H$. So we may write $f(cx)-f(x)=H\_c(cx)-H\_c(x)$ for a log-quasipolynomial $H\_c$. We may assume additionally $H\_c(b)=f(b)$. Note that $$H\_c(cx)-H\_c(x)=f(cx)-f(x)=\sum\_{j=0}^{m-1} f(c^{(j+1)/m}x)-f(c^{j/m}x)=H\_{c^{1/m}}(cx)-H\_{c^{1/m}}(x),$$
thus $p(x):=H\_c(x)-H\_{c^{1/m}}(x)$ satisfies $p(b)=0$ and $p(cx)=p(x)$ for $x\in [a/c,b]$. This yields $H\_c\equiv H\_{c^{1/m}}$. If $c\_1,c\_2$ are rational powers of 2, then there exist positive integers $m\_1,m\_2$ such that $c\_1^{1/m\_1}=c\_2^{1/m\_2}$. Thus $H\_{c}=:H$ does not depend on $c$ on a dense set $D\subset [a/b,1]$ of values of $c$'s. So the function $f(x)-H(x)=:F(x)$ satisfies $F(b)=0$, $F(x)=F(cx)$ for $c\in D$. Since $F$ is continuous, this yields $F\equiv 0$ and $f=H$. $\square$


Of course $\exp(x)$ is not log-quasipolynomial on an interval (for instance, because it grows faster at infinity, and if two functions which are analytic on the right half-plane coincide on an interval, they do coincide on the whole half-plane).


