Approximation of analytic function by a fixed number of monomials 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 $\exp$.
 A: 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).
A: Edit later - as per comments from @Eremenko and @Pinelis, there is a flaw in the argument and the solution works only if we assume that $\beta_{k,N}$ are bounded since that doesn't follow easily as I first thought (for example one can note that $|N\sqrt x- Nx^{1/2+1/N^2}| \le C\log N/N, 1 \ge x \ge 1/N^4$ and $|N\sqrt x- Nx^{1/2+1/N^2}| \le C/N, 0 \le x \le 1/N^4$, so even for the stronger approximation on the full $[0,1]$ interval, if there is one such, we can introduce two extra terms with the betas going to infinity that don't affect the approximation)
(Partial solution) I think the answer is no by a convergence argument. We will show that no such approximation is possible even on the subinterval $[1/2,1]$ and that will definitely imply the result
Let's assume there are sequences $f_N(x)=\sum_{k=1}^K \beta_{k,N}x^{\alpha_{k,N}}$ with parameters $\alpha_{k,N}, \beta_{k,N}, k=1,..K, 0 \le \alpha_{k,N} \le 1$, st $||f_N(x)-e^x||_{\infty} \le 1/N$ (where $||.||_{\infty}$ is the supremum norm on continuous functions on $[1/2,1]$ say).
Assuming $|\beta_{k,N}| \le C$ for all $k=1,..K, N \ge 1$, one can extract a subsequence for which all respective parameters converge to some parameters $\alpha_{k}, \beta_{k}, k=1,K$ and then (by renumbering) we need to show:
1: If $\alpha_{k,N}\to \alpha_k, \beta_{k,N} \to \beta_k, N \to \infty, k=1,..K$ then $||f_N-f||_{\infty} \to 0$ where $f$ is constructed with data $\alpha_k, \beta_k$
2: $\exp x \ne f$ for any data as above
For 1, the only thing which is not immediately obvious is to show that $||x^{\alpha_n}-x^{\alpha}||_{\infty} \to 0$ when $\alpha_n \to \alpha, 0 \le \alpha_n \le 1$ and by the mean value theorem for fixed $1/2 \le x<1$ we have $|x^{\alpha_n}-x^{\alpha}|=x^{c_n}|\alpha_n-\alpha||\log x| $ for some $c_n \in [0,1]$. But now since we are on $[1/2,1], |\log x|$ is uniformly bounded in $x$ as is obviously $x^c, 0 \le c \le 1$ so the norm convergence is clear.
For 2, assume $\exp x=\sum_{k=1}^K \beta_kx^{\alpha_k}, x \in [1/2,1]$. Then we can for example extend analytically both sides on say $\Re z >0$ using the principal branch of the logarithm for RHS and then by analytic continuation the identity is valid everywhere on $\Re z >0$ which is obviously not possible when $x \to \infty$
