Chebyshev rational approximation of $e^{x}, x >0$: does it exist?

Question

It's well known that the scalar function $e^x$, for $x \in (-\infty,0]$ can be approximated by Chebyshev rational approximation. In practice, one wants to use a partial fraction decomposition form like the following: $$R_{\nu}(x)= \alpha_0 + \sum_{i=1}^{\nu} \frac{\alpha_i}{x-\theta_i}$$

where for $\nu = 14$ we have "good" accuracy and the coefficients are listed in several papers. See e.g. this paper.

Now, it would be nice to find a rational approximation of $e^x$ , for $x>0$, similar to the one above. Honestly, I don't think that such an approximation can exist. The main reason is that for $x \rightarrow \infty$ such an expression is bounded, while it shouldn't.

Does anyone know if there exist a similar approximation, or if it is indeed impossible?

Answer 1

If $|e^x - R_\nu(x)| < C_\nu$ for $x\in (-\infty,0]$, then $|e^{x-a} - R_\nu(x-a)| < C_\nu$ for $x\in (-\infty,a]$. Moving around some scalar factors, you get $$|e^x - e^a R_\nu(x-a)| < e^a C_\nu \quad \text{for} \quad x \in [0,a].$$ You are right that you can't make the approximation uniform on the whole positive real axis, because $e^x$ is unbounded. But uniform absolute approximation on the whole negative real axis is only possible because both $e^x$ and $R_\nu(x) \to 0$ as $x\to -\infty$. The relative error, on the other hand, becomes unbounded for larger and larger negative $x$. After all, for sufficiently large $|x|$, $R_\nu(x) \sim \alpha_0$ approaches a non-zero constant, while $e^x$ continues to fall to zero.

Answer 2

It is easy to see that $e^x-R(x)$ is unbounded on $[0,+\infty)$ for every rational function $R$. Therefore, Chebyshev (=uniform) approximation cannot exist.