Approximating Exp[-x] in partial fraction form

Question

I'm looking to obtain order-$k$ approximation of $\exp(-z)$ for real-valued $z$. $$R_k(z)\approx \exp(-z)$$

The constraint is that I need the result in partial fraction form, ie:

$$ \begin{equation} R_k(z)=\alpha_0 + \sum_{i=1}^k \frac{\alpha_i}{z-\lambda_i} \label{1} \tag{1} \end{equation} $$

Where $\{\alpha_i\}$ and $\{\lambda_i\}$ are either complex-valued or real-valued. Restricting to real-valued allows doubling $k$ without increasing computational cost.

There's this answer for Chebychev approximation and PadeApproximant, but how do I turn those rational functions into partial fraction form $\ref{1}$?

All of the examples I tried have imaginary roots, so Apart fails to factor some terms.

Appendix B of this paper gives one such set of coefficients, but that's not easy to experiment with in Mathematica.

Here's a naive attempt at order=4 approximation with Taylor series.

Clear[x, t, i];
k = 4;
poly = Normal@Series[Exp[t], {t, 0, k}];
dpoly = Function @@ {{t}, D[poly, t]};
roots = t /. {ToRules@Roots[poly == 0, t]};
alpha[0] = 0;
alpha[i_] := dpoly[roots[[i]]]^-1;
lambda[i_] := roots[[i]];
nexp[z_] := alpha[0] + Sum[alpha[i]/(z - lambda[i]), {i, 1, k}];
LogPlot @@ {{Exp[-z], nexp[z]}, {z, 1, 10}, 
  PlotLegends -> {"exp[-z]", "approx"}, AxesLabel -> {"z"}, 
  PlotLabel -> StringForm["Order `` approximation", k]}
Table[{alpha[i], lambda[i]} // N, {i, 0, k}] // 
 TableForm[#, TableHeadings -> {None, {"alpha", "lambda"}}] &

The linked paper considers only real lambdas, but I see complex lambdas in the table from your question. Is it OK? — user64494, Mar 23 '23 at 20:12
Here's Table 5 in that paper, it uses complex numbers. Real-valued would work too, didn't know that was possible — Yaroslav Bulatov, Mar 23 '23 at 20:14
If $z=10$ how do you get the "value" being -10? $e^{-10}\approx 0.0000453999$. — JimB, Mar 23 '23 at 21:32
value in the graph is the logarithm of the function in question — Yaroslav Bulatov, Mar 23 '23 at 21:37
Understood. So you're trying to estimate $\log(e^{-z})=-z$ with a complicated function? Obviously I must be missing the point. — JimB, Mar 23 '23 at 21:40
If you need the coefficients of a particular model ($R_k(z)$), just generate a bunch of points for $z$ values of interest and perform a regression. — JimB, Mar 23 '23 at 21:43
@JimB The issue is that I don't have a way to compute $\exp(-z)$, but I do have a method to compute 1/(z-x). So the above approximation gives a way to express exponential in terms of inverses. Section 3 of this paper gives background — Yaroslav Bulatov, Mar 23 '23 at 21:48
If you have a Pade approximation you can use Apart to get a PFD. — Daniel Lichtblau, Mar 23 '23 at 22:32
@DanielLichtblau I need linear factors, but Apart gives me quadratic factors because of complex roots — Yaroslav Bulatov, Mar 23 '23 at 22:36
See https://mathematica.stackexchange.com/questions/68824/apart-for-complex-roots and linked questions. There's also a resource function for this. — Michael E2, Mar 23 '23 at 23:42

Approximating Exp[-x] in partial fraction form

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