Is there a way to express the following transform (and perhaps its inverse) in terms of known discrete transforms? For $a_i>0$ and $\sum_i a_i=1$ we have
$$f(x)=\sum_i^n a(i) \exp(-x a(i))$$
Equivalently, with $x=-\log y$
$$f(y)=\sum_i^n a(i) y^{a(i)}$$
The continuous version can be represented in terms of Laplace transform as follows
$$f(t)=\int_0^\infty g(i) \exp(-t g(i))=\mathcal{L}(yg^{-1}(y)')$$
However, this involves functional inverse of $g$ which is undefined in the discrete case.
I'm looking for something I can leverage in practice to invert this for a vector $(b_0,b_1,\ldots,b_n)$ with $n\approx 10000$. Find $(t_0,t_1,\ldots,t_n)$ such that $f(t_0),f(t_1),\ldots,f(t_n)=(b_0,b_1,\ldots,b_n)$
Motivation: this transform maps eigenvalues of quadratic Q to loss trajectory of gradient descent on Q, where $a(i)$ gives $i$th eigenvalue of $Q$ and $f(t)$ gives loss after $t$ steps.