I'm trying to find an alternative to the Taylor series which represents some function $f(x)$ as a sum of exponentials.
$$f(x) = a_0 e^{0x} + a_1 e^{1x} + a_2 e^{2x} + a_3 e^{3x} + ...$$
The coefficients $a_0, a_1, a_2...$ can be found by equating the derivatives: $$a_0 + a_1 + a_2 + a_3 + ... = f(0)$$ $$0 + a_1 + 2a_2 + 3a_3 + ... = f'(0)$$ $$0 + a_1 + 4a_2 + 9a_3 + ... = f''(0)$$
In general you can write out $n$ equations to solve for $n$ coefficients. You can solve for the coefficients by solving the following matrix equation $[M] a = f$
$$\begin{bmatrix} 1 & 1 & 1 & 1 & .. & 1\\ 0 & 1 & 2 & 3 & .. & (n-1)^{n-4}\\ 0 & 1 & 4 & 9 & .. & (n-1)^{n-3}\\ 0 & .. & .. & .. & .. & (n-1)^{n-2}\\ 0 & 1 & (n-4)^{n-1} & (n-3)^{n-1} & (n-2)^{n-1} & (n-1)^{n-1} \end{bmatrix} \begin{pmatrix} a_0\\ a_1\\ a_2\\ ..\\ a_{n-1} \end{pmatrix} = \begin{pmatrix} f(0)\\ f'(0)\\ f''(0)\\ ..\\ f^{(n-1)}(0) \end{pmatrix}$$
Now to find all coefficients I need to solve $[M]^{-1}f$ as $n$ approaches infinity. But how can I find and simplify the inverse of this matrix? Can someone help me find the closed analytical expression for the i'th coefficient $a_i$?
Note that if $u=e^x$ then $(f\circ\ln)(u)=f(x)=\sum_{n=0}^\infty a_ne^{nx}=\sum_{n=0}^\infty a_nu^n$. Hence, provided $f\circ\ln$ has a $C^\infty$ extension $g$ to a neighborhood of zero, $$a_n=\frac{g^{(n)}(0)}{n!}.$$ For the case where $g$ has no such extension, you will have to do more work. I would delete this answer if I could, but you will have to un-accept it first.