I was thinking of the following problem. Let $f$ be a Taylor expansion and $a_k$ the associated coefficients,
$$\forall x\in\mathbb{R},~f(x)\triangleq\sum_{k=0}^\infty a_kx^k.$$
Let suppose that we have:
$$\forall x\in\mathbb{R},~f(x)>0.$$
Is it possible to find another expansion such that:
$$\forall x\in\mathbb{R},~f(x) = \sum_{k=0}^\infty b_k\beta_k(x)$$
with $b_k$ positive real numbers and $\beta_k$ (exponential?) positive functions ?
Thank you very much!
I believe it is not possible. Here is an argument for this:
Disclaimer: All inequalities hereafter are meant elementwise
Let's consider a discrete version of this problem $x\in\{0,1,\dots,N-1\}$. Then we want to find an invertible $N\times N$ matrix $\beta$ (different from identity), such that for every elementwise non-negative vector $\boldsymbol{f} = (f(0), f(1), \dots f(N-1))^T \geq 0$ we have $$ \boldsymbol{b} = \beta^{-1} \boldsymbol{f} \geq 0\,. $$ The necessary and sufficient condition for this is that $\beta^{-1}\geq 0$. Moreover, you require that $\beta_k(x)\geq 0$, which in discrete case translates to $\beta\geq 0$.
However, citing Nonnegative matrix - Wikipedia:
The inverse of a non-negative matrix is usually not non-negative. The exception is the non-negative monomial matrices: a non-negative matrix has non-negative inverse if and only if it is a (non-negative) monomial matrix.
Where the monomial matrix is defined as:
In mathematics, a generalized permutation matrix (or monomial matrix) is a matrix with the same nonzero pattern as a permutation matrix, i.e. there is exactly one nonzero entry in each row and each column.
Thus the only allowed transformation $\beta$ is reshuffling (and possibly rescaling) the elements of $\boldsymbol{f}$.
