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!



Other expansion for positive Taylor expansion
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:](https://en.wikipedia.org/wiki/Nonnegative_matrix)



> 
> 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](https://en.wikipedia.org/wiki/Generalized_permutation_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 $\boldsymbol{f}$.


