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Aug 27, 2023 at 13:30 comment added Alexandre Eremenko The signs of $a_k$ do not matter, what does matter in my remark is $b_k>0$.
Aug 26, 2023 at 21:10 vote accept NancyBoy
Aug 26, 2023 at 18:38 history became hot network question
Aug 26, 2023 at 13:03 vote accept NancyBoy
Aug 26, 2023 at 21:10
Aug 26, 2023 at 12:00 vote accept NancyBoy
Aug 26, 2023 at 13:02
Aug 26, 2023 at 11:59 comment added NancyBoy Thabk you Alexander but here there is only $f>0$, we don't know the sign of the $a_k$.
Aug 26, 2023 at 11:53 comment added Alexandre Eremenko The answer is trivial: take $b_0=1, b_k=0$ for $k\geq 2$ and $\beta_0=f$. You have to specify what functions $\beta_k$ are allowed. With exponential functions this is in general not possible, since in this case we will have $f^{(n)}>0$ for all $n$, so not all positive functions can be expanded into a series of positive exponentials.
Aug 26, 2023 at 11:20 answer added Michał Jan timeline score: 5
Aug 26, 2023 at 10:23 comment added NancyBoy Arh, thank you for these information !
Aug 26, 2023 at 10:21 comment added Michał Jan Ah, sadly it won't. Inverse Laplace transform does not preserve positivity mathoverflow.net/questions/383996/…
Aug 26, 2023 at 10:17 comment added NancyBoy Thank you for your answer ! My problem is that the $a_k$ are not necessary positive but $f$ is positive. I want to have another expression with only positive terms. Do you think that Laplace transform will work ?
Aug 26, 2023 at 10:14 comment added Michał Jan If it's enough for this to work for $x>0$, and you change the sum for an integral, then this becomes Laplace transform $f(x)=\int_0^\infty dk b(k) \exp(-kx)$. For positive $b(k)$, $f(x)$ is of course positive, but I don't know if converse is true.
Aug 26, 2023 at 9:03 history asked NancyBoy CC BY-SA 4.0