Timeline for Hadamard-like product on infinitely differentiable functions
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 7, 2020 at 10:56 | comment | added | Igor Khavkine | Dividing the coefficients of a series by $n!$ is the Borel transform $\mathcal{B}$. Your product is $f * g = \mathcal{B}(\mathcal{B}^{-1}(f) \cdot \mathcal{B}^{-1}(g))$ where I wrote the usual Hadamard product as $\cdot$. So all the properties of the Hadamard product translate to yours via $\mathcal{B}$ and $\mathcal{B}^{-1}$. | |
| Feb 7, 2020 at 8:36 | history | edited | Jake Lai | CC BY-SA 4.0 |
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| Feb 6, 2020 at 20:46 | comment | added | Jake Lai | I was thinking of that example, I guess I didn't know the name "formal power series". Will make the edit, thank you! | |
| Feb 6, 2020 at 19:50 | comment | added | Pietro Majer | note that in general this series is not convergent, even with analytic functions, e.g. $f=g=\frac{1}{1-x}$ gives $(f*g)(x)=\sum_{n=0}^\infty n!x^n$. Maybe you want to consider it as an operation for formal power series. | |
| Feb 6, 2020 at 18:30 | review | First posts | |||
| Feb 6, 2020 at 18:36 | |||||
| Feb 6, 2020 at 18:25 | history | asked | Jake Lai | CC BY-SA 4.0 |