Timeline for Invertible linear transformation on the space of polynomials
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| May 28, 2022 at 3:59 | comment | added | LSpice |
TeX note: please use $\operatorname{char}\mathbb{F}$ \operatorname{char}\mathbb{F}, not $\mathrm{char}\mathbb{F}$ \mathrm{char}\mathbb{F}; note the difference in spacing. I have edited accordingly.
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| May 28, 2022 at 3:58 | history | edited | LSpice | CC BY-SA 4.0 |
`\operatorname`; deleted "Thank you"
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| May 28, 2022 at 3:23 | answer | added | Jorge Zuniga | timeline score: 4 | |
| May 26, 2022 at 3:59 | comment | added | Jorge Zuniga | @Federico_Poloni OK. Please hold on a couple of days. | |
| May 26, 2022 at 3:23 | comment | added | Jorge Zuniga | Using formal expansion of operators, $\mathscr{A}^{−1}=\frac{1}{2}\sum_{n=0}^∞(−\frac{1}{2})^nΔ^n$, where $Δ^n$ is the iterated forward difference operator [$Δ^n$ annihilates polynomials of degree lower than $n$]. The upper sum limit is finite for polynomials. Expanding $Δ^n$ in binomials gives an algorithm to find the polynomial inverse transformation you are looking for. | |
| May 25, 2022 at 22:36 | comment | added | Federico Poloni | @JorgeZuniga This seems a good answer, but please post it as an answer, not as a comment. | |
| May 25, 2022 at 16:02 | answer | added | loup blanc | timeline score: 7 | |
| May 24, 2022 at 14:11 | comment | added | Francesco Polizzi | Anyway, this should be somehow related to the inverse of the Pascal matrix: math.stackexchange.com/questions/1852826/… | |
| May 24, 2022 at 9:30 | comment | added | Francesco Polizzi | Just a quick remark. Since $\mathscr{A}^{-1}$ is linear, computing $\mathscr{A}^{-1}(x^k)$ for all positive integers $k$ is enough. Perhaps you could try to guess a formula after computing some of them, and the prove it by induction. | |
| May 24, 2022 at 7:31 | history | asked | Zhihua Chang | CC BY-SA 4.0 |