Timeline for Polynomial approximation of continuous function with constraints
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| May 19, 2021 at 2:23 | comment | added | Iiro Ullin | You are totally right of course, I am just commenting that it's not necessarily totally obvious for someone who's asking this question in the first place. | |
| May 19, 2021 at 0:25 | comment | added | Piotr Hajlasz | @IiroUllin It does. If $0\leq f\leq 1$, then the Bernstein polynomials satisfy the same estimate. $p\geq 0$ is obvious, because all terms in the definition of $p$ are nonnegative and then replacing $f(k/n)$ by the upper bound of $1$ you get the binomial formula for $((1-x)+x)^n$ which equals $1$ which is the upper bound for $p$. If I find time, I will add details. | |
| May 18, 2021 at 23:50 | comment | added | Iiro Ullin | The answer is fitting, but perhaps does not fully address the $0\leq p\leq1$ constraint. So I would add that the reason that $p$ lies within the same bounds as $f$ can be most clearly seen via De Casteljau's construction of the corresponding polynomial approximation (convex hull property of Bezier curves). | |
| May 18, 2021 at 21:35 | vote | accept | Rahul Sarkar | ||
| May 18, 2021 at 21:28 | history | answered | Piotr Hajlasz | CC BY-SA 4.0 |