*Where B₀(x)=(1−x)³, B₁(x)=3x(1−x)², B₂(x)=3x²(1−x), B₃(x)=x³ are the cubic Bernstein polynomials. https://en.wikipedia.org/wiki/Bernstein_polynomial …
-
-
Prikaži ovu nitHvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi
-
-
-
But given the coefficients of a higher degree polynomial how do you check if it’s strictly increasing? (My guess is this is way worse, smells like Positivstellensatz/SOS hierarchy to me!)
-
Yes, things work out nicely here because it's only cubic…
- Još 1 odgovor
Novi razgovor -
-
-
how do you pick coefficients such that (c₁²+c₂²+c₀(c₃−c₂)−c₁(c₂+c₃))/(c₀−3c₁+3c₂−c₃) > 0 ?
-
I used rejection sampling. :-)
- Još 2 druga odgovora
Novi razgovor -
-
-
What high school did you go to?
Hvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi
-
-
-
another method is to use bernstein polynomials, and pick each coefficient to be greater than the one before, which, when reduced to the power function surely is going to end up in the same solution as yours for cubic beziers.
-
yours is nice tho in that when you already have a polynomial in power form, you can quickly check if it qualifies.
Kraj razgovora
Novi razgovor -
-
-
I like coloured spaghetti.
Hvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi
-
-
-
It seems there is more pattern to "c₁²+c₂²+c₀(c₃−c₂)−c₁(c₂+c₃))/(c₀−3c₁+3c₂−c₃)", can you collect the roots? the denominator smells like Pascal triangle (of course, because it's Bezier).
Hvala. Twitter će to iskoristiti za poboljšanje vaše vremenske crte. PoništiPoništi
-
Čini se da učitavanje traje već neko vrijeme.
Twitter je možda preopterećen ili ima kratkotrajnih poteškoća u radu. Pokušajte ponovno ili potražite dodatne informacije u odjeljku Status Twittera.