oh no are people following me because i did math?
-
@eniko Ohhhh. Nice approximation though! I wonder if there's something deep behind it or it's just a coincidence
@lisyarus i dunno! i started with `lerp(x, pi/2, pow(x, 8))` but the distortion wasn't strong enough near the edges cause the curve wasn't steep enough
then i remembered that there's a circle ease where the graph, well, looks distinctly like a circle, including a very steep angle at the end
so i decided to switch `pow(x,8)` with that term and sure enough, if i made it even steeper by raising it to the 2nd power it was almost identical
-
oh no are people following me because i did math?
oh no
@eniko mandatory bird meme
-
undefined oblomov@sociale.network shared this topic
-
@eniko Ohhhh. Nice approximation though! I wonder if there's something deep behind it or it's just a coincidence
-
-
oh no are people following me because i did math?
oh no
@eniko yes 🤣
-
@j_bertolotti @eniko Taylor is typically only good around expansion point, but this approximation is good across the whole range
-
@eniko yes 🤣
@BrunoLevy01 prepare to be dazzled by my lack of any actual knowledge! :'D
-
@j_bertolotti @eniko Taylor is typically only good around expansion point, but this approximation is good across the whole range
@lisyarus @j_bertolotti @eniko I was thinking that the c factor could be computed with Taylor. Given that the actual best results are achieved with 1.95 and not 2, I was wondering how that would affect it.
-
@j_bertolotti @eniko Taylor is typically only good around expansion point, but this approximation is good across the whole range
-
@j_bertolotti @eniko It will still struggle towards the end of the range. Here's the expansion up to x^11: https://www.desmos.com/calculator/1bai5gkuay
Tbh my first idea would be to least-squares fit a polynomial
-
@BrunoLevy01 prepare to be dazzled by my lack of any actual knowledge! :'D
@eniko asking good questions is at least as interesting as giving good answers !
-
@j_bertolotti @eniko It will still struggle towards the end of the range. Here's the expansion up to x^11: https://www.desmos.com/calculator/1bai5gkuay
Tbh my first idea would be to least-squares fit a polynomial
@lisyarus @j_bertolotti @eniko can you do the same but only for the c^2 part? (I would do it myself but currently impossibilitated)
-
@lisyarus @j_bertolotti @eniko can you do the same but only for the c^2 part? (I would do it myself but currently impossibilitated)
@oblomov @j_bertolotti @eniko What do you mean by "only for the c^2 part"?
-
@oblomov @j_bertolotti @eniko What do you mean by "only for the c^2 part"?
@lisyarus @j_bertolotti in @eniko's formula there's a linear interpolation by c^2 with c = 1 - sqrt(1 - x^2) and even better results are achieved with c^1.95, so I was wondering what happens with c^2 replaced by its Taylor expansion to order 8. Or given that it should only have even terms (so order 8 would be 4 terms) maybe even up to 6 or 7 terms
