oh no are people following me because i did math?

[oblomov@sociale.network]

@j_bertolotti @lisyarus @eniko hey

https://sociale.network/@oblomov/115811316048332524

[brunolevy01@fosstodon.org]

@eniko yes 🤣

[lisyarus@mastodon.gamedev.place]

@j_bertolotti @eniko Taylor is typically only good around expansion point, but this approximation is good across the whole range

[eniko@mastodon.gamedev.place]

@BrunoLevy01 prepare to be dazzled by my lack of any actual knowledge! :'D

[oblomov@sociale.network]

@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@mathstodon.xyz]

@lisyarus
For an arcsin you only need it to be good in the (-1,1) interval, and for a smooth analytic function Taylor converges pretty quickly. Again, might not be optimal, but it would have been my first attempt.
@eniko

[lisyarus@mastodon.gamedev.place]

@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@fosstodon.org]

@eniko asking good questions is at least as interesting as giving good answers !

[oblomov@sociale.network]

@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@mastodon.gamedev.place]

@oblomov @j_bertolotti @eniko What do you mean by "only for the c^2 part"?

[oblomov@sociale.network]

@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

[diegor@social.gl-como.it]

@oblomov @j_bertolotti @lisyarus @eniko Taylor expansion will be problematic at -1 or 1, because the derivative goes to infinity.

Even if you get a good approximation in value, discontinuities in first order derivative are usually visually appalling on screen.

[oblomov@sociale.network]

@diegor @j_bertolotti @lisyarus

oh damn, good point. And that's actually also probably the reason why @eniko's works so well for the asin (the derivative of asin being rsqr(1-x^2))

[oblomov@sociale.network]

@diegor @j_bertolotti @lisyarus @eniko

BTW as I'm now feeling better, I toyed around with the exponent to c and apparently the lowest error is found with exponent 1.9296 but yeah having to resort to the power function is URGH

[eniko@mastodon.gamedev.place]

@oblomov @diegor @j_bertolotti @lisyarus looks like that's more accurate for the ends so i can edit that in

[oblomov@sociale.network]

@eniko @diegor @j_bertolotti @lisyarus

It's more accurate basically everywhere 8-D