arcsinc( 1-y ), for small y

[Albert Chan]

Rearange Acton fromula in terms of arcsinc(k = 1-y):

x = arcsinc(k) ~ sqrt(6y / (1 - 0.3 y))

this is much better than my estimate of sqrt(6y) / (1 - 0.15y)
For small y, errors cut down by almost 33% !

To push the idea a bit further, below cut down error (small y) by 97% !

x = arcsinc(k) ~ sqrt(6y / sqrt(1 - 0.6 y))

For y > 0.5, asymptotic formula is better: x ~ Pi / (1 + k + 1.244 k^3)

For 0<y<1, above combined setup kept relative error to 0.1% or less.

For more accuracy, use either form of Newton's method:

f = sin(x) - kx      → x = x - (sin(x) - k*x) / (cos(x) - k)
f = sin(x)/x - k     → x = x + x * (z - k) / (z - cos(x)), where z = sin(x)/x

Update: More accurate Asinc (small y), but more complicated formula:
http://www.hpmuseum.org/forum/thread-291...ight=Asinc

Update2: instead of asymptotic formula, correction also work:
For y >= 0.4, asinc(1-y) ~ sqrt(6y / sqrt(1 - 0.6 y)) * (1.001 + y^6/53)