arcsinc( 1-y ), for small y
07-05-2018, 11:43 PM
Post: #1
 Albert Chan Senior Member Posts: 2,103 Joined: Jul 2018
arcsinc( 1-y ), for small y
Come across a fun civil engineering problem from HP archive [1]
"Five minute challenge" essentially evaluate arcsinc( 1-y )

sin(x) / x = ( 1-y ) -- y = 1/5281 = rail expansion effect
x = arcsinc( 1-y ) -- then, rail warp e = 2640 tan(x/2)

From Gerson W. Barbosa (msg #21), sinc(x) ~ cos(x / sqrt(3))

x = arcsinc( 1-y ) ~ sqrt(3) arccos( 1-y )
= 0.03370733324 -- using my Casio fx-115ms

Estimate is good, exact x = 0.033707758809879429 ...

I wanted a simpler, and more accurate arcsinc( 1-y )

arccos( 1-y ) taylor series = sqrt(2y) (1 + y/12 + ...) [2]

x = arcsinc( 1-y ) ~ sqrt(6y) = 0.03370680134

For better accuracy, apply Newton's method, x0 = sqrt(6y):

f(x) = sinc(x) - (1-y)
= y - x^2/6 + x^4/120 - x^6/5040 + ...
f'(x) = -x/3 + x^3/30 - (3/70) x^5 + ...

f(x0) = y - y + 0.3 y^2 - (3/70) y^3 + ... ~ 0.3 y^2
f'(x0) = x0 (-1/3 + y/5 - (54/35) y^2 + ...) ~ -x0 / 3

x = x0 - f(x0) / f'(x0)
~ x0 + (0.9 y^2 / x0) * (x0/x0)
= x0 (1 + 0.9 y^2 / (6y))
= sqrt(6y) (1 + 0.15 y)

x = arcsinc( 1-y ) ~ sqrt(6y) (1 + 0.15 y) = 0.03370775874

with x this good, f(x) = 0.0 on my Casio :-)