Re: ArcSin
- To: mathgroup at christensen.cybernetics.net
- Subject: [mg1570] Re: ArcSin
- From: bob Hanlon <hanlon at pafosu2.hq.af.mil>
- Date: Sun, 2 Jul 1995 00:02:33 -0400
Needs["Algebra`SymbolicSum`"]
Expanding the ArcSin about x=0
Series[ArcSin[x], {x, 0, 13}]
3 5 7 9 11 13
x 3 x 5 x 35 x 63 x 231 x 14
x + -- + ---- + ---- + ----- + ------ + ------- + O[x]
6 40 112 1152 2816 13312
See Abramowitz and Stegun, 4.4.40; Hansen, 5.22.14
SymbolicSum[Pochhammer[1/2, n]/(n! (2n+1)) x^(2n+1),
{n, 0, Infinity}]
ArcSin[x]
An equivalent expression without the Pochhammer symbol
SymbolicSum[2 (2n)!/((n!)^2 (2n+1)) (x/2)^(2n+1),
{n, 0, Infinity}]
ArcSin[x]
If by a recursive relation you mean how to calculate the n-th term of the
series from the (n-1) term, the ratio of the terms is given by
r = ( 2 (2n)!/((n!)^2 (2n+1)) (x/2)^(2n+1) ) /
( ( 2 (2n)!/((n!)^2 (2n+1)) (x/2)^(2n+1) ) /.
n -> n-1 ) // Simplify
2 2
(-1 + 2 n) x (-1 + n)! (2 n)!
-------------------------------
2
4 (1 + 2 n) (2 (-1 + n))! n!
r = r /. { n! -> n (n-1)!, (2n)! -> (2n)(2n-1) (2(n-1))! }
2 2
(-1 + 2 n) x
--------------
2 n (1 + 2 n)
t[0, x_] := x
t[n_Integer?Positive, x_] := t[n, x] =
(2n-1)^2 x^2 / (2n (2n+1)) t[n-1, x]
Table[t[n, x], {n, 0, 6}]
3 5 7 9 11 13
x 3 x 5 x 35 x 63 x 231 x
{x, --, ----, ----, -----, ------, -------}
6 40 112 1152 2816 13312
> From: njachimi at glibm10.cen.uiuc.edu (Nathan Jachimiec)
> Newsgroups: comp.soft-sys.math.mathematica
> Subject: arcsin
> Date: 23 Jun 1995 02:39:29 GMT
>
> Is there a recurrence relation that one can use to compute an arcsin???
> How can one numerically compute the value of arcsin with a recurrence
> relation (restated for clarity)? Can someone direct me to some
resource
> or formula for numerical solutions of this function...
>
> njachimi at uiuc.edu