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

```

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