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