Arctangent approximation
- To: mathgroup at smc.vnet.net
- Subject: [mg107560] Arctangent approximation
- From: sidey <sidey.p.timmins at census.gov>
- Date: Wed, 17 Feb 2010 07:02:10 -0500 (EST)
A professor (Herbert Medina) came up with this remarkable
approximation.
Since I don't have Mathematica, could someone run it for
numberofdigits=9 and 12 please? (and send me the result?)
here is a short result
h2(x) = x - x^3/3 + x^5/5 - x^7/7 + 5x^9/48..
Thank you.
PS reference is: http://myweb.lmu.edu/hmedina/Papers/Arctan.pdf
Clear[h, m, a, numberofdigits]
numberofdigits=12;
m=Floor[-(1/5) Log[4,5*0.1^(numberofdigits+1)]]+1;
Do[a[2j]=(-1)^(j+1) Sum[Binomial[4m,2k] (-1)^k, {k,j+1,2m}];
a[2j-1]=(-1)^(j+1) Sum[Binomial[4m,2k+1] (-1)^k, {k,j,2m-1}],
{j,0,2m-1}];
h[m,x_]:=Sum[(-1)^(j+1) / (2j-1) x^(2j-1), {j,1,2m}] +
Sum[a[j]/((-1)^(m+1) 4^m (4m+j+1)) x^(4m+j+1), {j,0, 4m-2}];
Print["h[",m,",x] given below computes ArcTan[x] with ",
numberofdigits," digits of accuracy for x in [0,1]."]
Print["h[m,x] = ", h[m,x]]
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