binomial fraction subharmic functions-3rd, and 4th

• To: mathgroup at smc.vnet.net
• Subject: [mg52602] binomial fraction subharmic functions-3rd, and 4th
• From: Roger Bagula <tftn at earthlink.net>
• Date: Fri, 3 Dec 2004 03:54:41 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```I got the subharmonics for three and four to work too.
I used binomial fractions of (x+1)^n, n=3,4
Triplet:
{ 2*n/(1+n)2,1/(1+n)2,n2/(1+n)2}
Quartet:
{3*n/(1+n)3, 3*n2/(1+n)3,1/(1+n)3,n3/(1+n)3}

(* using a set of functions that give a 3rd level subharmonic functions
:  *)
(*triplet { 2*x/(1+x)^2},x^2/(1+x)^2,1/(1+x)^2*)
(* resulting sum of functions is a circle*)
f1[n_]:=2*n/(n+1)^2/;Mod[n,3]==1
f1[n_]:=n^2/(1+n)^2/;Mod[n,3]==2
f1[n_]:=1/(1+n)^2/;Mod[n,3]==0

f2[n_]:=2*n/(n+1)^2/;Mod[n,3]==2
f2[n_]:=n^2/(1+n)^2/;Mod[n,3]==0
f2[n_]:=1/(1+n)^2/;Mod[n,3]==1

f3[n_]:=2*n/(n+1)^2/;Mod[n,3]==0
f3[n_]:=n^2/(1+n)^2/;Mod[n,3]==1
f3[n_]:=1/(1+n)^2/;Mod[n,3]==2

f1sin[x_]:=Sum[(-1)^(n)*f1[n]*x^(2*n+1)/((2*n+1)!),{n,0,digits}]
f1cos[x_]:=Sum[(-1)^(n)*f1[n]*x^(2*n)/((2*n)!),{n,0,digits}]

f2sin[x_]:=Sum[(-1)^(n)*f2[n]*x^(2*n+1)/((2*n+1)!),{n,0,digits}]
f2cos[x_]:=Sum[(-1)^(n)*f2[n]*x^(2*n)/((2*n)!),{n,0,digits}]

f3sin[x_]:=Sum[(-1)^(n)*f3[n]*x^(2*n+1)/((2*n+1)!),{n,0,digits}]
f3cos[x_]:=Sum[(-1)^(n)*f3[n]*x^(2*n)/((2*n)!),{n,0,digits}]

Plot[f1sin[x],{x,-Pi,Pi}]
Plot[f1cos[x],{x,-Pi,Pi}]
ParametricPlot[{f1sin[x],f1cos[x]},{x,-Pi, Pi}]

Plot[f2sin[x],{x,-Pi,Pi}]
Plot[f2cos[x],{x,-Pi,Pi}]
ParametricPlot[{f2sin[x],f2cos[x]},{x,-Pi, Pi}]

Plot[f3sin[x],{x,-Pi,Pi}]
Plot[f3cos[x],{x,-Pi,Pi}]
ParametricPlot[{f3sin[x],f3cos[x]},{x,-Pi, Pi}]

Plot[f1sin[x]+f2sin[x]+f3sin[x],{x,-Pi,Pi}]
Plot[f1cos[x]+f2cos[x]+f3cos[x],{x,-Pi,Pi}]
ParametricPlot[{f1sin[x]+f2sin[x]+f3sin[x],f1cos[x]+f2cos[x]+f3cos[x]},{x,-Pi,Pi}]

Respectfully, Roger L. Bagula