Re: Evaluation of 1/(1/a + 1/b + 1/r)
- To: mathgroup at smc.vnet.net
- Subject: [mg70165] Re: [mg70129] Evaluation of 1/(1/a + 1/b + 1/r)
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Fri, 6 Oct 2006 01:58:22 -0400 (EDT)
- Reply-to: hanlonr at cox.net
Clear[f]; f[a_,b_,r_]:=0/;a==b==r==0; f[a_,b_,r_]:=1/Total[ 1/Select[{a,b,r},#=!=0&&#=!=0.&]]; {f[a,b,r],f[0,b,r],f[a,0,r],f[a,b,0],f[a,0.,0],f[0,b,0],f[0,0,r],f[0,0,0]} {1/(1/b + 1/r + 1/a), 1/(1/r + 1/b), 1/(1/r + 1/a), 1/(1/b + 1/a), a, b, r, 0} Prior to version 5 use Clear[f]; f[a_,b_,r_]:=0/;a==b==r==0; f[a_,b_,r_]:=1/Plus@@ (1/Select[{a,b,r},#=!=0&&#=!=0.&]); Bob Hanlon ---- Diana <diana.mecum at gmail.com> wrote: > Folks, > > I am trying to write a program to evaluate the sum of two continued > fractions, written in polynomials of T. > > I would like a short routine to evaluate 1/(1/a + 1/b + 1/r), just > working with a, b, and r if they are not equal to zero. > > So, if a = 0, then evaluate 1/(1/b + 1/r) > > If a, b, = 0, then evaluate 1/r > > If all of a, b, r = 0, give me 0 as output. > > If (1/a + 1/b + 1/r) = 0, give me 0 as output. > > Assume that a, b, r are arbitrary polynomials in T. > > Can someone help? > > Thanks, > > Diana > -- Bob Hanlon hanlonr at cox.net