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)

```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

```