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Re: Evaluation of 1/(1/a + 1/b + 1/r)


Bob, thanks so much. Diana

On 10/5/06, Bob Hanlon <hanlonr at cox.net> wrote:
>
> 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
>
>
>


-- 
"God made the integers, all else is the work of man."
L. Kronecker, Jahresber. DMV 2, S. 19.



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