From:Quentin Grady <quentin@inhb.co.nz>Newsgroups:sci.mathSubject:Egyptian Fractions & Parallel Resistors.Date:10 Jun 1996 10:12:45 GMTOrganization:Ram Management Ltd

G'day G'day Folks, I teach electrical theory a fairly introductory level. When resistors are added in parallel, the recipocals add 1/R(total) = 1/R1 + 1/R2. In writing test questions it is often desirable to have the answers as integers. As it happens I have read that Egyptian mathematicians expressed all fractions as the sum of reciprocals. eg 2/3 would be written as 1/2 + 1/6 I believe they compiled tables of these sums of reciprocals and one such table was found on an historic monument. Does anyone have or know how to generate a set of integral solutions for 1/z = 1/y + 1/x that I could use for writing simple parallel resistor problems? Please email if possible. Quentin Grady ^ ^ __/ New Zealand, >#,#< [ / \ /\ "... and the blind dog was leading."

To:quentin@inhb.co.nzSubject:Egyptian fractionsDate:Mon, 10 Jun 1996 10:35:02 -0700From:David Eppstein <eppstein@ICS.UCI.EDU>

Quentin Grady <quentin@inhb.co.nz> wrote: > I teach electrical theory a fairly introductory level. > When resistors are added in parallel, the recipocals add > > 1/R(total) = 1/R1 + 1/R2. > > Does anyone have or know how to generate a set of integral solutions > for 1/z = 1/y + 1/x that I could use for writing simple parallel resistor > problems? I have probably more than you wanted to know about generating Egyptian fractions, online at my web page http://www.ics.uci.edu/~eppstein/numth/egypt/ However, regarding your particular equation 1/z = 1/y + 1/x: let g=gcd(x,y), p=x/g, q=y/g, then your equation can be rewritten 1/z = 1/g (1/p + 1/q) with gcd(p,q)=1. But now 1/p+1/q= (p+q)/pq must be in lowest terms already, since any divisor of pq must divide either p or q, and we know that gcd(p+q,p)=gcd(p+q,q)=1. So if z is to be an integer, g must cancel the factor p+q in the numerator. Suppose g=(p+q)r. To generate fractions of the form you describe, simply choose p, q, and r, then you have 1/p(p+q)r + 1/q(p+q)r = 1/pqr E.g. p=2, q=3, r=7 gives 1/70 + 1/105 = 1/42. The analysis above shows that all solutions can be generated in this way. -- David Eppstein UC Irvine Dept. of Information & Computer Science eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

To:quentin@inhb.co.nzSubject:Egyptian fractionsDate:Mon, 10 Jun 1996 18:08:51 -0700From:David Eppstein <eppstein@ICS.UCI.EDU>

I assume the technique for Egyptian fractions can be extended to the form, 1/z = 1/y + 1/x + 1/w This would be really useful for setting up two-mesh circuit problems. I think the following procedure generates all the solutions: choose five numbers w,x,y,z,r, with w<x and gcd(x,w)=gcd(y,z)=1 let f = gcd(xw(y+z),yz(x-w)) let g = yz(x-w)/f let h = xw(y+z)/f then 1/wgr = 1/xgr + 1/yhr + 1/zhr e.g. w,x,y,z=2,5,4,11,1 f = gcd(150,132)=6 g = 132 / 6 = 22 h = 150 / 6 = 25 1/44 = 1/110 + 1/100 + 1/275 -- David Eppstein UC Irvine Dept. of Information & Computer Science eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

From:Quentin Grady <quentin@inhb.co.nz>Newsgroups:sci.mathSubject:Re: Egyptian Fractions & Parallel Resistors.Date:11 Jun 1996 16:37:18 GMTOrganization:Ram Management Ltd

Thanks all you people who came up with an answer and emailed it to me. You have been most helpful. It will make the marking more certain as rounding errors are eliminated in multipart questions. One simple method sent me, is to factorise the total resistance thus, 1/pq = 1/p(p + q) + 1/q(p +q) Thanks again. Quentin Grady ^ ^ __ / New Zealand, >#,#< [ / \ /\ "... and the blind dog was leading."

From:ksbrown@ksbrown.seanet.com (Kevin Brown)Newsgroups:sci.mathSubject:Unit Fractions and FibonacciDate:Tue, 18 Jun 1996 23:26:13 GMTOrganization:Seanet Online Services, Seattle WA

Quentin Grady <quentin@inhb.co.nz> wrote: > When resistors are added in parallel, the recipocals add > 1/R(total) = 1/R1 + 1/R2 > Does anyone have or know how to generate a set of integral > solutions for 1/z = 1/y + 1/x ...? Quentin Grady <quentin@inhb.co.nz> wrote: > ...Thanks all you people who came up with an answer and emailed it > to me. You have been most helpful. One simple method sent me is to > factorise the total resistance thus: 1/pq = 1/p(p+q) + 1/q(p+q) This can be regarded as a special case of a more general expansion related to the Fibonacci numbers. Let s[j], j=0,1,2,... be a sequence of integers that satisfy the recurrence s[k] = s[k-1] + s[k-2] with arbitrary initial values s[0] and s[1]. It can be shown that for any integers m,n with m>n we have 1 1 m 1 ----------- = ----------- + SUM ------------- (1) s[n-1] s[n] s[m] s[m+1] j=n s[j-1] s[j+1] For example, setting s[0]=s[1]=1 and n=5, m=10 gives 1 1 1 1 1 1 1 1 --- = --- + --- + --- + ---- + ---- + ---- + ----- 40 65 168 442 1155 3026 7920 12816 In general, to expand 1/D into a sum of unit fractions, the method is to split D into two factors, D = pq. Then we can set s[0]=p and s[1]=q and generate the s sequences as follows k s[k] s[k] s[k-1] s[k] s[k-2] --- ------- ----------- ----------- 0 p 1 q pq 2 p+q q(p+q) p(p+q) 3 p+2q (p+q)(p+2q) q(p+2q) 4 2p+3q (p+2q)(2p+3q) (p+q)(2p+3q) 5 3p+5q (2p+3q)(3p+5q) (p+2q)(3p+5q) 6 5p+8q (3p+5q)(5p+8q) (2p+3q)(5p+8q) 7 8p+13q (5p+8q)(8p+13q) (3p+5q)(8p+13q) etc etc etc We can now express 1/pq as the sum of the inverses of the numbers in the third column down to the mth row, plus the inverse of the mth number in the second column. Thus we have 1/pq = 1/p(p+q) + 1/q(p+q) = 1/p(p+q) + 1/q(p+2q) + 1/(p+q)(p+2q) = 1/p(p+q) + 1/q(p+2q) + 1/(p+q)(2p+3q) + 1/(p+2q)(2p+3q) etc. Of course, we can let m in equation (1) go to infinity, giving the infinite unit fraction expansion 1 inf 1 ----------- = SUM ------------- (2) s[n-1] s[n] j=n s[j-1] s[j+1] This can also be generalized to higher order recurrences. For example, if we define the sequence s[j] to satisfy the 3rd order recurrence s[k] = s[k-2] + s[k-3] with the initial values a,b,c, then we can generate the following sequences k s[k] s[k]s[k-1]s[k-2] s[k]s[k-1]s[k-3] --- ------ -------------------- ---------------------- 0 a 1 b 2 c abc 3 a+b bc(a+b) ac(a+b) 4 b+c c(a+b)(b+c) b(a+b)(b+c) 5 a+b+c (a+b)(b+c)(a+b+c) c(b+c)(a+b+c) 6 a+2b+c (b+c)(a+b+c)(a+2b+c) (a+b)(a+b+c)(a+2b+c) etc etc etc so we have 1/abc = 1/ac(a+b) + 1/bc(a+b) = 1/ac(a+b) + 1/b(a+b)(b+c) + 1/c(a+b)(b+c) = 1/ac(a+b) + 1/b(a+b)(b+c) + 1/c(b+c)(a+b+c) + 1/(a+b)(b+c)(a+b+c) and so on. To illustrate, with a=3,b=7,c=11 this last formula gives 1/231 = 1/330 + 1/770 = 1/330 + 1/1260 + 1/1980 = 1/330 + 1/1260 + 1/3780 + 1/4158 and with a=23,b=c=1 it gives 1/23 = 1/24 + 1/552 = 1/48 + 1/50 + 1/552 + 1/1200 ===================================================================== MathPages at --> http://www.seanet.com/~ksbrown/ =====================================================================