```From:           Quentin Grady  <quentin@inhb.co.nz>
Newsgroups:     sci.math
Subject:        Egyptian Fractions & Parallel Resistors.
Date:           10 Jun 1996 10:12:45 GMT
Organization:   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.nz
Subject:        Egyptian fractions
Date:           Mon, 10 Jun 1996 10:35:02 -0700
From:           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.nz
Subject:        Egyptian fractions
Date:           Mon, 10 Jun 1996 18:08:51 -0700
From:           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.math
Subject:        Re: Egyptian Fractions & Parallel Resistors.
Date:           11 Jun 1996 16:37:18 GMT
Organization:   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.math
Subject:        Unit Fractions and Fibonacci
Date:           Tue, 18 Jun 1996 23:26:13 GMT
Organization:   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 and s.  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=s=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=p and
s=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/
=====================================================================
```