Re: A lost treasure (Series within Parallel resistor combinations.)

Quentin Grady wrote :
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On Thu, 19 Jul 2007 14:22:41 +0200, "Philippe 92"
<nospam@xxxxxxxxxxxx> wrote:

Quentin Grady wrote :
On Wed, 18 Jul 2007 18:22:50 +0200, "Philippe 92" wrote:
Quentin Grady wrote :
...
[ find integer solutions to parallel/series combinations ]
[snip]

the full list of resistors r1, r2, r3, r4 with
rx = (r1+r2)//(r3+r4)
ry = a + b = (r1//r3) + (r2//r4)
r1, r2, r3, r4, a, b, rx, ry integers < 100
and |rx-ry| > 10

[snip]

20 solutions

[snip]

If it's OK with you I'd like to post a link to your website if you
would care to post them there or even post them on my own website with
an acknowledgement to you and a link to your website.

Hi,

As you like.
You could post the table on your site right now, may be as an extent of
your "reciprocal sums" page, or just wait for I post it on my site.
I'm in a problem of triangle construction now.
When I'm finished, I intend to extend my former problem of two parallel
resistors, and give a javascript to dynamically calculate solutions.
Or maybe a java applet for faster execution.

For more than two resistors in parallel, there is a relation with
unitary fractions (your page about Sylvester and Farey methods)
I allready have a script at
http://chephip.free.fr/pba_en/exe003.html
It can give all solutions of 1/a + 1/b + 1/c ... = 1/r
Number of solutions increase very fast with number of resistors.
The program is useless with 4 or more resistors in parallel.
I should add an upper bound for values.

Several minutes for 4 resistors to get an arithmetic overflow and more
than 8000 solutions, for instance :
1/108 + 1/110 + 1/132 + 1/135 = 1/30 (the simplest)
but also some 1/52 + 1/73 + 1/2476 + 1/17622930 = 1/30 !

Regards.

--
Philippe C., mail : chephip+news@xxxxxxx
site : http://chephip.free.fr/ (recreational mathematics)


.

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