From help-request at octave dot org Mon Aug 2 16:39:40 2004 Subject: RE: Mathematical Question From: "grumpy steve" To: "'Henry F. Mollet'" , "'Octave_post'" Date: Mon, 2 Aug 2004 22:37:47 +0100 Another nice way of doing it is to replace x by P/(1+P), this time = taking the limit as P tends to infinity. Put everything over a common = denominator, and use the fact that (1+P)^C =3D P^C + CP^(C-1) + C(C-1)P^(C-2) ...... wherever you find it. The answer comes out as expected. Steve -----Original Message----- From: Henry F. Mollet [mailto:mollet at pacbell dot net]=20 Sent: 02 August 2004 21:20 To: Octave_post Subject: Mathematical Question I have nowhere else to turn, I'm hoping the octave help list can provide a tip. Many thanks, Henry y =3D x/(1-x) - cx^c/(1-x^c), where c is a positive integer constant I need to know y in the limit as x approaches 1. I even have the result by biological reasoning and it is (c-1)/2 but how do I prove it mathematically? For example using x =3D 0.999 and c =3D 100 we have: y =3D 999 - 950.4 =3D 48.6 whereas (c-1)/2 =3D 49.5; Using x =3D 0.9999 and same c =3D 100, we have: y =3D 9999 - 9949.6 =3D 49.4 whereas (c-1)/2 =3D 49.5, close enough. =20 Somehow I have to expand the second term in the expression into a series where the first term of the series will be the same as the first term of the expression (i.e. x/(1-x)) and they will cancel out. In my application, the constant c is a positive integer, and that's what I used for empirical checks of the result but I have not checked if that's a requirement. ------------------------------------------------------------- Octave is freely available under the terms of the GNU GPL. Octave's home on the web: http://www.octave.org How to fund new projects: http://www.octave.org/funding.html Subscription information: http://www.octave.org/archive.html ------------------------------------------------------------- ------------------------------------------------------------- Octave is freely available under the terms of the GNU GPL. Octave's home on the web: http://www.octave.org How to fund new projects: http://www.octave.org/funding.html Subscription information: http://www.octave.org/archive.html -------------------------------------------------------------