RE: Mathematical Question

[grumpy steve]

Sorry - typed too quickly!
(1+P)^C = P^C + CP^(C-1) + (C(C-1)P^(C-2))/2 .....
I'd omitted the factor of 1/2 in the last term, which is a pity, since
that's where the final result comes from!

Steve

-----Original Message-----
From: grumpy steve [mailto:address@hidden 
Sent: 02 August 2004 22:38
To: 'Henry F. Mollet'; 'Octave_post'
Subject: RE: Mathematical Question

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 = 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:address@hidden 
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 = 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 = 0.999 and c = 100 we have:
y = 999 - 950.4 = 48.6 whereas (c-1)/2 = 49.5;
Using x = 0.9999 and same c = 100, we have:
y = 9999 - 9949.6 = 49.4 whereas (c-1)/2 = 49.5, close enough.
 
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.



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Octave is freely available under the terms of the GNU GPL.

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