Re: Mathematical Question

[Henry F. Mollet]

Wow, three answers within seconds.  I'd better get reacquainted with
L'Hopital's rule.
Many Thanks, Henry

on 8/2/04 2:04 PM, Hall, Benjamin at address@hidden wrote:

> I also cheated and used Mathematica (I haven't had a chance to learn maxima
> syntax yet...).  If you apply L'Hopital's rule twice to your equation, you
> get the right answer
> 
> 
> 
> -----Original Message-----
> From: John W. Eaton [mailto:address@hidden
> Sent: Monday, August 02, 2004 4:52 PM
> To: Henry F. Mollet
> Cc: Ada Cheng; Octave_post
> Subject: Re: Mathematical Question
> 
> 
> On  2-Aug-2004, Ada Cheng <address@hidden> wrote:
> 
> | You can write (1-x^c)=(1-x)(1+x+...+x^{c-1}), this allows you to rewrite
> | y as (sum_{k=1}^c x^k-x^c)/(1-x^c). y is clearly now in indeterminate
> | form as x approaches 1.  Apply L'Hopital rule, do some algebra and you
> | should get the answer.  (to make life easy on yourself, you will want to
> | use finite sum of arithmetic series formula).
> | 
> | Hope this help.
> 
> And at least Maxima thinks your result is the right one:
> 
> $ maxima
> GCL (GNU Common Lisp)  2.6.2 CLtL1   Jun 29 2004 18:53:13
> Source License: LGPL(gcl,gmp), GPL(unexec,bfd)
> Binary License:  GPL due to GPL'ed components: (READLINE BFD UNEXEC)
> Modifications of this banner must retain notice of a compatible license
> Dedicated to the memory of W. Schelter
> 
> Use (help) to get some basic information on how to use GCL.
> Maxima restarted.
> (C1) limit (x/(1-x)-c*x^c/(1-x^c), x, 1);
> 
>       C - 1
> (D1)                                 -----
> 2
> 
> jwe
> 
> 
> 
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