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Re: How to solve this type of equation in Mathematica?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg59999] Re: How to solve this type of equation in Mathematica?
  • From: Jeffrey Lyons <not-me at nospam.net>
  • Date: Sun, 28 Aug 2005 03:07:35 -0400 (EDT)
  • References: <dehk3h$c9t$1@smc.vnet.net> <dep7f7$eur$1@smc.vnet.net>
  • Reply-to: not-me at nospam.net
  • Sender: owner-wri-mathgroup at wolfram.com

On Sat, 27 Aug 2005 08:15:35 +0000 (UTC), robert.dodier at gmail.com
wrote:

>It looks to me like this isn't strong enough
>to yield a unique solution (or even a solution in
>a well-known class).
>
>If you replace f by A f where A is a constant
>you get A times \int_P^{\infty} (x + P) f(x) dx.
>So let f be any function s.t. the integral is defined
>and not equal to zero. (I believe that's a fairly
>large class.) If the integral I is not equal to 1,
>replace f by f/I and that's a solution,
>if I'm not mistaken.
>
>The equation is an example of an integral equation,
>so you might be able to find more info under that
>heading. Sorry that I can't be more helpful.
>
>Robert Dodier

Robert,
thanks so much for your interest in this problem. Your solution,
replacing f(x) with  f(x)/I if f is not equal to 1, still does not
prove there is a solution for f(x).  For all the solutions of this
type that I can think of the new f(x) is now a function of both x and
P, ie f(x, P). For example:
 f(x,P) =0.6666 * x^ (-3) * P^(-1)
is a solution.  What I need is a an f that is only a function of the
variable "x".

Can it be proven that there is/(is not) a solution to my equation if f
is only a function of the variable x and not a function of "P"?

Again any assistance would be appreciated.


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