Re: How to solve this type of equation in Mathematica?
- To: mathgroup at smc.vnet.net
- Subject: [mg59964] Re: How to solve this type of equation in Mathematica?
- From: Jeffrey Lyons <not-me at nospam.net>
- Date: Fri, 26 Aug 2005 04:54:19 -0400 (EDT)
- References: <dehk97$cbs$1@smc.vnet.net> <dek81g$a7m$1@smc.vnet.net>
- Reply-to: not-me at nospam.net
- Sender: owner-wri-mathgroup at wolfram.com
On Thu, 25 Aug 2005 10:54:40 +0000 (UTC), Ronald Bruck <bruck at math.usc.edu> wrote: >In article <dehk97$cbs$1 at smc.vnet.net>, Jeffrey Lyons ><not-me at nospam.net> wrote: > >> Hi, I would like to be able to solve the equation: >> >> Equal[Integrate[Times[Plus[P, x], f[x]], List[x, P, >> DirectedInfinity[1]]], 1] >> >> in Mathematica. Is there any way to do this? >> >> That is I would like to solve for f(x) when we know that >> integrating (x + P) f(x) from P to Infinity over x equals 1. >> > >Nothing beats mathematics. Not even Mathematica. > >Your equation (in LaTeX) is, as I make it out, > > \int_P^\infty (x+P)f(x) dx = 1. > >Assuming f is continuous (!), we can differentiate both sides wrt P to >obtain > > -2P f(P) + \int_P^\infty f(x) dx = 0. > >This implies f is differentiable (except maybe at 0); so we can >differentiate again wrt P, > > -2f(P) - 2P f'(P) - f(P) = 0. > >Thus f'(P) = -3/2 f(P)/P. > >It follows that f(P) = k P^(-3/2). But now you have a problem, >because (x+P)f(x) is NOT integrable at infinity (asymptotically, it's >like k/sqrt(x)). > >I don't think there is such a function. However, I haven't examined >what happens at points of discontinuity. > >--Ron Bruck Ron, are you saying there is no solution to my equation, ie there is no f(x) for which my equation holds?? Just to clarify P>0.