Re: How to solve this type of equation in Mathematica?
- To: mathgroup at smc.vnet.net
- Subject: [mg59921] Re: How to solve this type of equation in Mathematica?
- From: Ronald Bruck <bruck at math.usc.edu>
- Date: Thu, 25 Aug 2005 06:34:29 -0400 (EDT)
- References: <dehk97$cbs$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <dehk97$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