Re: Hi, I would like to solve for f(x)

On 25 Aug 2005 15:22:24 -0700, Marc.Mercurio@xxxxxxxxx wrote:

>Yes, int_P^infinity (x+P) f(x) dx = 1 or all P.
>
>Taking the derviative of both sides yields:
>
>f(x) = x^(-1.5)

Does it? If I take the derivative of both sides
(with respect to P) I get

-2P f(P) + int_P^infinity f(x) dx = 0

Differentiating again I get

-2 f(P) - 2P f'(P) -f(P) = 0,

or

(*) f'(x) = -3 f(x)/(2x).

Since f(x) = x^(-1.5) is one solution to (*) it
seems likely that we've both done the calculus
correctly so far. But (*) has many _other_
solutions...

>But plugging this result back into the equation does not equal 1.
>
>Can anyone tell me if there is a solution to this equation or is there
>simply no f(x) that satisfies this equation??
>Note: f(x) can not be a function of P. It should be a function of "x"
>only.

Huh? Yes, f(x) is a function of x only. And f(t) is a function
of t only, f(q) is a function of q only. The way you set up
the notation makes it natural to derive the equation for f(P).


************************

David C. Ullrich
.

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