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From: steve@siemens.UUCP
Newsgroups: net.puzzle
Subject: Re: More on Interview Questions.
Message-ID: <23600001@siemens.UUCP>
Date: Wed, 27-Mar-85 10:02:00 EST
Article-I.D.: siemens.23600001
Posted: Wed Mar 27 10:02:00 1985
Date-Received: Thu, 28-Mar-85 02:47:01 EST
References: <379@cavell.UUCP>
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Nf-ID: #R:cavell:-37900:siemens:23600001:000:2546
Nf-From: siemens!steve    Mar 27 10:02:00 1985


/* Written  2:40 am  Mar 23, 1985 by panos@utcsri in siemens:net.puzzle */
> Here are the answers to the following questions:
> 
> > From: mouli@cavell.UUCP (Bopsi Chandramouli)
> > 
> >  2) what is the derivative of factorial x?
> 
> Factorial x or  x! for short, is  in general defined as
> 
> 	x! = G (x+1) =  integral from 0 to inf  (t to the power of x
> 			times e to the power of -t) dt
> 
> where G(x) is the Gamma function.
> 
> The derivative of x! is, therefore, the derivative of the integral
> with respect to x. Because the function g(t,x) (t to the power ... -t)
> which is to be integrated is continuous w.r.t. t and x, and 
> because its derivative dg(t,x)/dx is also continuous, we can
> interchange the integral and the derivative, i.e. integrate
> the derivative of g(t,x) w.r.t. x  (Leibnitz?? thrm??)
> Doing this we get:
> 
****** WARNING!! ****** MISTAKE (see below) AT THIS STEP ******
> 	dx!/dx = integral from 0 to inf ( x * t to the power of (x-1) *
> 			e to the power -t) dt
> 	       = x * integral from 0 to inf ( t to the power of (x-1) *
> 			e to the power -t) dt
> 	       = x * G (x)
> 	       = x * (x-1)!
> 	       = x!
> 
> That is, the derivative of x! is x!
> 
> I wasn't expecting this result, since the one function you can easily
> get that equals its derivative is the exponential. A question one
> could ask is can you really find ALL functions, defined in any way,
> that equal their derivatives and  which are they?
> -- 
> 					Panos Economopoulos

Panos, I'm afraid you made a mistake.  First of all,
the exponential function can be defined/discovered by looking for the
set of functions whose derivatives are equal to themselves.  (You get
the set of all functions re^x where r is any real constant.)  One learns
more than one ever wanted to know about this sort of thing in Differential
Equations.  Secondly, the reason you got your erroneous answer was that,
once you correctly got the derivative operator inside the integral, you
took the derivative with respect to t instead of x.  I'm surprised you
did this since you fluent enough with calculus to know better.

	dx!/dx	= d/dx [integral from 0 to inf (t to the power of x *
			e to the power -t) dt]
		= integral from 0 to inf (d/dx [t to the power of x] *
			e to the power -t) dt
	d/dx [t^x] is NOT equal to x * t^(x-1)
	dx!/dx	= integral from 0 to inf (t to the power of x * log t *
			e to the power -t) dt

I will leave the solution of this integral as an exercise for the reader.

-Steve Clark

{princeton | astrovax}!siemens!steve