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From: msb@lsuc.UUCP (Mark Brader)
Newsgroups: net.puzzle
Subject: Re: More on Interview Questions...derivative of x!
Message-ID: <541@lsuc.UUCP>
Date: Sun, 24-Mar-85 04:37:29 EST
Article-I.D.: lsuc.541
Posted: Sun Mar 24 04:37:29 1985
Date-Received: Sun, 24-Mar-85 06:47:23 EST
References: <379@cavell.UUCP> <917@utcsri.UUCP>
Reply-To: msb@lsuc.UUCP (Mark Brader)
Distribution: net
Organization: Law Society of Upper Canada, Toronto
Lines: 58
Summary: derivative of x! is not x!

panos@utcsri.UUCP (Panos Economopoulos) writes:

> >  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.

Wrong.  x! is a function of whole numbers (nonnegative integers) only.
It is therefore discontinuous and has no derivative.

However, the Gamma function is the generalization of factorial to real
numbers, so the question of its derivative is of some interest.
Let's continue, pretending that the above was correct...

> 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:
> 
> 	dx!/dx = integral from 0 to inf ( x * t to the power of (x-1) *
> 			e to the power -t) dt

No we don't.  "x * t to the power (x-1)" is the derivative of "t to the
power x" with respect to t, not x.  The derivative with respect to x is
"ln t * t to the power (x - 1)".

Therefore

	dx!/dx = integral from 0 to inf (ln t * t to the power (x-1) *
			e to the power -t) dt

This is beyond me.

> That is, the derivative of x! is x!

This conclusion from the incorrect expression above could have been rejected
by a plausibility check.  Remember that 0! = 1! = 1.  Since we're talking about
a continuous function, its derivative must therefore be zero somewhere on this
interval (Rolle's theorem).  But then between 0! = 1 and the first a! = 0,
the function must have a negative derivative somewhere since it is decreasing,
yet must be positive since the first a!=0 has not been reached.  Therefore
the function can't be its own derivative.

> 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?

I think the exponential is the only one, but I don't know.

Mark Brader