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Path: utzoo!watmath!watdaisy!mvramakrishn
From: mvramakrishn@watdaisy.UUCP (Rama)
Newsgroups: net.puzzle
Subject: Re: More on Interview Questions.
Message-ID: <7115@watdaisy.UUCP>
Date: Sun, 24-Mar-85 12:22:11 EST
Article-I.D.: watdaisy.7115
Posted: Sun Mar 24 12:22:11 1985
Date-Received: Mon, 25-Mar-85 01:47:04 EST
References: <379@cavell.UUCP> <917@utcsri.UUCP>
Distribution: net
Organization: U of Waterloo, Ontario
Lines: 71

> > 
> >
> 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:
> 
> 	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?
> 
> -- 
   There seem to be slight error in the above.
	I am not an expert on Maths but from fundamental high school calculus,
		Let    y(x) = x!
			y(x+1) = (x+1)!

			(let  deltax = 1)
			dy/dx  =  limit as deltax -> 0 of 
					(y(x+deltax)-y(x)) / (deltax)
				= { y(x+1) - y(x) } / {(x+1)-x}
				= { (x+1)! - x! } / 1
				= (x+1) x! - x!
				= x x! + x! - x!
				= x * x!
	That is,  the derivative of  x!  is  x * x!
	
	Well, I do realise that the above is not a mathematical proof.
	The correct procedure seems to be the one given by <379@cavell.UUCP>
--------------------------------------------------------------------------------
	May I take this opportunity to REQUEST the people on the net to try
	to avoid posting something on a problem on which already scores
	of people have posted (unless it is very important).
	An example is obout manhole covers, a topic which has degenerated into
	personal fights (into talks about personal holes !).  
--------------------------------------------------------------------------------
    								Rama

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Mail:  M.V.Ramakrishna, Dept of Computer Science, University of Waterloo,
       Waterloo Ont., N2L 3G1 Canada