Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.2 9/18/84; site utcsri.UUCP
Path: utzoo!utcsri!panos
From: panos@utcsri.UUCP (Panos Economopoulos)
Newsgroups: net.puzzle
Subject: Re: More on Interview Questions.
Message-ID: <917@utcsri.UUCP>
Date: Sat, 23-Mar-85 02:40:05 EST
Article-I.D.: utcsri.917
Posted: Sat Mar 23 02:40:05 1985
Date-Received: Sat, 23-Mar-85 04:16:55 EST
References: <379@cavell.UUCP>
Distribution: net
Organization: CSRI, University of Toronto
Lines: 81

> 
>
Here are the answers to the following questions:

> From: mouli@cavell.UUCP (Bopsi Chandramouli)
> Subject: More on Interview Questions.
> Message-ID: <379@cavell.UUCP>
> Date: 14 Mar 85 06:00:58 GMT
> Organization: U. of Alberta, Edmonton, AB
> 
> 
> Here are two more 'good' interview questions. Actually I was asked
> the first question in a job interview.
> 
> 1) Hope you agree that
> 
>               x**2 = x + x + x + ... x such terms. 
> 
>    Then consider these two derivatives.
> 
>               d(x**2)
>              ----     = 2*x.
>               dx
> 
>               d(x + x + x + x .. x such terms)
>              ____                              = 1 + 1 + 1 + 1 .. x such terms
>               dx
>                       = x.
> 
>    Why is this anamoly?

The problem appears because differentiation is a linear operator,
i.e.  d(af(x)+bg(x))/dx = a df(x)/dx + b dg(x)/dx for a,b constants.
Therefore, d(x+x+x+..+x)/dx = dx/dx + dx/dx + ... + dx/dx ONLY
for a constant number of additions. However, the way the problem
is presented, we have x additions at the righthand side, i.e. a
function of the variable and the above property does NOT hold,
i.e., we cannot get rid of the brackets and differentiate each term
independently.


> 
>  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?

-- 

					Panos Economopoulos

UUCP:   {decvax,linus,ihnp4,uw-beaver,allegra,utzoo}!utcsri!panos
CSNET:  panos@toronto