From: wilson@web.ctron.com (David Wilson)
Newsgroups: sci.math
Subject: Re: Strange Constant
Date: 13 Aug 92 14:09:47 GMT
Reply-To: wilson@web.ctron.com (David Wilson)
Organization: Cabletron Systems INc.
Originator: wilson@web


 
> I have some questions about a constant, I don't know it's name,
> but here's how it is obtained.
> 
>                        1     1     1           1     1
>   <constant> =  lim   --- + --- + --- + ... + --- + --- - ln n
>                n->oo   1     2     3          n-1    n
> 
> <constant> is approximately equal to the following
> 
>   0.577216164900715256180774304084479808807373046875...
> 
> My questions are :
> 
>   1.  What is this constant?
>   2.  Who first obtained it?
>   3.  Why is the way it is?
>   4.  How is it used (If in fact it is used anywhere)?
> 
> Thanks for the information.
> Could you please post a respsonse, mail isn't working at this end.


    1.  As has been posted, this constant is known as Euler's constant,
        is generally denoted by a lower case gamma, and has the value

            0.57721 56649 01532 86060 6512

	Euler's constant is hard to approximate by naive methods.  I
	don't know if sophisticated methods are known for its accurate
	computation. 

    2.  Euler.

    3.  I assume you are asking why the limit approaches a constant
	value.  If f(x) is the greatest integer less than or equal to x,
	we have

	    f(x) <= x < f(x)+1

	If x >= 1, then f(x) >= 1 and

	    1/(f(x)+1) < 1/x <= 1/(f(x))

	    ==>  0 < 1/x - 1/(f(x)+1) <= 1/(f(x)) - 1/(f(x)+1)

	    ==>  0 < 1/x - 1/(f(x)+1) <= 1/(f(x)(f(x)+1))

	Letting n be a positive integer, we integrate the above between
	1 and n, getting

                         n-1              n-1
	    0 <= ln n -  sum  1/(k+1) <=  sum  1/(k(k+1))
			k = 1            k = 1

	From here, a few algebraic manipultions bound

	      n
	     sum  1/k - ln n
	    k = 1

	by a constant on both sides, for sufficient n.  Finally, noting
	that the above expression is monotonically decreasing in n, we
	deduce that a limit exists.

	The precise value of that limit is harder to explain.  It
	involves analysis of the digamma function, generally denoted
	by the lowercase greek letter psi, which is the first
	derivative of log of the gamma function.  The important property
	of the digamma function is the recurrence:

	    digamma(n+1) = digamma(n) + 1/n

	From which we deduce that

	      n
	     sum  1/k = digamma(n+1) - digamma(1).
	    k = 1

	The Euler-Maclaurin sum formula gives us an asymptotic expansion
	for digamma(n) for large n, which is

	    digamma(n) = ln n + 0(1/n)

	That is, digamma(n) approachs ln(n) asymptotically for large n.
	Thus, for large integer n, we have


	      n
	     sum  1/k = digamma(n+1) - digamma(1)
	    k = 1
	    = ln(n+1) + 0(1/n) - digamma(1)
	    = (ln(n) + 0(1/n)) + 0(1/n) - digamma(1)
	    = ln(n) + 0(1/n) - digamma(1)

	And so

	      n
	     sum  1/k - ln(n) = -digamma(1) + 0(1/n)
	    k = 1

	So that the limit of the left side approaches -digamma(1) as
	n grows without bound.  Playing with the definition of digamma,
	reveals that digamma(1) is the derivative of the gamma function
	at x = 1.  Therefore, Euler's constant is the negative of the
	derivative of the gamma function at x = 1.

	4.  Euler's constant shows up in various definite integrals,
        which pop up from time to time, e.g.:

                      / inf  -x
            gamma = - |     e   log x dx
                      / 0

	gamma pops up from time to time in unexpected places. I would
	go so far as to say that gamma is the third most pervasive
	"transcendental" constant after e (and its clan, such as
	log 2), and pi (and its clan, such as sqrt(2/pi)), although
	it is still unknown whether gamma is transcendental (or
	irrational, for that matter, though you know where my money
	falls.)


-- 
David W. Wilson (wilson@ctron.com)

Disclaimer: "Truth is just truth...You can't have opinions about truth."
- Peter Schikele, introduction to P.D.Q. Bach's oratorio "The Seasonings."