Zelaron Gaming Forum  
Stats Arcade Portal Forum FAQ Community Calendar Today's Posts Search
Go Back   Zelaron Gaming Forum > The Zelaron Nexus > The Lounge

 
 
Thread Tools Display Modes

 
Simple math problem. (Drug related)
Reply
Posted 2007-11-14, 12:57 AM
The Euler-Mascheroni constant gamma, sometimes also called the Euler constant (but not to be confused with the constant e==2.718281...) is defined as the limit of the sequence
gamma = lim_(n->infty)(sum_(k==1)^(n)1/k-lnn)
(1)
= lim_(n->infty)(H_n-lnn),
(2)

where H_n is a harmonic number (Graham et al. 1994, p. 278). It was first defined by Euler (1735), who used the letter C and stated that it was "worthy of serious consideration" (Havil 2003, pp. xx and 51). The symbol gamma was first used by Mascheroni (1790).

gamma has the numerical value
gamma==0.577215664901532860606512090082402431042.. .
(3)

(Sloane's A001620), and is implemented in Mathematica as EulerGamma.

It was calculated to 16 digits by Euler in 1781 and to 32 decimal places by Mascheroni (1790), although only the first 19 decimal places were correct. It was subsequently computed to 40 correct decimal placed by Soldner in 1809 and verified by Gauss and Nicolai in 1812 (Havil 2003, pp. 89-90). No quadratically converging algorithm for computing gamma is known (Bailey 1988). X. Gourdon and P. Demichel computed a record 108 million digits of gamma in October 1999 (Gourdon and Sebah). On Dec. 8, 2006, Alexander J. Yee computed 116580041 decimal digits in 38.5 hours (Yee 2006; Uunited Press International 2007). S. Kondo has computed gamma to 2 billion digits, which is apparently the current world record.

It is not known if this constant is irrational, let alone transcendental (Wells 1986, p. 28). The famous English mathematician G. H. Hardy is alleged to have offered to give up his Savilian Chair at Oxford to anyone who proved gamma to be irrational (Havil 2003, p. 52), although no written reference for this quote seems to be known. Hilbert mentioned the irrationality of gamma as an unsolved problem that seems "unapproachable" and in front of which mathematicians stand helpless (Havil 2003, p. 97). Conway and Guy (1996) are "prepared to bet that it is transcendental," although they do not expect a proof to be achieved within their lifetimes. If gamma is a simple fraction a/b, then it is known that b>10^(10000) (Brent 1977; Wells 1986, p. 28), which was subsequently improved by T. Papanikolaou to b>10^(242080) (Havil 2003, p. 97).

The Euler-Mascheroni constant arises in many integrals
gamma = -int_0^inftye^(-x)lnxdx
(4)
= -int_0^1lnln(1/x)dx
(5)
= int_0^infty(1/(1-e^(-x))-1/x)e^(-x)dx
(6)
= int_0^infty1/x(1/(1+x)-e^(-x))dx
(7)

(Whittaker and Watson 1990, p. 246). Integrals that give gamma in combination with other simple constants include
int_0^inftye^(-x^2)lnxdx = -1/4sqrt(pi)(gamma+2ln2)
(8)
int_0^inftye^(-x)(lnx)^2dx = gamma^2+1/6pi^2.
(9)

Double integrals include
gamma==int_0^1int_0^1(x-1)/((1-xy)ln(xy))dxdy
(10)

(Sondow 2003, 2005; Borwein et al. 2004, p. 49). An interesting analog of equation (10) is given by
ln(4/pi) = sum_(n==1)^(infty)(-1)^(n-1)[1/n-ln((n+1)/n)]
(11)
= int_0^1int_0^1(x-1)/((1+xy)ln(xy))dxdy
(12)
= 0.241564...
(13)

(Sloane's A094640; Sondow 2005).

gamma is also given by Mertens theorem
e^gamma==lim_(n->infty)1/(lnp_n)product_(i==1)^n1/(1-1/(p_i)),
(14)

where the product is over primes p. By taking the logarithm of both sides, an explicit formula for gamma is obtained,
gamma==lim_(x->infty)[sum_(p<=x)ln(1/(1-1/p))-lnlnx].
(15)

It is also given by series
gamma==sum_(k==1)^infty[1/k-ln(1+1/k)]
(16)

due to Euler, which follows from equation (1) by first replacing lnn by ln(n+1), which works since
lim_(n->infty)[ln(n+1)-lnn]==lim_(n->infty)ln(1+1/n)==0,
(17)

and then substituting the telescoping sum
sum_(k==1)^nln(1+1/k)
(18)

for ln(n+1), which is its sum since again
ln(1+1/k)==ln(k+1)-lnk,
(19)

obtaining
gamma = lim_(n->infty)[sum_(k==1)^(n)1/k-sum_(k==1)^(n)ln(1+1/k)]
(20)
= lim_(n->infty)sum_(k==1)^(n)[1/k-ln(1+1/k)]
(21)

which equals equation (◇).

Other series include
gamma = sum_(n==2)^(infty)(-1)^n(zeta(n))/n
(22)
= ln(4/pi)+sum_(n==1)^(infty)((-1)^(n-1)zeta(n+1))/(2^n(n+1))
(23)

(Gourdon and Sebah 2003, p. 3), where zeta(z) is the Riemann zeta function, and
gamma==sum_(n==1)^infty(-1)^n(|_lgn_|)/n
(24)

(Vacca 1910, Gerst 1969), where lg is the logarithm to base 2 and |_x_| is the floor function. Nielsen (1897) earlier gave a series equivalent to (24),
gamma==1-sum_(n==1)^inftysum_(k==2^(n-1))^(2^n-1)n/((2k+1)(2k+2)).
(25)

To see the equivalence of (25) with (24), expand
1/((2k+1)(2k+2))==1/(2k+1)-1/(2k+2)
(26)

and add
0==-1/2+1/4+1/8+1/(16)+...
(27)

to Nielsen's equation to get Vacca's formula.

The sums
gamma = sum_(n==1)^(infty)sum_(k==2^n)^(infty)((-1)^k)/k
(28)
= sum_(k==1)^(infty)1/(2^(k+1))sum_(j==0)^(k-1)(2^(k-j)+j; j)^(-1)
(29)

(Gosper 1972, with k-j replacing the undefined i; Bailey and Crandall 2001) can be obtained from equation (24) by rewriting as a double series, then applying Euler's series transformation to each of these series and adding to get equation (29). Here, (n; k) is a binomial coefficient, and rearranging the conditionally convergent series is permitted because the plus and minus terms can first be grouped in pairs, the resulting series of positive numbers rearranged, and then the series ungrouped back to plus and minus terms.

The double series (28) is equivalent to Catalan's integral
gamma==int_0^11/(1+x)sum_(n==1)^inftyx^(2^n-1)dx.
(30)

To see the equivalence, expand 1/(1+x) in a geometric series, multiply by x^(2^n-1), and integrate termwise (Sondow and Zudilin 2003).

Other series for gamma include
gamma==3/2-ln2-sum_(m==2)^infty(-1)^m(m-1)/m[zeta(m)-1]
(31)

(Flajolet and Vardi 1996), and
gamma==(2^n)/(e^(2^n))sum_(m==0)^infty(2^(mn))/((m+1)!)sum_(t==0)^m1/(t+1)-nln2+O(1/(2^ne^(2^n))),
(32)

(Bailey 1988), which is an improvement over Sweeney (1963).

A rapidly converging limit for gamma is given by
gamma = lim_(n->infty)[(2n-1)/(2n)-lnn+sum_(k==2)^(n)(1/k-(zeta(1-k))/(n^k))]
(33)
= lim_(n->infty)[(2n-1)/(2n)-lnn+sum_(k==2)^(n)1/k(1+(B_k)/(n^k))],
(34)

where B_k is a Bernoulli number (C. Stingley, pers. comm., July 11, 2003).

Another limit formula is given by
gamma==-lim_(n->infty)[(Gamma(1/n)Gamma(n+1)n^(1+1/n))/(Gamma(2+n+1/n))-(n^2)/(n+1)]
(35)

(P. Walker, pers. comm., Mar. 17, 2004). An even more amazing limit is given by
gamma==lim_(x->infty)zeta(zeta(z))-2^x+(4/3)^x+1
(36)

(B. Cloitre, pers. comm., Oct. 4, 2005), where zeta(z) is the Riemann zeta function.

Another connection with the primes was provided by Dirichlet's 1838 proof that the average number of divisors d(n)==sigma_0(n) of all numbers from 1 to n is asymptotic to
(sum_(k==1)^(n)d(k))/n∼lnn+2gamma-1
(37)

(Conway and Guy 1996). de la Vallée Poussin (1898) proved that, if a large number n is divided by all primes <=n, then the average amount by which the quotient is less than the next whole number is gamma.

An elegant identity for gamma is given by
gamma==(S_0(z)-K_0(z))/(I_0(z))-ln(1/2z),
(38)

where I_0(z) is a modified Bessel function of the first kind, K_0(z) is a modified Bessel function of the second kind, and
S_0(z)=sum_(k==0)^infty((1/2z)^(2k)H_k)/((k!)^2),
(39)

where H_n is a harmonic number (Borwein and Borwein 1987, p. 336; Borwein and Bailey 2003, p. 138). This gives an efficient iterative algorithm for gamma by computing
B_k = (B_(k-1)n^2)/(k^2)
(40)
A_k = 1/k((A_(k-1)n^2)/k+B_k)
(41)
U_k = U_(k-1)+A_k
(42)
V_k = V_(k-1)+B_k
(43)

with A_0==-lnn, B_0==1, U_0==A_0, and V_0==1 (Borwein and Bailey 2003, pp. 138-139).

Reformulating this identity gives the limit
lim_(n->infty)[sum_(k==0)^infty(((n^k)/(k!))^2H_k)/(sum_(k==0)^(infty)((n^k)/(k!))^2)-lnn]==gamma
(44)

(Brent and McMillan 1980; Trott 2004, p. 21).

Infinite products involving gamma also arise from the Barnes G-function with positive integer n. The cases G(2) and G(3) give
product_(n==1)^(infty)e^(-1+1/(2n))(1+1/n)^n = (e^(1+gamma/2))/(sqrt(2pi))
(45)
product_(n==1)^(infty)e^(-2+2/n)(1+2/n)^n = (e^(3+2gamma))/(2pi).
(46)

The Euler-Mascheroni constant is also given by the expressions
gamma = -Gamma^'(1)
(47)
= -psi_0(1),
(48)

where psi_0(x) is the digamma function (Whittaker and Watson 1990, p. 236),
gamma==lim_(s->1)[zeta(s)-1/(s-1)]
(49)

(Whittaker and Watson 1990, p. 271), the antisymmetric limit form
gamma==lim_(s->1^+)sum_(n==1)^infty(1/(n^s)-1/(s^n))
(50)

(Sondow 1998), and
gamma==lim_(x->infty)[x-Gamma(1/x)]
(51)

(Le Lionnais 1983).

The difference between the nth convergent in equation (◇) and gamma is given by
sum_(k==1)^n1/k-lnn-gamma==int_n^infty(x-|_x_|)/(x^2)dx,
(52)

where |_x_| is the floor function, and satisfies the inequality
1/(2(n+1))<sum_(k==1)^n1/k-lnn-gamma<1/(2n)
(53)

(Young 1991).

The symbol gamma is sometimes also used for
gamma^'=e^gamma approx 1.781072
(54)

(Sloane's A073004; Gradshteyn and Ryzhik 2000, p. xxvii).

There is a the curious radical representation
e^gamma==(2/1)^(1/2)((2^2)/(1.3))^(1/3)((2^3.4)/(1.3^3))^(1/4)((2^4.4^4)/(1.3^6.5))^(1/5)...,
(55)

which is related to the double series
gamma==sum_(n==1)^infty1/nsum_(k==0)^(n-1)(-1)^(k+1)(n-1; k)ln(k+1)
(56)

and (n; k) a binomial coefficient (Ser 1926, Sondow 2004a, Guillera and Sondow 2005). Another proof of product (55) as well as an explanation for the resemblance between this product and the Wallis formula-like "faster product for pi"
pi/2==(2/1)^(1/2)((2^2)/(1.3))^(1/4)((2^3.4)/(1.3^3))^(1/8)((2^4.4^4)/(1.3^6.5))^(1/16)...
(57)

(Guillera and Sondow 2005, Sondow 2005), is given in Sondow (2004). (This resemblance which is made even clearer by changing n->n+1 in (57).) Both these formulas are also analogous to the product for e given by
e==(2/1)^(1/2)((2^2)/(1.3))^(1/2)((2^3.4)/(1.3^3))^(1/3)((2^4.4^4)/(1.3^6.5))^(1/4)...
(58)

due to Guillera (Sondow 2005).
EulerMascheroniSondow

The values r(n) obtained after inclusion of the first n terms of the product for e^gamma are plotted above.

A curious sum limit converging to gamma is given by
lim_(n->infty)1/nsum_(k==1)^(n-1)([n/k]-n/k)==gamma

Because only someone on drugs would read through all of that before giving up
Old
Profile PM WWW Search
Xenn shows clear signs of ignorance and confidence; the two things needed to succeed in lifeXenn shows clear signs of ignorance and confidence; the two things needed to succeed in life
 
Xenn
 



 
Reply
Posted 2007-11-14, 01:09 AM in reply to Xenn's post "Simple math problem. (Drug related)"
Copypasta
Old
Profile PM WWW Search
!King_Amazon! simplifies with no grasp of the basics!King_Amazon! simplifies with no grasp of the basics!King_Amazon! simplifies with no grasp of the basics!King_Amazon! simplifies with no grasp of the basics!King_Amazon! simplifies with no grasp of the basics!King_Amazon! simplifies with no grasp of the basics!King_Amazon! simplifies with no grasp of the basics
 
 
!King_Amazon!
 
 

Bookmarks

« Previous Thread | Next Thread »

Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)
 

Posting Rules [Forum Rules]
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts
BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Forum Jump

Similar Threads
Thread Thread Starter Forum Replies Last Post
Simple math problem. (Drug related) -Spector- The Lounge 10 2007-11-14 10:07 AM
Drug to Make You Forget Demosthenes Science and Art 27 2007-10-22 03:52 PM
Math Problem of the Day slaynish The Lounge 6 2004-09-07 10:26 PM
problEm with battle layout dr. Musashi RPGMaker 2 2004-05-07 03:26 PM
A simple math problem Ganga The Lounge 36 2003-12-07 10:17 AM


All times are GMT -6. The time now is 12:03 PM.
'Synthesis 2' vBulletin 3.x styles and 'x79' derivative
by WetWired the Unbound and Chruser
Copyright ©2002-2008 zelaron.com
Powered by vBulletin® Version 3.8.2
Copyright ©2000 - 2024, Jelsoft Enterprises Ltd.
This site is best seen with your eyes open.