11 Series Expansion of Reciprocal of Gamma Function 11.1 Functions with Integers as Roots Function fz with negative integers as roots can be described as follows. z z z z f()z = 1+ 1+ 1+ = Π 1+ 1 2 3 n =1 n However, this infinite product diverges. That is, such a function cannot exist alone. So, let us consider the following function which added compensation to each factor. z z - Π 1+ e n n =1 n z Then, this infinite product uniformly converges for arbitrary z and results in function e /1+z . This was shown by Weierstrass. If the functions with various integers as roots are shown, it is as follows. In addition, these left sides can be expanded to series of z . So, we call the formulas factorization around 0. Formula 11.1.1 (factorization around 0) When is Euler-Mascheroni constant and z is the Gamma function, the following expressions hold. z z - - z n e Π 1+ e = (1,1-) n =1 n ()1+z z z z n e Π 1- e = (1,1+) n =1 n ()1-z z z - - z 2n 1 2 Π 1+ e = e (1,2-) n =1 2n ()1+z/2 z z z 2n 1 2 Π 1- e = e (1,2+) n =1 2n ()1-z/2 z - - +log2 z z 2n-1 2 Π 1+ e = e (1,3-) n =1 2n -1 ()1+z /2 z +log2 z z 2n-1 2 Π 1- e = e (1,3+) n =1 2n -1 ()1-z /2 z z - - z 2n 2 Π 1+ e = e (1,4-) n =1 2n -1 ()1+z /2 z z z 2n 2 Π 1- e = e (1,4+) n =1 2n -1 ()1-z /2 Proof Weierstrass expression on the gamma function was as follows. z - 1 z z n = e z Π 1+ e ()z n =1 n - 1 - From this, (1,1-) is obtained immediately. Reversing the sign of z in (1,1-) , we obtain (1,1+) . Replacing z with z /2 in (1,1-) and (1,1+) , we obtain (1,2-) and (1,2+) . Next, dividing (1,1-) by (1,2-) , z z - - z - z 2n-1 ()1+z/2 e ()z/2 2 Π 1+ e = z = e n =1 2n -1 ()1+z e- /2 2()z Here, using the Legendre duplication formula, ()z/2 1 e-zlog2 = = 2()z 2z ()1+z /2 ()1+z /2 Substituting this for the above, z - - +log2 z z 2n-1 2 Π 1+ e = e (1,3-) n =1 2n -1 ()1+z /2 And, reversing the sign of z , we obtain (1,3+) . Next, z z 1 1 1 - z 1- + - +- n n zlog Πe 2 -1 e 2 = e 2 3 4 = e 2 n =1 Multiplying each side of (1,3-) by this, z z - - z 2n 2 Π 1+ e = e (1.4-) n =1 2n -1 ()1+z /2 And, reversing the sign of z , we obtain (1,4+) . Giving z =1to the formulas, we obtain the following special values. Special Values 1 - 1 n - 1 Π 1+ e = e = 0.56145948 (1.1- ) n =1 n 1 - - 1 2n 2 2 1 Π 1+ e = e = 0.84550128 (1.2- ) n =1 2n 1 1 2n 1 2 1 Π 1- e = e = 0.75294950 (1.2+) n =1 2n 1 - - 1 2n-1 2 1 Π 1+ e = e = 0.66405515 (1.3- ) n =1 2n -1 2 1 - - 1 2n 2 1 Π 1+ e = e = 1.32811030 (1.4- ) n =1 2n -1 From Formula 11.1.1 , the following corollary is derived - 2 - Corollary 11.1.1 z z 1 sin z Π 1+ 1- = = (1.1) n =1 n n ()1+z ()1-z z z z 1 2 z Π 1+ 1- = = sin (1.2) n =1 2n 2n z z z 2 1+ 1- 2 2 z z z Π 1+ 1- = = cos (1.3) n =1 2n -1 2n -1 1+z 1-z 2 2 2 z z Π 1+ 1- = (1-2+) n =1 2n -1 2n 1+z z 1- 2 2 z z Π 1- 1+ = (1+2-) n =1 2n -1 2n 1-z z 1+ 2 2 (1.1) is a function with all Integers as roots. (1.2) is a function with all even numbers as roots, and (1.3) is a function with all odd numbers as roots. These are all expressed with an elementary function. (1-2+) is a function with negative odd numbers and positive even numbers as froots. (1+2-) is a function with positive odd numbers and negative even numbers as froots. These cannot be expressed with an elementary function. If both sides of (1+2-) are illustrated, it is as follows. The product is calculated 500 terms. Since both sides are overlapped exactly, the left side (blue) is not visible. Compensation Term We discuss (1,1-) as an example. z z - - e z n = Π 1+ e (1,1-) ()1+z n =1 n - 3 - -z/n Although the compensation term is e , the compensation term is not necessarily limited to this. First of all, z -1 the complete compensation term for the right side is Π 1+ . However, if such a compensation term n =1 n is adopted, the function of the left side becomes a constant function and the utility value disappears. This is the -z/n reason why the compensation term like e is adopted. In fact, if the body and the compensation term are illustrated, it is as follows. The products are calculated 10000 terms respectively. Both are very close but they do not overlap. Even from this figure, it is hard to consider compensation term other than an exponential function. On the other hand, it seems that it may be other exponential functions. So let us consider the following exponential function. z z - f()z = Π 1+ c n c > 1 n =1 n The necessity for the condition c >1 is obvious. Anything is okay as long as this is satisfied. For example, if fz is illustrated for c = 2.4 , c =2.71828 , c = 3.2 , it is as follows. We can see that the maximum value of fz is minimum at c = e . Actually, If c deviates from e greatly, -z/n the maximum value of fz becomes large surprising. Considering this, e seems to be optimal as a compensation term. - 4 - 11.2 Maclaurin Expansion Since the reciprocal of the gamma function is holomorphic over the whole complex plane, the Maclaurin expansion should be possible. In this section, treating the divergent product and divergent series as numbers or functions, we aim at the Maclaurin expansion of the reciprocal of gamma function. Formula 11.1.2 When is Euler-Mascheroni constant , z is the Gamma function, Hn is a harmonic number and H is a harmonic series , the following expressions hold. 2 2 1 H Hn =1 + z + + -Σ z 2 1+z 2! 2 n =1 n 3 2 3 H H H 2H - n+1 -H 3 + + - +ΣHn - z + (2.1-) 3! 2 3 n =1 n +1 n 2 2 1 H Hn =1 - z + + -Σ z 2 1-z 2! 2 n =1 n 3 2 3 H H H 2H - n+1 -H 3 - + - +ΣHn - z +- (2.1+) 3! 2 3 n =1 n +1 n 2 2 1 1 H Hn =1 + z + + - z 2 2 Σ 1+z/2 2 2 2! 2 n =1 n 3 2 3 H H 1 H 2H - n+1 -H 3 + + - +ΣHn - z + (2.2-) 23 3! 2 3 n =1 n +1 n 2 2 1 1 H Hn =1 - z + + - z 2 2 Σ 1-z/2 2 2 2! 2 n =1 n 3 2 3 H H 1 H 2H - n+1 -H 3 - + - +ΣHn - z +- (2.2+) 23 3! 2 3 n =1 n +1 n Proof From the Weierstrass expression on the gamma function, z - 1 z z n = e Π 1+ e (2.1-) ()1+z n =1 n When Hn is a harmonic number and H is a harmonic series, the compensation term is z 1 - -zΣ Πe n = e n=1 n = e-Hz n =1 Then z - 2 3 z n Hz -H 1 -H 2 -H 3 e Πe = e - = 1 + z + z + z + (2.1e) n =1 1! 2! 3! Next, assume that the body is expanded as follows. z z z z z Π 1+ = 1+ 1+ 1+ 1+ n =1 n 1 2 3 4 1 2 3 = 1 + a1z + a2z + a3z + - 5 - Then, from Vieta's formulas, 1 1 1 a = + + + = H 1 1 2 3 1 1 1 1 1 1 1 1 1 1 1 1 a = + + + + + + + + + + + + 2 1 2 3 4 2 3 4 5 3 4 5 6 H-H1 H-H2 H-H3 H-Hn = + + + = Σ 1 2 3 n =1 n 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 a = + + + + + + + + + + + + 3 1 2 3 4 5 1 3 4 5 6 1 4 5 6 7 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 + + + + + + + + + + + + + 2 3 4 5 6 2 4 5 6 7 2 5 6 7 8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 + + + + + + + + + + + + + 3 4 5 6 7 3 5 6 7 8 3 6 7 8 9 + 1 1 1 1 1 1 = H-H + H-H + H-H + 1 2 2 1 3 3 1 4 4 1 1 1 1 1 1 + H-H + H-H + H-H + 2 3 3 2 4 4 2 5 5 1 1 1 1 1 1 + H-H + H-H + H-H + 3 4 4 3 5 5 3 6 6 + Rearranging this along the diagonal line, 1 1 1 1 1 1 1 1 1 a = H-H + + H-H + + + H-H + 3 1 2 2 1 2 3 3 1 2 3 4 4 1 1 1 HnH-Hn+1 = H1 H-H2 +H2 H-H3 +H3 H-H4 + = Σ 2 3 4 n =1 n +1 Thus, z 2 H-Hn 3 HnH-Hn+1 Π 1+ = 1 + zH + z Σ + z Σ + (2.1p) n =1 n n =1 n n =1 n +1 The right side of (2.1-) is a product of (2.1e) and (2.1p) .