## Variations on a Theme by James Stirling (2006)

### BibTeX

@MISC{Dominici06variationson,

author = {Diego Dominici},

title = {Variations on a Theme by James Stirling},

year = {2006}

}

### OpenURL

### Abstract

Dedicated to Donald Silberger on the occasion of his 76th birthday 1

### Citations

855 |
A Course of Modern Analysis
- Whittaker, Watson
- 1969
(Show Context)
Citation Context ...laurin formula. Knopp [24] and Wilf [44]. n∑ [ln (k) − ln (z + k)] k=1 A similar analysis was done by 14. Patin [34] used (11) and the Lebesgue Dominated Convergence Theorem. 715. Whittaker & Watson =-=[42]-=- used Binet’s second formula and ∫ ln [Γ(z)] = P(z) + 2 0 ∞ arctan ( ) t z e2πt − 1 dt, Re(z) > 0 ∞∑ (−1) arctan(x) = x k 2k + 1 x2k , |x| ≤ 1. k=0 Thus, there have been a huge variety of approaches t... |

574 |
Special Functions
- Andrews, Askey, et al.
- 1999
(Show Context)
Citation Context ...ods. In an effort to join such illustrious company, we present still another direction for deriving (9). Our starting point shall be the difference equation (6). A parallel approach was considered in =-=[3]-=-. For a different analysis of (6) using the method of controlling factors, see [5]. Extensions and other applications of the method used can be found in [10], [15] and [16]. 3 Asymptotic analysis 3.1 ... |

458 |
Asymptotics and Special Functions
- Olver
- 1974
(Show Context)
Citation Context ...= 0 for all n, then yn ∼ Cnα , n → ∞ for some non-zero constant C. he showed that n! ∼ C √ nn n e −n . 2. Bender & Orszag [5], Bleistein & Handelsman [6], Diaconis & Freedman [13], Dingle [14], Olver =-=[33]-=- and Wong [45] applied Laplace’s method to the Euler integral of the second kind ∫ Γ (z) = 0 ∞ t z−1 e −t dt, Re(z) > 0. (11) 3. Bender & Orszag [5] and Temme [39] used Hankel’s contour integral [1] 1... |

361 |
Special Functions and Their Applications
- Lebedev
- 1972
(Show Context)
Citation Context ...e(z) > 0. (11) 3. Bender & Orszag [5] and Temme [39] used Hankel’s contour integral [1] 1 Γ(z) ∫ 1 = 2πi and the method of steepest descent. (0+) −∞ t −z e t dt 4. Bleistein & Handelsman [6], Lebedev =-=[25]-=-, Sasvári [37] and Temme [39] used Binet’s first formula and ∫ ln [Γ(z)] = P(z) + 0 ∞ 1 t ( 1 1 1 1 − + t 2 t et ) − 1 ( 1 1 1 − + 2 t et ) e − 1 −tz dt, Re(z) > 0 = ∑ k≥1 B2k (2k)! t2k−2 , |t| < 2π. ... |

255 |
Advanced Mathematical Methods for Scientists and Engineers
- Bender, Orszag
- 1978
(Show Context)
Citation Context ...nn e −n n! . Using his lemma Lemma 3 If yn+1 = 1 + yn α n + O ( n −2) and yn ̸= 0 for all n, then yn ∼ Cnα , n → ∞ for some non-zero constant C. he showed that n! ∼ C √ nn n e −n . 2. Bender & Orszag =-=[5]-=-, Bleistein & Handelsman [6], Diaconis & Freedman [13], Dingle [14], Olver [33] and Wong [45] applied Laplace’s method to the Euler integral of the second kind ∫ Γ (z) = 0 ∞ t z−1 e −t dt, Re(z) > 0. ... |

201 |
Asymptotic Expansions of Integrals
- Bleistein, Handelsman
- 1975
(Show Context)
Citation Context ... Lemma 3 If yn+1 = 1 + yn α n + O ( n −2) and yn ̸= 0 for all n, then yn ∼ Cnα , n → ∞ for some non-zero constant C. he showed that n! ∼ C √ nn n e −n . 2. Bender & Orszag [5], Bleistein & Handelsman =-=[6]-=-, Diaconis & Freedman [13], Dingle [14], Olver [33] and Wong [45] applied Laplace’s method to the Euler integral of the second kind ∫ Γ (z) = 0 ∞ t z−1 e −t dt, Re(z) > 0. (11) 3. Bender & Orszag [5] ... |

132 |
editors, Handbook of mathematical functions
- Abramowitz, Stegun
- 1965
(Show Context)
Citation Context ... [33] and Wong [45] applied Laplace’s method to the Euler integral of the second kind ∫ Γ (z) = 0 ∞ t z−1 e −t dt, Re(z) > 0. (11) 3. Bender & Orszag [5] and Temme [39] used Hankel’s contour integral =-=[1]-=- 1 Γ(z) ∫ 1 = 2πi and the method of steepest descent. (0+) −∞ t −z e t dt 4. Bleistein & Handelsman [6], Lebedev [25], Sasvári [37] and Temme [39] used Binet’s first formula and ∫ ln [Γ(z)] = P(z) + 0... |

96 |
A remark on Stirling’s formula
- Robbins
- 1955
(Show Context)
Citation Context ...ed to show that rn ∼ R3(n). Michel [29], proved the inequality ∣ern − 1 − 1 1 − 12n 288n2 1 1 ∣ ≤ + 360n3 108n4, n = 3, 4 . . .. Nanjundiah [32], showed that R2(n) < rn < R1(n) = 1, 2, . . .. Robbins =-=[36]-=-, established the double inequality 1 12n + 1 < rn < 1 , n = 1, 2, . . .. 12n 611. Marsaglia & Marsaglia [27] derived from (11) the asymptotic expansion n! ∼ n n+1 e −n ∞∑ k=1 bk ( ) k 2 2 Γ n ( ) k ... |

76 |
Asymptotic expansions: their derivation and interpretation
- Dingle
- 1973
(Show Context)
Citation Context ...−2) and yn ̸= 0 for all n, then yn ∼ Cnα , n → ∞ for some non-zero constant C. he showed that n! ∼ C √ nn n e −n . 2. Bender & Orszag [5], Bleistein & Handelsman [6], Diaconis & Freedman [13], Dingle =-=[14]-=-, Olver [33] and Wong [45] applied Laplace’s method to the Euler integral of the second kind ∫ Γ (z) = 0 ∞ t z−1 e −t dt, Re(z) > 0. (11) 3. Bender & Orszag [5] and Temme [39] used Hankel’s contour in... |

75 |
Theory and Application of Infinite Series
- Knopp
- 1990
(Show Context)
Citation Context ... . . . m with m = 3. His results where extended by Deeba & Rodriguez in [12]. 13. Olver [33] used Euler’s definition (5) ln[Γ(z)] = − ln (z) + lim n→∞ z ln(n) + and the Euler-Maclaurin formula. Knopp =-=[24]-=- and Wilf [44]. n∑ [ln (k) − ln (z + k)] k=1 A similar analysis was done by 14. Patin [34] used (11) and the Lebesgue Dominated Convergence Theorem. 715. Whittaker & Watson [42] used Binet’s second f... |

53 |
The Gamma function
- Artin
- 1964
(Show Context)
Citation Context ...ohr-Mollerup theorem [8]: 2Theorem 1 The gamma function is the only function Γ : (0, ∞) → (0, ∞) which satisfies 1. Γ(1) = 1 2. Γ(x + 1) = xΓ(x) 3. ln [Γ(x)] is convex for all x ∈ (0, ∞). Proof. See =-=[4]-=- and [30]. Another complex-analytic characterization is due to Wielandt [43]: Theorem 2 The gamma function is the only holomorphic function in the right half plane A satisfying 1. Γ(1) = 1 2. Γ(z + 1)... |

53 |
A generalization of Stirling’s formula
- Hayman
(Show Context)
Citation Context ...− bk) + an, (12) ∫ k+ 1 2 There is a mistake in his Equation (2.4), where he states that instead of (12). ln (n!) − 1 ln(n) + I(n) − I 2 5 ( ) 1 ∑n−1 = 2 k=1 k ln ( ) t dt k (ak − bk) + an,9. Hayman =-=[19]-=- used the exponential generating function e z = ∞∑ 1 k! k=0 zk and his method for admissible functions. 10. Hummel [20], established the inequalities where 11 12 < rn + 1 ln(2π) < 1, n = 2, 3, . . ., ... |

24 |
Mathematics for the physical sciences
- Wilf
- 1962
(Show Context)
Citation Context ...m = 3. His results where extended by Deeba & Rodriguez in [12]. 13. Olver [33] used Euler’s definition (5) ln[Γ(z)] = − ln (z) + lim n→∞ z ln(n) + and the Euler-Maclaurin formula. Knopp [24] and Wilf =-=[44]-=-. n∑ [ln (k) − ln (z + k)] k=1 A similar analysis was done by 14. Patin [34] used (11) and the Lebesgue Dominated Convergence Theorem. 715. Whittaker & Watson [42] used Binet’s second formula and ∫ l... |

17 |
Euler’s integral: A historical profile of the gamma function
- Davis, Leonhard
- 1959
(Show Context)
Citation Context ...ts of De Moivre in a letter to Gabriel Cramer. In 1729 Leonhard Euler (15 April 1707 – 18 Sept. 1783) proposed a generalization of the factorial function from natural numbers to positive real numbers =-=[11]-=-. It is called the gamma function, Γ(z), which he defined as k≥1 n!n Γ(z) = lim n→∞ z , (5) z(z + 1) · · ·(z + n) and it is related to the factorial numbers by Γ (n + 1) = n!, n = 0, 1, 2, . . .. From... |

13 |
Some monotonicity properties and characterization of the gamma fuction
- MULDOON
- 1978
(Show Context)
Citation Context ...erup theorem [8]: 2Theorem 1 The gamma function is the only function Γ : (0, ∞) → (0, ∞) which satisfies 1. Γ(1) = 1 2. Γ(x + 1) = xΓ(x) 3. ln [Γ(x)] is convex for all x ∈ (0, ∞). Proof. See [4] and =-=[30]-=-. Another complex-analytic characterization is due to Wielandt [43]: Theorem 2 The gamma function is the only holomorphic function in the right half plane A satisfying 1. Γ(1) = 1 2. Γ(z + 1) = zΓ(z) ... |

7 |
Rigorous WKB for finite-order linear recurrence relations with smooth coefficients
- Costin, Costin
- 1996
(Show Context)
Citation Context ... (6). A parallel approach was considered in [3]. For a different analysis of (6) using the method of controlling factors, see [5]. Extensions and other applications of the method used can be found in =-=[10]-=-, [15] and [16]. 3 Asymptotic analysis 3.1 Stirling’s formula We begin with a derivation of (7), to better illustrate how the method works. We assume that ln [Γ(z)] ∼ f(z) + g(z), z → ∞ (14) with Usin... |

7 |
Stirling’s Series and Bernoulli Numbers
- Deeba, Rodriguez
- 1991
(Show Context)
Citation Context ...idered the triplication case, using Gauss’ multiplication formula Γ(mz) = (√ ) 1−m 2π m mz−1 2 m−1 ∏ Γ k=0 ( z + k ) , m = 2, 3, . . . m with m = 3. His results where extended by Deeba & Rodriguez in =-=[12]-=-. 13. Olver [33] used Euler’s definition (5) ln[Γ(z)] = − ln (z) + lim n→∞ z ln(n) + and the Euler-Maclaurin formula. Knopp [24] and Wilf [44]. n∑ [ln (k) − ln (z + k)] k=1 A similar analysis was done... |

7 |
WKB methods for difference equations
- Dingle, Morgan
- 1967
(Show Context)
Citation Context ...A parallel approach was considered in [3]. For a different analysis of (6) using the method of controlling factors, see [5]. Extensions and other applications of the method used can be found in [10], =-=[15]-=- and [16]. 3 Asymptotic analysis 3.1 Stirling’s formula We begin with a derivation of (7), to better illustrate how the method works. We assume that ln [Γ(z)] ∼ f(z) + g(z), z → ∞ (14) with Using (14)... |

6 |
Calculation of the Gamma Function by Stirling’s Formula
- Spira
- 1971
(Show Context)
Citation Context ...z)] ∼ P(z), z → ∞ (7) 1 ln (z) + ln (2π) (8) 2 ln [Γ(z)] ∼ P(z) + RN(z), z → ∞ (9) RN(z) = N∑ B2k 2k (2k − 1)z 2k−1, N ≥ 1. (10) k=1 Estimations of the remainder ln [Γ(z)]−P(z)−RN(z) were computed in =-=[38]-=-. 32 Previous results Over the years, there have been many different approaches to the derivation of (7) and (9), including: 1. Aissen [2] studied the sequence Vn = nn e −n n! . Using his lemma Lemma... |

5 |
Special Functions (Wiley-Interscience Publication
- Temme
- 1996
(Show Context)
Citation Context ... & Freedman [13], Dingle [14], Olver [33] and Wong [45] applied Laplace’s method to the Euler integral of the second kind ∫ Γ (z) = 0 ∞ t z−1 e −t dt, Re(z) > 0. (11) 3. Bender & Orszag [5] and Temme =-=[39]-=- used Hankel’s contour integral [1] 1 Γ(z) ∫ 1 = 2πi and the method of steepest descent. (0+) −∞ t −z e t dt 4. Bleistein & Handelsman [6], Lebedev [25], Sasvári [37] and Temme [39] used Binet’s first... |

5 |
Asymptotic approximations of integrals, volume 34
- Wong
- 2001
(Show Context)
Citation Context ... then yn ∼ Cnα , n → ∞ for some non-zero constant C. he showed that n! ∼ C √ nn n e −n . 2. Bender & Orszag [5], Bleistein & Handelsman [6], Diaconis & Freedman [13], Dingle [14], Olver [33] and Wong =-=[45]-=- applied Laplace’s method to the Euler integral of the second kind ∫ Γ (z) = 0 ∞ t z−1 e −t dt, Re(z) > 0. (11) 3. Bender & Orszag [5] and Temme [39] used Hankel’s contour integral [1] 1 Γ(z) ∫ 1 = 2π... |

4 |
Asymptotic analysis of the Hermite polynomials from their differentialdifference equation
- Dominici
(Show Context)
Citation Context ...l approach was considered in [3]. For a different analysis of (6) using the method of controlling factors, see [5]. Extensions and other applications of the method used can be found in [10], [15] and =-=[16]-=-. 3 Asymptotic analysis 3.1 Stirling’s formula We begin with a derivation of (7), to better illustrate how the method works. We assume that ln [Γ(z)] ∼ f(z) + g(z), z → ∞ (14) with Using (14) in (6), ... |

4 |
A very short proof of Stirling’s formula
- Patin
- 1989
(Show Context)
Citation Context ...3] used Euler’s definition (5) ln[Γ(z)] = − ln (z) + lim n→∞ z ln(n) + and the Euler-Maclaurin formula. Knopp [24] and Wilf [44]. n∑ [ln (k) − ln (z + k)] k=1 A similar analysis was done by 14. Patin =-=[34]-=- used (11) and the Lebesgue Dominated Convergence Theorem. 715. Whittaker & Watson [42] used Binet’s second formula and ∫ ln [Γ(z)] = P(z) + 2 0 ∞ arctan ( ) t z e2πt − 1 dt, Re(z) > 0 ∞∑ (−1) arctan... |

3 |
A note on easy proofs of Stirling's theorem
- Blyth, Pathak
- 1986
(Show Context)
Citation Context ...emme [39] used Binet’s first formula and ∫ ln [Γ(z)] = P(z) + 0 ∞ 1 t ( 1 1 1 1 − + t 2 t et ) − 1 ( 1 1 1 − + 2 t et ) e − 1 −tz dt, Re(z) > 0 = ∑ k≥1 B2k (2k)! t2k−2 , |t| < 2π. 45. Blyth & Pathak =-=[7]-=- and Khan [23] used probabilistic arguments, applying the Central Limit Theorem and the limit theorem for moment generating functions to Gamma and Poison random variables. 6. Coleman [9] defined cn = ... |

3 |
Mathematische Werke/Mathematical works. Vol. 1. Group theory, edited and with a preface by Bertram Huppert and Hans Schneider, Walter de Gruyter
- Wielandt
- 1994
(Show Context)
Citation Context ...on Γ : (0, ∞) → (0, ∞) which satisfies 1. Γ(1) = 1 2. Γ(x + 1) = xΓ(x) 3. ln [Γ(x)] is convex for all x ∈ (0, ∞). Proof. See [4] and [30]. Another complex-analytic characterization is due to Wielandt =-=[43]-=-: Theorem 2 The gamma function is the only holomorphic function in the right half plane A satisfying 1. Γ(1) = 1 2. Γ(z + 1) = zΓ(z) for all z ∈ A 3. Γ (z) is bounded in the strip 1 ≤ Re(z) < 2 Proof.... |

2 |
Correction to: “A direct proof of Stirling’s formula
- Feller
- 1968
(Show Context)
Citation Context ...by Aissen [2], using the concavity of ln(x). 7. Dingle [14] used Weierstrass’ infinite product 1 Γ(z) = zeγz ∞∏ [( n=1 1 + z n ) e z − n (where γ is Euler’s constant) and Mellin transforms. 8. Feller =-=[17]-=-, [18] proved the identity where ln (n!) − 1 ln(n) = I(n) − I 2 I(n) = ∫n 0 and showed that ln(t)dt, ak = ∫k k− 1 2 ∞∑ (ak − bk) − I k=1 ( ) 1 ∑n−1 + 2 ln k=1 ] , ( ) k dt, bk = t ( ) 1 = 2 1 ln (2π).... |

2 |
Stirling’s series made easy
- Impens
(Show Context)
Citation Context ... z = ∞∑ 1 k! k=0 zk and his method for admissible functions. 10. Hummel [20], established the inequalities where 11 12 < rn + 1 ln(2π) < 1, n = 2, 3, . . ., 2 ( ) n n!e rn = ln √ . 2πnnn Impens [21], =-=[22]-=-, showed that for x > 0 R2n(x) < ln [Γ(x)] − P(x) < R2m+1(x), n, m ≥ 0, where Rn(x) was defined in (10). Maria [26] showed that [ ] −1 3 12n + < rn, n = 1, 2, . . .. 2 (2n + 1) Mermin [28] proved the ... |

2 |
A probabilistic proof of Stirling’s formula
- Khan
- 1974
(Show Context)
Citation Context ...d Binet’s first formula and ∫ ln [Γ(z)] = P(z) + 0 ∞ 1 t ( 1 1 1 1 − + t 2 t et ) − 1 ( 1 1 1 − + 2 t et ) e − 1 −tz dt, Re(z) > 0 = ∑ k≥1 B2k (2k)! t2k−2 , |t| < 2π. 45. Blyth & Pathak [7] and Khan =-=[23]-=- used probabilistic arguments, applying the Central Limit Theorem and the limit theorem for moment generating functions to Gamma and Poison random variables. 6. Coleman [9] defined cn = and showed tha... |

2 |
Stirling’s formula
- Mermin
- 1984
(Show Context)
Citation Context ...mpens [21], [22], showed that for x > 0 R2n(x) < ln [Γ(x)] − P(x) < R2m+1(x), n, m ≥ 0, where Rn(x) was defined in (10). Maria [26] showed that [ ] −1 3 12n + < rn, n = 1, 2, . . .. 2 (2n + 1) Mermin =-=[28]-=- proved the identity e rn = ∞∏ k=n e −1 ( 1 + 1 k ) k+ 1 2 , which he used to show that rn ∼ R3(n). Michel [29], proved the inequality ∣ern − 1 − 1 1 − 12n 288n2 1 1 ∣ ≤ + 360n3 108n4, n = 3, 4 . . ..... |

2 |
A simple derivation of Stirling's asymptotic series
- Namias
- 1986
(Show Context)
Citation Context ... asymptotic expansion n! ∼ n n+1 e −n ∞∑ k=1 bk ( ) k 2 2 Γ n ( ) k k, 2 where the generating function G(z) = ∑ bkzk is defined by G(z) exp [1 − G(z)] = exp k≥0 ( − 1 2 z2 ) , G ′ (0) = 1. 12. Namias =-=[31]-=- introduced the function F(n) = Γ(n) , with P(n) defined in P(n) (8). From Legendre’s duplication formula Γ(2n) = 22n−1 √ π Γ(n)Γ he derived a functional equation for F(n) F(2n) F(n)F ( n − 1 2 ) = √ ... |

2 |
Note on Stirling’s formula
- Nanjundiah
- 1959
(Show Context)
Citation Context ...identity e rn = ∞∏ k=n e −1 ( 1 + 1 k ) k+ 1 2 , which he used to show that rn ∼ R3(n). Michel [29], proved the inequality ∣ern − 1 − 1 1 − 12n 288n2 1 1 ∣ ≤ + 360n3 108n4, n = 3, 4 . . .. Nanjundiah =-=[32]-=-, showed that R2(n) < rn < R1(n) = 1, 2, . . .. Robbins [36], established the double inequality 1 12n + 1 < rn < 1 , n = 1, 2, . . .. 12n 611. Marsaglia & Marsaglia [27] derived from (11) the asympto... |

1 |
Some remarks on Stirling’s formula
- Aissen
- 1954
(Show Context)
Citation Context ...ions of the remainder ln [Γ(z)]−P(z)−RN(z) were computed in [38]. 32 Previous results Over the years, there have been many different approaches to the derivation of (7) and (9), including: 1. Aissen =-=[2]-=- studied the sequence Vn = nn e −n n! . Using his lemma Lemma 3 If yn+1 = 1 + yn α n + O ( n −2) and yn ̸= 0 for all n, then yn ∼ Cnα , n → ∞ for some non-zero constant C. he showed that n! ∼ C √ nn n... |

1 |
Laerebog i matematisk Analyse : Afsnit III. Funktioner af flere reelle Variable. Jul. Gjellerups Forlag
- Bohr, Mollerup
- 1922
(Show Context)
Citation Context ...uler’s definition (5), we immediately obtain the fundamental relation Γ(z + 1) = zΓ(z) (6) and the value Γ(1) = 1. In fact, the gamma function is completely characterized by the Bohr-Mollerup theorem =-=[8]-=-: 2Theorem 1 The gamma function is the only function Γ : (0, ∞) → (0, ∞) which satisfies 1. Γ(1) = 1 2. Γ(x + 1) = xΓ(x) 3. ln [Γ(x)] is convex for all x ∈ (0, ∞). Proof. See [4] and [30]. Another co... |

1 |
Classroom Notes: A Simple Proof of Stirling’s Formula
- Coleman
- 1951
(Show Context)
Citation Context ...lyth & Pathak [7] and Khan [23] used probabilistic arguments, applying the Central Limit Theorem and the limit theorem for moment generating functions to Gamma and Poison random variables. 6. Coleman =-=[9]-=- defined cn = and showed that cn → 1 − 1 2 ( n + 1 ) ln(n) − n + 1 − ln(n!) , 2 ln (2π) as n → ∞. A similar result was proved by Aissen [2], using the concavity of ln(x). 7. Dingle [14] used Weierstra... |

1 |
Notes: A Note on Stirling’s Formula
- Questions
- 1940
(Show Context)
Citation Context ...n(n) + I(n) − I 2 5 ( ) 1 ∑n−1 = 2 k=1 k ln ( ) t dt k (ak − bk) + an,9. Hayman [19] used the exponential generating function e z = ∞∑ 1 k! k=0 zk and his method for admissible functions. 10. Hummel =-=[20]-=-, established the inequalities where 11 12 < rn + 1 ln(2π) < 1, n = 2, 3, . . ., 2 ( ) n n!e rn = ln √ . 2πnnn Impens [21], [22], showed that for x > 0 R2n(x) < ln [Γ(x)] − P(x) < R2m+1(x), n, m ≥ 0, ... |

1 |
Stirling’s formula for n! made easy. Real Anal
- Impens
- 2002
(Show Context)
Citation Context ...tion e z = ∞∑ 1 k! k=0 zk and his method for admissible functions. 10. Hummel [20], established the inequalities where 11 12 < rn + 1 ln(2π) < 1, n = 2, 3, . . ., 2 ( ) n n!e rn = ln √ . 2πnnn Impens =-=[21]-=-, [22], showed that for x > 0 R2n(x) < ln [Γ(x)] − P(x) < R2m+1(x), n, m ≥ 0, where Rn(x) was defined in (10). Maria [26] showed that [ ] −1 3 12n + < rn, n = 1, 2, . . .. 2 (2n + 1) Mermin [28] prove... |

1 |
A remark on Stirling’s formula
- Maria
- 1965
(Show Context)
Citation Context ... 12 < rn + 1 ln(2π) < 1, n = 2, 3, . . ., 2 ( ) n n!e rn = ln √ . 2πnnn Impens [21], [22], showed that for x > 0 R2n(x) < ln [Γ(x)] − P(x) < R2m+1(x), n, m ≥ 0, where Rn(x) was defined in (10). Maria =-=[26]-=- showed that [ ] −1 3 12n + < rn, n = 1, 2, . . .. 2 (2n + 1) Mermin [28] proved the identity e rn = ∞∏ k=n e −1 ( 1 + 1 k ) k+ 1 2 , which he used to show that rn ∼ R3(n). Michel [29], proved the ine... |

1 |
On Stirling’s formula
- Michel
(Show Context)
Citation Context ... in (10). Maria [26] showed that [ ] −1 3 12n + < rn, n = 1, 2, . . .. 2 (2n + 1) Mermin [28] proved the identity e rn = ∞∏ k=n e −1 ( 1 + 1 k ) k+ 1 2 , which he used to show that rn ∼ R3(n). Michel =-=[29]-=-, proved the inequality ∣ern − 1 − 1 1 − 12n 288n2 1 1 ∣ ≤ + 360n3 108n4, n = 3, 4 . . .. Nanjundiah [32], showed that R2(n) < rn < R1(n) = 1, 2, . . .. Robbins [36], established the double inequality... |

1 |
Wielandt’s theorem about the Γ-function
- Remmert
- 1996
(Show Context)
Citation Context ...orem 2 The gamma function is the only holomorphic function in the right half plane A satisfying 1. Γ(1) = 1 2. Γ(z + 1) = zΓ(z) for all z ∈ A 3. Γ (z) is bounded in the strip 1 ≤ Re(z) < 2 Proof. See =-=[35]-=-. In terms of Γ(z), we can re-write (1) as with P(z) = z ln (z) − z − 1 2 and (3) in the form where R0(z) = 0 and ln[Γ(z)] ∼ P(z), z → ∞ (7) 1 ln (z) + ln (2π) (8) 2 ln [Γ(z)] ∼ P(z) + RN(z), z → ∞ (9... |

1 | An elementary proof of Binet’s formula for the gamma function
- Sasvári
- 1999
(Show Context)
Citation Context ... 3. Bender & Orszag [5] and Temme [39] used Hankel’s contour integral [1] 1 Γ(z) ∫ 1 = 2πi and the method of steepest descent. (0+) −∞ t −z e t dt 4. Bleistein & Handelsman [6], Lebedev [25], Sasvári =-=[37]-=- and Temme [39] used Binet’s first formula and ∫ ln [Γ(z)] = P(z) + 0 ∞ 1 t ( 1 1 1 1 − + t 2 t et ) − 1 ( 1 1 1 − + 2 t et ) e − 1 −tz dt, Re(z) > 0 = ∑ k≥1 B2k (2k)! t2k−2 , |t| < 2π. 45. Blyth & P... |

1 |
Approximating n!: historical origins and error
- Tweddle
- 1984
(Show Context)
Citation Context ...four terms” of the series ( n + 1 ) ( log n + 2 1 ) ( − a n + 2 1 ) + 2 1 log (2π) (2) 2 a 7a − ) + ) 3 − · · · , 24 ( n + 1 2 2880 ( n + 1 2 where log means the base-10 logarithm and a = [ln(10)] −1 =-=[40]-=-. In 1730 Stirling wrote to Abraham De Moivre (26 May 1667 – 27 Nov. 1754) pointing out some errors that he had made in a table of logarithms of factorials in the book and also telling him about (2). ... |