ISSN: 1017-060X (Print) ISSN: 1735-8515 (Online) Special Issue of the Bulletin of the Iranian Mathematical Society in Honor of Professor Heydar Radjavi's 80th Birthday Vol. 41 (2015), No. 7, pp. 99{106 . Title: A Haar wavelets approach to Stirling's formula Author(s): M. Ahmadinia and H. Naderi Yeganeh Published by Iranian Mathematical Society http://bims.ims.ir Bull. Iranian Math. Soc. Vol. 41 (2015), No. 7, pp. 99{106 Online ISSN: 1735-8515 A HAAR WAVELETS APPROACH TO STIRLING'S FORMULA M. AHMADINIA∗ AND H. NADERI YEGANEH (Communicated by Peter Rosenthal) Dedicated to Professor Heydar Radjavi on his 80th birthday Abstract. This paper presents a proof of Stirling's formula using Haar wavelets and some properties of Hilbert space, such as Parseval's identity. The present paper shows a connection between Haar wavelets and certain sequences. Keywords: Haar wavelets, Parseval's identity, Stirling's formula. MSC(2010): Primary: 15-xx; Secondary: 40-xx. 1. Introduction n! Stirling's formula (lim !1 p = 1) plays an important role in statis- n nne−n 2πn tics and probability; its main use is estimating the value of n!. The first proofs of Stirling's formula were presented by de Moivre and Stirling (see [4, 8]). Es- timates of n! have been obtained via upper and lower bounds of n! by various authors (for example see [3], [5] and [6]). Some papers prove Stirling's formula by different techniques as well (for example see [2]). The present paper in- troduces a new method of proving Stirling's formula by using Haar wavelets. Although there are some proofs of Stirling's formula that are simpler than the present one, we feel that it is of interest to know that this formula can be established using Haar wavelets. The outline of the paper is as follows: In Section 2, we introduce some notation and present two lemmas related to Parseval's identity and the Haar wavelet basis. In Section 3, we present an application of the two lemmas, and in the last section we give a proof of Stirling's formula and of a lower and upper bound for n!. Article electronically published on December 31, 2015. Received: 22 November 2014, Accepted: 18 April 2015. ∗Corresponding author. ⃝c 2015 Iranian Mathematical Society 99 Haar wavelets and Stirling's formula 100 2. Preliminaries and Haar Wavelets This section introduces Haar wavelets and some related notation. The following notation is fixed throughout the paper: Z∗ := N [ f0g; (N is the set of natural numbers). The bijective function P and the surjective function Q are defined as follows: P : S ! N; P (n; k) := 2n + k; and Q : W ! Z∗;Q(r; n) := n; where S := f(n; k) 2 Z∗ × Z∗j0 ≤ k ≤ 2n − 1g; & W := f(r; n) 2 Z∗ × Z∗jr ≤ ng: Consider the following functions ( 1; 0 ≤ x < 1; ϕ(x) := 0; otherwise; and 8 > ≤ 1 <1; 0 x < 2 ; (x) := −1; 1 ≤ x < 1; :> 2 0; otherwise; and also n n n;k(x) := 2 2 (2 x − k); (n; k) 2 S: R 2 h i 1 The real Hilbert space L ([0; 1]) with inner product f; g := 0 f(x)g(x)dx, has the following orthonormal basis, which is called \the Haar wavelets basis": (2.1) fϕ(x)g [ f n;k(x) j (n; k) 2 Sg : Parseval's identity implies that X1 2Xn−1 (2.2) hf; gi = hf; ϕihg; ϕi + hf; n;kihg; n;ki; n=0 k=0 (for details see [1] and [7]). For convenience, we employ the following notation: Z Z k+1 2n f := f(x)dx: k n;k 2n 101 Ahmadinia and Naderi Yeganeh Lemma 2.1. If f and g are in L2([0; 1]), then 2Xr −1 Z Z X1 2Xn−1 r (2.3) hf; gi = 2 f g + hf; n;kihg; n;ki; k=0 r;k r;k n=r k=0 for all r 2 Z∗. Proof. This lemma is proved by induction. If r = 0, then (2.3) and (2.2) are the same. If the equation (2.3) holds for r, then (2.4) 2Xr −1 Z Z 2Xr −1 X1 2Xn−1 r hf; gi = 2 f g + hf; r;kihg; r;ki + hf; n;kihg; n;ki: k=0 r;k r;k k=0 n=r+1k=0 Let h be an arbitrary function in L2([0; 1]). Then Z Z Z (2.5) h = h + h; r;k r+1;2k r+1;2k+1 and (Z Z ) r (2.6) hh; r;ki = 2 2 h − h : r+1;2k r+1;2k+1 We can derive the following equation by using (2.5) and (2.6) in the left side of the equation: 2Xr −1 Z Z 2Xr −1 2rX+1−1Z Z r r+1 (2.7) 2 f g + hf; r;kihg; r;ki = 2 f g: k=0 r;k r;k k=0 k=0 r+1;k r+1;k Substitution of (2.7) into (2.4) yields (2.3) for r + 1. □ Lemma 2.2. If f and g are continuous functions on [0; 1] that are differentiable 0 0 on (0; 1), and if also jf (x)j < M1 and jg (x)j < M2 for all x 2 (0; 1), then the following assertions are true: jh ij M1 jh ij M2 2 (i) f; n;k < 3n +1 and g; n;k < 3n +1 ; (n; k) S; R R2 2 R 2 2 j − r j M1M2 2 (ii) r;k fg 2 r;k f r;k g < 3·23r ; (r; k) S. Proof. The mean value theorem for integrals and (2.6) imply (Z Z ) n f(c1) f(c2) h i 2 − − f; n;k = 2 f f = n +1 n +1 ; n+1;2k n+1;2k+1 2 2 2 2 2 2k 2k+1 2 2k+1 2k+2 for some c1 ( 2n+1 ; 2n+1 ) and c2 ( 2n+1 ; 2n+1 ). The mean value theorem for derivatives yields 0 − h i f (c)(c1 c2) f; n;k = n +1 ; 2 2 Haar wavelets and Stirling's formula 102 2 j − j 1 for some c (c1; c2). Note that c1 c2 < 2n . Then jf 0(c)j M jhf; ij ≤ < 1 : n;k 3n +1 3n +1 2 2 2 2 The proof of statement (i) is completed by applying a similar proof for g. To prove statement (ii), consider the following functions on [0; 1] with (r; k) 2 S arbitrary and fixed: x + k x + k f (x) := f( ); & g (x) := g( ): 1 2r 1 2r Then Z Z Z − r 1 h i − 1 h ih i fg 2 f g = r f1 ; g1 r f1 ; ϕ g1 ; ϕ r;k r;k r;k 2 2 1 n− 1 X 2X1 (2.8) = hf ; ihg ; i: 2r 1 n;l 1 n;l n=0 l=0 (The first equation is clear and the second equation is obtained from (2.2).) Since M1 and M2 are the bounds for f 0 and g0 respectively, statement (i) 2r 2r 1 1 implies M M (2.9) jhf ; ij < 1 & jhg ; ij < 2 : 1 n;k 3n +1+r 1 n;k 3n +1+r 2 2 2 2 Using (2.8) and (2.9) yields Z Z Z n X1 2X−1 r 1 M1M2 M1M2 fg − 2 f g < = : 23n+2 23r 3 · 23r r;k r;k r;k n=0 l=0 □ 3. An Application: An Infinite Product An infinite product can be obtained by applying the above two lemmas, in 1 the special case where f = 1+x and g = 1 + x. Theorem 3.1. The following equation holds: Y1 e m 2m+1 p (3.1) p e( ) 2 = π: m + 1 2 m=1 Proof. Consider the following functions on [0; 1]: 1 f(x) := ; g(x) := 1 + x: 1 + x The inner product hf; gi can be written as follows, for all r 2 Z∗: 2Xr −1 Z (3.2) hf; gi = fg; k=0 r;k 103 Ahmadinia and Naderi Yeganeh Lemma 2.1 and (3.2) imply that r ( ) n 2X−1 Z Z Z X1 2X−1 r (3.3) fg − 2 f g = hf; n;kihg; n;ki; k=0 r;k r;k r;k n=r k=0 The following equation is clear, by (3.3), r ( ) n X1 2X−1 Z Z Z X1 X1 2X−1 r r r (3.4) 2 fg − 2 f g = 2 hf; n;kihg; n;ki; r=0 k=0 r;k r;k r;k r=0 n=r k=0 Note that the left hand side of (3.4) is convergent by part (ii) of Lemma 2.2. Also, part (i) of Lemma 2.2 implies that the right hand side of (3.4) is absolutely convergent. Thus this series can be rearranged using the surjective function Q as follows: Q(r;n) X1 X1 2Xn−1 X1 X1 2 X−1 r h ih i r h ih i 2 f; n;k g; n;k = 2 f; Q(r;n);k g; Q(r;n);k r=0 n=r k=0 r=0 n=r k=0 X1 X 2Xi−1 t = 2 hf; i;kihg; i;ki i=0 Q(t;u)=i k=0 ! X1 Xi 2Xi−1 t = 2 hf; i;kihg; i;ki i=0 t=0 k=0 X1 2Xi−1 i+1 = (2 − 1) hf; i;kihg; i;ki: i=0 k=0 Hence, (3.4) can be written as (3.5) r ( ) r X1 2X−1 Z Z Z X1 2X−1 r r r+1 2 fg − 2 f g = (2 − 1) hf; r;kihg; r;ki: r=0 k=0 r;k r;k r;k r=0 k=0 Together, (2.2) and (3.5) yield: (3.6) hf; gi − hf; ϕihg; ϕi + A = B; where r ( ) X1 2X−1 Z Z Z A = 2r fg − 2r f g ; r=0 k=0 r;k r;k r;k and X1 2Xr −1 r+1 B = 2 hf; r;kihg; r;ki: r=0 k=0 Haar wavelets and Stirling's formula 104 Parts (ii) and (i) of Lemma 2.2 show that the series A and B are absolutely convergent.