American Journal of Signal Processing 2019, 9(1): 1-5 DOI: 10.5923/j.ajsp.20190901.01
Derivation of Euler's Formula and ζ(2k) Using the Nyquist-Shannon Sampling Theorem
Er'el Granot
Department of Electrical and Electronics Engineering, Ariel University, Ariel, Israel
Abstract There are multiply approaches to prove Euler's well-known formula, however, none of them is trivial or simple. In this paper we utilize Nyquist-Shannon sampling theorem to prove Euler's formula in particular and Riemann's ζ (2k ) in general in a straightforward approach. The presented method allows calculating the summation of several other similar series, ∞ n −3 ∞ such as −1 2n += 1π 3 / 32 , and even more surprising ones such as sin pn n−3 =−− pππ p2 p /12 . ∑n=0( ) ( ) ∑n=1 ( ) ( )( ) Keywords Euler's theorem, Sampling theorem, Riemann's zeta function, Basel problem, Nyquist-Shannon theorem
trigonometric identities, inverse trigonometric functions, 1. Introduction complex analysis, Fourier analysis, and even geometric reasoning, however, there is none, to the best my Euler is definitely one of the greatest mathematicians of knowledge, which is based on Signal Processing's Sampling all times. His contribution to science is incredible by any theorem. measure or standard. However, his first contribution that The objective of this paper is to show that with the aid of made him famous by the age of twenty eight was the Digital Signal Processing (DSP) analysis, using the identity, which is called after him. We are not referring to Sampling theorem, the proof of this mathematical identity the well-known identity eiπ +=10, which is erroneously becomes almost straightforward. called Euler’s identity, for it should have been related to For completeness, we will remind the reader of the Cotes, who presented it many years prior to Euler. The real Sampling theorem and present the original Euler’s Euler’s identity was the solution of, what was termed the derivation. Basel problem, which was first presented in 1644 by Pietro The Nyquist- Shannon Sampling Theorem: If an Mengoli and later by the Bernoulli brothers, who could not analog signal xt( ) is sampled at a rate fs (which means solve the problem. The problem was to calculate the sum of the infinite that only xn( / fs ) are known), then the original signal ζ series (also known as Riemann's (2) ) can be exactly recovered from its sample values xn( / fs ) by the discrete convolution 11∞ 1 S =++1? += = (1) 22∑ 2 ∞ sin π (tf− n) 23 n=1 n s xt( ) = ∑ xn( / fs ) π tf− n Euler first presented an advanced numerical method to n=−∞ ( s ) calculate the sum, which he later used to validate the exact provided xt( ) is spectrally bounded by the frequency closed-form solution, which came later on. Euler proved ff< /2. that this series sums to π 2 /6 [1]. After the publications max s of the first proof, many equivalent proofs followed, which Euler's Derivation: Euler used the Taylor expansion of illustrates the fact that this identity appears in many places the sinc function [16] in mathematics and science [2-15]. n 246 22 One can find proofs that are based on multiple integrals, sin (πx) ( πππ xxx) ( ) ( ) ∞ (−π x ) =−1 + + += (2) ∑ π xn3! 5! 7! = (2+ 1) ! * Corresponding author: n 0 [email protected] (Er'el Granot) Then, Euler made a brilliant idea, which took him about a Published online at http://journal.sapub.org/ajsp decade to prove in a more rigorous method, and in fact, it Copyright © 2019 The Author(s). Published by Scientific & Academic Publishing This work is licensed under the Creative Commons Attribution International took about an additional century to cement the idea with License (CC BY). http://creativecommons.org/licenses/by/4.0/ Weierstrass factorization theorem [17].
2 Er'el Granot: Derivation of Euler's Formula and ζ(2k) Using the Nyquist-Shannon Sampling Theorem
Euler idea was to implement Newton factorization It is well-known that any spectrally bounded signal can be method to infinite products, i.e., it was shown by Newton written as a superposition of delayed sinc functions [19, 20]. that any polynomial can be rewritten as a product of linear In particular (see Fig.1), factors, which vanish at its roots [18]. Euler idea was to ∞ generalize it even to cases where the number of roots is 1= ∑ sinc( xn− ) (7) infinite. n=−∞ Since the sinc function vanishes for n =±±±1,2,3, There are many ways to prove this identity, one of which then Euler assumed, but he had difficulties in proving, that is to use Nyquist- Shannon(NS) theorem (mostly known as the "sampling theorem", see above) [19, 20], which shows sin (π x) ∞ x2 xxx 222 =−=−−−1 111 (3) that since the constant "1" (left side of Eq.(1)) has a zero π ∏2 2 22 x n=1n 123 bandwidth, then if this identity is valid for any xm= , where m =0, ±± 1, 2, then it is valid for any x . Another Now, after opening all brackets and collecting all terms, option is to recognize the fact that the Fourier transform of sin (π x) ∞ x2 xxx 222 both sides of the equation are equal to the delta function, i.e. =−=−−−=∏1 111 π 2 2 22 ∞ x n=1n 123 (4) 2πδ( ω ) =F{ 1} = F∑ sinc( xn− ) 2411 1 =−∞ 1−xx +++ +( ) n 1 2322 ∞ = Fsinc( x) exp( inω ) . (8) 2 { } ∑ That is, the coefficient of x is the infinite series n=−∞ ∞ 11∞ 1 1+ + += (5) = rect(ω) 2 π δω( −k ) = 2 πδω( ) 22 ∑ 2 ∑ 23 n=1 n k=−∞ However, from (2) it is clear that the coefficient of x2 Since Eq.(7) is valid for any x , then clearly all its 2 derivatives must vanishes. In particular, the second is −π / 3!, therefore derivative must vanish ∞ 2 11 1π ∂∂22∞ 1+ + += ∑ = . (6) 0= 1 = sinc( xn− ) . (9) 2322n 26 22∑ n=1 ∂∂xxn=−∞ In the literature Now since, In what follows we will show that this relation can be 2 retrieved easily using Digital Signal Processing analysis. ∂ sinc(ξ ) 2 n+1 =( −1) (10) Moreover, it will be shown that the derivation is almost ∂ξ 22n straightforward. τ =n ≠0 ∂2 sinc(ξ ) π 2 and = − , (11) ∂ξ 2 3 2. The Sampling Theorem Approach τ =0 then the second derivative at any integer xm= must be 1 equal to ∞ n 2 0.8 11 (−1) π 1− + −=− ∑ = (12) 2322 n2 12 0.6 n=1 ) n -
x This is the first identity, but it is not Euler's formula yet. 0.4 However, it is very simple to show that Eq.(12) is exactly sinc(
n half Eq.(5) since 0.2 1111 1−+−++= 2222 0 2345 1111 11 12+++++− ++ = -0.2 23452222 24 22 (13)
-4 -2 0 2 4 6 8 10 1111 211 1+++++− ++ = x 2222 22 23454 12 Figure 1. Illustration of Eq.(7). "1" can be rewritten as an infinite sum of 1 1111 1+++++ "sinc" pulses (in the figure only eight are presented) 2 23452222
American Journal of Signal Processing 2019, 9(1): 1-5 3
Thus, Eq.(6) is retrieved. By comparing the coefficient of x Eq.(12) appears, and Similarly, by taking the 4th derivative of Eq.(7) and using by comparing the coefficient of x3 Eq.(16) appears. 4 The same approach can be applied to other spectrally ∂ sinc(ξ ) 46n 2 =−−( 1) π , bounded functions, such as (see Fig.2) ∂ξ 42 2 τ = ≠ nn n 0 π sin(πππ( xxx++− 3)) sin( ( 1)) sin( ( 1)) sin x = −+ = ∂4 sinc(ξ ) π 4 2πππ( xxx++− 311) ( ) ( ) = (14) ∂ξ 4 5 ∞∞n τ = n sinπ ( xn−− 2 1) (−1) 0 (−=1) sinπ ( x − 1) ∑∑ππ−− −− One finds nn=−∞ ( xn21) =−∞ ( xn21) ∞ 4 (22) 46n 2 π 21∑ (−) π − += 0 (15) which is equivalent to 225 n=1 nn π which yields π sin x ∞ n 2 π (−1) n =−=∑ . (23) ∞ (−1) ππ447 π 4 sinπ ( x − 1) π ( xn−−21) =−+ =− . (16) 2cosx n=−∞ ∑ 4 2 n=1 n 72 240 720 Now since Again, Eq.(22) agrees with the Sampling theorem since π n the spectrum of sin( x / 2) is bounded within the spectral ∞(−1) ∞∞1 1 71 ∞ =−+=−2 ,(17) ωπ< ∑44 ∑∑ 4 ∑ 4regime . n=1nn nn= 11 = (2n) 8 n= 1n then, finally 1 0.8 ∞ 1 π 4 = . (18) 0.6 ∑ ) 4 90 n=1 n 0.4 n-1 2
- 0.2
Clearly, this method can be used to derive any summation x ∞ − p 0
n sinc( ∑ = n n 1 -0.2
(-1) -0.4 for any even integer p, i.e., it can be used to calculate n
Riemann's ζ (2k ) for any integer k. -0.6
However, one can use even a simpler approach for this -0.8 problem: -1
Eq.(7) is actually 0 5 10 15 x sin(π( x+− 1)) sin( ππ( xx)) sin( ( 1)) 1 =+++ + π( x+−11) ππ( xx) ( ) Figure 2. Illustration of Eq.(22). sin(π x / 2) (solid curve) can be (19) n rewritten as an infinite sum of "sinc" pulses with alternating signs (in the ∞∞sin π ( xn− ) sin (π x) (−1) = ∑∑= figure only eight are presented) ππ( xn−−) ( xn) nn=−∞ =−∞ Again, one can expand the two sides of the equation to get Therefore, Eq.(7) can be rewritten n π ππ32xx5 π 54 ∞ (−1) ∞ n − =−+ + + = π 1 (−1) π ∑ −− = + 2 . (20) 2 16 768 n=−∞ ( xn2 1) π ∑ − 2cosx sin ( x) xn=1 ( xn) 2 (24) ∞ Then, by applying the Lauren expansion on the two sides n 1 x =∑ ( −1) + + of the equation, one obtains 21n + 2 n=−∞ ( ) (21n + ) π1 ππ2xx 7 43 31 π 65 x =+++ + 2 sin (π xx) 6 360 15120 By comparing the coefficient of the x term on both (21) sides of Eq.(24) one gets ∞ 2 11n xx = −21 − + + + 3 ∞ n ∑ ( ) 23 π (−1) xn= nn = (25) n 1 ∑ 3 16 n=−∞ (21n + )
4 Er'el Granot: Derivation of Euler's Formula and ζ(2k) Using the Nyquist-Shannon Sampling Theorem
and by comparing the x4 coefficients ∞ sin ( pn) p =−−ππpp2 (34) ∑ 3 ( )( ) ∞ n n 12 5π 5 (−1) n=1 = (26) ∑ 5 for m = 4 768 n=−∞ (21n + ) 22 (Clearly, the odd terms vanish). Therefore ∞ sin ( pn) pp(π−)(2 π −−+ pp)( 36 ππ p + 4) = (35) n ∑ 5 ∞ (−1) π 3 n=1 n 720 = (27) ∑ 3 n=0 (21n + ) 32 for m = 6 ∞ sin ( pn) and = ∑ 7 n n ∞ (−1) 5π 5 n=1 (36) = . (28) ∑ 5 p 43 3 4 = (21n + ) 1536 (π−−p)(2 π pp) 3 −++ 12 pππ 24 p 16 π n 0 30240 ( ) Clearly, this method can be applied to calculate the etc. generic summation 2 n ∞ (−1) 1.5 ∑ p n=0 (21n + ) 1 for any odd p . 3 0.5 n )/
pn 0
3. Generalization sin( n -0.5 One can use the original NS sampling theorem -1 ∞∞n sin π ( xn− ) (−1) fx( ) = fn( ) = sin (π x) fn( ) (29) ∑∑ππ−−-1.5 nn=−∞ (xn) =−∞ ( xn)
-2 for any function fx( ) , whose spectrum, i.e. its Fourier -2 -1 0 1 2 3 4 5 6 7 8 p transform is spectrally bounded in ωπ< , to derive a Figure 3. Illustration of Eq.(34). The solid curve represents the left side of generic relation the equation, while the dashed curve represents its right side. As can be seen ∞n ∞∞m−1 the two agree provided 02<
limit where the last equality is simply the Taylor expansion of the p → 0 is taken, the original Euler formula, i.e. Eq.(6), is former. retrieved m By comparing the coefficients of x ∞ sin ( pn) π 2 lim = . m ∞ n ∑ 3 π d π fx( ) fn( )(−1) p→0 = pn 6 = − ∑ . (31) n 1 mx!mm sin (π ) +1 dx x=0 n=−∞ n Similarly, when p = π /2 then Eq.(34) reduces to Eq.(27). In particular, if f( x) =sin ((π − px) ) for 02<
summations has a
American Journal of Signal Processing 2019, 9(1): 1-5 5
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