NOTES ON EULER-BOOLE SUMMATION (1) f (l 1) (n) f (l 1) (m) + ( 1)k 1 k! B k (y) f (k) (y) dy,

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1 NOTES ON EULER-BOOLE SUMMATION JONATHAN M BORWEIN, NEIL J CALKIN, AND DANTE MANNA Abstract We stuy a connection between Euler-MacLaurin Summation an Boole Summation suggeste in an AMM note from 196, which explains them as two cases in a general approach to approximation that also encompasses Taylor sums Here we give aitional etails of the construction j=n+1 m 1 Introuction The Euler-Maclaurin summation formula, m n k f(x + j) = f(x) x + ( 1) l B l l! (1) l=1 + ( 1)k 1 k! ( ) f (l 1) (n) f (l 1) (m) n m B k (y) f (k) (y) y, is a well-known formula from classical analysis giving a relation between the sum of a (m times ifferentiable) function f an its integral Notably, it can be use to prove Stirling s asymptotic for factorials In the formula above, the B l are the Bernoulli numbers an the B l (x) are the perioic Bernoulli polynomials The Bernoulli polynomials are given by the formula (2) te xt e t 1 = B n (x) tn The perioic Bernoulli Polynomials are efine as B n (x) = B n (x x ) Evaluating at the point x = gives the Bernoulli numbers: B l := B l () A similar, but lesser known formula comes from starting with a ifferent set of polynomials The Euler polynomials E n (x) are given by the generating function (3) 2e xt e t + 1 = E n (x) tn an the perioic Euler polynomials Ẽn(x) by Ẽn(x+1) = Ẽn(x) an Ẽn(x) = E n (x) for x < 1 Thirly, efine the Euler number by E n = E n (1/2); ie, ( ) t t n (4) sech = E n 2 The parallel of (1) is calle the Boole summation formula If f(t) is a function with m continuous erivatives, efine on the interval t [x, x + ω], then (5) f(x + hω) = m 1 k=1 ω k k! E k(h) 1 ( ) f (k) (x + ω) + f (k) (x) Ẽ m 1 (h t) 2 ωm f (m) (x + ωt)t (m 1)! 1

2 2 JONATHAN M BORWEIN, NEIL J CALKIN, AND DANTE MANNA In his 196 note, Strot inicate an operator-theoretic approach to proving both of these formulae in the same way, but with very few of the etails given The main goal of this note is to supply the etails of that suggeste argument 2 Details of Strot s Observation The following is a more etaile account of a note by Strot that escribes a connection between Euler-MacLaurin an Boole summation by comparing the erivations of formulas (1) an (5) Define, for a function f : [x, x + 1] R an integer n N, the operator I n by (6) L n (f)(x) := n 1 f(x + i/n) n This operator, being a sum of translations, will sen a polynomial in x to another one of the same egree Furthermore, it is an automorphism on the ring of egree-k polynomials Proposition 21 For all k N, let P k := { k i= a ix i, a i R} Then for all n N, given A P k there is a unique B P k such that I n (B) = A This property is a special case of one for a general probability istribution function on a finite interval Given a istinct set of points x i [, 1] an a weight function N w : {x i } N i=1 (, 1) so that w(x i ) = 1, efine the corresponing operator (7) Even more generally, efine (8) L w (f)(x) := i=1 N f(x + x i )w(x i ) i=1 L g (f)(x) := f(x + u)g(u) u where g is any probability weight function with finite moments: (9) g(u) u = 1 an u k g(u) u < for all k N Here the interval of integration is the entire real line, where g possibly has compact support Hence (7) can be seen to be an instance of (8) by way of the Dirac elta function : N (1) g(u) = w(x i )δ(u x i ) Now suppose f P k so that f := k f nx n, for f n R By efinition, k (11) h(x) := L g (f) = f n (x + u) n g(u) u i=1 an we see via the binomial theorem that h(x) is again a polynomial of egree k: k k ( ) n (12) h(x) = h j x j, where h j = f n M n j an M l := u l g(u) u j n=j If one chooses to represent the polynomials in P k as (k + 1) 1 coefficient matrices, then the restriction of the operator L g to P k is a (k + 1) 2 upper-triangular matrix with coefficients { ( j 1 L g [i, j] := i 1) Mj i for 1 i j k + 1 (13) otherwise

3 NOTES ON EULER-BOOLE SUMMATION 3 This allows us to prove the invertibility of this operator Proposition 22 Let g be a probability ensity function whose absolute moments exist For all h P k, there is a unique f P k so that L g (f) = h Proof The eterminant of the matrix in (13) is (14) et(l g ) = k t= ( ) t X t,t = 1 t an so h(x) is the unique solution of a linear system of equations: h = L 1 g f (15) The reaer will note that Proposition 21 follows as a special case of this We can recover both the Euler Polynomials an the Bernoulli Polynomials using this uniqueness property with ifferent weight functions Define The Euler operator as f(x) + f(x + 1) (16) L E (f)(x) := 2 an the Bernoulli operator by (17) L B (f)(x) := 1 ie, g(u) = δ(u)/2 + δ(u 1)/2, f(x + u) u ie, g(u) = χ [,1] We can efine the Euler Polynomials E n (x) as the unique set of polynomials that satisfy (18) L E (E n (x)) = x n for all n N an the Bernoulli Polynomials as the unique class of polynomials satisfying (19) L B (B n (x)) = x n for all n N One now sees the use of the efinition of L n in (6); the Euler operator is L 2 an the Bernoulli operator correspons to the limit of L n as n, hence the range of n provies an interpolation of sorts We now prove the stanar efinitions of the Bernoulli an Euler Polynomials follow from here as consequences Theorem 23 Let P n (x), n N be the polynomials associate with a given ensity g(x); that is, (2) Then (21) L g (P n (x)) = x n for all n N x P n(x) = np n 1 (x) for all n N Proof By the Lebesgue Dominate Convergence Theorem, we have that P n (x + u)g(u)u = lim ν x ν ( ) = lim ν P n (x + 1/ν + u) P n (x + u) g(u) u = ν Therefore, in view of (2), we have that ( ) (22) L g x P n np n 1 = x (xn ) nx n 1 = But L g is one-to-one by Proposition 22, hence the esire conclusion ( P n (x + 1/ν + u) P n (x + u) x P n(x + u)g(u)u ) g(u) u

4 4 JONATHAN M BORWEIN, NEIL J CALKIN, AND DANTE MANNA Theorem 24 Suppose a class of polynomials {C n (x)} n with real coefficients has an exponential generating function of the form (23) C n (x) tn = e xt R(t), where R(t) is a continuous function on the real line Then the exponential generating function of the polynomial sequence given by the image of the C n uner the operator L g is given by (24) L g (C n (x)) tn = e xt R(t)Q(t), where Q(t) := L g (e xt ); ie, Q is the Laplace transform of the weight function g Proof Assume that the parameter t is within the raius of uniform convergence of the formal power series (23) for an arbitrary fixe value of x Then we can integrate the series termwise to prouce: (25) C n (x + u) g(u) u tn = e (x+u)t R(t) g(u) u This is equivalent to (24) We can use this to verify our claim regaring the Bernoulli an Euler Polynomials Corollary 25 Formulae (18) an (19) are equivalent efinitions of the classical Euler Polynomials an Bernoulli Polynomials, respectively Proof We show that the conition (18) is equivalent to the generating function representation of the Euler Polynomials This is known to be (26) E n (x) tn = 2ext e t + 1 Assume that the polynomials E n (x) satisfy (18); that is, (27) L E (E n (x)) = x n for all n N Now if the E n (x) o have an exponential generating function, then it must be (28) because of Theorem 24 We verify that (29) E n (x) tn = ext Q(t) Q(t) = L g (e ut ) = et matching the generating function in (26) to that in (28) Therefore we conclue that (with appropriate bouns onx) the formula (18) implies that the E n (x) are the Euler Polynomials Reversing the irection of the steps proves the converse statement, The argument for Bernoulli Polynomials is similar, again appealing to Theorem 24

5 NOTES ON EULER-BOOLE SUMMATION 5 3 Consequences Another consequence of Theorem 24 is the relation it provies to a particular generalization of the Bernoulli Polynomials The Bernoulli Polynomials of the kth orer (or type, but not to be confuse with the Bernoulli Polynomials of the kth kin, which are something else!) are given by the generating function ( ) k B n (k) (x) tn t (3) = e t e xt 1 so that the first orer correspons to plain Bernoulli: B (1) n (x) = B n (x) Since the kth Bernoulli generating function is just e xt ivie by a power of Q(t), we can also efine these polynomials by iterations of L B, the Bernoulli operator Corollary 31 For a positive integer k, the Bernoulli Polynomials of the kth kin are exactly those for which (31) (L B ) k [B n (k) (x)] = x n for all n N, where L B is the Bernoulli operator given in (17) Proof We emulate the proof of Corollary 25 for Bernoulli Polynomials, using Theorem 24 in conjunction with the observation that the generating function in (3) is exactly e xt [Q(t)] k The Bernoulli operator also has the following property Proposition 32 Let V = L 1 [, 1] The Bernoulli operator L B : V V has norm equal to 1 but no continuous inverse Proof First we show that L B = 1 We have that L B (1) = 1, hence L B 1 But for any f V, 1 (32) L B (f) = f(x + u) χ [,1] (u) u x { } 1 max f(x + u) χ [,1] (u) u : x 1 f(u) u = f, an thus L B 1 For the secon statement of the proposition, consier the following sequence of functions: f n := Hence there is no real number C such that f C L B (f) for all f V 31 Laplace Transforms of Density Functions In the previous section we arrive at the generating functions of Euler an Bernoulli Polynomials by taking the Laplace transform of an appropriate ensity function Of course, one coul choose any ensity function an get a class of polynomials with similar properties Here we summarize the outcome for a selection of functions; the reaer is encourage to experiment with other ensities Gaussian Density Function (33) Here we take g(u) = 1 2π e u2 /2, < u < Its Laplace Transform is Q(t) = 1 2π e ut e u2 /2 u = e t2 /2, an so the generating function of our polynomial sequence is (34) P n (x) tn = e xt /Q(t) = e xt t2 /2

6 6 JONATHAN M BORWEIN, NEIL J CALKIN, AND DANTE MANNA Expan the right han sie to obtain (35) P n (x) = n 2 ( 1) j (n 2j)! xn 2j Then we have the following immeiate corollary of Theorem 24 an 23 Corollary 33 Let P n (x) be the polynomials efine by the formula (35) for n N Then the P n are the unique polynomials that satisfy (36) an (37) 1 2π P n (x + u) e u2 /2 u = x n x P n(x) = np n 1 (x) The symmetry of the function e u2 /2 about the origin leas to the following property Proposition 34 For g(u) = e u2 /2, the operator L g sens even polynomials to even polynomials an o polynomials to o polynomials Proof Examine the image of x n uner the Gaussian transformation We expan the integran using the binomial theorem; thus n ( n (38) L g (x n ) = )x j u n j e u2 /2 u j We see that the integrals in this finite sum vanish if n j is an o integer, for that causes the integran to be an o function Therefore, the resulting polynomial must have the same parity as the original function, the parity of n itself Since any even (o) polynomial is a linear combination of even (o) monomials, the proposition follows by linearity of the operator Poisson Distribution (39) The Poisson istribution X for a real parameter λ is given by the probability function This correspons to the weight function (4) P(X = j) = e λλj g(u) := for j N δ(u j)e λ λ u Γ(u + 1) We will evelop the main properties using a general value of λ as far as this is possible As before, we search for a close form of the polynomials which are inverses of x n uner the operator L g We first fin their generating function, which comes from the Laplace transform of the weight function: (41) Q(t) = e λ Thus the generating function is given by (42) P n,λ (x) tn jt λj e = t eλ(e 1) = e xt e λ(et 1),

7 NOTES ON EULER-BOOLE SUMMATION 7 where the polynomials this time epen on the value of the real parameter λ We expan the right-han sie of this equation in much the same way as the intermeiate step of (41) an bring the factor e xt within the sum Then expan in a Taylor series with respect to t to obtain (43) e xt e λ(et 1) = (j + x) n t n e λ( λ)j The sums on j an n are inepenent, an we switch their orer This yiels e xt e λ(et 1) t n (j + x) n ( λ) j (44) = eλ Now expan (j + x) n using the binomial theorem, an rearrange to prouce e xt e λ(et 1) t n n ( ) n = x n m e λ j m ( λ) j (45) m In the inner sum, replace j m using the following well-known formula: j m = m m (46) k= S(m, k)(j) m = S(m, k)j(j 1) (j m + 1), m= k= where the S(m, k) are the Stirling numbers of the secon kin This allows us to reach a close form for the inner sum e λ j m ( λ) j m = e λ S(m, k)( λ) m ( λ) j m m (47) = S(m, j)( λ) j (j m)! k= The polynomial on the right is commonly known as the mth Bell polynomial B m ( λ), but we will leave the notation to avoi confusion with the Bernoulli polynomials Finally, we have our esire close form: n ( ) n m (48) P n,λ (x) = S(m, j)( λ) j x n m m m= In the special cases λ = 1 an 1, the inner sum will reuce to the mth Bell number an complementary Bell number, respectively Again, we have a corollary of Theorems 24 an 23 Corollary 35 Let P n,λ (x) be the polynomials efine by the formula (48) for n N λ R Then the P nλ are the unique polynomials that satisfy (49) an (5) Exponential Distribution e λ j=m P n,λ (x + j) λj = x n x P n,λ(x) = np n 1,λ (x) We regar the ensity function associate with the exponential istribution as { λe λu for u > (51) g(u) = u so that its Laplace transform is (52) Q(t) = e ut λu u = λ λ t an fixe

8 8 JONATHAN M BORWEIN, NEIL J CALKIN, AND DANTE MANNA for real t < λ Therefore the generating function for L g (x n ) is given as e xt λ λ t We then introuce the scaling x λx an t t/λ in orer to remove the extra parameter Expaning a geometric series leas to the following property of the exponential operator Proposition 36 Given that g(u) is the exponential ensity, the operator L g applie to the monomial x n gives the Taylor series truncation of the exponential function: n L g (x n x m (53) ) = m! m= The inverse polynomials of x n uner L g are xn xn 1 (n 1)!, an Theorems 24 an 23 are again applicable 32 Summation Formulae For a continuously ifferentiable function g(x) on an interval incluing the value a, a partial Taylor sum approximation about a applies For all x in this interval an m N, m g (n) (a) (54) g(x) (x a) n with an error (remainer) term R m (x) given by the integral (55) R m (a, x) = 1 m! x This follows from an inuctive argument; by efinition (56) a R = g(x) g(a) = g (m+1) (y) (x y) m y x a g (y)y, an the inuctive step follows from an integration by parts The focus of Strot s very short note was to compare this summation formula to the those of Euler-MacLaurin an Boole The etails of the former are given within the boy of his note; now we fill the etails in the case of Boole Summation Recall the earlier efinitions of g(u) an L E given in Section 2 Let x = a + h for < h < 1 an for an integer m efine the remainer as the ifference between the function an the finite sum: m L E (f (k) (a) (57) R m (x) := f(x) E k (h) k! Start with m = Since g(u) u = 1, we have (58) R 1 (x) = f(a + h) f (a + u) g(u) u = [f(a + h) f(a + u)] g(u) u k= We recognize the integran as an integral in its own right, an switch the orer of integration to obtain h R (x) = f (a + s) s g(u) u = 1 (59) V (s, h) f (a + s) s, 2 u where we efine the piecewise function { s 2 (6) V (s, h) := g(u) u for s < h 2 s g(u) u 2 for s h

9 NOTES ON EULER-BOOLE SUMMATION 9 Therefore we have that (61) 1 for s < h V (s, h) = 1 for h s 1 otherwise which is the same as the constant (perioic) Euler polynomial on [, 1] Thus we have 1 (62) f(x) = L E (f(a)) 1 f (a + s)ẽ(h s) s 2 The general formula follows by inuction in the usual way One simply integrates by parts to check that 1 1 (63) Ẽ k 1 (h s)f (k) (a + s) s = 1 2 k L E(f (k) (a))ẽk(h) + 1 2k The resulting formula is equivalent to (5) with ω set to 1 1, Ẽ 1 (h s)f (k+1) (a + s) s Faculty of computer Science, Dalhousie University, Halifax, NS Canaa B3H 4R2 aress: jborwein@csalca Department of Mathematical Sciences, Clemson University, Clemson, SC aress: calkin@clemsoneu Department of Mathematics an Statistics, Dalhousie University, Halifax, NS Canaa B3K 1W5 aress: manna@mathstatalca