MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Deprtment 8.33: Reltivistic Quntum Field Theory I NOTES ON THE EULER-MACLAURIN SUMMATION FORMULA These notes re intended to supplement the Csimir effect problem of Problem Set 4. Tht clcultion depended crucilly on the Euler-Mclurin summtion formul, which ws stted without derivtion.here I will give self-contined derivtion of the Euler- Mclurin formul.for pedgogicl resons I will first derive the formul without ny reference to Bernoulli numbers, nd fterwrd I will show tht the nswer cn be expressed in terms of these numbers.an explicit expression will be obtined for the reminder tht survives fter finite number of terms in the series re summed, nd in n optionl ppendix I will show how to simplify this reminder to obtin the form given by Abrmowitz nd Stegun. The Euler-Mclurin formul reltes the sum of function evluted t evenly spced points to the corresponding integrl pproximtion, providing systemtic method of clculting corrections in terms of the derivtives of the function evluted t the endpoints. Consider first function defined on the intervl x 1, for which we cn imgine pproximting the sum of f() + f(1) by the integrl of the function over the intervl: f() + f(1) = dxf(x)+r 1, (1) where R 1 represents correction term tht we wnt to understnd.one cn find n exct expression for R 1 by pplying n integrtion by prts to the integrl: dxf(x) =f() + f(1) dxxf (x), () so R 1 = dxxf (x), (3) where prime denotes derivtive with respect to x.
EULER-MACLAURIN SUM FORMULA, 8.33, SPRING 003 p. 1. Expnsion by successive integrtions by prts: We wnt n pproximtion tht is useful for smooth functions f(x), nd smooth function is one for which the higher derivtives tend to be smll.therefore, if we cn extrct more terms in wy tht leves reminder term tht depends only on high derivtives of the function, then we hve mde progress.this cn be ccomplished by successively integrting by prts, ech time differentiting f(x) nd integrting the function tht multiplies it.we cn define set of functions V 0 (x) 1,V 1 (x) x, (4) nd V n (x) dxv n (x). (5) Eq.(5) is not quite well-defined, however, becuse ech indefinite integrl is defined only up to n rbitrry constnt of integrtion.regrdless of how these constnts of integrtion re chosen, however, one cn rewrite Eq.(1) by using Eq.(3) nd then successively integrting by prts: f() + f(1) = = = = + dxf(x)+ dxv 1 (x) f (x) dxf(x)+v (1)f (1) V ()f () dxf(x)+ ] V (1)f (1) V ()f () ] V 3 (1)f (1) V 3 ()f () + dxv (x) f (x) ] V 4 (1)f (1) V 4 ()f () +... ] V n (1)f n (1) V n ()f n () dxv n (x) f n (x) dxf(x)+ n l= () l ] V l (1) f l (1) V l () f l () dxv n (x) f n (x), where f n (x) denotes the n th derivtive of f with respect to x. (6)
EULER-MACLAURIN SUM FORMULA, 8.33, SPRING 003 p. 3. Elimintion of the odd l contributions: Eq.(6) is vlid for ny choice of integrtion constnts in Eq.(5), so we cn seek choice tht simplifies the result.note tht V 1 (x) is odd under x x.we cn therefore choose the integrtion constnts so tht { Vn (x) if n is even V n ( x) = V n (x) if n is odd. (7) This even/odd requirement uniquely fixes the integrtion constnt in Eq.(5) when n is odd, becuse the sum of n odd function nd constnt would no longer be odd.we re still free, however, to choose the integrtion constnts when n is even. Using the even/odd property, Eq.(6) cn be simplified to f() + f(1) = dxf(x)+ n l= ] () l V l (1) f l (1) () l f l () dxv n (x) f n (x). (8) Note tht the terms in V l (1) for even l involve the difference of f l t the two endpoints, while the terms for odd l involve the sum.eq.(8) describes single intervl, however, nd our gol is to obtin formul vlid for ny number of intervls.we will do this by first generlizing Eq.(8) to pply to n rbitrry intervl x + h, nd then pplying it to ech intervl in succession of evenly spced intervls.when this succession is summed, the even l terms involving the differences of the endpoints will cncel t ech interior point, but the odd l terms will dd.the odd l terms cn therefore mke considerbly more complicted contribution to the nswer, but we cn force them to vnish by using the remining freedom in the choice of integrtion constnts.when n is even in Eq.(5), we choose the integrtion constnt so tht dxv n (x) 0. (9) Eq.(9) is lwys true for odd functions, so it is true for ll n>0.it then follows for ll n>1tht V n (1) V n () = dxv n (x) =0. (10) If n is odd then Eq.(7) implies tht V n () = V n (1), nd so V n (1) = V n () = 0 for ll odd n>1, (11)
EULER-MACLAURIN SUM FORMULA, 8.33, SPRING 003 p. 4 s desired. The V n s re now uniquely defined.in our construction we used the ntisymmetry property of Eq.(5) to fix the constnt of integrtion for odd n, nd the vnishing of the integrl in Eq.(9) to fix the integrtion constnt for even n.eq.(9), however, holds lso for odd n, nd is sufficient to fix the integrtion constnt for the odd n cses.the functions V n (x) cn therefore be defined succinctly by V 0 (x) 1, V n (x) dxv n (x), nd (1) (1b) dxv n (x) 0 (for n>0). (1c) We cn use these properties to build tble for the lowest vlues of n: n V n (x) V n (1) 0 V 0 (x) =1 1 1 V 1 (x) =x 1 V (x) = x 1 6 3 V 3 (x) = x3 6 x 6 4 V 4 (x) = x4 4 x 1 + 7 360 5 V 5 (x) = x5 10 x3 36 + 7x 360 6 V 6 (x) = x6 70 x4 144 + 7x 70 31 1510 1 3 0 1 45 0 945 (13) Eq.(11) gurntees tht only the even-l terms contribute to Eq.(8), so we cn set
EULER-MACLAURIN SUM FORMULA, 8.33, SPRING 003 p. 5 l =k nd rewrite Eq.(8) s f() + f(1) = dxf(x)+ n k=1 dxv n (x) f n (x). ] V k (1) f k (1) f k () (14) 3. Appliction to n rbitrry intervl: To pply Eq.(14) to n rbitrry intervl <x<+ h, one needs only to chnge vribles.let f(x) = f( x), where x =(x +1) h +. (15) Rewriting Eq.(14) in terms of f( x), while dividing the whole eqution by for lter convenience, one hs 1 ] 1 +h f()+ f( + h) = h + 1 1 n k=1 d x f( x) ( ) k h V k(1) f k ( + h) f ()] k ( ) n h +h d xv n ( h ( x ) 1 ) f n ( x), (16) where f n ( x) denotes the n th derivtive of f with respect to its rgument x.now tht the originl f nd x hve been eliminted, we cn drop the tilde superscripts tht pper in Eq.(16). 4. Appliction to n rbitrry sum of intervls: The problem cn now be completed by extending Eq.(16) to n intervl <x<b, divided into m steps of size h =(b )/m.adding n expression of the form (16) for
EULER-MACLAURIN SUM FORMULA, 8.33, SPRING 003 p. 6 ech intervl of size h, one hs m f( + kh) = 1 h b + 1 1 dxf(x)+ 1 n k=1 ] f()+f(b) ( ) k h ] V k(1) f k (b) f k () ( ) n h +h ( ) m dxv n h (x ) 1 f n (x + kh). (17) Note tht when one dds up the left-hnd sides of the expressions of the form (16), ll the terms hve coefficient 1 except for the first nd lst term, ech of which hve coefficient 1.In Eq.(17), the sum is written with ll terms hving coefficient 1, nd the correction for the first nd lst term hs been moved to the right-hnd side. For ll prcticl purposes, including the ppliction to the Csimir effect in Problem Set 4, Eq.(17) is ll tht is necessry. For the Csimir ppliction =0,b =, nd h = 1, so Eq.(17) becomes f(k) = 0 dxf(x)+ 1 f(0) 1 1 f (0) + 1 70 f (0) 1 30, 40 f (0) +.... (18) On the problem set the sum on the left strted t 1 insted of 0, so the coefficient of f(0) on the right ws 1 insted of 1.(Note tht the coefficient of the f(0) term ws printed incorrectly in Hung s book nd in 8.33 problem sets from pst yers.) Before leving the subject, however, one might wnt to estblish the connection between Eq.(17) nd the usul expression of the Euler-Mclurin formul in terms of Bernoulli numbers, nd one might wish to find clener wy to express the finl term of Eq.(17), often clled the reminder term. One normlly does not evlute this term, but one wnts to use it to rgue tht the reminder is smll. 5. Connection to the Bernoulli numbers: The Euler-Mclurin summtion formul is stted in Abrmowitz nd Stegun,* herefter clled A&S, s follows: * M.Abrmowitz nd I.A.Stegun, Hndbook of Mthemticl Functions, With Formuls, Grphs, nd Mthemticl Tbles, p.806.
EULER-MACLAURIN SUM FORMULA, 8.33, SPRING 003 p. 7 Let F (x)hveitsfirstn derivtives continuous on n intervl (, b).divide the intervl into m equl prts nd let h =(b )/m.then for some θ,1>θ>0, depending on F (n) (x) on(, b), we hve m F ( + kh) = 1 h b + F (t)dt + 1 {F (b)+f()} n k=1 h k (k)! B k{f (k) (b) F (k) ()} + hn m (n)! B n F (n) ( + kh + θh). (19) Disregrding for now the reminder term (the 3rd term on the right), Eq.(19) grees with Eq.(17) provided tht B k (k)! = V k(1). (0) k According to A&S, p.804, the Bernoulli numbers B n re defined in terms of the Bernoulli polynomils B n (x), which re defined by the generting function te xt e t 1 = B n (x) tn n! (for t<π).(1) The Bernoulli numbers re given by B n = B n (0). () From these reltions one cn esily (i.e., esily with the help of computer lgebr progrm) clculte the lowest Bernoulli polynomils:
EULER-MACLAURIN SUM FORMULA, 8.33, SPRING 003 p. 8 n B n (x) B n = B n (0) 0 B 0 (x) =1 1 1 B 1 (x) =x 1 B (x) =x x + 1 6 1 1 6 3 B 3 (x) =x 3 3 x + 1 x 0 (3) 4 B 4 (x) =x 4 x 3 + x 1 30 1 30 5 B 5 (x) =x 5 5 x4 + 5 3 x3 1 6 x 0 6 B 6 (x) =x 6 3x 5 + 5 x4 1 x + 1 4 1 4 By compring Eqs.(3) with Eqs.(13), one cn conjecture the equlity V n (x) =Ṽ n(x), (4) where Ṽ n (x) n n! B n ( ) x +1. (5) To prove this equlity, it is sufficient to verify tht Ṽ n(x) stisfies the reltions (1), since these were the reltions tht defined V n (x).it is strightforwrd to determine the properties required for B n (x) sothtṽ n(x) obeys ech of the reltions (1): B 0 (x) =1, B n (x) =n dxb n (x), nd (6) (6b) 0 dxb n (x) =0 (forn>0). (6c)
EULER-MACLAURIN SUM FORMULA, 8.33, SPRING 003 p. 9 Eq.(6) ws lredy written within Eqs.(3). To verify Eq.(6b), differentite the generting eqution (1) with respect to x: t e xt e t 1 = db n (x) dx t n n!. (7) One cn obtin nother expnsion of the sme quntity by multiplying the generting eqution by t: t e xt e t 1 = B n (x) tn+1 n! (8) t n = B n (n 1)!. n=1 Compring like powers of t in expnsions (7) nd (8), one finds db n dx = nb n(x), (9) which is equivlent to Eq.(6b). Finlly, to verify Eq.(6c), one cn integrte the generting eqution over x from 0 to 1: 0 dx text e t 1 =1= t n n! 0 dxb n (x). (30) Agin by compring powers of t, one verifies Eq.(6c). Thus Ṽ n(x) obeysllofthe reltions (1), nd hence V n (x) =Ṽ n(x). Hving estblished Eq.(4), it follows immeditely tht V k (1) = V k () = k (k)! B k(0), (31) which is just wht is needed to verify Eq.(0), nd hence the greement of our series expnsion with tht of A&S. Appendix: Simplifiction of the reminder term: The reminder term is the finl term on the right-hnd side of Eq.(17), given by R = 1 ( ) n h +h ( ) m dxv n h (x ) 1 f n (x + kh). (3)
EULER-MACLAURIN SUM FORMULA, 8.33, SPRING 003 p. 10 The gol of this ppendix is to mnipulte the reminder into the form given by A&S, s shown in Eq.(19) of this document. I m including this ppendix for completeness, but it is not physiclly importnt nd you need not red it if you re not curious. First, notice tht our definition of n is different from A&S s, since the sum on the right-hnd side of our Eq.(17) extends up to n, while the sum in A&S s eqution (Eq.(19)) extends only up to n.thus, if our reminder term is to gree with A&S s, it should be expressed in terms of f n+,notf n s in Eq.(3). To ccomplish this chnge, we will integrte by prts twice.the surfce term vnishes for the first integrtion by prts, since V n+1 hs n odd subscript nd therefore vnishes t the endpoints ccording to Eq.(11). For the second integrtion by prts there is surfce term which must be kept.in detil, the two integrtions by prts yield R = 1 = 1 ( ) n h +h ( h 1 ) n+1 m V n+(1) ( ) n+1 h +h ( ) m dxv n+1 h (x ) 1 f n+1 (x + kh) f n+1( +(k +1)h ) f n+1 ( + kh) ( ) m dxv n+ h (x ) 1 ] f n+ (x + kh). (33) (33b) Now notice tht the first term on the right-hnd side of Eq.(33b) cn be rewritten using f n+1( +(k +1)h ) f n+1 ( + kh) = +h dxf n+ (x + kh), (34) which llows one to combine the two terms: R = 1 ( ) n+1 h +h V n+(1) dxw(x) G(x), (35) where nd w(x) =1 V ( n+ (x ) 1) h V n+ (1) G(x) = m (36) f n+ (x + kh). (37) Using Eqs.(4), (5) nd (31) to rewrite this expression in terms of B n+, one finds R = h n+1 B +h n+ dxw(x) G(x), (38) (n +)!
EULER-MACLAURIN SUM FORMULA, 8.33, SPRING 003 p. 11 where w(x) =1 B ( x ) n+ h. (39) B n+ To complete the rgument, one needs to use the fct tht B n > B n (x) (for n =1,,...,nd0<x<1), (40) which we will prove below.this implies tht the second term in Eq.(39) hs mgnitude less thn 1, nd hence w(x) > 0. (41) Furthermore, from Eq.(6c) the second term in Eq.(39) vnishes when integrted over x from to + h, so +h dxw(x) =h. (4) Eqs.(41) nd (4) imply tht we cn interpret w(x) s weight fctor in the computtion of weighted verge, with 1 +h G(x) dxw(x) G(x). (43) h If we ssume tht every term on the right-hnd side of Eq.(37) is continuous, then G(x) is continuous, nd we cn pply the men vlue theorem* to conclude tht, somewhere in the rnge of integrtion ( <x<+h), G(x) mustbeequltoitsmenvlue G(x). Thusthereexistssomenumberθ in the rnge 0 <θ<1 such tht G( + θh) = G(x) = 1 h +h dxw(x) G(x). (44) Using the bove reltion to replce the integrl in Eq.(38), one hs finlly R = h n+ B n+ G( + θh) (n +)! = h n+ B m n+ f n+ ( + kh + θh), (n +)! (45) which mtches the reminder term in A&S s eqution (Eq.(19)). * See, for exmple, Methods of Mthemticl Physics, Third Edition, by Sir Hrold Jeffreys nd Berth Swirles (Ldy Jeffreys), Cmbridge University Press, 196, p.50, section 1.13.
EULER-MACLAURIN SUM FORMULA, 8.33, SPRING 003 p. 1 We hve reched the end, but to complete the proof we must justify the inequlity (40).This inequlity ws necessry to ssure the positivity of w(x), which in turn ws necessry for the men vlue theorem. The only proof tht I could construct for this inequlity depends on showing tht the generl shpe of the Bernoulli polynomils B n (x) in the intervl (0, 1), for n 3, is lwys one of four possibilities, depending on n mod 4.Smple grphs illustrting these shpes re shown on the following pge.specificlly, n mod 4 = 0: B n (x) is symmetric bout x = 1.The mximum is t x = 1,where B n > 0, nd the minimum is t x =0ndx =1,whereB n < 0.The slope is negtive for 1 <x<1, nd vnishes t the endpoints of this region. n mod 4 = 1: B n (x) is ntisymmetric bout x = 1, nd vnishes t x =0nd x = 1.Between x = 1 nd x = 1 the function rises monotoniclly to mximum nd then flls monotoniclly. n mod 4 = : B n (x) is symmetric bout x = 1.The minimum is t x = 1,where B n < 0, nd the mximum is t x =0ndx =1,whereB n > 0.The slope is positive for 1 <x<1, nd vnishes t the endpoints of this region. n mod 4 = 3: B n (x) is ntisymmetric bout x = 1, nd vnishes t x =0nd x = 1.Between x = 1 nd x = 1 the function flls monotoniclly to minimum nd then rises monotoniclly. Note tht we hve lredy shown (by Eqs.(11), (4), nd (5)) tht B n (x) vnishest0 nd 1 for n odd.the remining properties listed bove cn be shown by induction: one verifies tht B 3 (x) is being correctly described, nd then one uses Eqs.(6b) nd (6c) to show tht the properties for ech vlue of n mod 4 imply the properties for (n +1) mod 4. (Note tht grphiclly it ppers tht the zeros of B 6 (x)ndb 10 (x) coincide, but this is not exctly true.the zeros of B 6 (x), B 10 (x), nd B 14 (x) lie t 0.475407, 0.498447, nd 0.499903, respectively.) From the description bove for B n (x) whenn is even, one cn see tht the mximum bsolute vlue of the function in the rnge 0 <x<1 must occur either t the endpoints or t x = 1.We cn determine which of these two it is by using the generting function (1) to derive n identity tht reltes them. Consider the generting eqution (1) for x = 0, but with t replced by t/: ( t n B n (0) ) n! = 1 t e (t/) 1. (46)
EULER-MACLAURIN SUM FORMULA, 8.33, SPRING 003 p. 13
EULER-MACLAURIN SUM FORMULA, 8.33, SPRING 003 p. 14 By mnipulting the right-hnd side, it cn be re-expressed in terms of Bernoulli functions by using the generting eqution: ( t n B n (0) ) n! = 1 t e (t/) 1 e(t/) +1 e (t/) +1 = 1 te (t/) + t e t 1 { = 1 ( B 1 ) t n n n! + } B n (0) tn n!. (47) Compring like powers of t on both sides of the eqution, one finds B n (0) = 1 { ( 1 ) Bn n + Bn (0) }, (48) which implies tht ( B 1 n ) = Bn (0) 1 1 ] n. (49) It follows tht B n (0) > ( 1 Bn, ) nd therefore Eq.(40) holds.our proof excluded the specil cse B (x), which differs from the cses of lrger n in tht its derivtive does not vnish t 0 nd 1.It cn esily be verified, however, tht the mximum bsolute vlue of B (x) for0<x<1 must occur t x = 1 or t the endpoints, so Eq.(49) cn gin be used to show tht Eq.(40) pplies in this cse s well.