Pi, , and Asymptotic Expansions Author(s): J. M. Borwein, P. B. Borwein and K. Dilcher Source: The American Mathematical Monthly, Vol. 96, No. 8 (Oct., 1989), pp. 681-687 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/2324715 Accessed: 09-03-2015 18:09 UTC

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J.M. BORWEIN,P. B. BORWEIN, and K. DILCHER1,Dalhousie University,Halifax, Canada

JONATHANM. BORWEINwas an OntarioRhodes Scholar (1971) at Jesus College,Oxford, where he completeda D. Phil.(1974) with Michael Demp- ster.Since 1974 he hasworked at DalhousieUniversity where he is professor of .He has also beenon facultyat Carnegie-MellonUniversity (1980-82).He was the1987 Coxeter-James lecturer of theCanadian Mathe- maticalSociety and was awardedthe Atlantic Provinces Council on the Sciences1988 Gold Medal for Research.His researchinterests include functionalanalysis, classical analysis, and optimizationtheory.

PETER B. BORWEINobtained a Ph.D. (1979) fromthe Universityof BritishColumbia, under the supervision ofDavid Boyd. He spent1979-80 as 9 a NATO researchfellow in Oxford.Since then he has been on facultyat Dalhousie(except for a sabbaticalyear at theUniversity of Toronto)and is now AssociateProfessor of Mathematics.His researchinterests include approximationtheory, classical analysis, and complexitytheory.

KARuLDILCHER receivedhis undergraduate education and a Dipl. Math. degreeat TechnischeUniversitat Clausthal in WestGermany. He completed his Ph.D (1983) at Queen'sUniversity in Kingston,Ontario with Paulo Ribenboim.Since 1984 he has beenteaching at Dalhousie,where he is now AssistantProfessor. His researchinterests include Bernoulli numbers and polynomials,and classicalcomplex analysis.

1. Introduction.Gregory's series for 'T, truncatedat 500,000terms, gives to forty places

500,000 (1)k-1 4 E 2k 1 = 3.141590653589793240462643383269502884197. k=1

The numberon theright is not ir to fortyplaces. As one wouldexpect, the 6th digitafter the decimalpoint is wrong.The surpriseis thatthe next 10 digitsare correct.In fact,only the 4 underlineddigits aren't correct. This intriguing observa- tionwas sentto us by R. D. North[10] of ColoradoSprings with a requestfor an explanation.The pointof thisarticle is to providethat explanation. Two related

'Researchof theauthors supported in partby NSERC ofCanada.

681

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examples,to fiftydigits, are

IT 50,000 (W)k-1 2 k-i 2k - 1

- 1.5707863267948976192313211916397520520985833147388

1 -1 5 - 61 and 50,000 (1) k+1 log2- = k k=1 = .69313718065994530939723212147417656804830013446572,

1 -1 2 - 16 272 whereall but theunderlined digits are correct.The numbersunder the underlined digitsare thenumbers that must be addedto correctthese. The numbers1, - 1, 5, - 61 are the firstfour Euler numbers while 1, - 1, 2, - 16, 272 are thefirst five tangentnumbers. Our processof discoveryconsisted of generatingthese and thenidentifying them with the aid of Sloane'sHandbook of Sequences [11].What one is observing,in eachcase, is an asymptoticexpansion of theerror in Eulersummation. The amusingdetail is thatthe coefficients of theexpansion are .All of thisis explainedby Theorem 1. The standardfacts we needabout the Euler numbers { E1}, thetangent numbers { T }, and theBernoulli numbers { B1 }, mayall be foundin [1]or in [6].The numbers are definedas thecoefficients of thepower series

?? n E2n 2n secz = E (-1) (2)(1.1) 2+ (0n+ 2n)! tanz = E ( 1) 2 and To 1, (1.2) n=O ~ (2 n+1)! ad01 12 z 0BE Znn z ~~~~~~~~~~~~~(1.3) e 1 n= n! Theysatisfy the relations

= E (2kE2k 0, E2n+1 = 0, (1.4)

Bn= 2n(2n"-1) n>1 15 and

af (k ?)Bk = (1.6)

These threeidentities allow for the easy generation of { En}, {Tn}, and {Bn}. The

This content downloaded from 158.135.191.86 on Mon, 09 Mar 2015 18:09:19 UTC All use subject to JSTOR Terms and Conditions 1989] PI, EULER NUMBERS, AND ASYMPTOTIC EXPANSIONS 683 firstfew values are recordedbelow.

n 0 1 2 3 4 5 6 7 8 n~ 1 0 - 00 -61013 165 T7 - 1 0 2 -16 0 272 0

Bn, 1 0 0

It is clearfrom (1.4) thatthe Euler numbers are integral.From (1.5) and (1.6) it followsthat the tangent numbers are integers. Also,

2(2n)! _E_n_12n__- ad_ I2nI .72n?1 ad I2nl ( v2 as followsfrom (5.1) and(5.2) below.The maincontent of thisnote is thefollowing theorem.The simpleproof we offerrelies on theBoole SummationFormula, which is a prettybut less well-knownanalogue of Euler summation.The detailsare containedin Sections2 and3 (exceptfor c] whichis a straightforwardapplication of Eulersummation). More complicated developments can be baseddirectly on Euler summationor on resultsin [9]. THEOREM1. Thefollowing asymptotic expansions hold: - a] 2~ E_ 2m?1 2 k-i 2k -i =0 N 1 1 5 61 -N NT3 VN5 N7 k1 N/2 (_1) 1 co Trn-i k m=1 b] log2- Y, - + 2 k=1 k N =N 1 1 2 16 272 -N N2 N4 N N and

m=O c] k 22N2 + N2m+1 6kF' k=1 1 1 N 2N 6N3 30N' 42N7 Fromthe asymptotics of { E~,} and { Bn} and (1.5) we see thateach of theabove infiniteseries is everywheredivergent; the correct interpretation oftheir asymptotics is 00E K E ((2K + 1)!

a'] ~~m=1N2 - rn- N2 (E7rN)2K 00 T_ K T_ ((2K + 1)!

2m b'] L~M=N m= + 0 (ElTN)2K-1

c'] NB2- K B2 (2K + f~0 m1 1)!\ m=1 N m_1 N2m1 + (2u,gN)2K?1l

This content downloaded from 158.135.191.86 on Mon, 09 Mar 2015 18:09:19 UTC All use subject to JSTOR Terms and Conditions 684 J. M. BORWEIN, P. B. BORWEIN, AND K. DILCHER [October wherein each case theconstant concealed by theorder symbol is independentof N and K. In fact,the constant 10 worksin all cases.

2. The Boole SummationFormula. The Euler polynomialsEn(x) can be defined by thegenerating function

2etx = t n - et = En(X))- (Itin < T); (2.1) (see [1,p. 804]).Each En(x) is a polynomialof degreen withleading coefficient 1. We also definethe periodic En(x) by

En(X + 1) = -En(X) forall x, and En(x) = En(x) for 0 < x < 1. It can be shownthat En(x) has continuousderivatives up to the(n - I)st order. The following is known as Boole's summationformula (see, for example, [9, p. 34]). LEMMA 1. Let f(t) be a functionwith m continuousderivatives, defined on the intervalx < t < x + w. Thenfor 0 < h < 1

m-1 W;k 1 f (x + hw) = -Ek(h) - - (f (k)(X + W) + f (k)(X)) + Rm, k=OklOk 2 where

RmI= Mj (E (h t) (m)(x + wt)dt

This summationformula is easyto establishby repeatedintegration by partsof theabove integral. It is remarkedin [9,p. 26] thatthis formula was knownto Euler, forpolynomial f and withoutthe remainder term. Also notethat Lemma 1 turns intoTaylor's formula with Lagrange's remainder term if we replaceh by h/l and let coapproach zero. To derivea convenientversion of Lemma 1 forthe applications we havein mind, we set w = 1 and imposefurther restrictions on f.

LEMMA 2. Let f be a functionwith m continuousderivatives, defined on t > x. Suppose thatf (k)(t) 0 as t oofor all k = 0,1, ..., m. Then for 0 < h < 1

(1)f(x + h + v) rn-Ek(h)f(k)(X) +2R v=O k=O where

R m= _| (h )(M)(xf + t)dt. 2 (M - ~1)!

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3. The Remainderfor Gregory'sSeries. The Euler numbers En may also be definedby thegenerating function 2 _ 0 tn et + et n= n!(31 Comparing(3.1) with(2.1), we see that

En= 2nEn( . (3.2)

The phenomenonmentioned in theintroduction is entirelyexplained by thenext proposition-ifwe set n = 500,000.It is also clearthat we willget similar patterns forn = 10/2 withany positive integer m. PROPOSITION 1. For positive integersn and M we have

oo (_) k M 2E2 4 - . - + R ~~~~~~~(3.3) k= 2k + 1 k=O(_I)n E (2n )2k?l + (M), where I %(M) ~ 21E2M I PI ___z I (2n)2M?l

Proof. Apply Lemma 2 withf(x) = l/x; thenset x = n and h = 1/2. We get 00 (1)v rn-i E1(1/2) (_l) kk!+R

v=0 n + v + 1/2 k-O 2k! nk?l m (3.4) with I ooEmi_(h - t) (-1) m!dt Rm dt. m= 2 JO (mr-1)! (x + t)m+l We multiplyboth sides of (3.4) by 2(-1)n. Thenthe left-hand side is seen to be identicalwith the left-hand side of (3.3). Afterreplacing m by 2M + 1 and taking intoaccount (3.2) and thefact that odd-index Euler numbers vanish, we see thatthe firstterms on theright-hand sides of (3.3) and (3.4) agree.To estimatethe error term,we use thefollowing inequality,

IE2M(x) I < 2 2MIE2MI forO < x < 1 (see, e.g.,[1, p. 805]).Carrying out theintegration now leads to theerror estimate givenin Proposition1. 0 4. An AnalogueFor log 2. Lemma2 can also be usedto derivea resultsimilar to Proposition1, concerningtruncations of theseries

( 1)~ ~ ~ ~~~~(41 log2= E . (4.1) k=1 k In thiscase thetangent numbers Tn will play the role of theEn in Proposition1. It

This content downloaded from 158.135.191.86 on Mon, 09 Mar 2015 18:09:19 UTC All use subject to JSTOR Terms and Conditions 686 J.M. BORWEIN,P. B. BORWEIN,AND K. DILCHER [October followsfrom the identity 1 2e21z tanz=Tez2iz+1 1) togetherwith (1.2) and (2.1) that Tn-(-1) 2nEn(1) (4.2) as in [9, p. 28]. The Tncan be computedusing the recurrencerelation To = 1 and

k)2 T-k + Tn= 0 forn > 1. k=

Other propertiescan be found,e.g., in [81or [9, Ch. 21. PROPOSITION 2. For positiveintegers n and M we have

= n(1 2 + E (22k ) + R2(M), (4.3) E ( (-1) 2 k=n+l1 k==1 (2n) k) where

JR J~E2MI AR2(M)I< (2n)2M+l

Proof. We proceed as in the proof of Proposition1. Here we take x = n and h = 1. Using (4.2) and the fact that To = 1 and T2k = 0 for k > 1, we get the summationon the right-handside of (4.3). The remainderterm is estimatedas in the proofof Proposition1. 0 Using Proposition2 withn = 10m/2one again gets manymore correct digits of log 2 than is suggestedby the errorterm of the .

5. Generalizations.Proposition 1 and 2 can be extendedeasily in two different directions. i). The well-knowninfinite series (see, e.g., [1, p. 8071)

kO (_)k+ ) = (2nI s.2n+1 (n = 0,1, **.), (5.1) k=O (2k + 1 2) and

oo __2_ = (1 -21-2n)(2n) k=1

= (22n1 - 1) IB2n7I 2n (n = 1,2, ) (5.2) can be consideredas extensionsof Gregory'sseries and of (4.1). These seriesadmit exact analogues to Propositions1 and 2; one only has to replace f(x) = l/x by f(x) = x- (2n+ ) respectivelyx-(2n), in theproofs.

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We notethat the Euler-MacLaurin summation formula leads to similarresults for

00 IB2n 122n- 1 k=1 (2n)! ' (5) wheremultiples of theBernoulli numbers B2n takethe place of theEn and Tnin Propositions1 and 2. ii). A generalizationof theEuler-MacLaurin and Boole summationformulas was derivedby Berndt[3]. This can be appliedto characteranalogues of the series (5.1)-(5.3). The rolesof theEn and Tnin Proposition1 and 2 are thenplayed by generalizedBernoulli numbers or by relatednumbers. 6. AdditionalComments. The phenomenonobserved in theintroduction results fromtaking N to be a powerof ten; takingN = 2 * io0 also leads to "clean" expressions.References [1], [5], [6], and [9] includethe basic materialon Bernoulli and Euler numbers,while [8] deals extensivelywith their calculation, and [2] describesan entertaininganalogue of Pascal'striangle. Much on thecalculation of pi and relatedmatters may be foundin [4]. Eulersummation is treatedin [5],[6], and [9],while Boole summationis treatedin [9]. Relatedmaterial on thecomputa- tionand accelerationof alternatingseries is givenin [7].

Added in Proof. A versionof thephenomeon was observedby M. R. Powell and variousexplanations were offered (see The MathematicalGazette, 66 (1982) 220-221,and 67(1983)171-188).

REFERENCES 1. M. Abramowitzand I. Stegun,Handbook of MathematicalFunctions, Dover, N.Y., 1964. 2. M. D. Atkinson,How to computethe series expansions of sec x and tanx, Amer.Math. Monthly, 93 (1986)387-388. 3. B. C. Bemdt,Character analogues of thePoisson and Euler-MacLaurinsummation formulas with applications,J. NumberTheory, 7 (1975 413-445. 4. J. M. Borweinand P. B. Borwein,Pi and theAGM - A Studyin AnalyticNumber Theory and ComputationalComplexity, Wiley, N.Y., 1987. 5. T. J.I'a Bromwich,An Introductiontothe Theory of InfiniteSeries, 2nd ed., MacMillan, London, 1926. 6. R. L. Graham,D. E. Knuth,and 0. Patashnik,Concrete Mathematics, Addison Wesley, Reading, Mass.,1989. 7. R. Johnsonbaugh,Summing an alternatingseries, this MONTHLY, 86 (1979)637-648. 8. D. E. Knuthand T. J.Buckholtz, Computation of Tangent,Euler, and Bernoullinumbers, Math. Comput.,21 (1967)663-688. 9. N. Norlund, Vorlesungeniuber Differenzenrechnung, Springer-Verlag, Berlin, 1924. 10. R. D. North,personal communications, 1988. 11. N. J.A. Sloane,A Handbookof Integer Sequences, Academic Press, New York, 1973.

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