Bernoulli numbers and the Euler-Maclaurin summation formula

Transcription

1 Physics 6A Witer 006 Beroulli umbers ad the Euler-Maclauri summatio formula I this ote, I shall motivate the origi of the Euler-Maclauri summatio formula. I will also explai why the coefficiets o the right had side of this formula ivolve the Beroulli umbers. First, we defie the Beroulli umbers B. These arise i the Taylor series expasios of x coth(x) ad x cot(x) about x = 0. It is coveiet to defie the ormalizatio of the Beroulli umbers via the Taylor expasio of (x/) coth(x/) as follows: x ( x ) coth = =0 B x, x < π. () ()! This formula oly defies Beroulli umbers with eve o-egative idices. The more commo defiitio is based o the observatio that x ( x ) ] coth = x () e x is a idetity. The, x e x = x B, x < π.! =0 defies all the Beroulli umbers with o-egative idices. Comparig the above formulae, it follows that B = ad B + = 0 for =,, 3,.... For the remaider of this ote, we will oly be cocered with Beroulli umbers of the form B, for o-egative itegers. For the record, we list the first six B here: B 0 =, B = 6, B 4 = 30, B 6 = 4, B 8 = 30, B 0 = 5 66, etc. I geeral, the sigs alterate begiig with B, so that B = ( ) + B, for =,, 3,.... As x 0, both coth(x) ad cot(x) behave as /x; hece we multiply by x i order to have a fuctio with a fiite limit as x 0. To prove this, recall that coth(x) = cosh(x)/ sih(x) where cosh(x) (ex + e x ) ad sih(x) (ex e x ).

2 Give the Taylor series for x coth(x) expaded about x = 0], oe ca immediately obtai the Taylor series for x cot x by usig x cot(x) = ix coth(ix). By usig the followig hyperbolic ad trigoometric idetities: tah(x) = coth(x) coth(x), ta(x) = cot(x) cot(x), csch(x) = coth(x) coth(x), csc(x) = cot(x) cot(x), oe fids that the Taylor expasios about x = 0 for cot(x), ta(x), csc(x) ad the correspodig hyperbolic fuctios all ivolve the Beroulli umbers. Note that there are o similar idetities for sech(x) ad sec(x) i terms of coth(x) ad cot(x), respectively. Hece, the Taylor expasios about x = 0 of these two fuctios do ot ivolve the Beroulli umbers. Oe property of the Beroulli umbers will be importat i what follows. We will eed to ow the behavior of the B as becomes very large. We ca determie this by usig the famous coectio betwee the Beroulli umbers ad the Riema zeta fuctio: B = ()! ζ() (π), (3) where ζ() =0 (/() ). Note that lim ζ() =, sice i this limit oly the first term i the series (which is equal to oe) survives. Hece, usig this result i eq. (3) ad employig Stirlig s approximatio for ()!, we ed up with ()! (4π) / () e, as, B 4(π) / ( πe), as. (4) With this bacgroud, we are ow ready to itroduce the Euler-Maclauri summatio formula. This formula arises i the followig cotext. Suppose we wish to umerically approximate a itegral of the form: I + f(x)dx, where is a iteger ad is a positive iteger. The simplest possible approximatio to the itegral correspods to dividig up the iterval x + i uits of oe, ad estimatig the value of the itegral by computig the area of all the rectagles of uit legth that approximate the area uder the curve. We illustrate this procedure with the followig graph at the top of the ext page. Istead, they require the itroductio of a ew type of umbers called the Euler umbers. The explicit defiitio of the Euler umbers was give at the ed of the hadout o the Riema zeta fuctio. We will ot eed the Euler umbers i this ote. This presetatio is ispired by Jo Mathews ad R.L. Waler, Mathematical Methods of Physics, Chapter 3.

3 Here, oe ca either compute the area bouded by the rectagles idicated by the solid lies or by the rectagles idicated by the dashed lies. (I this particular example, = ad = 5.) The former uderestimates the area uder the curve y = f(x), while the latter overestimates this area. That is, I < I < I where I = f( + ) + f( + ) + + f( + ), I = f() + f( + ) + + f( + ). (5) The trapezoidal rule for umerical itegratio taes the average of I ad I. So, we shall mae the approximatio I = (I + I ), which we ca write as: + f(x) dx f() + f( + )] + f( + j). The Euler-Maclauri sum formula arises whe we attempt to covert the above result ito a exact formula. That is, we see to determie a expressio, R, such that: + f(x) dx = f() + f( + )] + j= f( + m) + R. (6) I will show you a sophisticated, yet simple, method for determiig R. You should be forewared that this method is slic ad will gloss over some subtleties that I will metio later. The tric is to itroduce two operators called D ad E. These operators act o a fuctio f(x) ad have very simple defiitios: Df() f (), Ef() f( + ), (7) By act o, I mea that D ad E operate o fuctios. You ca thi of D ad E as little machies. You feed these machies a fuctio ad they will spit out a ew fuctio. 3

4 where as usual, f () (df/dx) x=. With this otatio, eq. (6) reads: + f(x) dx = + E + E + + E + E ]f() + R, sice, e.g., E f() = E Ef() = Ef( + ) = f( + ), etc. Now, for the slic part. We shall write: + E + E + + E + E = + E + E + + E ( E ) = E E ( E ) = (E ) + ], (8) E where we have summed a fiite geometric series i the usual way. But, E is a operator, so what does ( E) mea? The aswer is that we are actually usig a short-had otatio. Whe i doubt, a fuctio of a operator is always defied by its Taylor series. For example, i eq. (8), ( E) = + E + E +. This is a ifiite series, so we should really worry about covergece (what does it mea whe you have a ifiite coverget series of operators rather tha umbers?). For the momet, I will treat these power series expasios as formal objects, ad postpoe questios of covergece util later. So, if you are willig to go alog with this strategy, the we have the followig result: + f(x) dx = (E ) + ] f() + R. (9) E Our ext step is to cosider the Taylor expasio of f(x) about x = : f(x) = f() + f (m) () m! (x ) m. Agai, we should chec for which values of x this series coverges, but we will sidestep this issue agai. If we set x = + i the above expasio, we fid: f( + ) = m=0 f (m) () m! Thus, we ca use our operators D ad E to rewrite this as Ef() = m=0 D m m! f().. 4

5 Note that this formula would be true for ay fuctio f, so we ca coclude that we have a operator idetity: E = m=0 D m m!. (0) I hope you recogize the sum o the right had side of eq. (0). This is the power series expasio of e D. Thus, we coclude that E = e D. () Oce agai, we have itroduced a strage ew object the expoetial of a operator. As before, this is a formal defiitio, ad you should always thi of e D as beig equal to its Taylor series expasio eq. (0)]. The fial step of our aalysis itroduces the idefiite itegral of f(x). Let us call it g(x): g(x) = f(x) dx, or equivaletly f(x) = dg(x) dx. I particular, g () = Dg() = f(). The fudametal theorem of calculus the allows us to write: + f(x) dx = g( + ) g() = (E )g(). Now for the bold move. Sice Dg() = f(), we shall write: This will allow us to write + g() = D f(). f(x) dx = (E ) f(). () D We ow have two differet expressios for + f(x) dx give by eqs. (9) ad (). Sice oly oe of these expressios ivolves R, this meas that we ca ow solve for R. Settig eqs. (9) ad () equal to each other, we obtai: R = (E ) D ] f() E = (E ) D D ( + E )] f(). At this poit, we shall substitute E = e D eq. ()] iside the bracets to obtai R = (E ) ( D D + )] f(). e D 5

6 Usig the idetity give by eq. (), we ca write this last result i a very suggestive way: R = (E ) D D coth D ] f(). Oce agai, we have a fuctio of a operator, which we are istructed to iterpret as a power series. It is at this poit that the Beroulli umbers eter. Usig eq. (), we ca write: D D coth D ] = B m D m (m)!. (3) Notice that at this poit, we oly have o-egative powers of the operator D o the right had side of eq. (3), which we ca easily hadle. Thus, we coclude that: ] R = ( E D m ) B m f(). (m)! We ca write out this expressio more explicitly by usig the defiitios of the operators D ad E eq. (7)]: R = B m f (m ) ( + ) f (m ) () ]. (m)! Isertig this result bac ito eq. (6) yields the followig remarable formula: + f(x) dx = f(+m)+ f()+f(+)] B m f (m ) ( + ) f (m ) () ]. (m)! Notice that this is a exact result. Somehow, we have maaged to tur a formula that started out as a approximatio to a itegral ito a exact result. The fiite sum m f(+m) is also a iterestig object, ad we ca reiterpret the above result as providig a formula for this fiite sum. If we write: f( + m) + f() + f( + )] = f( + m) f() + f( + )], the we ed up with the Euler-Maclauri summatio formula: f( + m) = + f(x) dx + f() + f( + )] + B m f (m ) ( + ) f (m ) () ]. (m)! (4) 6

7 It is ow time to face up to the questio of covergece. The Euler-Maclauri summatio formula as preseted here ivolves a ifiite sum. Give the behavior of the Beroulli umbers B m as m see eq. (4)], it is ot surprisig to lear that i most cases of iterest this is a diverget series. Our derivatio has bee too slic, i that it igored questios of covergece. I fact, oe ca be more careful by replacig all ifiite sums ecoutered above by fiite sums plus remaider terms. By carefully eepig trac of these remaider terms, oe ca obtai a more robust versio of the Euler-Maclauri summatio formula with a remaider term explicitly icluded. This derivatio is beyod the scope of these otes. You ca fid (the more covetioal) derivatio of the Euler-Maclauri summatio formula with remaider term i the textboo by Arfe ad Weber, Mathematical Methods for Physicists. For completeess, I shall display the fial result here: f( + m) = + + f(x) dx + f() + f( + )] p + (p)! B m f (m ) ( + ) f (m ) () ] (m)! 0 B p (x) m=0 f (p) (x + + m) dx, (5) where B p (x) is the Beroulli polyomial of order p (defied o p. 5 of Mc- Quarrie). I may applicatios, the Euler-Maclauri summatio formula provides a asymptotic expasio, i which case the diverget ature of the series i eq. (4) is ot problematical. I other cases, the ifiite sum turs out to be fiite. We shall ed this ote with a few applicatios. For our first example, we tae f(x) = x p ad = 0 i eq. (4). The ifiite sum o the right had side of eq. (4) is i fact fiite i this case, sice f (m ) (x) = 0 for m p +. Evaluatig the derivatives o the right had side of eq. (4), we ca cast the resultig formula ito the followig form: m p = p + p + p/] m=0 ( ) p + B m p+ m, m where p/] is the iteger part of p. For example, if p = the m = = ( + )( + ), 6 6 which reproduces a formula I derived i the first lecture. 7

8 Our secod applicatio is the derivatio of the asymptotic series for l! as. Here, we tae f(x) = l x, with = ad i eq. (4). The itegral i eq. (4) is easily computed: l x dx = l +, ad the derivatives f (m ) (x) are give by: f (m ) (m )! (x) =, m =,, 3,... x Notig that the summatio o the left had side of eq. (4) taes the followig form: we ca write eq. (4) as l( + m) = l + l + + l = l( 3 ) = l! l! = ( + ) l + C + B m m(m ), (6) m where the costat C represets all the remaiig terms of eq. (4) that are idepedet of : B m C = m(m ). (7) Ufortuately, due to the asymptotic behavior of B m as m eq. (4)], the sum i eq. (7) is diverget. However, this is ot surprisig sice we are usig the form of the Euler-Maclauri summatio formula without the remaider term. If we would have icluded the remaider term, the summatios o the right had sides of eqs. (6) ad (7) would have bee fiite sums. I additio, we would have icluded the -idepedet part of the remaider term i the defiitio of C above. I this case, the resultig expressio for C would have bee perfectly well-defied ad fiite. I fact, oe ca aalyze that resultig form for C ad evaluate this costat. However, this requires a umber of additioal trics that lie beyod the scope of these otes. Fially, the -depedet part of the remaider term would appear i eq. (6). By examiig its form c.f. eq. (5)], oe ca prove that the remaider term is of O(/ p ). This meas that eq. (6) is ideed a asymptotic expasio as. Here we shall tae the simpler approach. Namely, we shall simply assume that eq. (6) is a asymptotic expasio as. Comparig this result with Stirlig s approximatio, we coclude that C = l π. Thus, eq. (6) ow reads: l! ( + ) l + l π + 8 B m m(m ),. (8) m

9 This is called Stirlig s asymptotic series. Although the proof give here strictly applies oly for the case of positive iteger, there is a geeralizatio of this derivatio that ca yield the full asymptotic series for l Γ(x + ) for ay real umber x. Not surprisigly, the resultig asymptotic series is idetical to eq. (8) with replaced by x. For our third example, we choose f(x) = /x, with = ad i eq. (4). Agai, the itegral ad (m )-fold derivatives are easily computed: Thus, where dx x = l, f (m ) (m )! (x) =. x m m = l + C + C = + B m m. B m m, (9) m Oce agai, C appears to be diverget. However, as i the previous example, a more careful aalysis icludig the remaider term would produce a fiite expressio for C ad prove that eq. (9) is a asymptotic series. Thus, if we proceed uder this assumptio, we ca compute the costat C by taig the limit of eq. (9): C = lim ( ) m l = γ. (0) We recogize this limit as Euler s costat. Thus, we have derived the followig asymptotic series: m l + γ + B m m m,. By the way, we ca tur this equatio aroud ad use it for a accurate umerical computatio of γ. May other fiite series, summed from m = to, ca be expressed i the form of a asymptotic series as. I will leave it to you as a exercise to wor out the asymptotic series for: m p ζ(p) + p p + + O ( )],, where p > ad ζ(p) is the Riema zeta fuctio. The O(/) term above ca be expressed i terms of a asymptotic series with coefficiets proportioal to the Beroulli umbers, usig the same techiques employed above. 9

10 I our fial example, we choose f(x) = (l x)/x, with = ad. It follows that l m ( ) m l x l l dx + C + x + O C + l + l + O ( ) l, () where C represets the -idepedet pieces of eq. (4), ad the O(l / ) remaider correspods to a sum with Beroulli umber coefficiets (which we do ot write out explicitly here). Thus, C = lim ( l m m l This result ca be used to evaluate the sum S ( ) l. = ). () Cosider the sum S N made up of the first N terms of the above series. We ca write: = S N = N = = ( ) l N = l + N = N = = l l + N + l N = = l, (3) where we have used l() = l + l i obtaiig the secod lie of eq. (3). However, eqs. () ad () ad eqs. (9) ad (0) imply respectively that: N ( ) l N ( ) = C + l N + O, N = γ + l N + O. N Isertig these result ito eq. (3), the costat C drops out, ad we fid: ( ) S N = l (N) + l N + l (γ + l N) + O N = l ( ( ) γ l ) + O. N Taig the limit of N, we obtai the desired sum S = lim N S N : ( ) l = l ( γ l ). = This result (the = term does ot cotribute to the sum) was previously oted i the Riema zeta fuctio hadout. 0 =