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Johann Faulhaber and Sums of Powers

arXiv:math/9207222v1 [math.CA] 27 Jul 1992Johann Faulhaber and Sums of PowersDonald E. KnuthComputer Science Department, Stanford UniversityAbstract. Early 17th-century mathematical publications of Johann Faulhabercontain some remarkable theorems, such as the fact that the r-fold summationof 1m, 2m, .

. ., nm is a polynomial in n(n + r) when m is a positive odd number.The present paper explores a computation-based approach by which Faulhabermay well have discovered such results, and solves a 360-year-old riddle that Faul-haber presented to his readers.

It also shows that similar results hold when weexpress the sums in terms of central factorial powers instead of ordinary powers.Faulhaber’s coefficients can moreover be generalized to factorial powers of non-integer exponents, obtaining asymptotic series for 1α + 2α + · · · + nα in powersof n−1(n + 1)−1.1

Johann Faulhaber of Ulm (1580–1635), founder of a school for engineers early in the 17thcentury, loved numbers. His passion for arithmetic and algebra led him to devote a consid-erable portion of his life to the computation of formulas for the sums of powers, significantlyextending all previously known results.

He may well have carried out more computing thananybody else in Europe during the first half of the 17th century. His greatest mathematicalachievements appear in a booklet entitled Academia Algebræ (written in German in spiteof its latin title), published in Augsburg, 1631 [2].

Here we find, for example, the followingformulas for sums of odd powers:11 + 21 + · · · + n1 = N ,N = (n2 + n)/2 ;13 + 23 + · · · + n3 = N 2 ;15 + 25 + · · · + n5 = (4N 3 −N 2)/3 ;17 + 27 + · · · + n7 = (12N 4 −8N 3 + 2N 2)/6 ;19 + 29 + · · · + n9 = (16N 5 −20N 4 + 12N 3 −3N 2)/5 ;111 + 211 + · · · + n11 = (32N 6 −64N 5 + 68N 4 −40N 3 + 5N 2)/6 ;113 + 213 + · · · + n13 = (960N 7 −2800N 6 + 4592N 5 −4720N 4 + 2764N 3−691N 2)/105 ;115 + 215 + · · · + n15 = (192N 8 −768N 7 + 1792N 6 −2816N 5 + 2872N 4−1680N 3 + 420N 2)/12 ;117 + 217 + · · · + n17 = (1280N 9 −6720N 8 + 21120N 7 −46880N 6 + 72912N 5−74220N 4 + 43404N 3 −10851N 2)/45 .Other mathematicians had studied Σn1, Σn2, . .

., Σn7 and he had previously gotten as faras Σn12; but the sums had always previously been expressed as polynomials in n, not N.Faulhaber begins his book by simply stating these novel formulas and proceeding toexpand them into the corresponding polynomials in n. Then he verifies the results whenn = 4, N = 10. But he gives no clues about how he derived the expressions; he states onlythat the leading coefficient in Σn2m−1 will be 2m−1/m, and that the trailing coefficientswill have the form 4αmN 3 −αmN 2 when m ≥3.Faulhaber believed that similar polynomials in N, with alternating signs, would con-tinue to exist for all m, but he may not really have known how to prove such a theorem.

Inhis day, mathematics was treated like all other sciences; it was sufficient to present a largebody of evidence for an observed phenomenon. A rigorous proof of Faulhaber’s assertion2

was first published by Jacobi in 1834 [6]. A. W. F. Edwards showed recently how to obtainthe coefficients by matrix inversion [1], based on another proof given by L. Tits in 1923 [8].But none of these proofs use methods that are very close to those known in 1631.Faulhaber went on to consider sums of sums.

Let us write Σrnm for the r-fold sum-mation of mth powers from 1 to n; thus,Σ0nm = nm ;Σr+1nm = Σr1m + Σr2m + · · · + Σrnm .He discovered that Σrn2m can be written as a polynomial in the quantityNr = (n2 + rn)/2 ,times Σrn2. For example, he gave the formulasΣ2n4 = (4N2 −1) Σ2n2/5 ;Σ3n4 = (4N3 −1) Σ3n2/7 ;Σ4n4 = (6N4 −1) Σ4n2/14 ;Σ6n4 = (4N6 + 1) Σ6n2/15 ;Σ2n6 = (6N 22 −5N2 + 1) Σ2n2/7 ;Σ3n6 = (10N 23 −10N3 + 1) Σ3n2/21 ;Σ4n6 = (4N 24 −4N4 −1) Σ4n2/14 ;Σ2n8 = (16N 32 −28N 22 + 18N2 −3) Σ2n2/15 .He also gave similar formulas for odd exponents, factoring out Σrn1 instead of Σrn2:Σ2n5 = (8N 22 −2N2 −1)Σ2n1/14 ;Σ2n7 = (40N 32 −40N 22 + 6N2 + 6)Σ2n1/60 .And he claimed that, in general, Σrnm can be expressed as a polynomial in Nr times eitherΣrn2 or Σrn1, depending on whether m is even or odd.Faulhaber had probably verified this remarkable theorem in many cases includingΣ11n6, because he exhibited a polynomial in n for Σ11n6 that would have been quitedifficult to obtain by repeated summation.

His polynomial, which has the form6n17 + 561n16 + · · · + 1021675563656n5 + · · · −96598656000n2964061900800,turns out to be absolutely correct, according to calculations with a modern computer. (The denominator is 17!/120.

One cannot help thinking that nobody has ever checkedthese numbers since Faulhaber himself wrote them down, until today. )3

Did he, however, know how to prove his claim, in the sense that 20th century math-ematicians would regard his argument as conclusive? He may in fact have known how todo so, because there is an extremely simple way to verify the result using only methodsthat he would have found natural.Reflective functions.

Let us begin by studying an elementary property of integer func-tions. We will say that the function f(x) is r-reflective iff(x) = f(y)wheneverx + y + r = 0 ;and it is anti-r-reflective iff(x) = −f(y)wheneverx + y + r = 0 .The values of x, y, r will be assumed to be integers for simplicity.

When r = 0, reflectivefunctions are even, and anti-reflective functions are odd. Notice that r-reflective functionsare closed under addition and multiplication; the product of two anti-r-reflective functionsis r-reflective.Given a function f, we define its backward difference ∇f in the usual way:∇f(x) = f(x) −f(x −1) .It is now easy to verify a simple basic fact.Lemma 1.

If f is r-reflective then ∇f is anti-(r −1)-reflective. If f is anti-r-reflectivethen ∇f is (r −1)-reflective.Proof.

If x + y + (r −1) = 0 then x + (y −1) + r = 0 and (x −1) + y + r = 0. Thusf(x) = ±f(y −1) and f(x −1) = ±f(y) when f is r-reflective or anti-r-reflective.Faulhaber almost certainly knew this lemma, because [2, folio D.iii recto] presents atable of n8, ∇n8, .

. .

, ∇8n8 in which the reflection phenomenon is clearly apparent. Hestates that he has constructed “grosse Tafeln,” but that this example should be “allesgnugsam vor Augen sehen und auf h¨ohere quantiteten [exponents] continuiren k¨onde.”The converse of Lemma 1 is also true, if we are careful.

Let us define Σ as an inverseto the ∇operator:Σf(n) = C + f(1) + · · · + f(n) ,if n ≥0;C −f(0) −· · · −f(n + 1) ,if n < 0.Here C is an unspecified constant, which we will choose later; whatever its value, we have∇Σf(n) = Σf(n) −Σf(n −1) = f(n)for all n.4

Lemma 2. If f is r-reflective, there is a unique C such that Σf is anti-(r + 1)-reflective.If f is anti-r-reflective, then Σf is (r + 1)-reflective for all C.Proof.

If r is odd, Σf can be anti-(r + 1)-reflective only if C is chosen so that we haveΣf−(r + 1)/2= 0. If r is even, Σf can be anti-(r + 1)-reflective only if Σf(−r/2) =−Σf(−r/2 −1) = −Σf(−r/2) −f(−r/2); i.e., Σf(−r/2) = 12f(−r/2).Once we have found x and y such that x + y + r + 1 = 0 and Σf(x) = −Σf(y), itis easy to see that we will also have Σf(x −1) = −Σf(y + 1), if f is r-reflective, sinceΣf(x) −Σf(x −1) = f(x) = f(y + 1) = Σf(y + 1) −Σf(y).Suppose on the other hand that f is anti-r-reflective.

If r is odd, clearly Σf(x) = Σf(y)if x = y = −(r + 1)/2. If r is even, then f(−r/2) = 0; so Σf(x) = Σf(y) when x = −r/2and y = −r/2 −1.Once we have found x and y such that x + y + r + 1 = 0 andΣf(x) = Σf(y), it is easy to verify as above that Σf(x −1) = Σf(y + 1).Lemma 3.

If f is any even function with f(0) = 0, the r-fold repeated sum Σrf is r-reflective for all even r and anti-r-reflective for all odd r, if we choose the constant C = 0in each summation. If f is any odd function, the r-fold repeated sum Σrf is r-reflectivefor all odd r and anti-r-reflective for all even r, if we choose the constant C = 0 in eachsummation.Proof.

Note that f(0) = 0 if f is odd. If f(0) = 0 and if we always choose C = 0, it iseasy to verify by induction on r that Σrf(x) = 0 for −r ≤x ≤0.

Therefore the choiceC = 0 always agrees with the unique choice stipulated in the proof of Lemma 2, whenevera specific value of C is necessary in that lemma.When m is a positive integer, the function f(x) = xm obviously satisfies the conditionof Lemma 3. Therefore we have proved that each function Σrnm is either r-reflective oranti-r-reflective, for all r > 0 and m > 0.

And Faulhaber presumably knew this too.His theorem can now be proved if we supply one small additional fact, specializing fromarbitrary functions to polynomials:Lemma 4. A polynomial f(x) is r-reflective if and only if it can be written as a polynomialin x(x + r); it is anti-r-reflective if and only if it can be written as (x + r/2) times apolynomial in x(x + r).Proof.

The second statement follows from the first, because we have already observedthat an anti-r-reflective function must have f(−r/2) = 0 and because the function x +r/25

is obviously anti-r-reflective.Furthermore, any polynomial in x(x + r) is r-reflective,because x(x + r) = y(y + r) when x + y + r = 0. Conversely, if f(x) is r-reflective wehave f(x −r/2) = f(−x −r/2), so g(x) = f(x −r/2) is an even function of x; henceg(x) = h(x2) for some polynomial h. Then f(x) = g(x + r/2) = hx(x + r) + r2/4is apolynomial in x(x + r).Theorem (Faulhaber).

There exist polynomials gr,m for all positive integers r and msuch thatΣrn2m−1 = gr,2m+1n(n + r)Σrn1 ,Σrn2m = gr,2mn(n + r)Σrn2 .Proof. Lemma 3 tells us that Σrnm is r-reflective if m + r is even and anti-r-reflective ifm + r is odd.Note that Σrn1 =n+rr+1.

Therefore a polynomial in n is a multiple of Σrn1 if andonly if it vanishes at −r, . .

., −1, 0. We have shown in the proof of Lemma 3 that Σrnmhas this property for all m; therefore Σrnm/Σrn1 is an r-reflective polynomial when m isodd, an anti-r-reflective polynomial when m is even.

In the former case, we are done, byLemma 4. In the latter case, Lemma 4 establishes the existence of a polynomial g suchthat Σrnm/Σrn1 = (n + r/2)gn(n + r).

Again, we are done, because the identityΣrn2 = 2n + rr + 2 Σrn1is readily verified.A plausible derivation. Faulhaber probably didn’t think about r-reflective and anti-r-reflective functions in exactly the way we have described them, but his book [2] certainlyindicates that he was quite familiar with the territory encompassed by that theory.In fact, he could have found his formulas for power sums without knowing the theoryin detail.

A simple approach, illustrated here for Σn13, would suffice: Suppose14Σn13 = n7(n + 1)7 −S(n) ,where S(n) is a 1-reflective function to be determined. Then14n13 = n7(n + 1)7 −(n −1)7n7 −∇S(n)= 14n13 + 70n11 + 42n9 + 2n7 −∇S(n) ,6

and we haveS(n) = 70Σn11 + 42n9 + 2Σn7 .In other wordsΣn13 = 647 N 7 −5Σn11 −3Σn9 −17Σn7,and we can complete the calculation by subtracting multiples of previously computedresults.The great advantage of using polynomials in N rather than n is that the new formulasare considerably shorter. The method Faulhaber and others had used before making thisdiscovery was most likely equivalent to the laborious calculationΣn13 =114n14 + 132 Σn12 −26Σn11 + 1432 Σn10 −143Σn9 + 4292 Σn8+ 17167 Σn7 + 4292 Σn6 −143Σn5 + 1432 Σn4 −26Σn3 + 132 Σn2 −Σn1 + 114n ;the coefficients here are1141412, −1141411, .

. .

,114140.To handle sums of even exponents, Faulhaber knew thatΣn2m = n + 122m + 1 (a1N + a2N 2 + · · · + amN m)holds if and only ifΣn2m+1 = a12 N 2 + a23 N 3 + · · · +amm + 1N m+1 .Therefore he could get two sums for the price of one [2, folios C.iv verso and D.i recto]. It isnot difficult to prove this relation by establishing an isomorphism between the calculationsof Σn2m+1 and the calculations of the quantities S2m =(2m + 1)Σn2m(n + 12); forexample, the recurrence for Σn13 above corresponds to the formulaS12 = 64N 6 −5S10 −3S8 −17S6 ,which can be derived in essentially the same way.

Since the recurrences are essentiallyidentical, we obtain a correct formula for Σn2m+1 from the formula for S2m if we replace N kby N k+1/(k + 1).Faulhaber’s cryptomath. Mathematicians of Faulhaber’s day tended to conceal theirmethods and hide results in secret code.

Faulhaber ends his book [2] with a curious exercise7

of this kind, evidently intended to prove to posterity that he had in fact computed theformulas for sums of powers as far as Σn25 although he published the results only up toΣn17.His puzzle can be translated into modern notation as follows: LetΣ9n8 = a17n17 + · · · + a2n2 + a1ndwhere the a’s are integers having no common factor and d = a17 + · · · + a2 + a1. LetΣn25 = A26n26 + · · · + A2n2 + A1nDbe the analogous formula for Σn25.

LetΣn22 = (b10N 10 −b9N 9 + · · · + b0)b10 −b9 + · · · + b0Σn2 ,Σn23 = (c10n10 −c9N 9 + · · · + c0)c10 −c9 + · · · + c0Σn3 ,Σn24 = (d11n11 −d10N 10 + · · · −d0)d11 −d10 + · · · + d0Σn2 ,Σn25 = (e11n11 −e10N 10 + · · · −e0)e11 −e10 + · · · + e0Σn3 ,where the integers bk, ck, dk, ek are as small as possible so that bk, ck, dk, ek are mul-tiples of 2k. (He wants them to be multiples of 2k so that bkN k, ckN k, dkN k, ekN kare polynomials in n with integer coefficients; that is why he wrote, for example, Σn7 =(12N 2 −8N + 2)N 2/6 instead of (6N 2 −4N + 1)N 2/3.

See [2, folio D.i verso].) Thencomputex1 = (c3 −a12)/7924252 ;x2 = (b5 + a10)/112499648 ;x3 = (a11 −b9 −c1)/2945002 ;x4 = (a14 + c7)/120964 ;x5 = (A26a11 −D + a13 + d11 + e11)/199444 .These values (x1, x2, x3, x4, x5) specify the five letters of a “hochger¨uhmte Nam,” if we usefive designated alphabets [2, folio F.i recto].8

It is doubtful whether anybody solved this puzzle during the first 360 years after itspublication, but the task is relatively easy with modern computers. We havea10 = 532797408 , a11 = 104421616 , a12 = 14869764 , a13 = 1526532 , a14 = 110160 ;b5 = 29700832 ,b9 = 140800 ;c1 = 205083120 ,c3 = 344752128 ,c7 = 9236480 ;d11 = 559104 ;e11 = 86016 ;A26 = 42 ;D = 1092 .The fact that x2 = (29700832 + 532797408)/112499648 = 5 is an integer is reassuring: Wemust be on the right track!

But alas, the other values are not integral.A bit of experimentation soon reveals that we do obtain good results if we divide allthe ck by 4. Then, for example, x1 = (344752128/4 −14869764)/7924252 = 9, and we alsofind x3 = 18, x4 = 20.

It appears that Faulhaber calculated Σ9n8 and Σn22 correctly, andthat he also had a correct expression for Σn23 as a polynomial in N; but he probably neverwent on to express Σn23 as a polynomial in n, because he would then have multiplied hiscoefficients by 4 in order to compute c6N 6 with integer coefficients.The values of (x1, x2, x3, x4) correspond to the letters I E S U, so the concealed namein Faulhaber’s riddle is undoubtedly I E S U S (Jesus).But his formula for x5 does not check out at all; it is way out of range and not aninteger. This is the only formula that relates to Σn24 and Σn25, and it involves only thesimplest elements of those sums—the leading coefficients A26, D, d11, e11.

Therefore wehave no evidence that Faulhaber’s calculations beyond Σn23 were reliable. It is temptingto imagine that he meant to say ‘A26a11/D’ instead of ‘A26a11 −D’ in his formula for x5,but even then major corrections are needed to the other terms and it is unclear what heintended.All-integer formulas.

Faulhaber’s theorem allows us to express the power sum Σnmin terms of about12m coefficients. The elementary theory above also suggests anotherapproach that produces a similar effect: We can write, for example,n =n1;n3 = 6n+13+n1;n5 = 120n+25+ 30n+13+n1;(It is easy to see that any odd function g(n) of the integer n can be expressed uniquely asa linear combinationg(n) = a1n1+ a3n+13+ a5n+25+ · · ·9

of the odd functionsn1,n+13,n+25, . .

. , because we can determine the coefficientsa1, a3, a5, .

. .

successively by plugging in the values n = 1, 2, 3, . .

.. The coefficients ak willbe integers iffg(n) is an integer for all n.) Once g(n) has been expressed in this way, weclearly haveΣg(n) = a1n+12+ a3n+24+ a5n+36+ · · · .This approach therefore yields the following identities for sums of odd powers:Σn1 =n+12;Σn3 = 6n+24+n+12;Σn5 = 120n+36+ 30n+24+n+12;Σn7 = 5040n+48+ 1680n+36+ 126n+24+n+12;Σn9 = 362880n+510+ 151200n+48+ 17640n+36+ 510n+24+n+12;Σn11 = 39916800n+612+ 19958400n+510+ 3160080n+48+ 168960n+36+ 2046n+24+n+12;Σn13 = 6227020800n+714+ 3632428800n+612+ 726485760n+510+ 57657600n+48+ 1561560n+36+ 8190n+24+n+12.And repeated sums are equally easy; we haveΣrn1 =n+r1+r,Σrn3 = 6n+1+r3+r+n+r1+r,etc.The coefficients in these formulas are related to what Riordan [R, page 213] has calledcentral factorial numbers of the second kind.

In his notationxm =mXk=1T(m, k)x[k] ,x[k] = xx + k2 −1x + k2 −2. .

.x + k2 + 1,when m > 0, and T(m, k) = 0 when m −k is odd; hencen2m−1 =mXk=1(2k −1)! T(2m, 2k)n + k −12k −1,Σn2m−1 =mXk=1(2k −1)!

T(2m, 2k)n + k2k.The coefficients T(2m, 2k) are always integers, because the identity x[k+2] = x[k](x2−k2/4)implies the recurrenceT(2m + 2, 2k) = k2T(2m, 2k) + T(2m, 2k −2) .10

The generating function for these numbers turns out to becosh2x sinh(y/2)=∞Xm=0 mXk=0T(2m, 2k)x2k y2m(2m)! .Notice that the power-sum formulas obtained in this way are more “efficient” thanthe well-known formulas based on Stirling numbers (see [5, (6.12)]):Σnm =Xkk!mkn + 1k + 1=Xkk!mk(−1)m−kn + kk + 1.The latter formulas give, for example,Σn7 = 5040n+18+ 15120n+17+ 16800n+16+ 8400n+15+ 1806n+14+ 126n+13+n+12= 5040n+78−15120n+67+ 16800n+56−8400n+45+ 1806n+34−126n+23+n+12.There are about twice as many terms, and the coefficients are larger.

(The Faulhaberianexpression Σn7 = (6N 4 −4N 3 + N 2)/3 is, of course, better yet. )Similar formulas for even powers can be obtained as follows.

We haven2 = nn1= U1(n) ,n4 = 6nn+13+ nn1= 12U2(n) + U1(n) ,n6 = 120nn+25+ 30nn+13+ nn1= 360U3(n) + 60U2(n) + U1(n) ,etc., whereUk(n) = nkn + k −12k −1=n + k2k+n + k −12k.HenceΣn2 = T1(n) ,Σn4 = 12T2(n) + T1(n) ,Σn6 = 360T3(n) + 60T2 + T1(n) ,Σn8 = 20160T4(n) + 5040T3(n) + 252T2(n) + T1(n) ,Σn10 = 1814400T5(n) + 604800T4(n) + 52920T3(n) + 1020T2(n) + T1(n) ,Σn12 = 239500800T6(n) + 99792000T5(n) + 12640320T4(n)+ 506880T3(n) + 4092T2(n) + T1(n) ,11

etc., whereTk(n) =n + k + 12k + 1+ n + k2k + 1= 2n + 12k + 1n + k2k.Curiously, we have found a relation here between Σn2m and Σn2m−1, somewhat anal-ogous to Faulhaber’s relation between Σn2m and Σn2m+1: The formulaΣn2m2n + 1 = a1n + 12+ a2n + 24+ · · · + amn + m2mholds if and only ifΣn2m−1 = 31a1n + 12+ 52a2n + 24+ · · · + 2m + 1mamn + m2m.4. Reflective decomposition.

The forms of the expressions in the previous section leadnaturally to useful representations of arbitrary functions f(n) defined on the integers. Itis easy to see that any f(n) can be written uniquely in the formf(n) =Xk≥0akn + ⌊k/2⌋k,for some coefficients ak; indeed, we haveak = ∇kf(⌊k/2⌋) .

(Thus a0 = f(0), a1 = f(0) −f(−1), a2 = f(1) −2f(0) + f(−1), etc.) The ak are integersifff(n) is always an integer.

The ak are eventually zero ifff is a polynomial. The a2k areall zero ifff is odd.

The a2k+1 are all zero ifff is 1-reflective.Similarly, there is a unique expansionf(n) = b0T0(n) + b1U1(n) + b2T1(n) + b3U2(n) + b4T2(n) + · · · ,in which the bk are integers ifff(n) is always an integer. The b2k are all zero ifff is evenand f(0) = 0.

The b2k+1 are all zero ifff is anti-1-reflective. Using the recurrence relations∇Tk(n) = Uk(n) ,∇Uk(n) = Tk−1(n −1) ,we findak = ∇kf(⌊k/2⌋) = 2bk−1 + (−1)kbkand thereforebk =kXj=0(−1)⌈j/2⌉+⌊k/2⌋2k−jaj .In particular, when f(n) = 1 for all n, we have bk = (−1)⌊k/2⌋2k.

The infinite series isfinite for each n.12

Theorem. If f is any function defined on the integers and if r, s are arbitrary integers,we can always express f in the formf(n) = g(n) + h(n)where g(n) is r-reflective and h(n) is anti-s-reflective.

This representation is unique, exceptwhen r is even and s is odd; in the latter case the representation is unique if we specifythe value of g or h at any point.Proof. It suffices to consider 0 ≤r, s ≤1, because f(x) is (anti)-r-reflective ifff(x + a) is(anti)-(r + 2a)-reflective.When r = s = 0, the result is just the well known decomposition of a function intoeven and odd parts,g(n) = 12f(n) + f(−n),h(n) = 12f(n) −f(−n).When r = s = 1, we have similarlyg(n) = 12f(n) + f(−1 −n),h(n) = 12f(n) −f(−1 −n).When r = 1 and s = 0, it is easy to deduce that h(0) = 0, g(0) = f(0), h(1) =f(0) −f(−1), g(1) = f(1) −f(0) + f(−1), h(2) = f(1) −f(0) + f(−1) −f(−2), g(2) =f(2) −f(1) + f(0) −f(−1) + f(−2), etc.And when r = 0 and s = 1, the general solution is g(0) = f(0) −C, h(0) = C,g(1) = f(−1) + C, h(1) = f(1) −f(−1) −C, g(2) = f(1) −f(−1) + f(−2) −C, h(2) =f(2) −f(1) + f(−1) −f(−2) + C, etc.When f(n) = Pk≥0 akn+⌊k/2⌋k, the case r = 1 and s = 0 corresponds to the decom-positiong(n) =∞Xk=0a2kn + k2k,h(n) =∞Xk=0a2k+1 n + k2k + 1.Similarly, the representation f(n) = Pk≥0 b2kTk(n) + Pk≥0 b2k+1Uk+1(n) corresponds tothe case r = 0, s = 1, C = f(0).Back to Faulhaber’s form.

Let us now return to representations of Σnm as polynomialsin n(n + 1). Setting u = 2N = n2 + n, we haveΣn = 12u= 12A(1)0 uΣn3 = 14u2= 14A(2)0 u2 + A(2)1 uΣn5 = 16u3 −12u2= 16A(3)0 u3 + A(3)1 u2 + A(3)2 uΣn7 = 18u4 −43u3 + 23u2= 18A(4)0 u4 + A(4)1 u3 + A(4)2 u2 + A(4)3 u13

and so on, for certain coefficients A(m)k.Faulhaber never discovered the Bernoulli numbers; i.e., he never realized that a singlesequence of constants B0, B1, B2, . .

. would provide a uniform formulaΣnm =1m+1B0nm+1 −m+11B1nm +m+12B2nm−1 −· · · + (−1)mm+1mBmnfor all sums of powers.

He never mentioned, for example, the fact that almost half ofthe coefficients turned out to be zero after he had converted his formulas for Σnm frompolynomials in N to polynomials in n. (He did notice that the coefficient of n was zerowhen m > 1 was odd. )However, we know now that Bernoulli numbers exist, and we know that B3 = B5 =B7 = · · · = 0.

This is a strong condition. Indeed, it completely defines the constants A(m)kin the Faulhaber polynomials above, given that A(m)0= 1.For example, let’s consider the case m = 4, i.e., the formula for Σn7: We need to findcoefficients a = A(4)1 , b = A(4)2 , c = A(4)3such that the polynomialn4(n + 1)4 + an3(n + 1)3 + bn2(n + 1)2 + cn(n + 1)has vanishing coefficients of n5, n3, and n. The polynomial isn8 + 4n7 + 6n6 + 4n5+ n4+ an6 + 3an5 + 3an4 + an3+ bn4 + 2bn3 + n2+ cn2 + cn ;so we must have 3a + 4 = 2b + a = c = 0.

In general the coefficient of, say, n2m−5 in thepolynomial for 2mΣn2m−1 is easily seen to bem5A(m)0+m−13A(m)1+m−21A(m)2.Thus the Faulhaber coefficients can be defined by the rulesA(w)0= 1 ;kXj=0w −j2k + 1 −2jA(w)j= 0 ,k > 0 . (∗)(The upper parameter will often be called w instead of m, in the sequel, because we willwant to generalize to noninteger values.) Notice that (∗) defines the coefficients for each14

exponent without reference to other exponents; for every integer k ≥0, the quantity A(w)kis a certain rational function of w. For example, we have−A(w)1= w(w −2)/6 ,A(w)2= w(w −1)(w −3)(7w −8)/360 ,−A(w)3= w(w −1)(w −2)(w −4)(31w2 −89w + 48)/15120 ,A(w)4= w(w −1)(w −2)(w −3)(w −5)(127w3 −691w2 + 1038w −384)/6048000 ,and in general A(w)kis wk = w(w −1) . .

. (w −k + 1) times a polynomial of degree k, withleading coefficient equal to (2 −22k)B2k/(2k)!

; if k > 0, that polynomial vanishes whenw = k + 1.Jacobi mentioned these coefficients A(m)kin [6], although he did not consider therecurrence (∗), and he tabulated them for m ≤6. He observed that the derivative of Σnmwith respect to n is m Σnm−1 + Bm; this follows because power sums can be expressed interms of Bernoulli polynomials,Σnm =1m+1Bm+1(n + 1) −Bm+1(0),and because B′m(x) = mBm−1(x).Thus Jacobi obtained a new proof of Faulhaber’sformulas for even exponentsΣn2 = 13 24A(2)0 u + 14A(2)1(2n + 1) ,Σn4 = 15 36A(3)0 u2 + 26A(3)1 u + 16A(3)2(2n + 1) ,Σn6 = 17 48A(4)0 u3 + 38A(4)1 u2 + 28A(4)2 u + 18A(4)3(2n + 1) ,etc.

(The constant terms are zero, but they are shown explicitly here so that the patternis plain.) Differentiating again gives, e.g.,Σn5 =16 · 7 · 8(4 · 3 A(4)0 u2 + 3 · 2 A(4)1 u + 2 · 1 A(4)2 )(2n + 1)2+ 2(4A(4)0 u3 + 3A(4)1 u2 + 2A(4)2 u + 1A(4)3 )−16B6=16 · 7 · 88 · 7 A(4)0 u3 + (6 · 5 A(4)1+ 4 · 3 A(4)0 )u2 + (4 · 3 A(4)2+ 3 · 2 A(4)1 )u+ (2 · 1 A(4)3+ 2 · 1 A(4)2 )−16B6 .This yields Jacobi’s recurrence(2w −2k)(2w −2k −1)A(w)k+ (w −k + 1)(w −k)A(w)k−1 = 2w(2w −1)A(w−1)k,(∗∗)15

which is valid for all integers w > k + 1 so it must be valid for all w. Our derivation of(∗∗) also allows us to conclude thatA(m)m−2 =2m2B2m−2 ,m ≥2 ,by considering the constant term of the second derivative of Σn2m−1.Recurrence (∗) does not define A(m)m , except as the limit of A(w)mwhen w →m. Butwe can compute this value by setting w = m + 1 and k = m in (∗∗), which reduces to2A(m+1)m−1= (2m + 2)(2m + 1)A(m)mbecause A(m+1)m= 0.

ThusA(m)m= B2m ,integer m ≥0 .Solution to the recurrence. An explicit formula for A(m)kcan be found as follows: WehaveΣn2m−1 =12mB2m(n + 1) −B2m=12m(A(m)0um + · · · + A(m)m−1u) ,and n + 1 = (p1 + 4u + 1)/2; hence, using the known values of A(m)m , we obtain∞Xk=0A(m)kum−k = B2m p1 + 4u + 12!= B2m 1 −p1 + 4u2!,a closed form in terms of Bernoulli polynomials.We have used the fact that A(m)m+1 =A(m)m+2 = · · · = 0, together with the identity Bn(x + 1) = (−1)nBn(−x) .Expanding theright side in powers of u givesXl2ml 1 −p1 + 4u2!lB2m−l=Xj,l2ml2j + ljl2j + l(−u)j+lB2m−l ,using equation (5.70) of [5].

Setting j + l = m −k finally yieldsA(m)k= (−1)m−k Xj2mm −k −jm −k + jjm −k −jm −k + j Bm+k+j ,0 ≤k < m .16

This formula, which was first obtained by Gessel and Viennot [4], makes it easy to confirmthat A(m)m−1 = 0 and A(m)m−2 =2m2B2m−2, and to derive additional values such asA(m)m−3 = −22m2B2m−2 = −2A(m)m−2 ,m ≥3 ;A(m)m−4 =2m4B2m−4 + 52m2B2m−2 ,m ≥4 .The author’s interest in Faulhaber polynomials was inspired by the work of Ed-wards [1], who resurrected Faulhaber’s work after it had been long forgotten and underval-ued by historians of mathematics. Ira Gessel responded to the same stimulus by submittingproblem E3204 to the Math Monthly [3] regarding a bivariate generating function for Faul-haber’s coefficients.

Such a function is obtainable from the univariate generating functionabove, using the standard generating function for Bernoulli polynomials: SinceXB2mx + 12 z2m(2m)! = 12XBmx + 12 zmm!

+ 12XBmx + 12 (−z)mm!= z e(x+1)z/22(ez −1) −z e−(x+1)z/22(e−z −1) = z cosh(xz/2)2 sinh(z/2) ,we haveXk,nA(m)kum−k z2m(2m)! =XmB2m p1 + 4u + 12!z2m(2m)!

= z2cosh (p1 + 4u z/2)sinh (z/2);Xk,wA(m)kuk z2m(2m)! = zpu cosh (pu + 4 z/2)2 sinh (zpu /2).The numbers A(m)kare obtainable by inverting a lower triangular matrix, as Edwardsshowed; indeed, recurrence (∗) defines such a matrix.

Gessel and Viennot [4] observed thatwe can therefore express them in terms of a k × k determinant,A(w)k=1(1 −w) . .

. (k −w)w−k+13w−k+110.

. .0w−k+25w−k+23w−k+21.

. .0......... w−12k−1 w−12k−3 w−12k−5.

. .w−11w2k+1w2k−1w2k−3.

. .w3.17

When w and k are positive integers, Gessel and Viennot proved that this determinant isthe number of sequences of positive integers a1a2a3 . .

. a3k such thata3j−2 < a3j−1 < a3j ≤w −k + j ,for1 ≤j ≤k ,a3j−2 < a3j+1 ,a3j−1 < a3j+3 ,for1 ≤j < k .In other words, it is the number of ways to put positive integers into a k-rowed triplestaircase such aswith all rows and all columns strictly increasing from left to right and from top to bottom,and with all entries in row j at most w −k + j.

This provides a surprising combinatorialinterpretation of the Bernoulli number B2m when w = m + 1 and k = m −1 (in whichcase the top row of the staircase is forced to contain 1, 2, 3).The combinatorial interpretation proves in particular that (−1)kA(m)k≥0 for allk ≥0. Faulhaber stated this, but he may not have known how to prove it.Denoting the determinant by D(w, k), Jacobi’s recurrence (∗∗) implies that we have(w −k)2(w −k + 1)(w −k −1)D(w, k −1)= (2w −2k)(2w −2k −1)(w −k −1)D(w, k) −2w(2w −1)(w −1)D(w −1, k) ;this can also be written in a slightly tidier form, using a special case of the “integer basis”polynomials discussed above:D(w, k −1) = T1(w −k −1)D(w, k) −T1(w −1)D(w −1, k) .It does not appear obvious that the determinant satisfies such a recurrence, nor that thesolution to the recurrence should have integer values when w and k are integers.

But,identities are not always obvious.Generalization to noninteger powers. Recurrence (∗) does not require w to be apositive integer, and we can in fact solve it in closed form when w = 3/2:Xk≥0A(3/2)ku3/2−k = B3 p1 + 4u + 12!= u2p1 + 4u = u3/2 Xk≥01/2k(4u)−k .18

Therefore A(3/2)k=1/2k4−k is related to the kth Catalan number. A similar closed formexists for A(m+1/2)kwhen m is any nonnegative integer.For other cases of w, our generating function for A(w)kinvolves Bn(x) with nonintegersubscripts.

The Bernoulli polynomials can be generalized to a family of functions Bz(x),for arbitrary z, in several ways; the best generalization for our present purposes seems toarise when we defineBz(x) = xz Xk≥0zkx−kBk ,choosing a suitable branch of the function xz. With this definition we can develop theright-hand side ofXk≥0A(w)ku−k = B2w p1 + 4u + 12!u−w= p1 + 4u + 12pu!2w Xk≥02wk p1 + 4u + 12!−kBk(∗∗∗)as a power series in u−1 as u →∞.The factor outside the P sign is rather nice; we have p1 + 4u + 12pu!2w=Xj≥0ww + j/2w + j/2ju−j/2 ,because the generalized binomial series B1/2(u−1/2) [5, equation (5.58)] is the solution tof(u)1/2 −f(u)−1/2 = u−1/2 ,f(∞) = 1 ,namelyf(u) = p1 + 4u + 12pu!2.Similarly we find p1 + 4u + 12!−k=Xj−kj −kj/2 −k/2ju−k/2−j/2= u−k/2 −Xj≥1k2jj/2 −k/2 −1j −1u−k/2−j/2 .19

So we can indeed expand the right-hand side as a power series with coefficients that arepolynomials in w. It is actually a power series in u−1/2, not u; but since the coefficientsof odd powers of u−1/2 vanish when w is a positive integer, they must be identically zero.Sure enough, a check with computer algebra on formal power series yields 1 + A(w)1u−1 +A(w)2u−2 + A(w)3u−3 + O(u−4), where the values of A(w)kfor k ≤3 agree perfectly withthose obtained directly from (∗). Therefore this approach allows us to express A(w)kas apolynomial in w, using ordinary Bernoulli number coefficients:A(w)k=2kXl=0ww + l/2w + l/2l× 2w2k −lB2k−l −122k−l−1Xj=12wjj2k −l −jk −l/2 −j −12k −l −j −1Bj.The power series (∗∗∗) we have used in this successful derivation is actually divergentfor all u unless 2w is a nonnegative integer, because Bk grows superexponentially whilethe factor2wk= (−1)kk −2w −1k= (−1)k Γ(k −2w)Γ(k + 1) Γ(−2w) ∼(−1)kΓ(−2w)k−2w−1does not decrease very rapidly as k →∞.Still, (∗∗∗) is easily seen to be a valid asymptotic series as u →∞, because asymptoticseries multiply like formal power series.

This means that, for any positive integer p, wehave2pXk=02wk p1 + 4u + 12!2w−kBk =pXk=0A(w)kuw−k + O(uw−p−1) .We can now apply these results to obtain sums of noninteger powers, as asymptoticseries of Faulhaber’s type. Suppose, for example, that we are interested in the sumH(1/3)n=nXk=11k1/3 .Euler’s summation formula [5, exercise 9.27] tells us thatH(1/3)n−ζ( 13) ∼32n2/3 + 12n−1/3 −136n−4/3 −· · ·= 32Xk≥02/3kn2/3−kBk + n−1/3,20

and the parenthesized quantity is what we have called B2/3(n + 1). And when u = n2 + nwe have B2/3(n + 1) = B2/3(p1 + 4u + 1)/2; henceH(1/3)n−ζ( 13) ∼32Xk≥0A(1/3)ku1/3−k= 32 u1/3 + 536 u−2/3 −171215 u−5/3 + · · ·as n →∞.

(We can’t claim that this series converges twice as fast as the usual one,because both series diverge! But we do get twice as much precision in a fixed number ofterms.

)In general, the same argument establishes the asymptotic seriesnXk=1kα −ζ(−α) ∼1α + 1Xk≥0A((α+1)/2)ku(α+1)/2−k ,whenever α ̸= −1. The series on the right is finite when α is a positive odd integer; it isconvergent (for sufficiently large n) if and only if α is a nonnegative integer.The special case α = −2 has historic interest, so it deserves a special look:nXk=11k2 ∼π26 −A(−1/2)0u−1/2 −A(−1/2)1u−3/2 −· · ·= π26 −u−1/2 + 524 u−3/2 −1611920 u−5/2 + 4017168 u−7/2 −32021491520 u−9/2 + · · · .These coefficients do not seem to have a simple closed form; the prime factorization 32021 =11 · 41 · 71 is no doubt just a quirky coincidence.Acknowledgments.

This paper could not have been written without the help providedby several correspondents. Anthony Edwards kindly sent me a photocopy of Faulhaber’sAcademia Algebræ, a book that is evidently extremely rare: An extensive search of printedindexes and electronic indexes indicates that no copies have ever been recorded to existin America, in the British Library, or the Biblioth`eque Nationale.Edwards found itat Cambridge University Library, where the volume once owned by Jacobi now resides.

(I have annotated the photocopy and deposited it in the Mathematical Sciences Libraryat Stanford, so that other interested scholars can take a look. )Ivo Schneider, who iscurrently preparing a book about Faulhaber and his work, helped me understand some ofthe archaic German phrases.

Herb Wilf gave me a vital insight by discovering the first21

half of Lemma 4, in the case r = 1. And Ira Gessel pointed out that the coefficients in theexpansion n2m+1 = P ak n+k2k+1are central factorial numbers in slight disguise.References[1] A. W. F. Edwards, “A quick route to sums of powers,” American Mathematical Monthly93 (1986), 451–455.

[2] Johann Faulhaber, Academia Algebræ, Darinnen die miraculosische Inventiones zu denh¨ochsten Cossen weiters continuirt und profitiert werden. Augspurg, bey Johann Ul-rich Sch¨onigs, 1631.

(Call number QA154.8 F3 1631a f MATH at Stanford UniversityLibraries. )[3] Ira Gessel and University of South Alabama Problem Group, “A formula for powersums,” American Mathematical Monthly 95 (1988), 961–962.

[4] Ira M. Gessel and G´erard Viennot, “Determinants, paths, and plane partitions,” pre-print, 1989. [5] Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics(Reading, Mass.

: Addison-Wesley, 1989). [6] C. G. J. Jacobi, “De usu legitimo formulae summatoriae Maclaurinianae,” Journal f¨urdie reine und angewandte Mathematik 12 (1834), 263–272.

[7] John Riordan, Combinatorial Identities (New York: John Wiley & Sons, 1968). [8] L. Tits, “Sur la sommation des puissances num´eriques,” Mathesis 37 (1923), 353–355.22

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