Sums of Integer Powers--The Faulhaber Expansion



Sums of Integer Powers--The Bernoulli, Binomial, Stirling and Faulhaber Expansions

Norman Brenner

George Washington University

DRAFT VERSION

1960; 2002 Sep 12, 20, Oct 8; 2003 Apr 15 (shift to 1..n); Apr 18 (add binom); May 21; May 22 (shift back to 0..n-1); 2009 Sep 8

Perhaps publish in College Math Monthly (edited by Lowell Leinecke, known to jw@cc)

I. Statement of the problem

Consider computing the sum of consecutive integers to a given power; we define for all values of p:

Sp(n) := [pic]ip , p>0

Every schoolchild has seen the formula for the sum of the integers themselves:

S1(n) = n(n-1)/2

But the formulas for higher powers are unfamiliar to most people. In this paper, we give four explicit formulas for such sums. One formula may be original, and all the proofs are. The purpose of this paper, then, is pedagogic exposition.

A special definition for the case p=0 is necessary because of the indeterminacy of 00; taking it equal to 1, we define:

S0(n) := n

This value is chosen to be consistent with the recursion and differential relations below.

[Alternately, we could work with the sum from i=1 through n; but the range as indicated tends to yield simpler formulas in the rest of this work.]

II. Recursion relation

We can easily derive a recursion relation among these sums. Consider Sp(n+1). From the definition for p>0,

Sp(n+1) = [pic]ip

Rewrite this sum in two different ways. First, split off the largest term:

Sp(n+1) = np + [pic]ip

= np + Sp(n)

Alternately, replace the summation index by defining i:=j+1 :

Sp(n+1) = [pic](j+1)p

= 0p + [pic](j+1)p

The kernel of this latter summation can be expanded by the binomial theorem, and then the order of the two summations exchanged:

Sp(n+1) = [pic][pic][pic] jk

= [pic][pic][pic]jk

The limits of the summation variable k here are defaulted to –[pic] to +[pic]; since the binomial coefficient is zero outside the range k=0..p, the latter becomes the de facto range. However, it is simpler to write index k bare, meaning “over all possible values of k”.

The inner summation is exactly the definition of Sk(n), so:

Sp(n+1) = [pic][pic] Sk(n)

Combining the alternate expansions,

np + Sp(n) = [pic][pic] Sk(n)

Rearranging,

np = [pic][pic] Sk(n) - Sp(n)

A helpful notation is to define an abstract symbol "S", whose p'th power is defined to be Sp(n); then we may write more concisely,

np = (S+1)p - Sp, p>0

A similar formula was published by X. Bernoulli in 17??. [Look it up!]

III. Table of explicit expressions

With this recursion relation, we can now build a table of explicit formulas for the Sp(n). Start by substituting p=1 in it:

n1 = (S+1)1 – S1

i.e.

n = (S1+S0) – S1

or

n = S0

Working upward:

S0(n) = n

S1(n) = n2/2 - n/2

S2(n) = n3/3 - n2/2 + n/6

S3(n) = n4/4 - n3/2 + n2/4

S4(n) = n5/5 - n4/2 + n3/3 - n/30

S5(n) = n6/6 - n5/2 + 5 n4/12 - n2/12

S6(n) = n7/7 - n6/2 + n5/2 - n3/6 + n/42

S7(n) = n8/8 - n7/2 + 7 n6/12 – 7 n4/24 + n2/12

It is a striking result (Nicomachus's theorem) that

S3(n) = [S1(n)]2

[Author’s aside. An elegant geometric proof of this formula exists by unpacking each cube into slabs, which are then attached to a square to form a larger square. An algebraic proof is, of course, elementary, by performing mathematical induction on n.]

We note also, for future reference, that

3S5(n) + S3(n) = 4[S1(n)]3

and

S7(n) + S5(n) = 2[S1(n)]4

which will fall out below from another recursion relation.

IV. Numerical checks

Simple numerical checks on the correctness of these formulas are available by noting that, from the definition,

Sp(1) = 0

Sp(2) = 1

Sp(3) = 1 + 2p

Check e.g. S7(1) = 1/8 - 1/2 + 7/12 – 7/24 + 1/12 = 0. This is correct.

V. Worpitzky’s Expansion in terms of Binomial Coefficients

Rather than express Sp(n) as a sum of powers of n, it may be useful to express them as a sum over binomial coefficients of comparable magnitude. In lieu of proof (cf. Worpitzky, J., J. reine angew. Math. 94, 203-232, 1883), we simply give the first few here.

S0(n) = [pic]

S1(n) = [pic]

S2(n) = [pic] + [pic]

S3(n) = [pic] + 4 [pic] + [pic]

S4(n) = [pic] + 11 [pic] + 11 [pic] + [pic]

S5(n) = [pic] + 26 [pic] + 66 [pic] + 26 [pic] + [pic]

By substituting n=2 and n=3, and using the numerical values of Sp(n) given above, we solve the linear equations for different n and easily determine the leading coefficients to be:

Sp(n) = [pic] + (2p – p – 1) [pic] + [pic]3p - 2p(p+1) + [pic][pic] [pic] + ...

Also, the coefficients within one expression for Sp(n) are symmetric. The general expression is

Sp(n) = [pic][pic][pic](-1)k (j+1-k)p [pic][pic] [pic]

and the inner sum is the same if j be replaced by p-1-j (i.e. symmetric coefficients). This is an attractively simple and easily computed formula.

The coefficients are called Eulerian numbers (not to be confused with Euler numbers). Their row sum for fixed p is (p+1)!

Notice that all binomial coefficients in the expansion for Sp(n) are terms of order of magnitude np+1, and can be readily computed in sequence by a single multiplication and division. Also, the Eulerian coefficients can be directly calculated without recurrence, making these formulas easier to find, even though longer than ones below. They are not listed in the standard handbook by Abramowitz & Stegun, but are described in Concrete Mathematics, by Graham, Knuth & Patashnik (1994).

VI. Stirling’s Expansion in terms of Binomial Coefficients

An alternative expression of Sp(n) as a sum of binomial coefficients of increasing magnitude, for which the coefficients are the Stirling coefficients of the second kind, will be found in Abramowitz & Stegun, Handbook of Mathematical Functions, formula 24.1.4 C(2):

[Stirling “numbers”?]

Sp(n) = [pic]k![pic]p(k) [pic]

[We use a script T here because we do not have A&S’s symbol, a script S.]

E.g. here is S4(n):

S4(n) = [pic] + 14 [pic] + 36 [pic] + 24 [pic]

It will be seen that the lower indices of these binomial coefficients differ; therefore, they represent ascending powers of n, from order of n2 to order of n5. Also, the Stirling coefficients are not symmetric within a single Sp(n), unlike the coefficients in the previous expansion.

Both of these expansions, in terms of binomial coefficients, are rather more bulky than the more compact expansions that we now turn to.

VII. Differentiation formula and Bernoulli’s Expansion

We may try differentiating Sp(n) as given in the table above, and discover empirically that

dSp(n)/dn = p Sp-1(n) + Bp

where Bp are a series of constants to be determined. This differentiation formula is easily proved by mathematical induction using the recursion formula above.

[Insert proof by induction]

The Bernoulli numbers may be defined by the recursion formula remaining after Sp(n) is removed from the derivative of the basic recursion formula. Here is a table of the first few Bernoulli numbers:

B0 = 1

B1 = -1/2

B2 = 1/6

B3 = 0

B4 = -1/30

B5 = 0

B6 = 1/42

B7 = 0

B8 = -1/30

B9 = 0

B10 = 5/66

B11 = 0

B12 = -691/2730

B13 = 0

B14 = 7/6

B15 = 0

B16 = -3617/510

Note that, except for B1, all odd-indexed B2q+1 are 0, while the even-indexed ones alternate in sign. Further, the denominator of any Bernoulli number is readily computable, whereas a formula for the numerator is not known to me.

[Insert Bernoulli denominator formula]

An explicit formula for the coefficients of the nk in any Sp(k) is derivable, dependent on the Bernoulli numbers:

Sp(n) = (p+1)-1 [pic][pic] Bi np+1-i

Using the symbol “B”, whose j’th power is a notation for Bj, we may write this more compactly as

Sp = [(n+B)p+1 – Bp+1]/(p+1)

In effect, this is the „matrix inverse” to the recursion equation shown above for the Sp. Note that this can be written suggestively as a definite integral:

⌠n

Sp = │ (x+B)p dx

⌡0

Incidentally, it is not possible to compute S-1(n) (called the harmonic sums) viaa the above differentiation formula, since substituting p=0 therein yields

dS0(n)/dn = 0 S-1(n) + B0

i.e.

1 = 0 S-1(n) + 1

which tells us nothing about S-1(n), except that it is finite.

VIII. Generator function

A generator function for the sums may be defined by

G(n,t) := [pic]Sp(n) tp/p!

Here, t is a dummy placeholder variable, whose sole purpose is to mark and separate the coefficients. (As before, the bare index p is to be interpreted as ranging over all possible values of p, in this case, for p=0..[pic].) Inserting the definition of Sp(n), and exchanging summations,

G(n,t) = (ent - 1)/(et - 1)

By taking the partial derivative of G with respect to n, we may again prove the formula for the derivative dSp(n)/dn.

The generator formula for the Bernoulli numbers is similar; first define the Bernoulli polynomials Bp(x) by

text/(et - 1) := [pic]Bp(x) tp/p!

Then define

Bp := Bp(0)

That is,

t/(et - 1) = [pic]Bp tp/p!

If we replace the dummy variable t in G(n,t) by it, where i is the imaginary unit, and extract the pure imaginary part, we have:

Imag(G(n,it)) = [G(n,it) - G(n,-it)]/(2i)

= [(enit - 1)/(eit - 1) - (e-nit - 1)/(e-it - 1)]/(2i)

= [enit + e(1-n)it - eit - 1] / [2i(eit - 1)]

and, from the power series definition,

Imag(G(n,it)) = [pic][pic]Sp(n) (it)p/p! - [pic]Sp(n) (-it)p/p![pic] / (2i)

Only the odd-indexed terms do not cancel. Substituting p:=2q+1,

= [pic]S2q+1(n) (-1)q t2q+1/(2q+1)!

A recursion equation for the odd-indexed sums will appear below. Meantime, notice that in the left-hand side, if n is replaced by 1–n, there is no change in the expression. This suggests defining a variable which is also invariant when n is replaced by 1–n, e.g.

M := n(n-1)

In this case,

n = (1 + [pic])/2

and

1-n = (1 - [pic])/2

Now, the exponential functions on the left-hand side are expansible in terms of powers of n and 1-n. The odd-numbered powers of the square root will cancel out, leaving only integer powers of 4M+1. Therefore, the left-hand side becomes a (very complicated) polynomial in terms of M, and not in terms of n directly. We will see explicit formulas in terms of M, below.

[We have found no further need for the generator function.]

[Another possible approach, suggested by Balaji, is to apply a Fourier transform.]

IX. Explicit factoring and Faulhaber’s expansion

A possibly fruitful path to explore lies in factoring the explicit expressions above:

S0(n) = n

S1(n) = n (n-1) / 2

S2(n) = n (n-1) (2n-1) / 6

S3(n) = n2 (n-1)2 / 4

S4(n) = n (n-1) (2n-1) (3n2 - 3n - 1) / 30

S5(n) = n2 (n-1)2 (2n2 - 2n - 1) / 12

S6(n) = n (n-1) (2n-1) (3n4 - 6n3 + 3n + 1) / 42

S7(n) = n2 (n-1)2 (3n4 - 6n3 - n2 + 4n + 2) / 24

By direct substitution in this table, it appears that

Sp(1-n) = (-1)p+1 Sp(n), p>0

As before, this suggests defining a variable which is also invariant when n is replaced by 1–n. On further examination, we see that the parameter

M := n(n-1)

appears as a factor in every Sp(n) for p>0, and its derivative

M' := dM/dn = 2n - 1

appears in every S2q(n) for q>0. Therefore, rewriting the table in terms of M, we have more compact expressions:

S0(n) = M'/2 + 1/2

S1(n) = M/2

S2(n) = (M'/2) M/3

S3(n) = M2/4

S4(n) = (M'/2)(M2 - M/3)/5

S5(n) = (M3 - M2/2)/6

S6(n) = (M'/2)(M3 - M2 + M/3)/7

S7(n) = (M4 - 4 M3/3 + 2 M2/3)/8

It will be helpful to give the odd-indexed sums a new name; define

Fq(M) := S2q+1(n)

and call these the Faulhaber polynomials. The even-indexed sums would thence be obtainable by differentiation:

S2q(n) = M' [ dFq(M)/dM ] /(2q+1) – B2q+1

Of course, B2q+1 = 0 except for B1 = -1/2.

These compact expansions in terms of M (with M’ written out as 2n-1) were published by Johann Faulhaber in 1631, proven by Jacobi in 1834, and rediscovered by D. E. Knuth in 1993: “Johann Faulhaber and Sums of Powers”, Mathematics of Computation 61 (1993), 277-294. A short biography of Faulhaber is on the web site



[I have not actually read any of these last three papers.]

X. Recursion relation for the Faulhaber polynomials

It is elementary to derive a recursion formula for the Faulhaber polynomials. [This proof is original with me, so far as I know.] From its definition,

Mq = nq (n-1)q

We may expand this immediately with the binomial theorem:

Mq = [pic][pic] (-n)q+i (*)

For later use, substitute i:=j-q so that

Mq = [pic][pic] (-n)j (**)

We will come back to these expansions later, so we give them markers. Alternately, we can first rewrite Mq as

Mq = [(n-1)+1]q (n-1)q

and expand this with respect to n-1 by the binomial theorem: [Proof needed?]

Mq = [pic][pic] (n-1)q+i

Next, expand (n-1)q+I by the binomial theorem:

Mq = [pic][pic](-1)q+i [pic][pic] (-n)j

Exchanging summations here and equating with coefficients of (-n)j in the previous expansion (**), we prove a useful identity for binomial coefficients:

[pic] = [pic](-1)q+i [pic][pic] (***)

Return now to equation (*). If we now restrict q>0, then a fortiori also q+i>0, and so we may substitute for nq+i therein the recursion identity for the Sp(n) above:

Mq = [pic][pic] (-1)q+i [pic][pic][pic]Sj(n) - Sq+i(n)[pic]

Apply the outer summation to each term in the bracketed sum and exchange summations in the first term:

Mq = [pic][pic][pic] (-1)q+i [pic][pic]Sj(n)[pic] - [pic][pic](-1)q+i Sq+i(n)

= [pic]Sj(n) [pic][pic](-1)q+i [pic][pic][pic] - [pic][pic] (-1)q+i Sq+i(n)

The bracketed expression in the central term appears in the identity (***) we derived above. Inserting it,

Mq = [pic][pic]Sj(n) - [pic][pic] (-1)q+i Sq+i(n)

Substitute i:=j-q in the last term:

Mq = [pic][pic]Sj(n) - [pic][pic](-1)j Sj(n)

Notice that terms with even-valued j cancel out, leaving only the odd j:=2k+1 terms:

Mq = 2 [pic][pic]S2k+1(n)

Substitute k:=q–p, and the definition of Faulhaber polynomials, and this becomes:

Mq = 2 [pic][pic]Fq-p(M), q>0

This is the desired recursion relation. Again, use an abstract symbol "F", whose p'th power is Fp(M), and we may write more concisely,

Mq = F(q-1)/2 [ (F1/2+1)q – (F1/2-1)q ], q>0

This formula can be rewritten as a matrix multiplication:

M = (P + P-1)·F

Where the matrix Pjk is the Pascal triangle extended with 0’s (i.e. the entries are the binomial coefficients), and its inverse matrix has the same entries but with alternating signs. M and F are of course the vectors of the powers of M and the symbol F. To express F in terms of M would be equivalent to inverting the matrix multiplier shown.

XI. Numerical checks for the Faulhaber recursion formula

When q=1, this is

M1 = F0/2 [ (F1/2+1)1 – (F1/2-1)1 ]

or

M1 = 2F0

i.e.

M = 2F0(M)

When q=2, this is

M2 = F1/2 [ (F1/2+1)2 – (F1/2-1)2 ]

or

M2 = 4F1

i.e.

M2 = 4F1(M)

which we saw earlier in the form

S3(n) = M2/4

When q=3, this is

M3 = F [ (F1/2+1)3 – (F1/2-1)3 ]

or

M3 = 6F2 + 2F

i.e.

M3 = 6F2(M) + 2F1(M)

which we saw earlier in the form

3S5(n) + S3(n) = 4[S1(n)]3

When q=4, this is

M4 = F3/2 [ (F1/2+1)4 – (F1/2-1)4 ]

or

M4 = 8F3 + 8F2

i.e.

M4 = 8F3(M) + 8F2(M)

We saw this same equation earlier in the form

S7(n) + S5(n) = 2[S1(n)]4

When q=5, this is

M5 = F2 [ (F1/2+1)5 – (F1/2-1)5 ]

or

M5 = 10F4 + 20F3 + 2F2

i.e.

M5 = 10F4(M) + 20F3(M) + 2F2(M)

XII. Differential equation for the Faulhaber polynomials

Since we have a differential equation for Sp(n) with respect to n, we may substitute the function Fq(M) and replace d/dn by (2n-1)d/dM. We find

2Fq’ + (4M+1)Fq” = (2q+1)(2q Fq-1 + B2q), q>-1

[Check!]

Possibly useful is

Fq’(M) = [(2q+1)S2q(n) + B2q+1]/(2n-1) , q>-1

Recall that B2q+1 = 0 except for B1 = -1/2.

[A generating function for the Faulhaber polynomials would be useful.]

XIII. Explicit expressions for the Faulhaber polynomials

Using the recursion formula, we may compute explicit expressions for the Fq(M), for larger values of q; the following table extends up to F8(M) (=S17(n)):

F0(M) = M/2

F1(M) = M2/4

F2(M) = (M3 - M2/2)/6

F3(M) = (M4 - 4 M3/3 + 2 M2/3)/8

F4(M) = (M5 - 5 M4/2 + 3 M3 - 3 M2/2)/10

F5(M) = (M6 - 4 M5 + 17 M4/2 - 10 M3 + 5 M2)/12

F6(M) = (M7 - 35 M6/6 + 287 M5/15 - 118 M4/3 + 691 M3/15 - 691 M2/30)/14

F7(M) = (M8 - 8 M7 + 112 M6/3 - 352 M5/3 + 718 M4/3 - 280 M3 + 140 M2)/16

F8(M) = (M9 - 21 M8/2 + 66 M7/3 - 293 M6/3 + 9114 M5/10 - 3711 M4/2 + 10851 M3/5 - 10851 M2/10)/18

[Hmm. F2(M) is evenly divisible by M-1/2 and F4(M) is evenly divisible by M-1. But F6(M) is divisible only by M-1.444844178. Bah!]

[The coefficients multiplying the powers of M form a triangular matrix which is the inverse of the matrix of binomial coeffients multiplying the F’s in the recursion relation. Is there a known way of inverting it?]

Note that the last two terms of any expansion (for q>1) may be written in terms of the Bernoulli numbers:

Fq(M) = .. + (-1)q (2q+1) B2q (M3 - M2/2)

This is easily shown by differentiating Fq(M) twice. [Check!]

XIV. Explicit expressions for the lower Faulhaber coefficients

Write the summation expansions in the format

Fq(M) = aq Mq+1/(q+1) - bq Mq/q + cq Mq-1/(q-1) - dq Mq-2/(q-2) + ..

[What about the denominator of 2(q+1) in all coefficients?] [Wha??]

For reference, here are the numerical values of some of the coefficients, taken from the explicit expressions above:

q aq/(q+1) -bq/q cq/(q-1) -dq/(q-2) eq/(q-3) -fq/(q-4) gq/(q-5) -hq/(q-6)

0 1/2 0 0 0 0 0 0 0

1 1/4 0 0 0 0 0 0 0

2 1/6 1/12 0 0 0 0 0 0

3 1/8 1/6 1/12 0 0 0 0 0

4 1/10 1/4 3/10 3/20 0 0 0 0

5 1/12 1/3 17/24 5/6 5/12 0 0 0

6 1/14 5/12 41/30 59/21 691/210 691/420 0 0

7 1/16 1/2 7/3 22/3 359/24 35/2 35/4 0

8 1/18 7/12 11/3 293/18 1519/30 1237/12 3617/30 3617/60

We find

aq = 1 / 2

bq = q(q-1) / 12

cq = q(q-1)(q-2) (7q-1) / 6!

dq = q(q-1)(q-2)(q-3) (31q2 - 27q - 10) / 6x7!

eq = q(q-1)(q-2)(q-3)(q-4) (127q3 - 310q2 + 37q + 90) / 30x8!

More suggestively, we may write in terms of the Bernoulli numbers:

aq = [pic] 1! (-B0/0!) [(2-1-1)q-1]

bq = [pic] 2! (B2/2!) [(21-1)q0]

cq = [pic] 3! (-B4/4!) [(23-1)q - 1]

dq = [pic] 4! (B6/6!) [(25-1)q2 - 27q – 10]

eq = [pic] 5! (-B8/8!) [(27-1)q3 - 310q2 + 37q + 90]

fq = [pic] 6! (B10/10!) [(29-1)q4 + (- 12674q3 + 14161q2 + 2486q - 3864)/5]

gq = [pic] 7! (-B12/12!) [(211-1)q5 + (- 11974437q4 + 31092673q3 - 22432587q2 - 7706534q + 6399960)/691]

[That |B2k| = (k-1)!(k-1)!/((k+1)!(2k+1)) for k=1,2,3,4 seems to just be a coincidence here. Because for k=5,6,7,8, B2k = 5/66, -691/2730, 7/6, -3617/510, while the formula yields 4/55, 20/91, 6/7, 70/17.]

A simple way to generate these coefficient expressions is to insert them into the recursion relation above, and solve for successively smaller powers of M. aq is determined first, then bq in terms of it, &c. Or the differential equation for Fq can be used; specifically, equating the coefficients of Mq-6:

-(q-5)fq + [2 + 4(q-6)]gq = 2q(2q+1)gq-1/(q-6)

Define the polynomials

FF(q) := 511q4 - ..

GG(q) := 2047q5 - ..

so that

fq = [pic] (6! B10/10!) FF(q)

gq = [pic] (7! (-B12)/12!) GG(q)

Then, given FF(q), the equation to be solved for GG(q) is

0 = (B1012!)/(B1210!)(q-5)FF(q)/2 + (2q-11)(q-6)GG(q) – (2q+1)(q-7)GG(q-1)

Set q to 6 different numerical values, and solve the resulting 6 simultaneous linear equations for the numerical coefficients of polynomial GG(q).

[Build a generator function for these expansions??]

[How confirm from these models that the diagonal coefficients are Bernoulli numbers?]

XV. A numerical check

When n=1, then M=2, and F6(1)=1, so

F6(1) ?= (2^7 - 35 2^6/6 + 287 2^5/15 - 118 2^4/3 + 691 2^3/15 - 691 2^2/30)/14

= 2^7/14 - 35 2^6/6x14 + 287 2^5/15x14 - 118 2^4/3x14 + 691 2^3/15x14 - 691 2^2/30x14

= 2^6/7 - 5 2^4/3 + 41 2^4/15 - 59 2^4/3x7 + 691 2^2/15x7 - 691/15x7

= 16 (4/7 - 5/3 + 41/15 - 59/3x7) + 691 3/15x7

= 16 (4x15 - 5x35 + 41x7 - 59x5)/3x5x7 + 691/5x7

= 16 (60 - 175 + 287 - 295)/3x5x7 + 691/5x7

= 16 (- 123)/3x5x7 + 691/5x7

= 16 (- 41)/5x7 + 691/5x7

= -656/5x7 + 691/5x7

= 35/5x7

= 1

Check.

XVI. Closed form for the Faulhaber coefficients

The Bernoulli expansion for S2q+1(n) is

Fq(M) [pic] S2q+1(n) = (n+B)2q+2/(2q+2) – B2q+2/(2q+2)

We may substitute for M herein. Expanding must produce terms which agree with the Faulhaber expansion in terms of M. Substitute

n = [pic] + 1/2

giving

Fq(M) = ([pic] + 1/2 + B)2q+2/(2q+2) – B2q+2/(2q+2)

Expand binomially:

Fq(M) = (2q+2)-1 [pic][pic] (M + 1/4)(2q+2-i)/2 (1/2 + B)i – B2q+2/(2q+2)

Since no square roots involving M appear in the Faulhaber expansion, we expect to find that the expressions when i is odd will be 0 when we expand. In fact, this is because the latter factor has a closed form value:

(1/2 + B)i = (21-i - 1) Bi

(cf. Abramowitz & Stegun, 23.?.?).

Since Bi is 0 for all odd i except i=1, and the multiplicative factor is 0 for i=1, only even values of i survive in the summation. Therefore, substitute i:=2j immediately:

Fq(M) = (2q+2)-1 [pic][pic] (M + 1/4)q+1-j (21-2j - 1) B2j – B2q+2/(2q+2)

Now expand the factor in M + 1/4 binomially:

Fq(M) = (2q+2)-1 [pic][pic][pic][pic] Mq+1-j-k (21-2j-2k - 4-k)B2j – B2q+2/(2q+2)

Exchange the inner and outer summation, and define coefficients Cq(i) by

Fq(M) := [pic]Cq(i) Mq+1-i – B2q+2/(2q+2)

where we substitute k:=i-j and we find a closed form:

Cq(i) := (2q+2)-1 [pic][pic][pic](21-2i – 4j-i)B2j

Let us compute some of these coefficients. The largest term in Fq(M) is where i=0. In this case, the summation definition of Cq(i) is over only j=0, so

Cq(0) = (2q+2)-1 (1)(1)(21 – 40)B0 = (2q+2)-1

which agrees with the formula for aq/(q+1) above. In the next smaller term, for i=1, the sum over j ranges over the indices 0 and 1, so

Cq(1) = (2q+2)-1 [pic][pic][pic](2-1 – 4-1)B0 + [pic][pic](2-1 – 40)B2[pic]

= (2q+2)-1 [(q+1)(1/4) + ((2q+2)(2q+1)/2)(–1/2)(1/6)]

= 1/8 + (2q+1)(–1/24)

= -(q-1)/12

which agrees exactly with -bq /q above. Next,

Cq(2) = (2q+2)-1 [pic][pic][pic](21-4 – 4j-2)B2j

= (2q+2)-1 [[pic][pic](2-3 – 40-2)B0

+ [pic][pic](2-3 – 41-2)B2

+ [pic][pic](2-3 – 42-2)B4]

= (2q+2)-1 [[pic](2-3 – 4-2)

+ [pic][pic](2-3 – 4-1)(1/6)

+ [pic](2-3 – 40)(-1/30)]

= (2q+2)-1 [[pic](1/16) + [pic]q(-1/8)(1/6) + [pic](-7/8)(-1/30)]

= q(q-2)(7q-1)/720

which agrees exactly with cq /(q-1) above.

This closed form expression for Cq(i) is not completely satisfactory, since it does not explicitly display the factors of binomial coefficients in q, which we have deduced experimentally above. Nor does it suggest a form for the polynomial in q which is a factor in each coefficient.

[Hint, hint. Prove at least the presence of factor (q above i). Or, try to compute the last coefficients in each Fq, especially that the last two have a ratio of -1/2.]

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