Recursive Formulae for Power Sums [expository notes]

4 Pages·2014·0.241 MB·English

by  Carl Wagner

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Recursive Formulae for Power Sums 1 April 2013
Carl Wagner
If r and n are nonnegative integers, the rthpower sum S (n) is defined by
r
n
(1) S (n):kr, where 00:1.
r
k0
In particular,
(2) S (n)1n.
0
Along with (2), the following recurrence provides a recursive formula for S (n):
r
1 r1r1
(3) S (n) {(n1)r1 S (n)}, for all r 1.
 
r r1  i  i
i0
Proof of (3): By the binomial theorem,
r r1
(4) (k1)r1kr1  ki,
 
 i 
i0
and so
n n r r1 r r1 n r r1
(5) [(k1)r1kr1] ki  ki  S (n).
     
 i   i   i  i
k0 k0 i0 i0 k0 i0
Since the left-most sum above telescopes to (n1)r1, this yields the implicit recurrence
r r1
(6)  S (n)(n1)r1,
 
 i  i
i0
which yields (3) upon solving for S (n). See Wagner [2] for a combinatorial proof of (6). □
r
From (2) and (3) it follows by induction on r that S (n) is a polynomial in n of degreer1.
r
Accordingly, let us write
r1
(7) S (n)a(r, j)nj.
r
j0
From (2) and the recurrence (3), we may compute
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1 1
(8) S (n) n n2,
1 2 2
1 1 1
(9) S (n) n n2  n3,
2 6 2 3
1 1 1
(10) S (n) n2  n3 n4, and
3 4 2 4
1 1 1 1
(11) S (n) n n3 n4  n5, etc.
4 30 3 2 5
Tabulating the coefficients of these polynomials clarifies the connection between the coefficients
of S (n) and those of S (n).
r r1
Table 1. a(r, j) for 0r4 and 0 j5
j 0 j 1 j 2 j=3 j 4 j 5
r 0 1 1
r 1 0 ½ ½
r 2 0 1/6 ½ 1/3
r 3 0 0 ¼ ½ ¼
r 4 0 -1/30 0 1/3 ½ 1/5
A careful examination of the above table leads us to conjecture the following theorem:
Theorem. For all r 1, a(r,0)0 and
r1 r1
(12) a(r, j)a(r, j)1, whence,
j0 j1
r1
(13) a(r,1)1 a(r, j).
j2
For all r 1 and j2,
r
(14) a(r, j) a(r1, j1).
j
3

r1
Proof (Owens [1]). Setting n1 in (7) yields a(r, j)S (1)0r 1r 1. Setting n0 in
r
j0
(7) for r 1 yields a(r,0)S (0)0r 0. Now consider the polynomial
r
r1
(15) S (x):a(r, j)xj.
r
j0
In what follows we use the easily established fact that, applied to a polynomial p(x), the finite
difference operator  and the differentiation operator D commute, i.e.,
(16) (Dp(x))D(p(x)).
Now by (1), it is the case that for fixed r 0 and all n0, S (n1)S (n)(n1)r, which
r r
implies the polynomial identity
(17)  S (x)S (x1)S (x)(x1)r.
r r r
From (16) and (17) it follows that
(18) (DS (x))D( S (x))D(x1)r= r(x1)r1 r S (x)rS (x).
r r r1 r1
Hence,
(19) (D S (x)rS (x)0,
r r1
from which it follows that the polynomial DS (x)rS (x)is equal to a constant1, namely, the
r r1
constant c DS (0) rS (0)= a(r,1)r . Hence,
r r1 r,1
(20) DS (x)rS (x)c, which implies that
r r1
r1 r r
(21)  ja(r, j)xj1 (j1)a(r, j1)xj  ra(r1, j)xj+ c.
j1 j0 j0
Comparing coefficients of xj for j1 yields
(22) (j1)a(r, j1)ra(r1, j) for all r, j1, i.e.,
(23) ja(r, j)ra(r1, j1) for all r 1 and all j2. □
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Remark. The numbers B defined by B a(0,1), B a(1,1), and B a(r,1) for all r 2,
r 0 1 r
are called Bernoulli numbers. We have
B 1,B 1/2,B 1/6,B 0,B 1/30,B 0,B 1/42,B 0,B 1/30,B 0,B 5/66,
0 1 2 3 4 5 6 7 8 9 10
B 0, B 691/2730, etc. It may be proved that B 0 for all odd r 3, and that B
11 12 r r
alternates in sign for r even. If we define a function : RR by
x x
(24) (x) if x0 and (0)1 (lim ) ,
ex 1 x0 ex 1
then
 xr  xr
(25) B (x) (and so a(r,1) (x)x) for |x|<2.
r r! r!
r0 r0

Notes
m x m  x 
1 If p(x)c , and p(x)c 0, then c c 0, since the set of
   
j j j j1 1 m
j0 j1
 x 
polynomials { , j 1,...,m} is a basis for the vector space of polynomials of degree
 
 j1
m1. So p(x)c .
0

References
1. R. Owens, Sums of powers of integers, Mathematics Magazine 65 (1992), 38-40.
2. C. Wagner, Combinatorial proofs of formulas for power sums, Archiv der Mathematik 68
(1997), 464-467.