p2(x) = x 2+x+. 4x + 6x 2 + 2x. P~(~) = 4. Then p~(z)dx = k and p~(z)dz = --1. 2 ... --1. fOn. = p2(x)dx = n(n + 1)(2n + 1). 6. Also we have. $3(n)=13+2 3 + -"+n 3. /0.
CLASSROOM N Marikannan and V Ravichandran Sri Venkateswara College of Engineering Sriperumbudur 602 105, India.
Introduction We know t h a t the sum of the first n n a t u r a l n u m b e r s is n(n + 1)/2. This can be proved easily in several ways. For example, if S denotes the s u m of first n n a t u r a l numbers, t h e n S = ~i=1 ~ i = ~ ='~1 ( n + l - ~ )" a n d therefore 2S = ~ l ( i + n + 1 - i ) = n(n+ 1) which gives S = n(n+ 1)/2. One can also use the principle of m a t h e m a t i c a l induction to prove the result. Here we present a proof of the result using integration. Let Sk(n) = ~ = 1 ik be the sum of k t h powers of first n n a t u r a l numbers. Let 1 p2(x)
=
1
x 2+x+
4x ~ + 6x 2 + 2x P~(~)
=
4
Then
p~(z)dx
= k
and
p~(z)dz =
--1
2
'
and
Sl(n)
=
l+2+...+n
/0
pl(x)dx + f~
=
--1
/2
pl(x)dx + . . . +
pl(x)dx
px(x)dx - n(n + 1) 2
Similarly we have Keywords Bernoulli polynomials, Vandermonde delerminanl.
80
~ k p2(x)dx = k 2
and
-1
RESONANCE ( February 2003
CLASSROOM
The sum of k th
f n p 2 ( x ) d x = n ( n + 1)(2n + 1) 6 Jo
powers of the first n natural numbers
Therefore we have
is a polynomial of
S2(n) = 12 + 2 2 + = =
.
.
.
/oI p 2 ( x ) d x fOn
degree k+l without
q-n 2
+...
+
L --1 V2(x)dx
constant term.
p 2 ( x ) d x = n ( n + 1)(2n + 1) 6
Also we have $3(n)=13+2
3 + = =
-"+n
3
/0 /o;
p3(x)dx+...+ v3(x)dx-
/n
_lpa(x)dx
[n(n + 1)] 2 4
T h e s e e x a m p l e s suggest t h e p r o b l e m of finding a function p k ( x ) such t h a t ik =
1
pk(x)dx.
T h i s f u n c t i o n pk(X) can be used to represent t h e s u m of t h e k t h powers of t h e first n n a t u r a l n u m b e r s as an integral: i k + 2k + ... + n k =
pk(x)dx.
In this article, we show t h a t such a p o l y n o m i a l pk(x) exists a n d we prove a recurrence relation satisfied by pk(x). It t u r n s o u t from this recurrence relation t h a t p'k(x) = k p k _ l ( x ) . Also, we prove t h a t t h e s u m of k t h powers of t h e first n n a t u r a l n u m b e r s is a p o l y n o m i a l of degree k + 1 w i t h o u t c o n s t a n t t e r m a n d o b t a i n a nice relation b e t w e e n t h e s u m of powers of n a t u r a l n u m b e r s and the Vandermonde determinant.
RE.SONANCE I February
2003
81
CLASSROOM
Sum of Powers by Integration Let us start with the assumption t h a t such a polynomial Pk (x) exists for each k and analyse w h a t conditions t h e y must satisfy. As we shall see, these conditions allow us to actually define the polynomials. Since
(i+l)k+~--ik+~= ( k ~ l ) ik+(k~l)ik-~+'"+(~++i) we have (k + 1) -1
(x + 1)kdx = ~_, r=l
r
-1
pk+l_~(x)dx. define the
Therefore, it is clear t h a t we m a y actually polynomials pk(x) recursively by
k+l
pa(x) = (x + 1) k - (k+l)pk-l(X) +'''+ (k+I)PO(X) k+l This recurrence relation along with po(x) = 1 can be used to find pk(x) for any k > 0. For example, we have
1 pl(X)
=
X A-
p2(x)
=
z 2+x+-6
p3(x)
=
4
p4(x)
=
X4 q - 2 x 3 - f - x 2 -
1
4x a + 6x 2 + 2x 1
3O 6x 5 + 15x 4 +
ps(x)
=
p6(x)
=
10x 3 --
x
6 x6+3x 5+~x 4
x 2
1
z
In fact, it follows t h a t Pk is a polynomial of degree k and has rational coefficients. Notice t h a t p'(x) = np,,_l (x) for n = 1, 2 , . . . , 6. In general we can easily prove t h a t it is true for any integer n by using induction.
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RESONANCE I February 2003
CLASSROOM
The polynomials Pn are closely related to what are known in the literature as Bernoulli polynomials Bn; indeed p,~(x) = B , , ( x + 1). Vandermonde Determinant and Sum of Powers
Consider the equations
for i = 1, 2 , . . . , n. On adding them, we get
n+l,k+l, =
1
2
1) S k - l ( n )
+ . . . . 4-
(k+l~
k+~/So (n).
As go(n) = n, this equality proves easily by induction on k that S k ( n ) is a polynomial in n of degree k + 1 without constant term. Also the coefficient of n k+l in S k ( n ) is 1 This can be proved from the fact that pk(x) is a k+l" polynomial of degree k with the coefficient of x k is one by using the integral representation Sk(n) = f ~ p k ( x ) d x . Since S k ( n ) is a polynomial in n of de~ee k + 1 without constant term, we can write S k ( n ) = a l n + a2n 2 + a3n 3 + . . . + ak+ln k+l.
One can find the coefficients ai by substituting the values n = 1 , . . . , k + 1 and solving the corresponding matrix equation 1 2 9
(k + 1)
... 999 .
9 .
...
/ (al 2 1+1 '
(k + 1) k+l/
SONANCE I February 2003
a2
sk(1) sk(2)
"
ak+l
Sk(k + 1)
83
CLASSROOM
Note t h a t the d e t e r m i n a n t of the coefficient m a t r i x M is (k + 1)! times the d e t e r m i n a n t of
1
1
..-
1 )
1
2
...
2k
:
9
1
...
(k+l)
...
:
(k+l) k
which is nonzero; the latter factor is the V a n d e r m o n d e determinant. As we have
(al) a2
=
M_1r
a k!+l this gives an interesting connection between the s u m of kth powers of the first n n a t u r a l n u m b e r s a n d the V a n d e r m o n d e determinant.
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RESONANCE I February 2003
