Why is ∑_(k=1)^n k^m a polynomial with degree m+1 in n?

Colten Andrade 2022-09-23 Answered
Why is k = 1 n k m a polynomial with degree m + 1 in n?
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Answers (2)

Solomon Fernandez
Answered 2022-09-24 Author has 5 answers
Let V be the space of all polynomials f : N 0 F (where F is a field of characteristic zero). Define the forward difference operator Δ f ( n ) = f ( n + 1 ) f ( n ). It is not hard to see that the forward difference of a polynomial of degree d is a polynomial of degree d 1, hence defines a linear operator V d V d 1 where V d is the space of polynomials of degree at most d. Note that dim V d = d + 1.
We want to think of Δ as a discrete analogue of the derivative, so it is natural to define the corresponding discrete analogue of the integral ( f ) ( n ) = k = 0 n 1 f ( k ). But of course we need to prove that this actually sends polynomials to polynomials. Since ( Δ f ) ( n ) = f ( n ) f ( 0 ) (the "fundamental theorem of discrete calculus"), it suffices to show that the forward difference is surjective as a linear operator V d V d 1 .
But by the "fundamental theorem," the image of the integral is precisely the subspace of V d of polynomials such that f ( 0 ) = 0, so the forward difference and integral define an isomorphism between V d 1 and this subspace.
More explicitly, you can observe that Δ is upper triangular in the standard basis, work by induction, or use the Newton basis 1 , n , ( n 2 ) , ( n 3 ) , . . . for the space of polynomials. In this basis we have Δ ( n k ) = ( n k 1 ) , and now the result is really obvious.
The method of finite differences provides a fairly clean way to derive a formula for n m for fixed m. In fact, for any polynomial f ( n ) we have the "discrete Taylor formula"
f ( n ) = k 0 Δ k f ( 0 ) ( n k )
and it's easy to compute the numbers Δ k f ( 0 ) using a finite difference table and then to replace ( n k ) by ( n k + 1 ) . I wrote a blog post that explains this, but it's getting harder to find.
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Thordiswl
Answered 2022-09-25 Author has 2 answers
The formula just drops right out if we use the Euler Maclaurin Summation Formula.
For f ( x ) = x m we have
k = 0 n f ( k ) = 0 n f ( x )   d x + n m 2 + j = 0 B 2 j ( 2 j ) ! ( f ( 2 j 1 ) ( n ) f ( 2 j 1 ) ( 0 ) )
Where B j are the Bernoulli numbers and f ( j ) ( x ) is the j t h derivative of f .
Since f ( x ) is polynomial, the terms in
j = 0 B 2 j ( 2 j ) ! ( f ( 2 j 1 ) ( n ) f ( 2 j 1 ) ( 0 ) )
all are zero after a point 2 j 1 > m and thus we get the formula for
as a polynomial in n , with degree m + 1.
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