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Topic:
1 + 2 + ... + n a polynomial how?
Replies:
5
Last Post:
Apr 2, 2013 11:55 AM




Re: 1 + 2 + ... + n a polynomial how?
Posted:
Apr 2, 2013 11:55 AM


On 01 Apr 2013 19:18:54 +0300, Jussi Piitulainen <jpiitula@ling.helsinki.fi> wrote:
>Is it obvious that 1 + 2 + ... + n is a polynomial of degree 2? How? > >I mean the sum of the first n positive integers. I would like to see >that it is a polynomial of degree 2 _without using_ the fact that it >is equal to n(n + 1)/2. Zeilberger (his new Opinion 129) says Gauss >could have used the polynomiality of the sum to support the equality, >rather than the other way around.
We'll write sequences using parentheses instead of subscripts, because we need subscripts to index sequences of sequences: If s is a sequence then s is the sequence
s = (s(0), s(1), ...) .
We'll say a sequence "is" a polynomial if there exists a polynomial p with s(n) = p(n).
Given a sequence s, we define the sequence Ds by
(Ds)(n) = s(n+1)  s(n).
Various people have replied that the reason your sequence s is a polynomial is that Ds is a polynomial. You _knew_ that Ds is a polynomial, so just saying that it follows that s is a polynomial is presumably no help, you want to know _why_ it follows that s is a polynomial.
Thought about it for a minute, came up with a proof with a reasonably high concept/calculation ratio.
Say P_N is the space of all polynomials of degree less than or equal to N. The result follows from this:
Prop. If s is in P_N then there exists q in P_{N+1} with Dq = s.
That's not exactly what you need, but you know how to fill the tiny gap.
Lemma. If p_j is a polynomial of degrree exactly j then p_0, ... , p_N are linearly independent.
The grubbing around with coefficients and systems of equations if contained in the proof of the lemma. But luckily there's no need to write that proof down, because I'm sure you agree the lemma is obvious.
Let b_j denote the monomial sequence
b_j(n) = n^j.
So in particular b_0, ... b_N is a basis for P_N.
It's clear that q in P_{N+1} implies Dq is in P_N.
So we fix N and regard D as an operator from P_{N+1} to P_N.
Now it's clear that the degree of Db_j is exactly j  1, at least for j >= 1. Hence the lemma shows that Db_1, ... , Db_{N+1} are independent.
So D(P_{N+1}) is a subspace of P_N, and the dimension of D(P_{N+1}) is at least N+1. Hence D(P_{N+1}) = P_N. QED.
> >Thanks for any insight.



