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Re: Derivation of the Sum of a Series
Posted:
Sep 25, 2001 12:43 PM


On Mon, 24 Sep 2001, Matthias Kawski wrote:
> > The more interesting qurestion is: why should it be a priori > clear that this has a polynomial solution? > > Matthias > *********************************************************
Matthias asks a good question. Its difficult to answer because there are functions which are not polynomials which agree with the desired polynomial at whole numbers. Whenever there is no unique solution to a problem its harder to show that a solution of the type you want exists. The proof that there is a polynomial comes from the construction. But there are plausibility arguments that such a polynomial exists.
If you are doing divided differences with your students then they know that if the first divided difference of a table is linear then a quadratic function fits the data of the original table. If they look at a table of values of the sum of the first n squares, then the divided differences are squares, so the second differences are linear, so its reasonable that a cubic fits the data of the original table.
If you encounter these sums in calculus, then they are the upper sums of the integral of x^k, from 0 to n , using the integers to partition the interval from 0 to n. Thus they are greater than (n^{k+1})/k+1. The geometric argument of my last post shows that they are less than n^{k+1}. This makes it reasonable to hope that we are looking for a polynomial of degree k+1.

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