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Topic: Derivation of the Sum of a Series
Replies: 4   Last Post: Sep 25, 2001 12:43 PM

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Terence Gaffney

Posts: 45
Registered: 12/6/04
Re: Derivation of the Sum of a Series
Posted: Sep 25, 2001 12:43 PM
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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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