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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
  
  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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  Date Subject Author
  9/23/01 Eric Abramovich
  9/24/01 Terence Gaffney
  9/24/01 Matthias Kawski
  9/25/01 Terence Gaffney
  9/24/01 me@talmanl1.mscd.edu