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

 Messages: [ Previous | Next ]
 Terence Gaffney Posts: 45 Registered: 12/6/04
Re: Derivation of the Sum of a Series
Posted: Sep 24, 2001 10:37 AM

I don't mean to encourage more questions like Eric's but he does raise an
interesting issue.

Consider the problem:

Find a polynomial in n whose value at n is the sum from i=1 to i=n of i^k,
k a positive integer.

There are a couple of nice points you can make.

By a geometric argument you can see that this polynomial is less than
n^{k+1}, for all n, hence must be a polynomial of degree less than or
equal to k+1. For example, if k=1, then look at an n by n square. The
terms of the sum are then the areas of squares of height i and base 1. If
k=2, then the terms of the sums are volumes with base dimensions 1 and i
and height i, and so on. This geometric argument can also be used to
illustrate the usual derivation when k=1.

So what do you know about this polynomial P? Well, you know that

P(n)-P(n-1)=n^k

This is very powerful; two polynomials are equal iff all of their
coefficients are equal.

You also know that P(0)=0. This means that the constant term of P is zero.

So look at a general polynomial of degree k+1, in n, evaluate the same
polynomial at n-1, subtract the second from the first, and set the
coefficients equal to the coefficients of n^k. You can then work out the
coefficients starting with the coefficient of n^{k+1}. You can do this
for k=2, and have your students do it for k=3.

If there are other points you like to make with your students about this
topic, Id be interested in hearing about it.

Best,
Terry Gaffney

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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