Here is a cute problem I first encountered many years ago:
A polynomial of degree satisfies for . Determine .
Recall that a polynomial of degree is uniquely determined by its value at points. Hence could have only one possible value. Is it ? Let’s see: here is the solution I saw at the time.
Consider the polynomial . Then is a polynomial of degree (the ‘th term has degree ). Furthermore, since for , we have for the following:
(Here we recall that the sum of a row of Pascal’s triangle is a power of 2.)
This shows that is our unique polynomial with the desired properties, and we find
Neat isn’t it that the polynomial almost climbs up to but ends up one short. 🙂 Let’s see how this result generalises. Modify the above question so that for (where is a constant not equal to 1). What is ? This time we let
As before is clearly a degree polynomial. Furthermore, for we apply the binomial theorem to obtain
This constructed polynomial has the desired properties, so by uniqueness we simply substitute to find :
The way such polynomials can be constructed (i.e. not pulled out of thin air!) is to write down a finite difference table – refer to an earlier blog post here: if the first diagonal of the table is where , then
In the first problem it is easy to verify that , in the generalisation .