Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: induction on finite set.
Replies: 2   Last Post: Feb 26, 2013 5:21 AM

 Messages: [ Previous | Next ] Topics: [ Previous | Next ]
 johnykeets Posts: 8 From: usa Registered: 1/1/13
Re: induction on finite set.
Posted: Feb 26, 2013 5:21 AM

The principle of finite induction can be derived from the fact that every nonempty set of natural numbers has a smallest element. This fact is known as the well-ordering principle for natural numbers. The finite induction therefore is not that much brader.
For any given positive integer N, we have 2N-1 real variables x_k \in [0, 1], with 1 \leq k \leq 2N-1.
We also know x_1 = 1 = x_{2N-1}, and for all other k, x_k < 1. Additionally, we may use the notation
S_k = \sum_{j = 1}^{k} x_j
We have a recurrence that is valid for 1 \leq k \leq 2N - 2:
S_k = x_k + (1 - x_k)S_{k+1}

We can manipulate that in a variety of ways, such as the following which are valid (if I haven?t made a mistake) when 1 < k < 2N - 1
If our aim is to find a general form of x_k as a function of k, for each positive integer N. A general form for S_k would be nice as well. My preliminary attempts suggest perhaps there will be symmetry, with x_{N-n} = x_{N+n}, and I would like to prove or disprove that. I also suspect that there will be N-1 polynomials p_k of degree N, such that for 1 < k \leq N we will have p_k(x_k) = 0. I?d like to know whether that is true or false.

algebra homework help
online maths tutoring

Date Subject Author
2/20/13 sean bruce
2/24/13 grei
2/26/13 johnykeets