Proof on this word problem

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October 16th 2012, 10:08 AM
gfbrd

Proof on this word problem

Hey there I need some help with this proof.

There are 2000 points on a circle, and each point is given a number that is equal to the average of its two nearest neighbors. Show that all the numbers must be equal.

I am not sure how to start this. I hope someone can guide me step by step on how to do this proof, so that I can understand this better.
October 16th 2012, 10:15 AM
MarkFL

Re: Proof on this word problem

Think of the sequence of numbers...what type of sequence must they be?
October 16th 2012, 10:24 AM
gfbrd

Re: Proof on this word problem

I'm not exactly sure what this sequence is called but this pattern is that the term your on is the average of the 2 closest neighbors, so it is the previous term plus the next term divided by 2?

Sorry if I'm way off, I take a longer time to learn than others
October 16th 2012, 11:21 AM
MarkFL

Re: Proof on this word problem

Let the sequence of numbers be denoted by $x_0,x_1,x_2,\cdots,x_{1999}$

We require:

$x_n=\frac{x_{n-1}+x_{n+1}}{2}$ where $1\le n\le1998$ and $n\in\mathbb{Z}$

This results in the homogeneous recursion:

$x_{n+1}=2x_n-x_{n-1}$

whose characteristic equation is:

$(r-1)^2=0$ hence:

$x_n=k_1+k_2n$

and so we see we must have an arithmetic sequence. We then may write:

$x_0=k_1$ and so we have:

$x_n=x_0+k_2n$

Now, we also know we must have:

$x_0=\frac{x_{1999}+x_1}{2}=\frac{x_0+1999k_2+x_0+k _2}{2}$

$x_0=\frac{2x_0+2000k_2}{2}=x_0+1000k_2$ hence:

$k_2=0$ and so we have:

$x_n=x_0$
October 16th 2012, 02:41 PM
gfbrd

Re: Proof on this word problem

Thank you so much for your help, I understand this a lot better now