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Math Help - Proof on this word problem

  1. #1
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    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.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Proof on this word problem

    Think of the sequence of numbers...what type of sequence must they be?
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  3. #3
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    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
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  4. #4
    MHF Contributor MarkFL's Avatar
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    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
    Thanks from gfbrd
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  5. #5
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    Re: Proof on this word problem

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