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Math Help - Function Help!

  1. #1
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    Function Help!

    The question is:

    f is a function defined on Z. Assume that f satisfies the following properties:

    f(0) doesn't equal 0
    f(1) equals 3
    f(x) * f(y) = f(x+y) + f(x-y)

    1. Determine f(7).
    2. Prove that f(n)=f(-n) for all integers n.

    ...I'm not sure even how to begin with this problem. Does anyone have any hints to get me started??? I'm sure after that I'll be fine. Thanks
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  2. #2
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    The second is not too hard. The key is that f(x) * f(y) = f(y) * f(x), so f(x + y) + f(x - y) = f(x + y) + f(y - x). Therefore, f(x - y) = f(y - x). Then substitute x = 0.

    To find f(7), start with small numbers. E.g., f(1) * f(0) = f(1) + f(1), which gives f(0).
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chubbs145 View Post
    The question is:

    f is a function defined on Z. Assume that f satisfies the following properties:

    f(0) doesn't equal 0
    f(1) equals 3
    f(x) * f(y) = f(x+y) + f(x-y)

    1. Determine f(7).
    2. Prove that f(n)=f(-n) for all integers n.

    ...I'm not sure even how to begin with this problem. Does anyone have any hints to get me started??? I'm sure after that I'll be fine. Thanks
    Forget \mathbb{Z} and just look at \mathbb{N} (you can do this because of what emakarov showed). Then we get one of the consequences (letting x=n,y=1) is that f(n)f(1)=f(n+1)+f(n-1)\implies f(n+1)=3f(n)-f(n-1). Then, using the basic facts about linear homogenous recurrence relations we can compute exactly what f is
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    Quote Originally Posted by chubbs145 View Post
    The question is:

    f is a function defined on Z. Assume that f satisfies the following properties:

    f(0) doesn't equal 0
    f(1) equals 3
    f(x) * f(y) = f(x+y) + f(x-y)

    1. Determine f(7).
    2. Prove that f(n)=f(-n) for all integers n.

    ...I'm not sure even how to begin with this problem. Does anyone have any hints to get me started??? I'm sure after that I'll be fine. Thanks


    So 2f(0)=f(0+0)+f(0-0)=f(0)^2\Longrightarrow f(0)=...? , as we're given it is not zero...

    After that, you get f(n)+f(-n)=f(0+n)+f(0-n)=f(0)f(n) ... , and this solves (2)

    Finally, f(2)+f(0)=f(1+1)+f(1-1)=f(1)^2=9\Longrightarrow f(2)=9-f(0) , and etc.
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