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Math Help - recursive function

  1. #1
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    recursive function

    I have a recursive function and I seem to be having a problem deducing a general solution for it:
    u_{n+2} = u_{n+1}^2 + u_{n}^2-2
    The initial conditions are u_{o} = 1, u_{1} = 1
    Some of the values are as followed:
    u_{o}=1, u_{1}=1,u_{2}=0,u_{3}=-1,u_{4}=-1,u_{5}=0,u_{6}=-1,u_{7}=-1,u_{8}=0
    My initial thought was that it was a periodic function. But I am as yet unable to begin to formulate a general solution,
    Can anyone help. Thank you.
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  2. #2
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    Quote Originally Posted by riptorn70 View Post
    I have a recursive function and I seem to be having a problem deducing a general solution for it:
    u_{n+2} = u_{n+1}^2 + u_{n}^2-2
    The initial conditions are u_{o} = 1, u_{1} = 1
    Some of the values are as followed:
    u_{o}=1, u_{1}=1,u_{2}=0,u_{3}=-1,u_{4}=-1,u_{5}=0,u_{6}=-1,u_{7}=-1,u_{8}=0
    My initial thought was that it was a periodic function. But I am as yet unable to begin to formulate a general solution,
    Can anyone help. Thank you.
    Hi

    Your initial thought is correct :
    u_{2}=0,u_{3}=-1,u_{4}=-1
    u_{5}=0,u_{6}=-1,u_{7}=-1
    u_{8}=0 etc ...
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  3. #3
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    Thank you running_gag for your input. I believe I may have the embroyo of a general solution, as followed: (do you think i'm on the right track)


    <br />
u_{n} = 1, \mbox{ }n\leq 1<br />

    <br />
u_{n}=\left\{\begin{array}{cc}<br />
0,&\mbox{ if } n\equiv 2 \bmod 3\\-1, & \mbox{ if } n\equiv 0 \bmod 3<br />
\\-1, & \mbox{ if } n\equiv 1 \bmod 3\end{array}\right.<br />

    But due to the fact that it is periodic do you think I should be looking more to trigonometric functions for a more elegant solution i.e the sine and cosine functions. Any input or ideas would be appreciated. Thank you
    Last edited by riptorn70; December 4th 2008 at 03:11 PM. Reason: Notice error in general solution
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  4. #4
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    Quote Originally Posted by riptorn70 View Post
    Thank you running_gag for your input. I believe I may have the embroyo of a general solution, as followed: (do you think i'm on the right track)


    <br />
u_{n} = 1, \mbox{ }n\leq 1<br />

    <br />
u_{n}=\left\{\begin{array}{cc}<br />
0,&\mbox{ if } n\equiv 2 \bmod 3\\-1, & \mbox{ if } n\equiv 0 \bmod 3<br />
\\-1, & \mbox{ if } n\equiv 1 \bmod 3\end{array}\right.<br />

    But due to the fact that it is periodic do you think I should be looking more to trigonometric functions for a more elegant solution i.e the sine and cosine functions. Any input or ideas would be appreciated. Thank you
    Maybe you could but I think that recursivity is a good way to demonstrate your idea
    Let P(n) be u_{3n+2} = 0, u_{3n+3} = -1, u_{3n+4} = -1

    P(0) is true since u_{2} = 0, u_{3} = -1, u_{4} = -1
    It is easy to demonstrate that if P(k) is true then P(k+1) is true
    Which demonstrates P(n) for every n
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  5. #5
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    Great!! thank you for your input. The inductive reasoning is very helpful thankyou.
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