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F(n)=(1/2)f(n+1)+(1/2)f(n-1)-1I got f(n)=n^2. I can not find

  1. Aug 25, 2012 #1

    dyh

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    f(n)=(1/2)f(n+1)+(1/2)f(n-1)-1

    I got f(n)=n^2.

    I can not find anymore solution except this.

    I just wonder there are some more solutions about this problem or not.

    I think there are more, but I don't know how to get to them.

    I wanna hear your opinion.

    Thanks
     
  2. jcsd
  3. Aug 25, 2012 #2

    Bacle2

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  4. Aug 25, 2012 #3

    dyh

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    Re: f(n)=(1/2)f(n+1)+(1/2)f(n-1)-1

    Thanks
    I know how to solve the general methods for solving recursions.

    But I think this is some special case of it because there is some contradiction in particular solution.
    I.e. when I put f(n)= c (constant) for non-negative integer n. then
    I can get

    c=(1/2)c+(1/2)c-1

    so it looks like 0=-1

    So, I need to try "cn" but there is also contradiction in this case similar to previous case.

    then I need to try "cn^2". In this case, if I choose c as 1 then it can be a solution.

    So, I just would like to know this is the only solution or not.

    If not, I wanna know how to get other solutions.
     
  5. Aug 25, 2012 #4

    dyh

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    Re: f(n)=(1/2)f(n+1)+(1/2)f(n-1)-1

    oh.. I got some other general solution form

    f(n) = a+bn+n^2, a,b are constant

    But I still would like to know there would be more solutions or not. hehe
     
  6. Aug 26, 2012 #5

    Bacle2

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    Re: f(n)=(1/2)f(n+1)+(1/2)f(n-1)-1

    Sorry, I don't see it. Don't you have to choose the first two terms to find out the

    value of f(n+1). Then, if you choose f(n)=f(n-1)=c , you get f(n+1)=2+c;

    I don't see any contradiction.
     
  7. Aug 26, 2012 #6

    dyh

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    Re: f(n)=(1/2)f(n+1)+(1/2)f(n-1)-1

    In my opinion,

    if I choose f(n)=f(n-1)=c for any non-negative integer n, then f(n+1) would be "c" too
     
  8. Aug 26, 2012 #7

    Bacle2

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    Re: f(n)=(1/2)f(n+1)+(1/2)f(n-1)-1

    O.K, let's see:

    f(n)=(1/2)f(n+1)+ (1/2)f(n-1)-1

    Set f(n)=c=f(n-1) . Then,

    c= (1/2)f(n+1)+c/2-1 , so:

    1+ c- c/2 = (1/2)f(n+1), so :

    1+c/2= (2+c)/2=(1/2)f(n+1) , so f(n+1)=2+c .

    That's what I get.
     
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