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

[dyh]

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 want to hear your opinion.

Thanks

 

[Bacle2]

Have you considered general methods for solving recursions:

http://en.wikipedia.org/wiki/Recurrence_relation ?

 

[dyh]

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 want to know how to get other solutions.

 

[dyh]

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

 

[Bacle2]

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 want to know how to get other solutions.

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.

 

[dyh]

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

 

[Bacle2]

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.