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

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Discussion Overview

The discussion revolves around the recurrence relation f(n)=(1/2)f(n+1)+(1/2)f(n-1)-1. Participants explore potential solutions, particularly focusing on whether f(n)=n^2 is the only solution or if other forms exist.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes f(n)=n^2 as a solution but expresses uncertainty about the existence of other solutions.
  • Another participant suggests that general methods for solving recursions may apply to this case.
  • A participant notes contradictions arise when testing constant solutions, leading them to try polynomial forms like "cn" and "cn^2".
  • One participant claims to have derived a general solution form f(n) = a + bn + n^2, where a and b are constants, but still questions if more solutions exist.
  • Some participants discuss the implications of setting f(n)=f(n-1)=c and derive f(n+1) in terms of c, leading to different interpretations of the results.
  • There is a contention regarding the validity of assuming constant values for f(n) and how it affects the recurrence relation.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the solutions, with some believing that f(n)=n^2 may not be the only solution, while others challenge the assumptions made in deriving solutions. The discussion remains unresolved regarding the completeness of the solution set.

Contextual Notes

Participants highlight contradictions encountered when testing specific forms of solutions, indicating that the assumptions made in their reasoning may affect the outcomes. The discussion does not resolve these contradictions.

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 want to hear your opinion.

Thanks
 
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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.
 


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
 


dyh said:
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.
 


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
 


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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