D dwsmith MHF Hall of Honor Mar 2010 3,093 582 Florida May 10, 2010 #1 Let f be a function such that \(\displaystyle f(n+1)=1-[f(n)]^2, \forall n\in\mathbb{Z}, n\geq0\). How would f(n+2) be expressed in terms of f(n)?

Let f be a function such that \(\displaystyle f(n+1)=1-[f(n)]^2, \forall n\in\mathbb{Z}, n\geq0\). How would f(n+2) be expressed in terms of f(n)?

Anonymous1 Nov 2009 517 130 Big Red, NY May 10, 2010 #2 dwsmith said: Let f be a function such that \(\displaystyle f(n+1)=1-[f(n)]^2, \forall n\in\mathbb{Z}, n\geq0\). How would f(n+2) be expressed in terms of f(n)? Click to expand... \(\displaystyle f(n+1)=1-[f(n)]^2\) \(\displaystyle \implies f(n+2) = f\left\{(n+1)+1\right\}=1-[f(n+1)]^2 = 1 - \left\{ 1-[f(n)]^2 \right\}^2\) Reactions: dwsmith

dwsmith said: Let f be a function such that \(\displaystyle f(n+1)=1-[f(n)]^2, \forall n\in\mathbb{Z}, n\geq0\). How would f(n+2) be expressed in terms of f(n)? Click to expand... \(\displaystyle f(n+1)=1-[f(n)]^2\) \(\displaystyle \implies f(n+2) = f\left\{(n+1)+1\right\}=1-[f(n+1)]^2 = 1 - \left\{ 1-[f(n)]^2 \right\}^2\)

D dwsmith MHF Hall of Honor Mar 2010 3,093 582 Florida May 10, 2010 #3 Ok that trumps the ... answer first posed.

Anonymous1 Nov 2009 517 130 Big Red, NY May 10, 2010 #4 dwsmith said: I am almost positive that isn't the answer but I could be wrong. Click to expand... Which step do you disagree with, particularly?

dwsmith said: I am almost positive that isn't the answer but I could be wrong. Click to expand... Which step do you disagree with, particularly?

D dwsmith MHF Hall of Honor Mar 2010 3,093 582 Florida May 10, 2010 #5 Anonymous1 said: Which step do you disagree with, particularly? Click to expand... I don't disagree with that but when your response was first .... I was in disagreement.

Anonymous1 said: Which step do you disagree with, particularly? Click to expand... I don't disagree with that but when your response was first .... I was in disagreement.

Anonymous1 Nov 2009 517 130 Big Red, NY May 10, 2010 #6 dwsmith said: I don't disagree with that but when your response was first .... I was in disagreement. Click to expand... Oh yeah that, haha. When I am fairly certain I know the answer, I like to secure my position. (Evilgrin) Cheers!

dwsmith said: I don't disagree with that but when your response was first .... I was in disagreement. Click to expand... Oh yeah that, haha. When I am fairly certain I know the answer, I like to secure my position. (Evilgrin) Cheers!