My Math Forum Intermediate Values

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 November 21st, 2011, 05:15 PM #1 Member   Joined: Oct 2011 Posts: 35 Thanks: 0 Intermediate Values Suppose that f is continuous on [0,1], and maps [0,1] into [0,1] Take any point c in [0,1] and define a sequence {Xn} inductively by X1= c, Xn+1=f(Xn). Suppose that {Xn} converges to the point p. Prove that p is a fixed point of f(x) I have no idea how this works.. smart people please HELP!
 November 21st, 2011, 05:44 PM #2 Member   Joined: Dec 2010 From: Miami, FL Posts: 96 Thanks: 0 Re: Intermediate Values Since $f$ is a continuous function and $x_{n} \rightarrow p$, the following folds: $\lim_{n \rightarrow \infty} f(x_{n})= f( \lim_{n \rightarrow \infty} x_{n})$
 November 21st, 2011, 06:08 PM #3 Member   Joined: Oct 2011 Posts: 35 Thanks: 0 Re: Intermediate Values any idea on how to prove p is a fixed point of f(x)?
 November 21st, 2011, 06:15 PM #4 Member   Joined: Dec 2010 From: Miami, FL Posts: 96 Thanks: 0 Re: Intermediate Values Consider $f(p)$ and apply what I mentioned above.
 November 21st, 2011, 06:28 PM #5 Member   Joined: Oct 2011 Posts: 35 Thanks: 0 Re: Intermediate Values i'm slowly trying to piece this together.. so if n=1 then Xn+1 = X2 = f(X1) if n=2 then Xn+1 = X3 = f(X2) = f(f(X1)) if n=3 then Xn+1 = X4 = f(X3) = f(f(f(X1))) and so on... maybe?
 November 21st, 2011, 06:43 PM #6 Member   Joined: Dec 2010 From: Miami, FL Posts: 96 Thanks: 0 Re: Intermediate Values We want to show $f(p)= p$. We know that $x_{n} \rightarrow p$, so that $f(p)= f(\lim_{n \rightarrow \infty} x_n)$, and $x_{n+1}= f(x_{n})$. We just have to piece it all together.
 November 21st, 2011, 07:00 PM #7 Member   Joined: Oct 2011 Posts: 35 Thanks: 0 Re: Intermediate Values i'm making a mess.. i have: Prove p is a fixed point. I want to show that f(p)=p I know that Xn->p so f(p)=f(lim n->oo Xn) and Xn+1=f(Xn) so f(lim n->oo Xn+1) = f(lim n->oo f(Xn))= f(f(lim n->oo Xn))= f(f(p))
 November 21st, 2011, 07:14 PM #8 Member   Joined: Dec 2010 From: Miami, FL Posts: 96 Thanks: 0 Re: Intermediate Values You are very close. Apply what I mentioned to you earlier before you apply $x_{n+1}=f(x_n)$
 November 21st, 2011, 07:57 PM #9 Member   Joined: Oct 2011 Posts: 35 Thanks: 0 Re: Intermediate Values i'm really not close.. i have: applying Xn+1 = f(Xn) so f(Xn+1)=f(f(Xn)) taking the limit of both we have lim n->oo f(Xn+1) = f(f(p)) and lim n->oo f(f(Xn))= f(f(p)) i'm not sure how you want me to apply Xn+1 = f(Xn)
November 21st, 2011, 08:34 PM   #10
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Re: Intermediate Values

Quote:
 Originally Posted by guynamedluis Since $f$ is a continuous function and $x_{n} \rightarrow p$, the following folds: $\lim_{n \rightarrow \infty} f(x_{n})= f( \lim_{n \rightarrow \infty} x_{n})$
No, apply this.

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