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November 21st, 2011, 05:15 PM   #1
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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!
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November 21st, 2011, 05:44 PM   #2
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Re: Intermediate Values

Since is a continuous function and , the following folds:
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November 21st, 2011, 06:08 PM   #3
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Re: Intermediate Values

any idea on how to prove p is a fixed point of f(x)?
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November 21st, 2011, 06:15 PM   #4
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Re: Intermediate Values

Consider and apply what I mentioned above.
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November 21st, 2011, 06:28 PM   #5
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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?
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November 21st, 2011, 06:43 PM   #6
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Re: Intermediate Values

We want to show .

We know that , so that , and .

We just have to piece it all together.
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November 21st, 2011, 07:00 PM   #7
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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))
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November 21st, 2011, 07:14 PM   #8
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Re: Intermediate Values

You are very close.

Apply what I mentioned to you earlier before you apply
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November 21st, 2011, 07:57 PM   #9
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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)
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November 21st, 2011, 08:34 PM   #10
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Re: Intermediate Values

Quote:
Originally Posted by guynamedluis
Since is a continuous function and , the following folds:
No, apply this.
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