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A fixed point theorem

  1. May 27, 2015 #1
    hm.png
    Here I do not perceive the a sequence generated by fixed-point iteration. First would you like to explain this. How can it be that if lim n->∞ pn=P, then lim n-> ∞ Pn+1 ?

    Source: Numerical Methods Using Matlab by Kurtis D. Fink and John Matthews.
     
    Last edited: May 27, 2015
  2. jcsd
  3. May 27, 2015 #2

    Svein

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    Your appendix does not say exactly that. But:
    1. pn+1 = g(pn) (definiton)
    2. Assume limn→∞pn=P. Then, of course, limn→∞pn+1=P
    3. g(P) =g(limn→∞pn)
    4. g is continuous (supposition). Therefore g(limn→∞pn) = limn→∞g(pn)
    5. By definition (see 1.) limn→∞g(pn) = limn→∞pn+1 =P (from 2.)
    6. 3. and 5. ⇒ g(P) = P.
     
  4. May 27, 2015 #3
    It seems that I have a little forgotten calculus. Would you like to give more information for step 2 and step 4 in your post. With what topic of calculus are they about and how do we know them. I will search for their proof to best understand.

    Thank you.
     
  5. May 27, 2015 #4

    Svein

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    Step 2: limn→∞pn=P means "given ε>0, there exists an N such that for all n>N, |pn - P|<ε". And if n>N, obviously (n+1)>N.
    Step 4: Again an ε-proof: Since g is continuous, there exists a δ>0 such that |g(P)-g(x)|<ε for all x such that |P-x|<δ. Also, due to step 2, there is an N such that for all n>N, |pn - P|<min(ε, δ). Therefore |g(P)-g(pn)|<ε for n>N, which means that limn→∞g(pn) = g(P) = g( limn→∞pn).
     
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