1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor

    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).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook