How Does a Fixed Point Theorem Explain Convergence in Iterative Methods?

[mech-eng]

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->∞ p_n=P, then lim n-> ∞ Pn+1 ?

Source: Numerical Methods Using Matlab by Kurtis D. Fink and John Matthews.

 

[Svein]

Your appendix does not say exactly that. But:

1. p_n+1 = g(p_n) (definiton)
2. Assume lim_n→∞p_n=P. Then, of course, lim_n→∞p_n+1=P
3. g(P) =g(lim_n→∞p_n)
4. g is continuous (supposition). Therefore g(lim_n→∞p_n) = lim_n→∞g(p_n)
5. By definition (see 1.) lim_n→∞g(p_n) = lim_n→∞p_n+1 =P (from 2.)
6. 3. and 5. ⇒ g(P) = P.

 

[mech-eng]

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.

 

[Svein]

Would you like to give more information for step 2 and step 4 in your post

Step 2: lim_n→∞p_n=P means "given ε>0, there exists an N such that for all n>N, |p_n - 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, |p_n - P|<min(ε, δ). Therefore |g(P)-g(p_n)|<ε for n>N, which means that lim_n→∞g(p_n) = g(P) = g( lim_n→∞p_n).