Numerical Analysis Question

Discussion in 'Math' started by numericalmadness, Apr 13, 2009.

  1. numericalmadness

    Thread Starter New Member

    Apr 11, 2009
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    This is another question which I seek some help for.
    The answers to this will be appreciated as soon as possible
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Here's a start

    Are you working on past papers towards an exam?



    En = √a - Pn............given

    Pn = √a - En...........rearrange....1

    Also E(n+1) = √a - P(n+1)

    P(n+1) = √a - E(n+1).....................2

    P(n+1) = β(a - Pn^2) +Pn........given

    So √a - E(n+1) = β(a - Pn^2) +Pn

    substitute for Pn from 1

    √a - E(n+1) = β(a - (√a - En)^2) +√a - En

    Collect terms and simplify to obtain required result
     
  3. numericalmadness

    Thread Starter New Member

    Apr 11, 2009
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    Thanks alot, i was able to put part a together though after giving it more thought.
    What I'm not sure bout now doe is part b...the conditions for β
    I already know what to do for part c) just dunno how to get part b)
     
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Here is an extract from a Stanford U def

    http://sepwww.stanford.edu/public/docs/sep97/paul1/paper_html/node5.html

    Ask yourself how can E(n+1) be proportional to En^2 in the formula you have just developed. That is what value of beta makes (1-2β√a)En = 0

    Is E(n+1) not then proportional to En, with β as the constant of proportionality?
     
    Last edited: Apr 14, 2009
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