# Numerical Analysis Question

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

Apr 11, 2009
7
0
This is another question which I seek some help for.
The answers to this will be appreciated as soon as possible

File size:
80.7 KB
Views:
29
2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
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

Apr 11, 2009
7
0
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
5,005
513
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