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The iterative formula \(x_{i+1}=a_0+a_1x_1^2\) \((a_0, a_1\) positive ) is being used to solve the equation \(x=a_0+a_1x^2.\) What is the condition of convergence?
 one year ago
 one year ago
Please help:) The iterative formula \(x_{i+1}=a_0+a_1x_1^2\) \((a_0, a_1\) positive ) is being used to solve the equation \(x=a_0+a_1x^2.\) What is the condition of convergence?
 one year ago
 one year ago

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ajprincessBest ResponseYou've already chosen the best response.0
sorry didn't get u?
 one year ago

mykoBest ResponseYou've already chosen the best response.0
couchy convergence test: \[x_{i+k}x_{i}<\epsilon\]
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
sorry i havnt learnt t.
 one year ago

mykoBest ResponseYou've already chosen the best response.0
basicly it says that the far enough terms have distance less than any positive number
 one year ago

perlBest ResponseYou've already chosen the best response.0
i think here you assume that there is a limit
 one year ago

perlBest ResponseYou've already chosen the best response.0
one sec, let me work on it
 one year ago

perlBest ResponseYou've already chosen the best response.0
if limit exists, that means lim xn = L , then x_i+1 is roughly the same as x_i for large i
 one year ago

perlBest ResponseYou've already chosen the best response.0
so substitute L for xi and xi+1 L = a0 + a1 L ^2, it is a quadratic
 one year ago

perlBest ResponseYou've already chosen the best response.0
use quadratic formula , a1 L^2  L + a0 = 0, where L = xi as i > oo,
 one year ago

perlBest ResponseYou've already chosen the best response.0
L = [(1) +  sqrt ( 1  4*a1*a0)] / ( 2*a1)
 one year ago

perlBest ResponseYou've already chosen the best response.0
this has real solutions only when the discriminant is at least positive
 one year ago

perlBest ResponseYou've already chosen the best response.0
so the condition for convergence is , when a1*a0 <= 1/4
 one year ago

perlBest ResponseYou've already chosen the best response.0
also you have a typo in your question
 one year ago

perlBest ResponseYou've already chosen the best response.0
so for instance, a0 = 1/2 and a1 = 1/3 converges for any x
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
Can u plz tell me what is the typo in the question? @perl
 one year ago

perlBest ResponseYou've already chosen the best response.0
it should say dw:1354436826022:dw
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
ya it does. Thanxx a lottt.
 one year ago

perlBest ResponseYou've already chosen the best response.0
I take that back that it converges for all x,
 one year ago

perlBest ResponseYou've already chosen the best response.0
your seed has to be close enough to the x intercept (zero) for the iteration to converge
 one year ago

perlBest ResponseYou've already chosen the best response.0
you know what, i dont think i answered the question. i answered what it will converge to in the case that it does converge
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
ohhh k. Thanksss a lottt for helping me.
 one year ago

perlBest ResponseYou've already chosen the best response.0
I think i have a condition in my book.
 one year ago

perlBest ResponseYou've already chosen the best response.0
the derivative has to be less than 1 . so 2a1*x < 1 , so x < 1/ (2*a1)
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1354453952416:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1354454207481:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1354454557959:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1354454728910:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1354454813862:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
dw:1354455134554:dw
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
Thanksss a lottttt.
 one year ago
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