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 2 years 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?
 2 years 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?

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myko
 2 years ago
Best ResponseYou've already chosen the best response.0couchy convergence test: \[x_{i+k}x_{i}<\epsilon\]

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0sorry i havnt learnt t.

myko
 2 years ago
Best ResponseYou've already chosen the best response.0basicly it says that the far enough terms have distance less than any positive number

perl
 2 years ago
Best ResponseYou've already chosen the best response.0i think here you assume that there is a limit

perl
 2 years ago
Best ResponseYou've already chosen the best response.0one sec, let me work on it

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

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

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

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

perl
 2 years ago
Best ResponseYou've already chosen the best response.0this has real solutions only when the discriminant is at least positive

perl
 2 years ago
Best ResponseYou've already chosen the best response.0so the condition for convergence is , when a1*a0 <= 1/4

perl
 2 years ago
Best ResponseYou've already chosen the best response.0also you have a typo in your question

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

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0Can u plz tell me what is the typo in the question? @perl

perl
 2 years ago
Best ResponseYou've already chosen the best response.0it should say dw:1354436826022:dw

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0ya it does. Thanxx a lottt.

perl
 2 years ago
Best ResponseYou've already chosen the best response.0I take that back that it converges for all x,

perl
 2 years ago
Best ResponseYou've already chosen the best response.0your seed has to be close enough to the x intercept (zero) for the iteration to converge

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

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0ohhh k. Thanksss a lottt for helping me.

perl
 2 years ago
Best ResponseYou've already chosen the best response.0I think i have a condition in my book.

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

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1354453952416:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1354454207481:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1354454557959:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1354454728910:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1354454813862:dw

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1354455134554:dw
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