ajprincess
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
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did you try couchy?
ajprincess
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sorry didn't get u?
myko
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couchy convergence test:
\[|x_{i+k}-x_{i}|<\epsilon\]
ajprincess
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sorry i havnt learnt t.
myko
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basicly it says that the far enough terms have distance less than any positive number
perl
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i think here you assume that there is a limit
perl
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one sec, let me work on it
perl
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if limit exists, that means lim xn = L , then x_i+1 is roughly the same as x_i for large i
perl
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so substitute L for xi and xi+1
L = a0 + a1 L ^2, it is a quadratic
perl
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use quadratic formula , a1 L^2 - L + a0 = 0, where L = xi as i -> oo,
perl
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L = [-(-1) + - sqrt ( 1 - 4*a1*a0)] / ( 2*a1)
perl
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this has real solutions only when the discriminant is at least positive
perl
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so the condition for convergence is , when a1*a0 <= 1/4
perl
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also you have a typo in your question
perl
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so for instance, a0 = 1/2 and a1 = 1/3 converges for any x
perl
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for any initial seed x
ajprincess
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Can u plz tell me what is the typo in the question? @perl
perl
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your iterative formula
perl
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it should say |dw:1354436826022:dw|
perl
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makes sense?
ajprincess
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ya it does. Thanxx a lottt.
perl
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I take that back that it converges for all x,
perl
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your seed has to be close enough to the x intercept (zero) for the iteration to converge
perl
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you know what, i dont think i answered the question. i answered what it will converge to in the case that it does converge
ajprincess
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ohhh k. Thanksss a lottt for helping me.
perl
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I think i have a condition in my book.
perl
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the derivative has to be less than 1 .
so
2a1*x < 1 , so x < 1/ (2*a1)
mahmit2012
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|dw:1354453952416:dw|
mahmit2012
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|dw:1354454207481:dw|
mahmit2012
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|dw:1354454557959:dw|
mahmit2012
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|dw:1354454728910:dw|
mahmit2012
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|dw:1354454813862:dw|
mahmit2012
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|dw:1354455134554:dw|
ajprincess
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Thanksss a lottttt.