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ajprincess

  • 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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  1. myko
    • 2 years ago
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    did you try couchy?

  2. ajprincess
    • 2 years ago
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    sorry didn't get u?

  3. myko
    • 2 years ago
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    couchy convergence test: \[|x_{i+k}-x_{i}|<\epsilon\]

  4. ajprincess
    • 2 years ago
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    sorry i havnt learnt t.

  5. myko
    • 2 years ago
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    basicly it says that the far enough terms have distance less than any positive number

  6. perl
    • 2 years ago
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    i think here you assume that there is a limit

  7. perl
    • 2 years ago
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    one sec, let me work on it

  8. perl
    • 2 years ago
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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

  9. perl
    • 2 years ago
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    so substitute L for xi and xi+1 L = a0 + a1 L ^2, it is a quadratic

  10. perl
    • 2 years ago
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    use quadratic formula , a1 L^2 - L + a0 = 0, where L = xi as i -> oo,

  11. perl
    • 2 years ago
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    L = [-(-1) + - sqrt ( 1 - 4*a1*a0)] / ( 2*a1)

  12. perl
    • 2 years ago
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    this has real solutions only when the discriminant is at least positive

  13. perl
    • 2 years ago
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    so the condition for convergence is , when a1*a0 <= 1/4

  14. perl
    • 2 years ago
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    also you have a typo in your question

  15. perl
    • 2 years ago
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    so for instance, a0 = 1/2 and a1 = 1/3 converges for any x

  16. perl
    • 2 years ago
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    for any initial seed x

  17. ajprincess
    • 2 years ago
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    Can u plz tell me what is the typo in the question? @perl

  18. perl
    • 2 years ago
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    your iterative formula

  19. perl
    • 2 years ago
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    it should say |dw:1354436826022:dw|

  20. perl
    • 2 years ago
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    makes sense?

  21. ajprincess
    • 2 years ago
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    ya it does. Thanxx a lottt.

  22. perl
    • 2 years ago
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    I take that back that it converges for all x,

  23. perl
    • 2 years ago
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    your seed has to be close enough to the x intercept (zero) for the iteration to converge

  24. perl
    • 2 years ago
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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

  25. ajprincess
    • 2 years ago
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    ohhh k. Thanksss a lottt for helping me.

  26. perl
    • 2 years ago
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    I think i have a condition in my book.

  27. perl
    • 2 years ago
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    the derivative has to be less than 1 . so 2a1*x < 1 , so x < 1/ (2*a1)

  28. mahmit2012
    • 2 years ago
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    |dw:1354453952416:dw|

  29. mahmit2012
    • 2 years ago
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    |dw:1354454207481:dw|

  30. mahmit2012
    • 2 years ago
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    |dw:1354454557959:dw|

  31. mahmit2012
    • 2 years ago
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    |dw:1354454728910:dw|

  32. mahmit2012
    • 2 years ago
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    |dw:1354454813862:dw|

  33. mahmit2012
    • 2 years ago
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    |dw:1354455134554:dw|

  34. ajprincess
    • 2 years ago
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    Thanksss a lottttt.

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