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mukushla Group Title

numerical analysis Bairstow method why the first initial guess doesn't lead us to answer always ???

  • 2 years ago
  • 2 years ago

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  1. experimentX Group Title
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    *

    • 2 years ago
  2. mukushla Group Title
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    see u man

    • 2 years ago
  3. mukushla Group Title
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    @Mikael see this plz

    • 2 years ago
  4. mukushla Group Title
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    which we use for finding all roots of a polynomial...help would be appreciated

    • 2 years ago
  5. mukushla Group Title
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    how our initial guess increases the rate of convergence

    • 2 years ago
  6. Mikael Group Title
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    Ok I will have to read more and think. Basically there should be a procedure that deals with quadratic root troubles. Initial guess is small change , forget it. I have some idea but have to test it. Imagine that qudr. root is NOT quadratic root. How ? Several ways are possible - you invent, and I invent....

    • 2 years ago
  7. Mikael Group Title
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    Then we talk.

    • 2 years ago
  8. Mikael Group Title
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    in 1-2 days.

    • 2 years ago
  9. mukushla Group Title
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    thank u...i'll check this :)

    • 2 years ago
  10. mukushla Group Title
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    @mahmit2012

    • 2 years ago
  11. vf321 Group Title
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    Wait, @mukushla I think I might know.

    • 2 years ago
  12. vf321 Group Title
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    Bairstow's method relies on newton's method to calculate the quadratic's coeffs, right?

    • 2 years ago
  13. mukushla Group Title
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    yeah we extract quadratics from a polynomial and then solve for all of its roots...

    • 2 years ago
  14. vf321 Group Title
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    But newton's method essentially makes the assumption that \[\Delta y = \Delta x dy\]

    • 2 years ago
  15. vf321 Group Title
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    So if I have some function where the tangent line (dy/dx) has a slope negative to the actual secant line (delta y/delta x) that it's approximating, then you'll get farther from where you started.

    • 2 years ago
  16. vf321 Group Title
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    Ex:|dw:1349025207545:dw|

    • 2 years ago
  17. mukushla Group Title
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    that makes sense ... lets say\[a_nx^n+....+a_1x+a_0=(b_{n-2}x^{n-2}+....+b_1x+b_0)(x^2-rx-s)+b_{-1}(x-r)+b_{-2}\]my initial guess will be \((r_1,s_1)\); how u explain it for this...

    • 2 years ago
  18. mukushla Group Title
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    i mean how u say relies on newton's method ??

    • 2 years ago
  19. vf321 Group Title
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    Well, if you were using Bairstow's method normally, how would you find r and s?

    • 2 years ago
  20. mukushla Group Title
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    starting with an initial guess and modify it \[r_2=r_1+\Delta r\]\[s_2=s_1+\Delta s\]

    • 2 years ago
  21. vf321 Group Title
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    Yes, how do you get delta r and delta s?

    • 2 years ago
  22. mukushla Group Title
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    system of equations which comes from \[b_{-1}(r+\Delta r,s+\Delta s)=0\]\[b_{-2}(r+\Delta r,s+\Delta s)=0\]

    • 2 years ago
  23. vf321 Group Title
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    yes, so if that system of equations happens to give you, for example, \(\Delta r = 0\) and \(\Delta s = 0\)?

    • 2 years ago
  24. mukushla Group Title
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    then i cant modify my r and s

    • 2 years ago
  25. vf321 Group Title
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    Exactly, and you're stuck in an infinite loop.

    • 2 years ago
  26. mukushla Group Title
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    sorry man...just one more question...under what conditions we get delta r and delta s =0

    • 2 years ago
  27. vf321 Group Title
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    Well, that's dependent on the actual polynomial and the r_1 and s_1 that you choose.

    • 2 years ago
  28. vf321 Group Title
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    Specifically, this occurs when \(s=-h/g\) and \(r=\frac{g}{h-g}\) Where g and h are the remainder coefficients after two divisions: \[a_nx^n+....+a_1x+a_0=\]\[(b_{n-2}x^{n-2}+....+b_1x+b_0)(x^2-rx-s)+ix+j=\]\[(c_{n-4}x^{n-4}...+ c_0)(x^2-rx-s)+gx+h\]

    • 2 years ago
  29. vf321 Group Title
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    I have to cite Wikipedia for the conditions in the last part though: http://upload.wikimedia.org/wikipedia/en/math/0/6/b/06bed931e9c9f2a198dc0b053f5fd43b.png

    • 2 years ago
  30. mukushla Group Title
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    thank u very much @vf321 ... i really appreciate your help :)

    • 2 years ago
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