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

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mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0which we use for finding all roots of a polynomial...help would be appreciated

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0how our initial guess increases the rate of convergence

Mikael
 2 years ago
Best ResponseYou've already chosen the best response.0Ok 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....

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0thank u...i'll check this :)

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1Wait, @mukushla I think I might know.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1Bairstow's method relies on newton's method to calculate the quadratic's coeffs, right?

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0yeah we extract quadratics from a polynomial and then solve for all of its roots...

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1But newton's method essentially makes the assumption that \[\Delta y = \Delta x dy\]

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1So 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.

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0that makes sense ... lets say\[a_nx^n+....+a_1x+a_0=(b_{n2}x^{n2}+....+b_1x+b_0)(x^2rxs)+b_{1}(xr)+b_{2}\]my initial guess will be \((r_1,s_1)\); how u explain it for this...

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0i mean how u say relies on newton's method ??

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1Well, if you were using Bairstow's method normally, how would you find r and s?

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0starting with an initial guess and modify it \[r_2=r_1+\Delta r\]\[s_2=s_1+\Delta s\]

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, how do you get delta r and delta s?

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0system of equations which comes from \[b_{1}(r+\Delta r,s+\Delta s)=0\]\[b_{2}(r+\Delta r,s+\Delta s)=0\]

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1yes, so if that system of equations happens to give you, for example, \(\Delta r = 0\) and \(\Delta s = 0\)?

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0then i cant modify my r and s

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1Exactly, and you're stuck in an infinite loop.

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0sorry man...just one more question...under what conditions we get delta r and delta s =0

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1Well, that's dependent on the actual polynomial and the r_1 and s_1 that you choose.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1Specifically, this occurs when \(s=h/g\) and \(r=\frac{g}{hg}\) Where g and h are the remainder coefficients after two divisions: \[a_nx^n+....+a_1x+a_0=\]\[(b_{n2}x^{n2}+....+b_1x+b_0)(x^2rxs)+ix+j=\]\[(c_{n4}x^{n4}...+ c_0)(x^2rxs)+gx+h\]

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1I have to cite Wikipedia for the conditions in the last part though: http://upload.wikimedia.org/wikipedia/en/math/0/6/b/06bed931e9c9f2a198dc0b053f5fd43b.png

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0thank u very much @vf321 ... i really appreciate your help :)
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