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

Mathematics
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see u man
@Mikael see this plz

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Other answers:

which we use for finding all roots of a polynomial...help would be appreciated
how our initial guess increases the rate of convergence
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....
Then we talk.
in 1-2 days.
thank u...i'll check this :)
Wait, @mukushla I think I might know.
Bairstow's method relies on newton's method to calculate the quadratic's coeffs, right?
yeah we extract quadratics from a polynomial and then solve for all of its roots...
But newton's method essentially makes the assumption that \[\Delta y = \Delta x dy\]
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.
Ex:|dw:1349025207545:dw|
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...
i mean how u say relies on newton's method ??
Well, if you were using Bairstow's method normally, how would you find r and s?
starting with an initial guess and modify it \[r_2=r_1+\Delta r\]\[s_2=s_1+\Delta s\]
Yes, how do you get delta r and delta s?
system of equations which comes from \[b_{-1}(r+\Delta r,s+\Delta s)=0\]\[b_{-2}(r+\Delta r,s+\Delta s)=0\]
yes, so if that system of equations happens to give you, for example, \(\Delta r = 0\) and \(\Delta s = 0\)?
then i cant modify my r and s
Exactly, and you're stuck in an infinite loop.
sorry man...just one more question...under what conditions we get delta r and delta s =0
Well, that's dependent on the actual polynomial and the r_1 and s_1 that you choose.
Specifically, this occurs when \(s=-h/g\) and \(r=\frac{g}{h-g}\) Where g and h are the remainder coefficients after two divisions: \[a_nx^n+....+a_1x+a_0=\]\[(b_{n-2}x^{n-2}+....+b_1x+b_0)(x^2-rx-s)+ix+j=\]\[(c_{n-4}x^{n-4}...+ c_0)(x^2-rx-s)+gx+h\]
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
thank u very much @vf321 ... i really appreciate your help :)

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