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

- anonymous

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

- katieb

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- anonymous

see u man

- anonymous

@Mikael see this plz

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## More answers

- anonymous

which we use for finding all roots of a polynomial...help would be appreciated

- anonymous

how our initial guess increases the rate of convergence

- anonymous

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....

- anonymous

Then we talk.

- anonymous

in 1-2 days.

- anonymous

thank u...i'll check this :)

- anonymous

- anonymous

Wait, @mukushla I think I might know.

- anonymous

Bairstow's method relies on newton's method to calculate the quadratic's coeffs, right?

- anonymous

yeah we extract quadratics from a polynomial and then solve for all of its roots...

- anonymous

But newton's method essentially makes the assumption that
\[\Delta y = \Delta x dy\]

- anonymous

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.

- anonymous

Ex:|dw:1349025207545:dw|

- anonymous

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...

- anonymous

i mean how u say relies on newton's method ??

- anonymous

Well, if you were using Bairstow's method normally, how would you find r and s?

- anonymous

starting with an initial guess and modify it \[r_2=r_1+\Delta r\]\[s_2=s_1+\Delta s\]

- anonymous

Yes, how do you get delta r and delta s?

- anonymous

system of equations which comes from \[b_{-1}(r+\Delta r,s+\Delta s)=0\]\[b_{-2}(r+\Delta r,s+\Delta s)=0\]

- anonymous

yes, so if that system of equations happens to give you, for example, \(\Delta r = 0\) and \(\Delta s = 0\)?

- anonymous

then i cant modify my r and s

- anonymous

Exactly, and you're stuck in an infinite loop.

- anonymous

sorry man...just one more question...under what conditions we get delta r and delta s =0

- anonymous

Well, that's dependent on the actual polynomial and the r_1 and s_1 that you choose.

- anonymous

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

- anonymous

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

- anonymous

thank u very much @vf321 ... i really appreciate your help :)

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