## anonymous 3 years ago numerical analysis Bairstow method why the first initial guess doesn't lead us to answer always ???

1. experimentX

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2. anonymous

see u man

3. anonymous

@Mikael see this plz

4. anonymous

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

5. anonymous

how our initial guess increases the rate of convergence

6. 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....

7. anonymous

Then we talk.

8. anonymous

in 1-2 days.

9. anonymous

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

10. anonymous

@mahmit2012

11. anonymous

Wait, @mukushla I think I might know.

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. 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.

16. anonymous

Ex:|dw:1349025207545:dw|

17. 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...

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. anonymous

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

22. 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$

23. anonymous

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

24. anonymous

then i cant modify my r and s

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. 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$

29. 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

30. anonymous

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