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mukushla
Group Title
numerical analysis
Bairstow method
why the first initial guess doesn't lead us to answer always ???
 one year ago
 one year ago
mukushla Group Title
numerical analysis Bairstow method why the first initial guess doesn't lead us to answer always ???
 one year ago
 one year ago

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mukushla Group TitleBest ResponseYou've already chosen the best response.0
see u man
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
@Mikael see this plz
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
which we use for finding all roots of a polynomial...help would be appreciated
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
how our initial guess increases the rate of convergence
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.0
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....
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.0
Then we talk.
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.0
in 12 days.
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
thank u...i'll check this :)
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
@mahmit2012
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Wait, @mukushla I think I might know.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Bairstow's method relies on newton's method to calculate the quadratic's coeffs, right?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
yeah we extract quadratics from a polynomial and then solve for all of its roots...
 one year ago

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

vf321 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Ex:dw:1349025207545:dw
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
that 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...
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
i mean how u say relies on newton's method ??
 one year ago

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

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

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Yes, how do you get delta r and delta s?
 one year ago

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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.0
then i cant modify my r and s
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Exactly, and you're stuck in an infinite loop.
 one year ago

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

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Well, that's dependent on the actual polynomial and the r_1 and s_1 that you choose.
 one year ago

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

vf321 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
thank u very much @vf321 ... i really appreciate your help :)
 one year ago
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