Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
mukushla
Group Title
numerical analysis
Bairstow method
why the first initial guess doesn't lead us to answer always ???
 2 years ago
 2 years ago
mukushla Group Title
numerical analysis Bairstow method why the first initial guess doesn't lead us to answer always ???
 2 years ago
 2 years ago

This Question is Closed

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

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

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

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

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

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

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

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

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

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

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

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

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

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

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Exactly, and you're stuck in an infinite loop.
 2 years 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
 2 years 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.
 2 years 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: \[a_nx^n+....+a_1x+a_0=\]\[(b_{n2}x^{n2}+....+b_1x+b_0)(x^2rxs)+ix+j=\]\[(c_{n4}x^{n4}...+ c_0)(x^2rxs)+gx+h\]
 2 years 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
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
thank u very much @vf321 ... i really appreciate your help :)
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.