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Consider the quadratic polynomial $f(x) = x^2 − 6x + 2$. The two roots of this function are $R_1 = 3 + \sqrt 7$, $R_2 = 3 − \sqrt 7$. Consider the following 4 different iterative processes.

a)$X_{n+1}=6-\frac{2}{X_n}$;

b)$X_{n+1}=\frac{1}{6}X_n^2+\frac{1}{3}$;

c)$X_{n+1}=\sqrt{6X_n-2}$;

d)$X_{n+1}=\frac{X_n^2-2} {2X_n-6}$;

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What is the question? –  André Nicolas May 31 '12 at 12:32
    
I edited the equations, I wasn't sure what b) is supposed to look like. –  Simon Markett May 31 '12 at 12:51
    
I have posed some edits of the post for formatting (please confirm that I haven't altered your intention), but I cannot interpret (b), at all. When the edit is approved (probably), you can look at how I did it, and format (b), yourself, in a similar fashion. Also, I second André...what are you actually asking? Please append the actual question. –  Cameron Buie May 31 '12 at 12:52
    
@Cameron looks like we edited at the same time... and for some reason mine was dominant and your's doesn't show up after all. Did you do anything different? –  Simon Markett May 31 '12 at 12:54
    
@Simon I don't have actual editing privileges, yet (halfway there), so that makes sense. I didn't do anything at all to (b), I didn't keep the parentheses in (c) or (d), and I removed the "$=0$" from the definition of $f$, but other than that, identical. –  Cameron Buie May 31 '12 at 13:06

1 Answer 1

I will take a leap here and assume that the question is why those processes should find a zero of the given polynomial. Well, if they converge then they converge to a zero of the polynomial since for the limit $x$ (and the case a)) we have

$$x=6-\frac 2x\Leftrightarrow x^2-6x+2=0$$

and similarly for the other cases. Whether they converge and to which zero they converge depends on your starting point.

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If you show $|\frac {dX_{n+1}}{dX_n}| \lt 1$ you know it converges if you start close enough and you can estimate the speed of convergence. –  Ross Millikan Jul 8 '12 at 22:11

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