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Newton's method
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georgech
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#1
Newton's method

Hello, i am having some difficulties solving the following problems using newton's method:

1)What does the sequence defined by
x_0=1,\;\;\;\;\;x_{k+1}=\frac{1}{2}x_k+\frac{1}{x_k}
converge to?
and

2)Devise a subroutine using only simple operations that finds, using newton's method, the cube root of some input number z.

Do u have any ideas?
Thanks

PostPosted: Sat Oct 31, 2009 10:13 am  Back to top 
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georgech
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#2
i only need some help to start..i know how to use newton to find roots of equations, but i haven't seen such problems before...

PostPosted: Sun Nov 01, 2009 1:37 am  Back to top 
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JBL
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#3
If I say that "a is a cube root of z," what's the corresponding equation?
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Joel
Hi Deeps! <3

PostPosted: Sun Nov 01, 2009 7:51 am  Back to top 
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georgech
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#4
Thanks, i have done the second question. But i am not sure how to proceed with the first one, which needs the convgence of the sequence. How can i do something like tha without using matlab? thanks

PostPosted: Sun Nov 01, 2009 1:11 pm  Back to top 
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atomicwedgie
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#5
georgech wrote:
Thanks, i have done the second question. But i am not sure how to proceed with the first one, which needs the convgence of the sequence. How can i do something like tha without using matlab? thanks


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Recall that Newton's method for finding roots of a function f(x) is given by

x_{k + 1} = x_k - \frac {f(x_k)}{f'(x_k)}.

Hence we must solve a differential equation to find the function f. We must have

x_k - \frac {f(x_k)}{f'(x_k)} = \frac {x_k}{2} + \frac {1}{x_k},

or

\frac {f(x_k)}{f'(x_k)} = \frac {x_k}{2} - \frac {1}{x_k} = \frac {x_k^2 - 2}{2x_k}.

Now we use the fact that \frac {f'(x)}{f(x)} = \frac {d}{dx}\left[\log f(x) \right]:

\frac {d}{dx_k}\left[ \log f(x_k) \right] = \frac {f'(x_k)}{f(x_k)} = \frac {2x_k}{x_k^2 - 2}.

Integrating,

\log f(x_k) = \int \frac {2x_k}{x_k^2 - 2} \, dx_k = \log |x_k^2 - 2|

f(x_k) = C(x_k^2 - 2)

where C is a constant of integration. Since the roots of f(x) = C(x^2 - 2) are the same for all nonzero values of C we know that the given recursion relationship, if it converges, must converge to either \sqrt {2} or - \sqrt {2}. But given x_0 = 1, it is obvious that x_k > 0 for all k, and therefore the sequence converges* to \sqrt {2}.

*Note. The only thing that remains is to show that the sequence is in fact convergent, which I leave as an exercise to the reader.


PostPosted: Sun Nov 01, 2009 1:48 pm  Back to top 
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JBL
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#6
That was a pretty impressive reverse-engineering of the question, atomicwedgie. The natural way to do that question has nothing to do with Newton's method: if x_k \to L then x_{k + 1} \to L and so L = \frac {1}{2}L + \frac {1}{L}.
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Joel
Hi Deeps! <3

PostPosted: Sun Nov 01, 2009 4:35 pm  Back to top 
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J.Y.Choi
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#7
Think about graph y = \frac {1}{2}x + \frac {1}{x} and y = x. The limit we want to find is the fixed point.

This is one well-known method, but I don't know this is the Newton's method. The book [Mathematical Methods for Physicists] by Arfken will help.

PostPosted: Sun Nov 01, 2009 4:54 pm  Back to top 
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