Dear all,
The question is in the link below. I have done part 1 already, it was a spiral that converges on alpha.
http://imageshack.us/photo/my-images/98/newtonraphson.jpg/
How do I do B? I have used f(x) and f'(x) and put them into the form x - f(x)/f'(x) for both forms. How would I explain how they converge?
Any help would be greatly appreciated.
Thank you in advance,
Jon
The question is in the link below. I have done part 1 already, it was a spiral that converges on alpha.
http://imageshack.us/photo/my-images/98/newtonraphson.jpg/
How do I do B? I have used f(x) and f'(x) and put them into the form x - f(x)/f'(x) for both forms. How would I explain how they converge?
Any help would be greatly appreciated.
Thank you in advance,
Jon
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You are not expected to use the Newton formula, but the simple
iteration formulas given in (a). x(n+1)=G(x(n))
These will converge to the root a if |G'(a)|<1, and use the given approximation
a=1. Find G'(x) where G(x)=7/(x^2+5) , giving -14x/(2x+5)^2
So G'(1) =-2/7 and |G'(1)|<1 so procedure converges to a.
Do the same with (7-x^3)/5 you get |G'(1)|=3/5
The one with the smaller G' converges more quickly.
If this is an MEI question look it up in the course book in Iterative Processes.
iteration formulas given in (a). x(n+1)=G(x(n))
These will converge to the root a if |G'(a)|<1, and use the given approximation
a=1. Find G'(x) where G(x)=7/(x^2+5) , giving -14x/(2x+5)^2
So G'(1) =-2/7 and |G'(1)|<1 so procedure converges to a.
Do the same with (7-x^3)/5 you get |G'(1)|=3/5
The one with the smaller G' converges more quickly.
If this is an MEI question look it up in the course book in Iterative Processes.
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this is not a Newton Raphson formula. it is an iteration.
x=G(x),
x_(n+1)=G(x_n)
For Newton Raphson method, let F(x) = x^3+5x-7, and do
x_(n+1)=x_n - f(x_n))/f'(x_n)
x=G(x),
x_(n+1)=G(x_n)
For Newton Raphson method, let F(x) = x^3+5x-7, and do
x_(n+1)=x_n - f(x_n))/f'(x_n)
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I know it