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Any suggestion? I have no idea what the question is asking. Thanks in advance.

Mark McClure
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Bob
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  • Just write down the formula $n(x)=x-f(x)/f'(x)$ in the case $f(x)=g(x)^2$ and compare to what you get when you write down $x-g(x)/g'(x)$ directly. You should find that you are adjusting your approximation in the former case by only half the amount that you do in the latter. – Mark McClure Mar 10 '14 at 12:38

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Hint: To see what's happening, calculate explicitly let $f(x)=x^2$. Start with say $x_1=1$. You will find that the convergence is somewhat slow, or at least a lot slower than Newton "usually" is.

Then generalize.

André Nicolas
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  • Well, in this case $g(x)=x$ and the corresponding Newton's method iteration function is $n(x)=0$ - i.e., you get to the exact root in one step from any starting value. :) – Mark McClure Mar 10 '14 at 12:43
  • Yes, but applying the Newton Method blindly to $x^2$ gets you the sequence of approximations $1,\frac{1}{2},\frac{1}{4},\dots$, instead of the "characteristic" roughly doubling the number of correct decimal digits for each iteration. – André Nicolas Mar 10 '14 at 12:47