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I have a problem understanding as to how does one set the initial value for newton's method to find reciprocal. After doing some analysis, the conclusion is that $0 < x_0 < \frac{2}{a}$

but isn't it a "chicken and egg problem"? We started out to determine $\frac{1}{a}$ and here we are using it to determine the solution... I'm confused.

for the uninitiated, http://www.derekroconnor.net/NA/Notes/RecSqRoot.pdf

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    Educated guesswork. Or you could look for an interval where your function changes sign...2012-12-07

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Hint: consider roots of the function $ f(x)=\frac{1}{x}-a.$ This has the analytical solution as $x=1/a$. Try using it with Newton's method.

For the initial iterate, if $10^{k-1}, then $1/10^k<1/a<1/10^{k-1}$. Then $1/10^k<1/a<2/a$ is a suitable initial iterate.

Added When you apply the Newton iterate to this function, you obtain the iterate $x_{k+1}=x_k\left(2-ax_k\right).$ The analysis you want should then be easily done to show the range of convergence of the initial estimate must be $0.

In regard to the initial iterate, $1/10^k=1/2^{k\log_2(10)}>1/2^{4k},$ which can be specified using bit shifts. Alternatively, you can express the initial iterate in scientific or hex notation.

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    but aren't you again using the '/' operator. I want to actually write a program where I do it using only * and + operators. But yea, That could have been a great starting point...2012-12-08