1
$\begingroup$

How is Newton's Iteration achieved?

I mean, can you please explain where does Newton's iterative formula

$x_{k+1}=\frac{1}{2}(x_k+\frac{N}{x_k})$

come from?

$\endgroup$
0
$\begingroup$

gt6989b has given a good general definition of Newton's method. Applying it to $f(x)=N-x^2$ we have the equation $f'(x)=-2x$. Then $f(x_{k+1}) =x_k-\frac{f(x_k)}{f'(x_k)}=x_k-\frac{N-x_k^2}{-2x_k}=\frac{x_k}2+\frac N{2x_k}$

$\endgroup$
2
$\begingroup$

Intuitively, Newton's method assumes that your function is a line at every point. So assume you are starting from some point $P_n = (x_n, f(x_n))$ and the slope of the Newton approximation line is given by $m = f'(x_n)$, which is given to you as well.

Now, the equation of the line that with slope $m$ that passes through $P_n$ must be $y - f(x_n) = m(x - x_n)$, or in other words $y = f(x_n) + f'(x_n)(x - x_n)$.

You are looking to find the root, i.e. the place where $y=0$. That line has a root when $0 = f(x_n) + f'(x_n)(x - x_n)$, or in other words, where

$x = x_n - \frac{f(x_n)}{f'(x_n)}$,

which is what you would choose as the next iteration point. In other words,

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.