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For the function $f(x)=x^2-a, a>0,$ the Newton iterative steps are given by?

$$ x_{n+1} = \frac 12 [x_n + \frac {a}{x_n}]$$ $$ x_{n+1} = 2[x_n + \frac {a}{x_n}]$$ $$ x_{n+1} = \frac 12 [x_n -ax_n]$$ $$ x_{n+1} = \frac 12 [x_n + ax_n]$$

Clueless regarding how to proceed. Here is my attempt for the sake of trying:

$$f(x) = x^2 - a$$ $$f'(x) = 2x$$ $$ x_{n+1} = x_n - \frac {f(x_n)}{f'(x_n)}$$

DanielWainfleet
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Splendid Digital Solutions
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    Where did you find that formula? What does $f(x)n$ even mean? – The correct one is $x{n+1} = x_n - \frac {f(x_n)}{f'(x_n)}$. Now substitute $f(x_n) = x_n^2-a$ and $f'(x_n) = 2x_n$ and simplify the expression. – Martin R Sep 01 '21 at 07:09
  • @MartinR Edited the formula. – Splendid Digital Solutions Sep 01 '21 at 07:15
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    Please proof-read it again: $x_n +1$ should be $x_{n+1}$ at several places, and $(x)_n - \frac {f(x)_n}{f'(x)_n}$ should be $x_n - \frac {f(x_n)}{f'(x_n)}$. – Martin R Sep 01 '21 at 07:20
  • My edit was cosmetic. \frac ax_n puts a small space between the x and the n (Displayed Lines 1 & 2) but \frac {a}{x_n} does not. – DanielWainfleet Sep 01 '21 at 13:31
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    Try a particular case. Let $a=1.$ The solutions to $f(x)=0$ are $x=\pm 1$. Let $x_1=2$. In the 2nd formula the sequence is increasing, so every $x_n\ge 2$ (for if $x_n>0$ then $x_{n+1}>2x_n>x_n>0$). In the 3rd formula $x_{n+1}=0$ for all $n\ge 1.$ In the 4th formula $x_n=2$ for every $n.$ – DanielWainfleet Sep 01 '21 at 13:51

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This is an iterative method where, starting from an initial approximation $x_0$, we form a sequence with each element, $x_{n+1}$, being computed from the previous, $x_n$, through the formula $$ x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}. $$

In this case, we have that $f(x) = x^2-a$ and therefore $$f(x_n)=x_n^2-a, \quad f'(x_n)= 2 x_n.$$

If you substitute this back into the formula and make some elementary simplifications you will get to one of the options.

PierreCarre
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