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$g(x) = \frac{1}{2}(x - \frac{a}{x^2})$, $a > 1$ and $a<100$, if $a>0$ then $1<10^{2k}a<100$. So that $\sqrt{a}=\sqrt{10^{-2k}10^{2k}a} = 10^{-k}\sqrt{10^{2k}a}$.
$x_0 = \min(a,10)$, $\forall a >0 $ $\exists k \in Z$ so that $10^{2k}a \in [1,100]$.

How can i find the number of iterations for Newton's method to find $\sqrt{a}$ with accuracy $10^{-8}$? I need some pointer as of where to start this.

Gerry Myerson
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Marko Taht
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  • You can have a look at here. – Maczinga Mar 05 '17 at 09:44
  • That link helped a bit, althoug i dont get where did he get those formulas in the beginning. And i have absolute accuaraccy he uses relative accuaraccy. – Marko Taht Mar 05 '17 at 10:32
  • The formulas are found solving a peculiar recurrence. Have a look here for example. You should find this in standard numerical calculus book (sorry but I've only pointers in Italien in mind at present). – Maczinga Mar 05 '17 at 11:35
  • but the part about (1+2^-b)sqrt(y) from what is this derived from? – Marko Taht Mar 05 '17 at 11:42
  • I don't see that... can you be more precise? – Maczinga Mar 05 '17 at 12:52
  • In the first link "if you want to achieve 2^−b relative accuracy, xn=(1+2^−b)sqrt(y)". This part. and can i use this for absolute accuaracy? – Marko Taht Mar 05 '17 at 13:36
  • No in this case, you should use $x_n-\sqrt{y}=2^{-b}$ – Maczinga Mar 05 '17 at 14:51
  • Making some advancments, but if i use the xn-sqrt(y) then i cant get rid of the sqrt(y) in the first element. And how should i calculate the second element if my a is between 1 and 100 and x0 is between 1 on 10? – Marko Taht Mar 05 '17 at 17:06

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