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I wanted to write a way of calculating a square root of a number in LaTeX. Unfortunately, as a beginner in this sphere, I do not know what math symbol means to repeat the process. To make it clear, here is my equation for finding a square root of a variable k

$$ \sqrt k = \frac{(n + \frac{k}{n})}{2} $$

After each calculation, a specific result is put instead of variable n, i.e: $$ \sqrt 5 = \frac{(5 + \frac{5}{5})}{2} = 3$$ $$ \sqrt 5 = \frac{(3 + \frac{5}{3})}{2} \approx 2.\overline{33} $$ $$ \sqrt 5 = \frac{(2.\overline{33} + \frac{5}{2.\overline{33}})}{2} \approx 2.23796 $$ $$ \sqrt 5 = \frac{(2.23796 + \frac{5}{2.23796})}{2} \approx 2.23606 $$ $$ \sqrt 5 = \frac{(2.23606797 + \frac{5}{2.23606797})}{2} \approx 2.23606797$$ $$ \sqrt 5 \approx 2.23606797... $$

FYI my English is not that good when explaining something, especially if it is Maths, so expect some weird or false vocabulary here and there...

Tian Vlašić
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Sarich
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1 Answers1

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Do not do that, instead define a sequence $(u_n)_n \in \mathbb{R}^\mathbb{N}$ with $$u_0 = x_0 = k$$ $$u_{n+1} = f(u_n) = f^n (u)$$ with $f(x) = \frac{x^2 + k}{2x}$.

Show that $u_n \rightarrow \sqrt{k}$ and it will be much cleaner.

For that show that $\sqrt{k}$ is a fixed point of $f$ and that your sequence is either decreasing or increasing wether you have $x_0 > \sqrt{k}$ or not

julio_es_sui_glace
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