Is there any simple method to calculate $\sqrt x$ without using logarithm

Solution 1:

For $y=\sqrt{x}$ there is a simple method: \begin{align} y &= 1 &&\text{initialize} \\ y &=\frac {(\frac{x}{y}+y)}{2} & &\text{repeat until convergence} \end{align} It can be modified for roots of higher orders.

Solution 2:

There is an old-fashioned digit-by-digit method that I learned when I was at school. The theory of it is explained here with a base 10 example here, and many old arithmetic books give more practical details for actually carrying out the calculations in a sensible manner.

I have a very old arithmetic textbook which does something similar for cube roots, but it gets more tedious, and I have never seen anything for 5th roots.

Solution 3:

$$f(x)=\sqrt [ n ]{ x }\Rightarrow f'(x)=\frac {x^{(1/n-1)}}{n}$$ $$f'(x_0)\approx\frac{f(x_0+h)-f(x_0)}{h}$$ $$\Rightarrow f(x_0+h)\approx f'(x_0)h+f(x_0)$$ Suppose you want to calculate $f(x)=\sqrt [ 3 ]{ x }$ at $x=7 $ then take $h=-1$ and $x_0=8$ $$f(7)\approx f'(8)(-1)+f(8)\approx-\frac{1}{12}+2\approx\frac{23}{12}$$

Solution 4:

The continued fraction method works like this: Suppose $x = a^2 + b$, where $a = \lfloor \sqrt x \rfloor$. Then

$$ \begin{align} x &= \sqrt{a^2 + b}\\ x-a &= \sqrt{a^2 + b} - a\\ \frac{1}{x-a} &= \frac{1}{\sqrt{a^2 + b} - a}\\ &= \frac{1}{\sqrt{a^2 + b} - a}\frac{\sqrt{a^2 + b} + a}{\sqrt{a^2 + b} + a}\\ &= \frac{\sqrt{a^2 + b} + a}{b}\\ &= \frac{x + a}{b} \end{align} $$

Substitute, and get:

$$ \begin{align} x &= a + (x-a)\\ &= a + \frac{b}{a+x}\\ %= a + \frac{b}{2a+\frac{b}{a+x}}\\ x &= a+\cfrac{b}{2a+\cfrac{b}{2a+\cfrac{b}{2a + \dots}}} \end{align} $$

Now, this is not a simple continued fraction. However, if one divides the numerator and denominator of $\frac{b}{2a+x}$ by $b$, then one can eventually get a periodic simple continued fraction, and one approximates by the convergents. The above expression turns out to be faster.

Solution 5:

If $x$ is an integer, then you can find the continued fraction expansion of $\sqrt x$ and get very close approximations with no division involved. If you want 6-place accuracy, for instance, continue till you get a convergent with denominator $>1000$.