Approximating square roots using binomial expansion. Through the binomial expansion of $(1 - 2x)^\frac{1}{2}$, I am required to find an approximation of $\sqrt2$.


  Binomial expansion
  
  $ (1 + x)^n = 1 + \frac{n}{1}x + \frac{n(n-1)}{1*2}x^2 + ... $


Thus, the expansion of $(1 - 2x)^\frac{1}{2}$: $$
 = 1 - x -\frac{1}{2}x^2 - \frac{1}{2}x^3 + ... $$

The suggested way, is to choose a value for $x$ so that $(1-2x)$ has the form $2*$'a perfect square'. This can be done by taking $x = 0.01$. Thus, $(1 - 2x)=(1-2*0.01) = 0.98 = 2*0.7^2$

And
$$
(1 - 2x)^\frac{1}{2} = 0.98^\frac{1}{2} = 0.7\sqrt2$$
Which is equal to the previously established expansion, so we can now go ahead and find $\sqrt2$. The problem I am facing, is that there was no mention of how the value of $x=0.01$ was arrived at.

Is there an easy way to determine an appropriate value for $x$?
Personally, I wouldn't have done it that way.  So here is how I would've done it:

Method 1:

$$\sqrt2=\sqrt{1+1}=1+\frac12-\frac18+\dots\approx1+\frac12-\frac18=\frac{11}8=1.375$$

which is much clearer to me, since it avoids having to take decimals raised to powers and gives you something you can easily do by hand.

$$1.375^2=1.890625$$

Obviously it approaches the correct value as you take more terms.



Method 2:

This is called fixed-point iteration/Newton's method, and it basically goes like this:

$$x=\sqrt2\implies x^2=2$$

$$2x^2=2+x^2$$

Divide both sides by $2x$ and we get

$$x=\frac{2+x^2}{2x}$$

Now, interestingly, I'm going to call the $x$'s on the left $x_{n+1}$ and the $x$'s on the right $x_n$, so

$$x_{n+1}=\frac{2+(x_n)^2}{2x_n}$$

and with $x_0\approx\sqrt2$, we will then have $x=\lim_{n\to\infty}x_n$.  For example, with $x_0=1$,

$x_0=1$

$x_1=\frac{2+1^2}{2(1)}=\frac32=1.5$

$x_2=\frac{2+(3/2)^2}{2(3/2)}=\frac{17}{12}=1.41666\dots$

$x_3=\dots=\frac{577}{408}=1.414215686$

And one can quickly check that $(x_3)^2=2.000006007\dots$, which is pretty much the square root of $2$.
In fact you can take any two numbers which can be added to get 2 (not nesserly 0.01 but at least you should know the root of one of them 

So for example $\sqrt{2} = {(1+1)^{1/2}}$ 

Know all what you need is to expand it using bio theorem and for 2 terms you ll get 1.5

$1+(0.5)\cdot 1= 1.5 $
Which is not too bad approximation for $\sqrt{2} = 1.4142135624 $