How to treat small number within square root guys.I am reading a math book. It has a equation shown as follows,
$\sqrt{(1+\Delta^2)}$
And then,since $\Delta$ is very small, it can be written as,
$\sqrt{(1+\Delta^2)} = (1+\frac12\Delta^2)$
What is the theory behind this equation?
 A: Let's use $\epsilon$ instead of $\Delta^2$, and write it as
$$\sqrt{1+\epsilon}$$
Then,
$$\sqrt{1+\epsilon}=1+\frac{1}{2}\epsilon$$
Squaring both sides, we find that
$$1+\epsilon=1+\epsilon+\frac{1}{4}\epsilon^2$$
This is true if $\epsilon$ is small enough that $\epsilon^2$ can be neglected.
The first two terms of the MacLaurin series for $\sqrt{1+x}$ are
$$\sqrt{1+x}\approx 1+\frac{1}{2}x$$
This is true for all small $x$, near $x=0$.
A: If we take a taylor expansion for representing equation with polynomials it gives us:
$$\sqrt {1+x^2} = 1 + \frac {1}{2} x + \frac {1}{4} x^2 + ... $$
As when $ x $ is very small $ x^2$ is very very small which can be neglected and also for $ x^{n} $ as $ n > 2$
For that 
$$\sqrt {1+x^2} \approx 1 + \frac {1}{2} x^2 $$
Edited: wrong expansion in prior formula.
A: Square both sides to get $1+\Delta^2\approx 1+\Delta^2+\frac{1}{4}\Delta^4$
which is a close approximation because when $\Delta$ is very small, $\frac{1}{4}\Delta^4$ is close to $0$.
A: What @Mohamed_Fadel noted is NOT true.
The correct taylor expansion for $\sqrt {1+x^2}$ at the point $x_0=0$ is:
$$\sqrt {1+x^2} = 1 + \frac {1}{2} x^2 - \frac {1}{8} x^4 + \frac {1}{16} x^6...$$
That you get by just going through the standard taylor expansion.
For getting $\sqrt{(1+\Delta^2)} = (1+\frac12\Delta^2)$, all you need to do is neglect every order following the quadratic.
@Mohamed_Fadel: You cannot treat $x^2$ like a single variable $x$ when expanding, because inner derivatives will arise when doing the Taylor expansion.
