How to treat small number within square root

Question

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?

Answer 1

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 $x=0$.

Answer 2

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 $0$.

Answer 3

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.