Sometimes we end up with a quantity of the form $(1+x)^n$, that is, the quantity $(1+x)$ is raised to the nth power. This can be written as an infinite sum of terms known as expansion:

$(1+x)^n = 1 + nx +\dfrac{n(n-1)}{2} x^2 + ....$

This formula is useful for us mainly when $x$ is very small compared to one $(x \ll 1)$. In this case, each successive term is much smaller than the preceding term.

For example, let $x=0.01$, and $n=2$. Then whereas the first term equals 1, the second term is $nx=(2)(0.01)=0.02$, and the third term is $\dfrac{n(n-1)}{2} x^2 = \dfrac{(2)(1)}{2} (0.01)^2=0.0001$ and so on.

Thus, when $x$ is small, we can ignore all but the first two (or three) terms and can write:

$$(1+x)^n \approx 1+nx$$

This approximation often allows us to solve an equation easily that otherwise might be very difficult.

Some Series Expansions

$(1+x)^n = 1 + nx + \dfrac{n(n-1)}{2!} x^2 + ....\\$

$(a+b)^n = a^n + \dfrac{n}{1!} a^{n-1} b + \dfrac{n(n-1)}{2!} a^{n-2} b^2 + ....\\$

$e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + ....\\$

$\ln (1 \pm x) = \pm x - \dfrac{1}{2} x^2 \pm \dfrac{1}{3} x^3 - ....$

For x in radians:

$\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - ....\\$

$\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - ....\\$

$\tan x = x + \dfrac{x^3}{3} + \dfrac{2x^5}{15} + .... \;\;\;\; |x| < \dfrac{\pi}{2}$

Some useful approximations

For $(x \ll 1)$:

$(1+x)^n \approx 1+nx$

$e^x \approx 1 + x$

$\ln (1 \pm x) \approx \pm x$

For $(x \leq 0.1 \;\mathrm{rad})$ [small-angle approximation]:

$\sin x \approx x$

$\cos x \approx 1$

$\tan x \approx x$

The small-angle approximation is excellent for $x<5^o=0.1 \;\mathrm{rad}$ and generally acceptable up to $x=10^o$.

So, let us evaluate $\sqrt{1.02}$:

$\sqrt{1.02}=(1.02)^{\frac{1}{2}}=(1+0.02)^{\frac{1}{2}} \approx 1+ \frac{1}{2} (0.02)=1.01 \;\;\;\;\;\;\;\;\;\;\; (x=0.02)$.

You can check that result with a calculator.