About Rayleigh's formula How to use $\displaystyle j_n(x)=(-1)^nx^n\left(\frac{1} {x} \frac{d} {dx}\right)^n \left(\frac{\sin x}{x}\right)$?
for example, to find $j_3(x)$ 

The formula you provided uses a compact way of writing operators.
For example, at $n = 2$, the term $\displaystyle \left(\frac{1}{x} \frac{d}{dx}\right)^2$ should be understood to mean the operator
$$\displaystyle \left(\frac{1}{x} \frac{d}{dx}\right)^2=\frac{1}{x} \frac{d}{dx}\frac1x \frac d{dx}.$$
You can see the values of Rayleigh's formula which are Spherical Bessel functions on the Wiki, but lets calculate some of them so you can get a feel for how this operator works.
We have (I am writing these as they are typically written - like on the Wiki page):
$$\displaystyle j_n(x)=(-1)^nx^n\left(\frac{1} {x} \frac{d} {dx}\right)^n \left(\frac{\sin x}{x}\right)$$
So:
$\displaystyle j_0(x) = (-1)^0x^0\left(\frac{1} {x} \frac{d} {dx}\right)^0 \left(\frac{\sin x}{x}\right) = \frac{\sin x}{x}$
$\displaystyle j_1(x) = (-1)^1x^1\left(\frac{1} {x} \frac{d} {dx}\right)^1 \left(\frac{\sin x}{x}\right) = (-1)^1x^1\left(\frac{1} {x} \frac{x \cos x - \sin x}{x^2}\right) = \frac{\sin x}{x^2} - \frac{\cos x}{x}$
$\displaystyle j_2(x) = (-1)^2x^2\left(\frac{1} {x} \frac{d} {dx}\right)^2 \left(\frac{\sin x}{x}\right)$
$\displaystyle = x^2 \left(\frac{1} {x} \frac{d} {dx} \frac{1} {x} \frac{d} {dx}\right) \left(\frac{\sin x}{x}\right)$
$\displaystyle = x^2 \left(\frac{1} {x} \frac{d} {dx} \frac{1} {x} \right) \left(\frac{x \cos x - \sin x}{x^2}\right)$
$\displaystyle = x^2 \left(\frac{1} {x} \frac{d} {dx} \right) \left(\frac{x \cos x - \sin x}{x^3}\right) $
$\displaystyle = x^2 \left(\frac{1} {x} \right) \left(\frac{(3-x^2) \sin x - 3x \cos x}{x^4}\right)$
$\displaystyle = \left(\frac{3}{x^2} -1 \right) \frac{\sin x}{x} - \frac{3 \cos x}{x^2}$
$\displaystyle j_3(x) = (-1)^3x^3\left(\frac{1}{x} \frac{d} {dx}\right)^3 \left(\frac{\sin x}{x}\right) = -x^3 \left(\frac{1}{x} \frac{d}{dx} \frac{1}{x} \frac{d}{dx} \frac{1}{x} \frac{d}{dx} \right)\left(\frac{\sin x}{x}\right) $
After some calculations, we arrive at:
$$j_3(x) = \left(\frac{15}{x^3} - \frac{6}{x} \right) \frac{\sin x}{x} -\left(\frac{15}{x^2} - 1 \right) \frac{\cos x}{x}$$
I will let you work that last one, but if you get stuck, just reply and I will add the details!
