About Rayleigh's formula

Question

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 $j_3(x)$

Answer 1

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!

Answer 2

Related problem:(I). You can use the new identity for Rayleigh's formula which computes the spherical Bessel functions $ j_n(z) $ of the first kind

$$ j_n(x)=\sum_{k=0}^{n}{ n\brack k} {2}^{n-k}\sum _{s=0}^{k} s! { k\brace s}\sum _{m=0}^{s}{\frac {\left( -1 \right)^{n+s-m}{x}^{-n-1+m}\sin \left( x+\frac{m\pi}{2} \right) }{ m! }}\quad n\in \mathbb{N} \cup {0}. $$

where $ n\brack k $ and $ n\brace k $ are Stirling numbers of the first and second kind respectively.

For instance, for $n=3,4$, you get

$$ -{\frac {-\cos \left( t \right) {t}^{3}+6\,\sin \left( t \right) {t}^{ 2}+15\,\cos \left( t \right) t-15\,\sin \left( t \right) }{{t}^{4}}},$$

$$ {\frac {\sin \left( t \right) {t}^{4}+10\,\cos \left( t \right) {t}^{3 }-45\,\sin \left( t \right) {t}^{2}-105\,\cos \left( t \right) t+105\, \sin \left( t \right) }{{t}^{5}}}.$$

Note: I'll appreciate if someone double check the formula.

Answer 3

The operator $\left(\frac{1}{x}\frac{d}{dx}\right)^{n}$ has the following expansion :

\begin{equation} \left(\frac{1}{x}\frac{d}{dx}\right)^{n}. = \sum_{k=0}^{n}b(n, k)\frac{1}{x^{2n-k}}\frac{d^{k}.}{dx^{k}} \end{equation}

where $b(n, k)$ is a Bessel number of the first kind, so Rayleigh's formula becomes:

$\displaystyle j_n(x)=(-1)^nx^n\left(\frac{1} {x} \frac{d} {dx}\right)^n \left(\frac{\sin x}{x}\right)=\frac{(-1)^n}{x^{n+1}}\sum_{k=0}^{n}(-1)^kk!b(n, k)\sum_{i=0}^{k}(-1)^{i}\sin \left(x+\frac{i\pi}{2}\right)\frac{x^i}{i!} $

Hence :

$\displaystyle j_3(x)=\frac{-1}{x^4}\sum_{k=0}^{3}(-1)^kk!b(3, k)\sum_{i=0}^{k}(-1)^{i}\sin \left(x+\frac{i\pi}{2}\right)\frac{x^i}{i!}=\frac{-1}{x^4}\left(-15\sin x+15x \cos x+6x^2\sin x-x^3\cos x \right) $