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10 Bessel FunctionsSpherical Bessel Functions

§10.49 Explicit Formulas

Contents
  1. §10.49(i) Unmodified Functions
  2. §10.49(ii) Modified Functions
  3. §10.49(iii) Rayleigh’s Formulas
  4. §10.49(iv) Sums or Differences of Squares

§10.49(i) Unmodified Functions

Define a_{k}(\nu) as in (10.17.1). Then

10.49.1 a_{k}(n+\tfrac{1}{2})=\begin{cases}\dfrac{(n+k)!}{2^{k}k!(n-k)!},&k=0,1,\dotsc%
,n,\\
0,&k=n+1,n+2,\dotsc.\end{cases}
10.49.3
\mathsf{j}_{0}\left(z\right)=\frac{\sin z}{z},
\mathsf{j}_{1}\left(z\right)=\frac{\sin z}{z^{2}}-\frac{\cos z}{z},
\mathsf{j}_{2}\left(z\right)=\left(-\frac{1}{z}+\frac{3}{z^{3}}\right)\sin z-%
\frac{3}{z^{2}}\cos z.
10.49.5
\mathsf{y}_{0}\left(z\right)=-\frac{\cos z}{z},
\mathsf{y}_{1}\left(z\right)=-\frac{\cos z}{z^{2}}-\frac{\sin z}{z},
\mathsf{y}_{2}\left(z\right)=\left(\frac{1}{z}-\frac{3}{z^{3}}\right)\cos z-%
\frac{3}{z^{2}}\sin z.

§10.49(ii) Modified Functions

Again, with a_{k}(n+\tfrac{1}{2}) as in (10.49.1),

10.49.8 {\mathsf{i}^{(1)}_{n}}\left(z\right)=\tfrac{1}{2}e^{z}\sum_{k=0}^{n}(-1)^{k}%
\frac{a_{k}(n+\frac{1}{2})}{z^{k+1}}+(-1)^{n+1}\*\tfrac{1}{2}e^{-z}\sum_{k=0}^%
{n}\frac{a_{k}(n+\frac{1}{2})}{z^{k+1}}.
10.49.9
{\mathsf{i}^{(1)}_{0}}\left(z\right)=\frac{\sinh z}{z},
{\mathsf{i}^{(1)}_{1}}\left(z\right)=-\frac{\sinh z}{z^{2}}+\frac{\cosh z}{z},
{\mathsf{i}^{(1)}_{2}}\left(z\right)=\left(\frac{1}{z}+\frac{3}{z^{3}}\right)%
\sinh z-\frac{3}{z^{2}}\cosh z.
10.49.10 {\mathsf{i}^{(2)}_{n}}\left(z\right)=\tfrac{1}{2}e^{z}\sum_{k=0}^{n}(-1)^{k}%
\frac{a_{k}(n+\frac{1}{2})}{z^{k+1}}+(-1)^{n}\tfrac{1}{2}e^{-z}\sum_{k=0}^{n}%
\frac{a_{k}(n+\frac{1}{2})}{z^{k+1}}.

\sum_{k=0}^{n}a_{k}(n+\tfrac{1}{2})z^{n-k} is sometimes called the Bessel polynomial of degree n. For a survey of properties of these polynomials and their generalizations see Grosswald (1978). See also §18.34, de Bruin et al. (1981a, b), and Dunster (2001c).

§10.49(iii) Rayleigh’s Formulas

§10.49(iv) Sums or Differences of Squares

Denote

10.49.17 s_{k}(n+\tfrac{1}{2})=\frac{(2k)!(n+k)!}{2^{2k}(k!)^{2}(n-k)!},k=0,1,\dotsc,n.

Then

10.49.19
{\mathsf{j}_{0}}^{2}\left(z\right)+{\mathsf{y}_{0}}^{2}\left(z\right)=z^{-2},
{\mathsf{j}_{1}}^{2}\left(z\right)+{\mathsf{y}_{1}}^{2}\left(z\right)=z^{-2}+z%
^{-4},
{\mathsf{j}_{2}}^{2}\left(z\right)+{\mathsf{y}_{2}}^{2}\left(z\right)=z^{-2}+3%
z^{-4}+9z^{-6}.
10.49.20 \left({\mathsf{i}^{(1)}_{n}}\left(z\right)\right)^{2}-\left({\mathsf{i}^{(2)}_%
{n}}\left(z\right)\right)^{2}=(-1)^{n+1}\sum_{k=0}^{n}(-1)^{k}\frac{s_{k}(n+%
\frac{1}{2})}{z^{2k+2}}.
10.49.21
\Big({\mathsf{i}^{(1)}_{0}}\left(z\right)\Big)^{2}-\Big({\mathsf{i}^{(2)}_{0}}%
\left(z\right)\Big)^{2}=-z^{-2},
\Big({\mathsf{i}^{(1)}_{1}}\left(z\right)\Big)^{2}-\Big({\mathsf{i}^{(2)}_{1}}%
\left(z\right)\Big)^{2}=z^{-2}-z^{-4},
\Big({\mathsf{i}^{(1)}_{2}}\left(z\right)\Big)^{2}-\Big({\mathsf{i}^{(2)}_{2}}%
\left(z\right)\Big)^{2}=-z^{-2}+3z^{-4}-9z^{-6}.