Bessel–Clifford function

Relation to Bessel functions

The Bessel function of the first kind can be defined in terms of the Bessel–Clifford function as

${\displaystyle J_n(z)={\left(\frac{z}{2}\right)}^n{\mathcal{C}}_n\left(-\frac{z^2}{4}\right);}$

when n is not an integer. We can see from this that the Bessel function is not entire. Similarly, the modified Bessel function of the first kind can be defined as

${\displaystyle I_n(z)={\left(\frac{z}{2}\right)}^n{\mathcal{C}}_n\left(\frac{z^2}{4}\right).}$

The procedure can of course be reversed, so that we may define the Bessel–Clifford function as

${\displaystyle {\mathcal{C}}_n(z)=z^{-n/2}{I}_n(2\sqrt{z});}$

but from this starting point we would then need to show ${\displaystyle \mathcal{C}}$ was entire.

Recurrence relation

From the defining series, it follows immediately that ${\displaystyle \frac{d}{dx}{\mathcal{C}}_n(x)={\mathcal{C}}_{n+1}(x).}$

Using this, we may rewrite the differential equation for ${\displaystyle \mathcal{C}}$ as

${\displaystyle x{\mathcal{C}}_{n+2}(x)+(n+1){\mathcal{C}}_{n+1}(x)={\mathcal{C}}_n(x),}$

which defines the recurrence relationship for the Bessel–Clifford function. This is equivalent to a similar relation for ₀F₁. We have, as a special case of Gauss's continued fraction

${\displaystyle \frac{{\mathcal{C}}_{n+1}(x)}{{\mathcal{C}}_n(x)}=\frac{1}{n+1+\frac{x}{n+2+\frac{x}{n+3+\frac{x}{\ddots }}}}.}$

It can be shown that this continued fraction converges in all cases.

The Bessel–Clifford function of the second kind

The Bessel–Clifford differential equation

${\displaystyle x{y}^{″}+(n+1)y^{\prime }=y}$

has two linearly independent solutions. Since the origin is a regular singular point of the differential equation, and since ${\displaystyle \mathcal{C}}$ is entire, the second solution must be singular at the origin.

If we set

${\displaystyle {\mathcal{K}}_n(x)=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\exp \left(-t-\frac{x}{t}\right)\frac{dt}{t^{n+1}}}$

which converges for ${\displaystyle \mathrm{\Re }(x)>0}$, and analytically continue it, we obtain a second linearly independent solution to the differential equation.

The factor of 1/2 is inserted in order to make ${\displaystyle \mathcal{K}}$ correspond to the Bessel functions of the second kind. We have

${\displaystyle K_n(x)={\left(\frac{x}{2}\right)}^n{\mathcal{K}}_n\left(\frac{x^2}{4}\right).}$

and

${\displaystyle Y_n(x)={\left(\frac{x}{2}\right)}^n{\mathcal{K}}_n\left(-\frac{x^2}{4}\right).}$

In terms of K, we have

${\displaystyle {\mathcal{K}}_n(x)=x^{-n/2}{K}_n(2\sqrt{x}).}$

Hence, just as the Bessel function and modified Bessel function of the first kind can both be expressed in terms of ${\displaystyle \mathcal{C}}$, those of the second kind can both be expressed in terms of ${\displaystyle \mathcal{K}}$.

Generating function

If we multiply the absolutely convergent series for exp(t) and exp(z/t) together, we get (when t is not zero) an absolutely convergent series for exp(t + z/t). Collecting terms in t, we find on comparison with the power series definition for ${\displaystyle {\mathcal{C}}_n}$ that we have

${\displaystyle \exp \left(t+\frac{z}{t}\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{t}^n{\mathcal{C}}_n(z).}$

This generating function can then be used to obtain further formulas, in particular we may use Cauchy's integral formula and obtain ${\displaystyle {\mathcal{C}}_n}$ for integer n as

${\displaystyle {\mathcal{C}}_n(z)=\frac{1}{2\pi i}{\oint }_C\frac{\exp (t+z/t)}{t^{n+1}}dt=\frac{1}{2\pi }{\int }_0^{2\pi }\exp (z\exp (-i\theta )+\exp (i\theta )-ni\theta )d\theta .}$