Generating function of Bessel polynomial and Recurrence relation

Generating function of Bessel polynomial:

The generating function of Bessel polynomial is given by

\( e^{\frac12 (t-1/t)}\) and is given by \( e^{\frac12 (t-1/t)}=\sum_{n=-\infty}^\infty t^n J_n(x)\)

Solution:

L.H.S

$$ e^{\frac12 (t-1/t)}$$

$$=e^{\frac{xt}{2}}\cdot e^{\frac{-x}{2t}}$$

$$=\biggl[1+\frac{(xt/2)^1}{1!}+\frac{(xt/2)^2}{2!}+…+\frac{(xt/2)^r}{r!}+…\biggr]\biggl[1+\frac{(-x/2t)^1}{1!}+\frac{(-x/2t)^2}{2!}+…+\frac{(-x/2t)^s}{s!}+…\biggr]$$

$$=\sum_{r=0}^\infty\frac{(xt/2)^r}{r!}\cdot \sum_{s=0}^\infty \frac{(-x/2t)^2}{s!}$$

$$=\sum_{r=0}^\infty \sum_{s=0}^\infty \frac{(-1)^s(x/2)^{r+s} t^{r-s}}{r!s!}$$

Put, r-s=n

$$\to r= n + s$$

$$= \sum_{n=0}^\infty \biggl[ \sum_{s=0}^\infty \frac{(-1)^s (x/2)^{n+2s}}{n+s)!s!}\biggr]t^n$$

$$=\sum_{n=0}^\infty J_n(x)t^n\dotsm(1)$$

Similarly, collecting the coefficient of \(t^{-n}\), we get

$$ e^{\frac12 x(t-1/t)}=\sum_{n=0}^\infty J_n(x) t^{-n}\dotsm(2)$$

Combining (1) and (2) we get,

$$ e^{\frac12 x(t-1/t)}= \sum_{n=-\infty}^\infty t^nJ_n(x)$$

Recurrence relation:

We know the generating function of Bessel polynomial

$$ e^{\frac12 x(t-1/t)}= \sum_{n=-\infty}^\infty t^nJ_n(x)$$

Now, differentiating equation (1) with resect to 't' we get

$$or,\;\; e^{\frac12 x(t-1/t)}\cdot \frac12 x(1+\frac{1}{t^2})= \sum_{n=-\infty}^\infty nt^{n-1} J_n(x)$$

$$or,\;\; \frac12 x \sum_{-\infty}^\infty t^n J_n(x) (1+\frac{1}{t^2}) = \sum_{-\infty}^\infty nt^{n-1} J_n(x)$$

$$or,\;\; x\sum_{-\infty}^\infty t^nJ_n(x)+x\sum_{-\infty}^\infty t^{n-2} J_n(x) = 2\sum_{n=-\infty}^\infty nt^{n-1}J_n(x)$$

Now, collecting the coefficient of \(t^n\) on both side, we get

$$ xJ_n(x) + xJ_{n+2}(x)= 2(n+1) J_{n+1} (x)$$

This is the first recurrence relation.

Again, differentiating equation (1) with respect to 'x' we get,

$$ e^{\frac12 x(t-1/t)}\cdot \frac12 (t-\frac1t)=\sum_{n=-\infty}^\infty t^n J_n'(x)$$

$$or,\;\; \sum_{n=--\infty} ^\infty t^n J_n(x)\cdot (t-\frac1t) = 2 \sum_{n=-\infty}^\infty t^n J_n'(x)$$

$$or,\;\; \sum_{n=-\infty}^\infty t^{n+1} J_n(x)-\sum_{n=-\infty}^\infty t^{n-1}J_n(x)=2\sum_{n=-\infty}^\infty t^n J_n'(x)$$

Collecting the coefficient of \(t^n\) on both side

$$J_{n-1} (x)= J_{n+1} (x)= 2J_n'(x)$$

This is second recurrence relation.

Proof:

$$\frac{d}{dx}(x^n J_n(x))= x^n J_{n-1}(x)$$

We know,

$$ J_n(x)=\sum_{j=0}^\infty\frac{(-1)^j (x/2)^{n+2j}}{j! (n+j)!}$$

$$=\sum_{j=0}^\infty \frac{(-1)^j x^{n+2j}}{2^{n+2j} j! (n+j)!}$$

Multiplying both side by \(x^n\) we get,

$$ x^n J_n(x)= \sum_{j=0}^\infty \frac{(-1)^j x^{2n+2j}}{2^{n+2j}j!(n+j)!}$$

Differentiating both side with respect to x, we get,

$$or, \frac{d}{dx}(x^n J_n(x))= \sum_{j=0}^\infty \frac{(-1)^j (2n+2)x^{2n+2j-1}}{2^{2n+2j}j! (n+j)(n+j-1)!}$$

$$=\sum_{j=0}^\infty \frac{(-1)^j 2x^{n+2j-1} x^n}{2^{n+2j}j! (n+j-1)!}$$

$$= x^n \sum_{j=0}^\infty \frac{(-1)^j (x/2)^{n+2j-1}}{j!(n-1+j)!}$$

$$\frac{d}{dx}(x^nJ_n(x))= x^nJ_{n-1}(x)$$

Hence proved.

Prove

$$J_1(x)= -J_0'(x)$$

We know the recurrence relation of Bessel is

$$2J_n'(x)= J_{n-1}(x)-J_{n+1}(x)$$

Put n=0

$$ 2J_0'(x)= J_{-1}(x)- J_1(x)$$

$$= -J_1(x) – J_1(x)$$

$$2J_0'(x)= -2J_1(x)$$

$$\therefore\;\;J_1(x)=-J_0'(x)$$ proved.

We know, $$2J_0'(x)= -2J_1(x)$$

Differentiating both side with respect to 'x' we get,

$$2J_0''(x)=-2J_1(x)$$

$$2J_0''(x)= J_0(x)-J_2(x)$$

$$2J_0''(x)= J_2(x)- J_0(x)$$

Reference:

1. Dass, H. K.Mathematical Physics. New Delhi: S. Chand, 2005.
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4. Carroll, Robert W.Mathematical Physics. Amsterdam: North-Holland, 1988.
5. Adhikari, Pitri Bhakta, and Dya Nidhi Chhatkuli.A Textbook of Physics. Third Revised Edition ed. Vol. III. Kathmandu: SUKUNDA PUSTAK BHAWAN, 2072.