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Generating function of bessel function

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Prove the generating function

\(\displaystyle e^{\frac{x}{2}\left(z-z^{-1}\right)}=\sum_{n=-\infty}^{\infty}J_n(x)z^n\)​
 

chisigma

Well-known member
Feb 13, 2012
1,704
$\displaystyle e^{\frac{x}{2}\ (z - z^{-1})} = \sum_{m=0}^{\infty} \frac{(\frac{x}{2})^{m}}{m!}\ z^{m} \ \sum_{k=0}^{\infty} (-1)^{k} \ \frac{(\frac{x}{2})^{k}}{k!}\ z^{k} = $

$\displaystyle = \sum_{n = - \infty}^{+ \infty} \{ \sum_{m - k = n} \frac{(-1)^{k}\ (\frac{x}{2})^{m + k}}{m!\ k!}\ \}\ z^{n} = \sum_{n = - \infty}^{+ \infty} \sum_{k=0}^{\infty}^{n} \{ \frac{(-1)^{k}}{(n+k)!\ k!}\ (\frac{x}{2})^ {2 k + n} \} z^{n} = \sum_{n = - \infty}^{+ \infty} J_{n} (x)\ z^{n}$


Kind regards

$\chi$ $\sigma$
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
$$2(n+1)\jmath_{n+1}(x) = x \jmath_{n+2}(x) + x \jmath_n(x)$$

Multiplying by $z^n$ and summing from $-\infty$ to $\infty$ both sides gives

$$\begin{aligned} \sum_{n = -\infty}^{\infty} 2n \jmath_{n}(x) z^{n-1} &= \sum_{n = -\infty}^{\infty} x \jmath_n(x) z^{n-2} + \sum_{n = -\infty}^{\infty} x \jmath_{n}(x) z^n \\ &= \sum_{n = -\infty}^{\infty} x \left (1 + \frac{1}{z^2} \right ) \jmath_n(x) z^n \end{aligned}$$

Hence, we have the differential equation :

$$ K'(z) = \frac{x}{2} \left (1 + \frac{1}{z^2} \right ) K(z) $$

where $K(z) = \sum_{n = -\infty}^{\infty} \jmath_n(x) z^n$. This results $K(z) = \bar{C} \exp \left (\frac{x}{2} \left ( z - \frac{1}{z} \right) \right )$ after resolving the ODE. Substituting $z = 1$ easily gives $\bar{C} = 1$. Thus, we have

$$\sum_{n = -\infty}^{\infty} \jmath_n(x) z^n = \exp \left (\frac{x}{2} \left ( z - \frac{1}{z} \right) \right ) \;\;\; \blacksquare$$


Balarka
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