Do the derivatives of the Bessel functions of the first kind with integer order have any singularities?

Question

Do the derivatives of $J_\nu(z)$, the Bessel functions of the first kind with integer order $\nu$, have any singularities?

I am interested in the case where $z$ is real and $\nu$ is a positive integer.

The definition is

$\mathop{J_{\nu}}\nolimits\!\left(z\right)=(\tfrac{1}{2}z)^{\nu}\sum_{k=0}^{% \infty}(-1)^{k}\frac{(\tfrac{1}{4}z^{2})^{k}}{k!\mathop{\Gamma}\nolimits\!% \left(\nu+k+1\right)}$

Since $\Gamma(x)$ is never zero I think that $J_\nu(z)$ has no singularities.

According to derivative formula

$2\mathop{J_{\nu}}\nolimits'\!% \left(z\right)=\mathop{J_{\nu-1}}\nolimits\!\left(z\right)-\mathop{J_{% \nu+1}}\nolimits\!\left(z\right)$

I can compute the first derivative of $J_\nu$ by means of $J_{\nu-1}$ and $J_{\nu+1}$ and so since $J_{\nu-1}$ and $J_{\nu+1}$ have no singularities also the first derivative of $J_\nu$ will not have singularities.

The case with $\nu=0$ is handled by this formula

$\displaystyle\mathop{J_{0}}\nolimits'\!\left(z\right)=-\mathop{J_{1}}% \nolimits\!\left(z\right)$

and so also the first derivative in case of order $\nu$ equal to zero is free of singularities.

For the second order derivative I think that I can apply two times the derivative formula and so I get again no singularities and the same happens with higher order derivatives.

Am I missing something?

What about this different formula for the derivatives? Does that $\frac{1}{z}$ give a singularity?

$\mathop{J_{\nu}}\nolimits'\!\left(z\right)=\mathop{J_{% \nu-1}}\nolimits\!\left(z\right)-(\nu/z)\mathop{J_{\nu}}% \nolimits\!\left(z\right)$

Answer 1

Bessel functions of the first kind with integer order are entire functions, i.e., are analytic in the entire complex plane. As such, they don't have any singularities, nor do their derivatives of any order. Any other expressions involving $J_\nu(z)$ and $1\over z$ (or whatever else, for that matter) can't change this fact.

You can verify that by checking the radius of convergence of the series in the definition. It is infinite.

BTW, $\Gamma(x)$ never being zero has nothing to do with $J_\nu(z)$ having or not having any singularities. Really, consider the series $\sum\limits_{k=0}^\infty{z^k}$: the denominator is never zero (for there is no denominator at all), yet the function behaves suspiciously as $z$ approaches $1$.

Answer 2

As far as I know, no singularity will appear. However, the exact answer for your question lies in knowing the process of deriving the formula

$${{J'}_\nu }(z) = {J_{\nu - 1}}(z) - {\nu \over z}{J_\nu }(z)$$

I think the firstly derived formula is

$$z{{J'}_\nu }(z) = z{J_{\nu - 1}}(z) - \nu {J_\nu }(z)$$

and then someone may divide it by $z$ to get what you mentioned so your mentioned formula is not valid for $z=0$.

Answer 3

Bessel functions of positive order, evaluated at zero, are equal to zero:¹ $$ {J_\nu}(0) = 0 \, \Big|_{\nu>0} $$

The derivatives of the Bessel functions with integer order greater than one, evaluated at zero, are also equal to zero: $$ {J_\nu}'(0) = 0 \, \Big|_{\nu>1} $$

Proof: Starting with the recurrence relation: $$ {J_\nu}'(z) = \frac12 \left(J_{\nu-1}(z) - J_{\nu+1}(z) \right) \Big|_{\nu >1} $$ $$ {J_\nu}'(0) = \frac12 \left(J_{\nu-1}(0) - J_{\nu+1}(0) \right) \Big|_{\nu >1} $$ $$ {J_\nu}'(0) = \frac12 \left(0 - 0 \right) = 0 \quad \blacksquare$$

Since the values of the derivatives at the origin, are zero, they are obviously not singularities.

The second term in the recurrence relation: $$ {J_\nu}'(z) = J_{ \nu-1}(z) - \nu \frac{{J_\nu}(z)}{z} $$

when $ z=0 $ and $ \nu>0 $; is not a singularity, but an indeterminate form of type $\! \tfrac00 $. $$ {J_\nu}'(0) = J_{ \nu-1}(0) - \nu \frac{0}{0} = 0 $$

The proper value of an indeterminate form can be determined by taking the limit $$ \lim\limits_{ z \rightarrow 0}\left(\frac{J_\nu(z)}{z} \right) = 0 $$

We now know that the indeterminate form must evaluate to zero ($\nu \tfrac00 = 0$). $$ {J_\nu}'(0) = J_{\nu-1}(0) - 0 = 0 $$

This is similar to the indeterminate form caused from lazily defining the sinc function as $\sin(x)/x$, instead of $\lim\limits_{ x \rightarrow z}(\sin(z)/z)$ $$ \operatorname{sinc}(0) = 1 = \lim\limits_{ x \rightarrow 0}\left(\frac{\sin(x)}{x}\right) \neq \frac{\sin(0)}{0} $$