For that reason the modified bessel function $I_q(\rho)$ defined as $$ I_{q}(\rho)=\sum_{m=0}^{\infty}\frac{(\rho/2)^{2m+q}}{m!\Gamma(m+q+1)},$$ where in the above $\rho$ is fixed, can be differentiated term by term in relation to $q$?
For fixed $\rho$ and considered as a function of $q$ the above is no more a power series in the variable $q$, but I see in the books that I'm studying that Bessel functions have this very good differentiability property in relation to $q$ too. How can I justify that Bessel functions are diferentiable in relation to $q$? Any help or hint, will be very useful, thanks
Let $$\rho=2x\qquad \text{and} \qquad x^{2m+q}=e^{(2m+q)\log(x)}$$ Now, $q$ can be treated as a continuous variable (because you used the gamma function instead of the factorial).
$$a_m=\frac{e^{(2m+q)\log(x)}}{m!\Gamma(m+q+1)}$$ $$\frac{\partial a_m}{\partial q}=\frac{x^{2 m+q} }{m! \Gamma (m+q+1)}(\log (x)-\psi (m+q+1))$$ $$\color{blue}{\frac{\partial I_q(2x)}{\partial q}=\sum_{m=0}^\infty \frac{\partial a_m}{\partial q}= (\log (x)-\psi (q+1))\,I_q(2 x)-}$$ $$\color{blue}{x^q\, \Gamma (q+1) \text{HypergeometricPFQRegularized}^{(\{1\},\{0,0\},0)}\left(\{q+1\},\{q+1,q+1\},x^2\right) }$$
Do it with numerical differentiation with $q=\pi$ and $x=4.567$ using central differences with $\Delta q= 10^{-6}\,\pi$ The above formula gives $-90.842073139191$ and numerical differentiation $-90.842073139159$