I got one of the proof that Bessel functions of the first kind of integer order are linearly dependent from proofwiki. However, it proves from the integral form of the Bessel function.
How can I prove
$${J_{-n}}(x) = (-1)^{n} \, J_n(x)$$
starting from
$$J_{\alpha }(x)=\sum _{m=0}^{\infty }{\frac {(-1)^{m}}{m!\Gamma (m+\alpha +1)}}{\left({\frac {x}{2}}\right)}^{2m+\alpha}$$
I can't get the equation to show linearly dependent by replacing $\alpha$ with $-\alpha$.
Using $$ J_{\nu }(x)=\sum_{k=0}^{\infty} \frac {(-1)^{k}}{k! \, \Gamma(k+\nu +1)} \, \left(\frac {x}{2}\right)^{2k+\nu} $$ and properties of the gamma function then for $\nu = n$: \begin{align} J_{-n}(x) &= \sum_{k=0}^{\infty} \frac {(-1)^{k}}{k! \, \Gamma(k-n+1)} \, \left(\frac {x}{2}\right)^{2k-n} \\ &= \sum_{k=0}^{n-1} \frac {(-1)^{k}}{k! \, \Gamma(k-n+1)} \, \left(\frac {x}{2}\right)^{2k-n} + \sum_{k=n}^{\infty} \frac {(-1)^{k}}{k! \, \Gamma(k-n+1)} \, \left(\frac {x}{2}\right)^{2k-n} \\ &= 0 + \sum_{k=0}^{\infty} \frac {(-1)^{k+n}}{(k+n)! \, \Gamma(k+1)} \, \left(\frac {x}{2}\right)^{2k+n} \\ &= (-1)^n \, \sum_{k=0}^{\infty} \frac{(-1)^k}{k! \, \Gamma(k+n+1)} \, \left( \frac{x}{2} \right)^{2 k +n} \\ &= (-1)^n \, J_{n}(x). \end{align}
