Question Number 220502 by SdC355 last updated on 14/May/25
$$\mathrm{each}\:{J}_{\nu} \left({z}\right),{Y}_{\nu} \left({z}\right)\:\mathrm{are}\:\mathrm{linear}\:\mathrm{independent}….?? \\ $$$${W}_{\mathrm{Ronskian}} \left\{{J}_{\nu} ^{\:} \left({z}\right),{Y}_{\nu} \left({z}\right)\right\}=\begin{vmatrix}{{J}_{\nu} \left({z}\right)}&{\:{Y}_{\nu} \left({z}\right)}\\{{J}_{\nu} '\left({z}\right)}&{{Y}_{\nu} '\left({z}\right)}\end{vmatrix} \\ $$$$={J}_{\nu} ^{\left(\mathrm{1}\right)} \left({z}\right){Y}_{\nu} \left({z}\right)−{J}_{\nu} \left({z}\right){Y}_{\nu} ^{\left(\mathrm{1}\right)} \left({z}\right) \\ $$$${J}_{\nu} ^{\left(\mathrm{1}\right)} \left({z}\right){Y}_{\nu} \left({z}\right)={J}_{\nu−\mathrm{1}} \left({z}\right){Y}_{\nu} \left({z}\right)−\frac{\nu}{{z}}{J}_{\nu} \left({z}\right){Y}_{\nu} \left({z}\right) \\ $$$${J}_{\nu} \left({z}\right){Y}_{\nu} ^{\left(\mathrm{1}\right)} \left({z}\right)={Y}_{\nu−\mathrm{1}} \left({z}\right){J}_{\nu} \left({z}\right)−\frac{\nu}{{z}}{J}_{\nu} \left({z}\right){Y}_{\nu} \left({z}\right) \\ $$$${J}_{\nu−\mathrm{1}} \left({z}\right){Y}_{\nu} \left({z}\right)−{J}_{\nu} \left({z}\right){Y}_{\nu−\mathrm{1}} \left({z}\right)…. \\ $$$$…..\mathrm{damn}….. \\ $$$$\mathrm{Result}\:\mathrm{is}\:\frac{\mathrm{2}}{\pi{z}}\:…… \\ $$
Answered by MrGaster last updated on 14/May/25
$$={J}_{\nu} \left({z}\right){Y}_{\nu} ^{'} \left({z}\right)−{J}_{\nu} ^{'} \left({z}\right){Y}_{\nu} \left({z}\right) \\ $$$${J}'_{\nu} \left({z}\right)={J}_{\nu−\mathrm{1}} \left({z}\right)−\frac{\nu}{{z}}{J}_{\nu} \left({z}\right) \\ $$$${Y}\underset{\nu} {'}\left({z}\right)={Y}_{\nu−\mathrm{1}} \left({z}\right)−\frac{\nu}{{z}}{Y}_{\nu} \left({z}\right) \\ $$$$\Rightarrow{W}={J}_{\nu} \left({z}\right)\left({Y}_{\nu−\mathrm{1}} \left({z}\right)−\frac{\nu}{{z}}{Y}_{\nu} \left({z}\right)\right)−{Y}_{\nu} \left({z}\right)\left({J}_{\nu−\mathrm{1}} \left({z}\right)−\frac{\nu}{{z}}{J}_{\nu} \left({z}\right)\right) \\ $$$$={J}_{\nu} \left({z}\right){Y}_{\nu−\mathrm{1}} \left({z}\right)−\frac{\nu}{{z}}{J}_{\nu} \left({z}\right){Y}_{\nu} \left({z}\right)−{Y}_{\nu} \left({z}\right){J}_{\nu−\mathrm{1}} \left({z}\right)+\frac{\nu}{{z}}{J}_{\nu} \left({z}\right){Y}_{\nu} \left({z}\right) \\ $$$$={J}_{\nu} \left({z}\right){Y}_{\nu−\mathrm{1}} \left({z}\right)−{Y}_{\nu} \left({z}\right){J}_{\nu−\mathrm{1}} \left({z}\right) \\ $$$$\because{J}_{\nu} \left({z}\right){Y}_{\nu−\mathrm{1}} \left({z}\right)−{J}_{\nu−\mathrm{1}} \left({z}\right){Y}_{\nu} \left({z}\right)=\frac{\mathrm{2}}{\pi{z}} \\ $$$$\therefore{W}_{\mathrm{Ronskian}} \left\{{J}_{\nu} ^{\:} \left({z}\right),{Y}_{\nu} \left({z}\right)\right\}=\frac{\mathrm{2}}{\pi{z}} \\ $$
Commented by SdC355 last updated on 14/May/25
$$\mathrm{I}\:\mathrm{mean}\:\mathrm{why}\:{J}_{\nu} \left({z}\right){Y}_{\nu−\mathrm{1}} \left({z}\right)−{J}_{\nu−\mathrm{1}} \left({z}\right){Y}_{\nu} \left({z}\right)=\frac{\mathrm{2}}{\pi{z}} \\ $$$$\mathrm{can}'\mathrm{t}\:\mathrm{Caculate}\:\mathrm{anymore} \\ $$