- #1
davidge
- 554
- 21
Hi
I want to show that ##V(x) = (1-x^2)^{m/2} P_l (x)## is a solution of the equation
$$\frac{d}{dx} \bigg[(1-x^2) \frac{dV(x)}{dx} \bigg] + \bigg[l(l+1) - \frac{m^2}{1-x^2} \bigg]V(x) = 0.$$ Because the equation for ##P_l (x)## is $$\frac{d}{dx} \bigg[(1-x^2) \frac{dP_l (x)}{dx}\bigg] + l(l+1) P_l (x) = 0,$$ my attempts have been consisting of trying to differentiate the last equation ##m## times. I did not realize how to use the Leibnitz formula for general derivative, so I have been trying to differentiate it a small number of times and by induction arguing what it looks like after the ##m \text{-th}## derivative. I'm having no success in by proceeding this way, though.
However, this is just an attempt. I would like equally well any other one, because I just want to show that ##V(x)## given as above is a solution for the equation shown.
I want to show that ##V(x) = (1-x^2)^{m/2} P_l (x)## is a solution of the equation
$$\frac{d}{dx} \bigg[(1-x^2) \frac{dV(x)}{dx} \bigg] + \bigg[l(l+1) - \frac{m^2}{1-x^2} \bigg]V(x) = 0.$$ Because the equation for ##P_l (x)## is $$\frac{d}{dx} \bigg[(1-x^2) \frac{dP_l (x)}{dx}\bigg] + l(l+1) P_l (x) = 0,$$ my attempts have been consisting of trying to differentiate the last equation ##m## times. I did not realize how to use the Leibnitz formula for general derivative, so I have been trying to differentiate it a small number of times and by induction arguing what it looks like after the ##m \text{-th}## derivative. I'm having no success in by proceeding this way, though.
However, this is just an attempt. I would like equally well any other one, because I just want to show that ##V(x)## given as above is a solution for the equation shown.
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