I'm having troubles with this demonstration. With the following expressión of the generating function of Legendre polynomials: $$e^{tx}J_0\left ( t\sqrt{1-x^2} \right )=\sum \frac {P_n(x)}{n!}t^n $$ I have to demonstrate that the equation is valid for $x=0$, so I evaluate both sides and i obtain: $$ e^{0}J_0\left ( t\sqrt{1-0} \right )= J_0(t)=\sum \frac{(-1)^n}{(n!)^2}\left(\frac{t}{2} \right)^{2n}$$ For odd n $$ P_n(0)=0$$ For pair n $$ P_n(0)=(-1)^n\frac{(2n)!}{2^{2n}(n!)^2}$$
If we put all this together in the first equation, the equations are not equal with a first look. $$\sum \frac{(-1)^n}{(n!)^2}\left(\frac{t}{2} \right)^{2n}=\sum(-1)^n\frac{(2n)!}{2^{2n}(n!)^2} \frac{t^n}{n!}$$
I know im doing something wrong, if someone can help me I really appreciate your help. Thanks!
This is an example of why it is important to write the subscripts in the Sigma notation - the discrepancy between summing over all terms versus just the even terms is what is causing the confusion here.
To be more precise, what you want to show is that $$ \sum_{n=0}^{\infty}\frac{(-1)^n}{(n!)^2}\left(\frac{t}{2}\right)^{2n}=\sum_{n=0}^{\infty}\frac{P_n(0)}{n!}t^n.\qquad (\star) $$ After substituting the values of $P_n(0)$ (as described here), we obtain that $$ \sum_{n=0}^{\infty}\frac{P_n(0)}{n!}t^n=\sum_{n=2k, k=0}^{\infty}\frac{(-1)^k(2k)!}{2^{2k}(k!)^2}\cdot\frac{1}{(2k)!}t^{2k}. $$ Notice how the $(2k)!$ cancels out, leaving $$ \sum_{n=0}^{\infty}\frac{P_n(0)}{n!}t^n=\sum_{n=2k, k=0}^{\infty}\frac{(-1)^k}{(k!)^2}\left(\frac{t}{2}\right)^{2k}. $$ Since we wrote everything in terms of $k$ instead of $n$, there is no reason to keep around the $n=2k$ anymore in the notation, so we obtain that $$ \sum_{n=0}^{\infty}\frac{P_n(0)}{n!}t^n=\sum_{k=0}^{\infty}\frac{(-1)^k}{(k!)^2}\left(\frac{t}{2}\right)^{2k}. $$ And this matches the expression on the left side of $(\star)$ after renaming the variables.