how to show that $\frac{d}{dr}J_0(\lambda r)= -\lambda J_1(\lambda r)$? as a bessel series expansion

Question

we already know that the Bessel series expansions is $$ J_n(\lambda r)=({\lambda r \over 2})^n \sum_m^{\infty}({\lambda r \over 2})^{2m} \frac{(-1)^m}{m!(n+m)!}$$ we want to prove that $$\frac{d}{dr}J_0(\lambda r)= -\lambda J_1(\lambda r)$$ we first notice that $$ \frac{d}{dr}J_0(\lambda r)= \lambda J_0^{'}(\lambda r)$$ so the Eq.(2)becomes $$\lambda J_0^{'}(\lambda r)=-\lambda J_1(\lambda r)$$ In other words we just need to prove that $$J_0^{'}(\lambda r)+J_1(\lambda r)=0$$ obviously $$J_0(\lambda r)=\sum_m^{\infty}({\lambda r \over 2})^{2m} \frac{(-1)^m}{m!m!} $$ so $$J_0^{'}(\lambda r)=2m\sum_m^{\infty}({\lambda r \over 2})^{2m-1} \frac{(-1)^m}{m!m!}$$ moreover $$J_1(\lambda r)=({\lambda r \over 2})^1\sum_m^{\infty}({\lambda r \over 2})^{2m} \frac{(-1)^m}{m!(1+m)!}$$ so that $$J_0^{'}(\lambda r)+J_1(\lambda r)=2m\sum_m^{\infty}({\lambda r \over 2})^{2m-1} \frac{(-1)^m}{m!m!}+({\lambda r \over 2})^1\sum_m^{\infty}({\lambda r \over 2})^{2m} \frac{(-1)^m}{m!(1+m)!}$$ after some simplification I notice I just need to prove that $$ (\frac{\lambda r}{2})^2 \frac{1}{m+1}+2m =0 $$ and I am stuck at here. I am just wondering, where am I wrong?

Answer 1

So first off, some errors. The start of the sum is $m= 0$. The derivative of $(\lambda r/2)^{2m}$ with respect to $\lambda r$ is $m(\lambda r/2)^{2m-1}$, not $2m(\lambda r/2)^{2m-1}$. Also you can't take $m$ out of the sum like that, since that's what's being summed over. Cleaning all that up and letting $u = \lambda r$ for brevity gives $$J_0^{'}(u)+J_1(u)=\sum_{m= 0}^{\infty}m\left({ u \over 2}\right)^{2m-1} \frac{(-1)^m}{m!m!}+\sum_{m=0}^{\infty}\left({u \over 2}\right)^{2m+1} \frac{(-1)^m}{m!(1+m)!} $$ To solve the problem, note that the first term in the $J_0'$ sum is $0$. Thus, you can have the sum start at $m=1$ instead. What do you notice about the $m=1$ term of the $J_0'$ sum and the $m=0$ term of the $J_1$ sum? What about the next terms? Can you make an argument that they all sum to $0$?

Answer 2

ok I finally got it. thank you very much @eyeballfrog. since the first sum $$\sum_{m=0}^{\infty} =\sum_{m=1}^{\infty}$$ I could just replace the second sum (m) as (m-1) started from 1
so the second sum becomes $$\sum_{m=1}^{\infty} \left({ u \over 2}\right)^{2m-1} \frac{(-1)^{m-1}}{(m-1)!m!}$$ so we add both up we get $$\sum_{m= 1}^{\infty}m\left({ u \over 2}\right)^{2m-1} \frac{(-1)^m}{m!m!}+\sum_{m=1}^{\infty} \left({ u \over 2}\right)^{2m-1} \frac{(-1)^{m-1}}{(m-1)!m!}$$ Then apparently $$m((-1)^m+(-1)^{m-1})) ==0$$ constantly. thus the entire equation becomes 0 all the time.
Again, Thank you very much @eyeballfrog!!!