Some verification of equation on the article of Bessel Function

[yungman]

I want to verify there are typos in page 11 of http://math.arizona.edu/~zakharov/BesselFunctions.pdf

1) Right below equation (51)
[tex]\frac{1}{2\pi}\left(e^{j\theta}-e^{-j\theta}\right)^{n+q}e^{-jn\theta}=\left(1-e^{-2j\theta}\right)^n\left(e^{j\theta}-e^{-j\theta}\right)^q[/tex]
There should not be ##\frac {1}{2\pi}## on the left hand side. That will not work.



2)Then in equation (52)
[tex]A_n(z)=\left(\frac z 2 \right)^n\sum_{n=0}^{\infty}\frac{1}{(n+2k)!}\left(\frac z 2 \right)^k I_{k,n} [/tex]
Should have a 2 in the power of k. Also, since ##p=n+2k##, It should be:
[tex]A_n(z)=\left(\frac z 2\right)^n\sum_{n+2k=0}^{\infty}\frac{1}{(n+2k)!}\left(\frac z 2\right)^{2k} I_{k,n} [/tex]
As ##p=n+q## where ##q=2k## for even order. Therefore ##p=n+2k##.
To further prove my assertion, if you look at the bottom of the page 11:
[tex]A_n(z)=\left(\frac z 2 \right)^n\sum_0^{\infty}\frac{(-1)^k}{k!(n+k)!}\left(\frac z 2 \right)^k ≠J_n(z)[/tex]
[tex]J_n(z)\;=\;\sum_{k=0}^{\infty}\frac{(-1)^k}{k!(n+k)!}\left(\frac z 2 \right)^{2k+n} [/tex]

I don't even know how to set the lower limit of the ##\sum## as I stated that the original was ##\sum_{p=0}^{\infty}##, all of a sudden, it becomes ##\sum_{n=0}^{\infty}## which should be ##\sum_{n+2k=0}^{\infty}##

 

[yungman]

[tex]A_n(z)=\left(\frac z 2 \right)^n\sum_0^{\infty}\frac{(-1)^k}{k!(n+k)!}\left(\frac z 2 \right)^k ≠J_n(z)[/tex]
[tex]J_n(z)\;=\;\sum_{k=0}^{\infty}\frac{(-1)^k}{k!(n+k)!}\left(\frac z 2 \right)^{2k+n} [/tex]

I don't even know how to set the lower limit of the ##\sum## as I stated that the original was ##\sum_{p=0}^{\infty}##, all of a sudden, it becomes ##\sum_{n=0}^{\infty}## which should be ##\sum_{n+2k=0}^{\infty}##

I was thinking, the reason ##\sum_{k=0}^{\infty}## is because even though ##p=n+2k##, n is really a constant and the ##\left(\frac{z}{2}\right)^n## can be moved totally out of the ##\sum##. So the ##\sum## is really for series representation of the exponential functions. Therefore, we start with k=0. What do you think?

Thanks