# Some verification of equation on the article of Bessel Function

1. Jul 14, 2013

### yungman

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

1) Right below equation (51)

$$\frac{1}{2\pi}\left(e^{j\theta}-e^{-j\theta}\right)^{n+q}e^{-jn\theta}=\left(1-e^{-2j\theta}\right)^n\left(e^{j\theta}-e^{-j\theta}\right)^q$$
There should not be $\frac {1}{2\pi}$ on the left hand side. That will not work.

2)Then in equation (52)
$$A_n(z)=\left(\frac z 2 \right)^n\sum_{n=0}^{\infty}\frac{1}{(n+2k)!}\left(\frac z 2 \right)^k I_{k,n}$$
Should have a 2 in the power of k. Also, since $p=n+2k$, It should be:
$$A_n(z)=\left(\frac z 2\right)^n\sum_{n+2k=0}^{\infty}\frac{1}{(n+2k)!}\left(\frac z 2\right)^{2k} I_{k,n}$$
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:
$$A_n(z)=\left(\frac z 2 \right)^n\sum_0^{\infty}\frac{(-1)^k}{k!(n+k)!}\left(\frac z 2 \right)^k ≠J_n(z)$$
$$J_n(z)\;=\;\sum_{k=0}^{\infty}\frac{(-1)^k}{k!(n+k)!}\left(\frac z 2 \right)^{2k+n}$$

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 $\sum_{n+2k=0}^{\infty}$

Last edited: Jul 14, 2013
2. Jul 15, 2013

### yungman

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