- #1
yungman
- 5,755
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I am reading the article Mirela Vinerean:
http://www.math.kau.se/mirevine/mf2bess.pdf
On page 6, I have a question about
[tex]e^{\frac{x}{2}t} e^{-\frac{x}{2}\frac{1}{t}}=\sum^{\infty}_{n=-\infty}J_n(x)e^{jn\theta}=\sum_{n=0}^{\infty}J_n(x)[e^{jn\theta}+(-1)^ne^{-jn\theta}][/tex]
I think there is a mistake at the last term. If you look at n=0, the equation will be:
[tex]J_0(x)[e^0+(-1)^0e^{-0}]\;=\;j_0(x)[1+1]\;=\;2J_0(x)[/tex]
Which is not correct. The problem is n=0 is being repeated in both the n=+ve and n=-ve.
The equation should be:
[tex]e^{\frac{x}{2}t} e^{-\frac{x}{2}\frac{1}{t}}=\sum_{n=0}^{\infty}[J_n(x)e^{jn\theta}]\;+\;\sum_{n=1}^{\infty}[(-1)^ne^{-jn\theta}][/tex]
With this, the first term covers the original n=+ve from 0 to ∞. The second term covers the original from -∞ to -1. Now you only have one term contain n=0. Am I correct?
Thanks
http://www.math.kau.se/mirevine/mf2bess.pdf
On page 6, I have a question about
[tex]e^{\frac{x}{2}t} e^{-\frac{x}{2}\frac{1}{t}}=\sum^{\infty}_{n=-\infty}J_n(x)e^{jn\theta}=\sum_{n=0}^{\infty}J_n(x)[e^{jn\theta}+(-1)^ne^{-jn\theta}][/tex]
I think there is a mistake at the last term. If you look at n=0, the equation will be:
[tex]J_0(x)[e^0+(-1)^0e^{-0}]\;=\;j_0(x)[1+1]\;=\;2J_0(x)[/tex]
Which is not correct. The problem is n=0 is being repeated in both the n=+ve and n=-ve.
The equation should be:
[tex]e^{\frac{x}{2}t} e^{-\frac{x}{2}\frac{1}{t}}=\sum_{n=0}^{\infty}[J_n(x)e^{jn\theta}]\;+\;\sum_{n=1}^{\infty}[(-1)^ne^{-jn\theta}][/tex]
With this, the first term covers the original n=+ve from 0 to ∞. The second term covers the original from -∞ to -1. Now you only have one term contain n=0. Am I correct?
Thanks