- #1
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- 10
Hey guys and gals,
While this technically isn't homework, I figured this is the place to post.
I am working over a problem and I am at a point in the solution that has me a bit stumped. Perhaps someone may provide some guidance.
In acoustics, we run into the problem of a radiating body in cylindrical coordinates. Essentially, after a bit of work, I have come to the point where I am stuck in equating a body's radial velocity and the radial derivative of the velocity potential. What I have is:
[tex]C cos(\theta) = \sum_{n=0}^\infty A_n \left(\frac{\omega}{a_o} \right) H_{n+1}^{(1)} \left(\frac{\omega}{a_o} R\right) cos(n \theta)[/tex]
Where C is a constant, An is what I am trying to solve for, and H is the Hankel function of the first kind.
Now, normally I don't have both sides as a function of theta and the solving for An is pretty straight forward. However, this time it is not the case. Is there a way to somehow come up with a general solution to An that does not include the summation? I'm thinking no, but I figured I'd ask.
Thanks!
While this technically isn't homework, I figured this is the place to post.
I am working over a problem and I am at a point in the solution that has me a bit stumped. Perhaps someone may provide some guidance.
In acoustics, we run into the problem of a radiating body in cylindrical coordinates. Essentially, after a bit of work, I have come to the point where I am stuck in equating a body's radial velocity and the radial derivative of the velocity potential. What I have is:
[tex]C cos(\theta) = \sum_{n=0}^\infty A_n \left(\frac{\omega}{a_o} \right) H_{n+1}^{(1)} \left(\frac{\omega}{a_o} R\right) cos(n \theta)[/tex]
Where C is a constant, An is what I am trying to solve for, and H is the Hankel function of the first kind.
Now, normally I don't have both sides as a function of theta and the solving for An is pretty straight forward. However, this time it is not the case. Is there a way to somehow come up with a general solution to An that does not include the summation? I'm thinking no, but I figured I'd ask.
Thanks!