Find the first 3 terms of the asymptotic expansion of Jn(x)

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Homework Statement


The bessel function Jn(x) is defined by the integral
Jn(x)=1/(inπ)∫0πeixcosφcos(nφ)dφ

From this formula, find the first 3 terms of the asymptotic expansion of Jn(x) when x=n and n is a large positive integer.

Homework Equations

The Attempt at a Solution


I tried combining the cos and exp term together
Jn(x) = ∫π ei(xcosφ+nφ)
 
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uaefame said:

Homework Statement


The bessel function Jn(x) is defined by the integral
Jn(x)=1/(inπ)∫0πeixcosφcos(nφ)dφ

From this formula, find the first 3 terms of the asymptotic expansion of Jn(x) when x=n and n is a large positive integer.

Homework Equations

The Attempt at a Solution


I tried combining the cos and exp term together
Jn(x) = ∫π ei(xcosφ+nφ)
How is this justified? ##e^{\cos(mx)} \cdot \cos(ny) \ne e^{\cos(mx + ny)}##.
 
Mark44 said:
How is this justified? ##e^{\cos(mx)} \cdot \cos(ny) \ne e^{\cos(mx + ny)}##.

Aside from the missing factor outside the integral, and a possible factor of 2 or 1/2, what he wrote is OK because he switched from ##\int_{\phi=0}^{\pi} \cdots## to ##\int_{\phi=-\pi}^{\pi} \cdots##, and used the evenness of the ##\cos(\phi)## in the exponential.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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