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

In summary: It is not clear why he switched from ##\cos(n\phi)## to ##\cos(ny)##. I would have used ##\cos(n\phi)##, and taken ##y=\phi##.
  • #1
uaefame
4
0

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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  • #2
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)}##.
 
  • #3
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.
 

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

1. What is the asymptotic expansion of Jn(x)?

The asymptotic expansion of Jn(x) is a mathematical series that approximates the Bessel function of the first kind, Jn(x), for large values of x. It is used to calculate the values of the Bessel function with high precision.

2. How is the asymptotic expansion of Jn(x) calculated?

The asymptotic expansion of Jn(x) is calculated using the Laplace method, which involves expanding the Bessel function into a power series and then applying the method of steepest descent to determine the dominant term in the series.

3. What are the first 3 terms of the asymptotic expansion of Jn(x)?

The first 3 terms of the asymptotic expansion of Jn(x) are given by:

Jn(x) ~ (1/√2πx)cos(x-πn/2) + (1/8πx^2)sin(x-πn/2) - (1/8πx^3)cos(x-πn/2)

4. How accurate is the asymptotic expansion of Jn(x)?

The accuracy of the asymptotic expansion of Jn(x) depends on the value of x and the number of terms included in the expansion. For large values of x, including more terms in the expansion can improve the accuracy of the approximation.

5. What are the practical applications of the asymptotic expansion of Jn(x)?

The asymptotic expansion of Jn(x) has many practical applications in physics, engineering, and other fields. It is commonly used in the analysis of wave phenomena, such as diffraction and scattering, and in solving differential equations that involve Bessel functions. It is also used in the study of heat transfer and electrical engineering problems.

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