What did I do wrong on this Bessel expansion?

In summary, the conversation discusses the use of Bessel functions to represent a given function, with the book providing the answer of -2 times the sum of a specific formula involving the Bessel function. However, the person has trouble reducing the order of 3 in the denominator and seeks help in finding a way to do so. They attempt a solution using a substitution and eventually arrive at an expression involving the Bessel function again. The conversation ends with a request for assistance and a holiday greeting.
  • #1
yungman
5,718
240

Homework Statement



I cannot get the answer given by the book. The question is:

Using Bessel function of order = 2 to represent f(x):
f(x)=0 for 0<x<1/2 and f(x)=1 for 1/2<x<1.


The Answer given by the book is [tex]-2\sum_{j=1}^{\infty} \frac{J_{1}(\alpha_{2,j})-2J_{1}(\frac{\alpha_{2,j}}{2})}{\alpha_{2,j}J_{1}(\alpha_{2,j})^{2}}J_{2}(\alpha_{2,j}x)[/tex]

I got everything correct except the deminator where I have

[tex]\alpha_{2,j}J_{3}(\alpha_{j})^{2}[/tex]

I know there is a way to reduce the order if it start with order of 1, I cannot reduce the order of 3 to 1.


Homework Equations


[tex]\int x^{-p+1}J_{p}(x)dx=-x^{-p+1}J_{p-1}(x)+C[/tex] for [tex]p=2\Rightarrow \int x^{-1}J_{2}(x)dx=-x^{-1}J_{1}(x)+C[/tex]


The Attempt at a Solution


We let [tex] s=\alpha_{j}x \Rightarrow \frac{ds}{\alpha_{j}}=dx, x=\frac{s}{\alpha_{j}},a=1[/tex]

[tex]A_{j}=\frac{\int_{0}^{a}xf(x)J_{2}(\lambda_{j}x)dx}{\int_{0}^{a}xJ_{2}(\lambda_{j}x)^{2}dx}=\frac{\int_{\frac{1}{2}}^{1}x \frac{1}{x^{2}} J_{2}(\alpha_{j}x)dx}{\frac{a^{2}}{2}J_{3}(\alpha_{j})^{2}dx}=\frac{2\int_{\frac{\alpha_{j}}{2}}^{\alpha_{j}}\alpha_{j}s^{-1}J_{2}(s)\frac{ds}{\alpha_{J}}}{J_{3}(\alpha_{j})^{2}} = \frac{-2}{J_{3}(\alpha_{j})^{2}}[\frac{J_{1}(s)}{s}]_{\frac{\alpha_{j}}{2}}^{\alpha_{j}}[/tex]

[tex]A_{j}=\frac{-2}{J_{3}(\alpha_{j})^{2}}[\frac{J_{1}(\alpha_{j})}{\alpha_{j}}-\frac{2J_{1}(\frac{\alpha_{j}}{2})}{\alpha_{j}}] = \frac{-2[J_{1}(\alpha_{j})-2J_{1}(\frac{\alpha_{j}}{2})]}{\alpha_{j}J_{3}(\alpha_{j})^{2}}[/tex]

[tex] f(x)=\sum_{j=1}^{\infty}A_{j}J_{2}(\lambda_{j}x) = -2\sum_{j=1}^{\infty}\frac{[J_{1}(\alpha_{j})-2J_{1}(\frac{\alpha_{j}}{2})]}{\alpha_{j}J_{3}(\alpha_{j})^{2}}J_{2}(\alpha_{j}x)[/tex]


Thanks for your time and Merry Christmas

Alan
 
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  • #2
Anyone please!
 

1. What is a Bessel expansion?

A Bessel expansion is a mathematical method for representing a function as a sum of simpler functions. It is often used in physics and engineering to solve problems involving functions with cylindrical symmetry.

2. How do I know if I have made a mistake in my Bessel expansion?

If your Bessel expansion does not accurately represent the original function, it is likely that you have made a mistake. This could be due to an error in your calculations or an incorrect choice of parameters.

3. Can I use a Bessel expansion for any type of function?

No, Bessel expansions are only suitable for functions with cylindrical symmetry. If your function does not have this type of symmetry, a different method may be more appropriate.

4. How do I improve my Bessel expansion if it is not accurate enough?

One way to improve the accuracy of a Bessel expansion is to increase the number of terms in the expansion. This means including more simpler functions in the sum and can result in a better representation of the original function.

5. Are there any common mistakes that scientists make when using Bessel expansions?

Yes, some common mistakes include choosing the wrong parameters, using an incorrect formula, or making a calculation error. It is important to double check your work and ensure that all steps are correct to avoid these mistakes.

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