# What did I do wrong on this Bessel expansion?

• yungman
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.
yungman

## 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 $$-2\sum_{j=1}^{\infty} \frac{J_{1}(\alpha_{2,j})-2J_{1}(\frac{\alpha_{2,j}}{2})}{\alpha_{2,j}J_{1}(\alpha_{2,j})^{2}}J_{2}(\alpha_{2,j}x)$$

I got everything correct except the deminator where I have

$$\alpha_{2,j}J_{3}(\alpha_{j})^{2}$$

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

$$\int x^{-p+1}J_{p}(x)dx=-x^{-p+1}J_{p-1}(x)+C$$ for $$p=2\Rightarrow \int x^{-1}J_{2}(x)dx=-x^{-1}J_{1}(x)+C$$

## The Attempt at a Solution

We let $$s=\alpha_{j}x \Rightarrow \frac{ds}{\alpha_{j}}=dx, x=\frac{s}{\alpha_{j}},a=1$$

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

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

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

Thanks for your time and Merry Christmas

Alan

## 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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