# Exploring the Bessel Function Expansion

• MHB
• Another1
In summary, the use of the Bessel function g(x,t)=g(u+v,t)=g(u,t)g(v,t) allows us to show that J_{0}(u+v)=J_{0}(u)J_{0}(v)+2\sum_{s=1}^{\infty}J_{s}(u)J_{-s}(v). This is achieved by manipulating the power series representation of the function and using the properties of Bessel functions to simplify the expression.
Another1
Bessel function

using $$\displaystyle g(x,t)=g(u+v,t)=g(u,t)g(v,t)$$

to show that $$\displaystyle J_{0}(u+v)=J_{0}(u)J_{0}(v)+2\sum_{s=1}^{\infty}J_{s}(u)J_{-s}(v)$$

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my solution

$$\displaystyle g(u+v,t)=e^{\frac{u+v}{2}(t-\frac{1}{t})}$$
$$\displaystyle g(u+v,t)=e^{\frac{u}{2}(t-\frac{1}{t})}\cdot e^{\frac{v}{2}(t-\frac{1}{t})}$$
$$\displaystyle g(u+v,t)=\sum_{n=-\infty}^{\infty}J_{n}(u)t^{n}\sum_{n=-\infty}^{\infty}J_{n}(v)t^{n}$$

$$\displaystyle J_{n}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!(n+s)!}(\frac{u+v}{2})^{n+2s}$$
$$\displaystyle J_{0}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}(\frac{u}{2}+\frac{v}{2})^{2s}$$
$$\displaystyle J_{0}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}\left\{(\frac{u}{2}+\frac{v}{2})^{2s} \right\}$$
$$\displaystyle J_{0}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}\left\{ \sum_{k=0}^{2s}{2s\choose k}\left(\frac{u}{2} \right)^{2s -k}\left(\frac{v}{2} \right)^{k} \right\}$$
$$\displaystyle J_{0}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}\left\{ \left(\frac{u}{2}\right)^{2s}+\left(\frac{v}{2}\right)^{2s}+ \sum_{k=1}^{2s-1}{2s\choose k}\left(\frac{u}{2} \right)^{2s -k}\left(\frac{v}{2} \right)^{k} \right\}$$
$$\displaystyle J_{0}(u+v)=J_{0}(u)+J_{0}(v)+\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}\left\{\sum_{k=1}^{2s-1}{2s\choose k}\left(\frac{u}{2} \right)^{2s -k}\left(\frac{v}{2} \right)^{k} \right\}$$

this is wrong
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Last edited:
Another said:
using $$\displaystyle g(x,t)=g(u+v,t)=g(u,t)g(v,t)$$

to show that $$\displaystyle J_{0}(u+v)=J_{0}(u)J_{0}(v)+2\sum_{s=1}^{\infty}J_{s}(u)J_{-s}(v)$$

___________________________________________________________________________________________
my solution

$$\displaystyle g(u+v,t)=e^{\frac{u+v}{2}(t-\frac{1}{t})}$$
$$\displaystyle g(u+v,t)=e^{\frac{u}{2}(t-\frac{1}{t})}\cdot e^{\frac{v}{2}(t-\frac{1}{t})}$$
$$\displaystyle g(u+v,t)=\sum_{n=-\infty}^{\infty}J_{n}(u)t^{n}\sum_{n=-\infty}^{\infty}J_{n}(v)t^{n}$$

$$\displaystyle J_{n}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!(n+s)!}(\frac{u+v}{2})^{n+2s}$$
$$\displaystyle J_{0}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}(\frac{u}{2}+\frac{v}{2})^{2s}$$
$$\displaystyle J_{0}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}\left\{(\frac{u}{2}+\frac{v}{2})^{2s} \right\}$$
$$\displaystyle J_{0}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}\left\{ \sum_{k=0}^{2s}{2s\choose k}\left(\frac{u}{2} \right)^{2s -k}\left(\frac{v}{2} \right)^{k} \right\}$$
$$\displaystyle J_{0}(u+v)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}\left\{ \left(\frac{u}{2}\right)^{2s}+\left(\frac{v}{2}\right)^{2s}+ \sum_{k=1}^{2s-1}{2s\choose k}\left(\frac{u}{2} \right)^{2s -k}\left(\frac{v}{2} \right)^{k} \right\}$$
$$\displaystyle J_{0}(u+v)=J_{0}(u)+J_{0}(v)+\sum_{s=0}^{\infty}\frac{(-1)^{s}}{s!s!}\left\{\sum_{k=1}^{2s-1}{2s\choose k}\left(\frac{u}{2} \right)^{2s -k}\left(\frac{v}{2} \right)^{k} \right\}$$

this is wrong
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now i can solve it thankkkk !

$$\displaystyle g(u+v,t)=g(u,t)g(v,t)$$

$$\displaystyle e^{\frac{u+v}{2}(t-\frac{1}{t})}=e^{\frac{u}{2}(t-\frac{1}{t})}e^{\frac{v}{2}(t-\frac{1}{t})}$$

$$\displaystyle \sum_{n=-\infty}^{\infty}J_{n}(u+v)t^n=\sum_{l=-\infty}^{\infty}J_{l}(u)t^l\sum_{m=-\infty}^{\infty}J_{m}(v)t^m$$

let n = 0
$$\displaystyle J_{0}(u+v)= \left(...+J_{-1}(u)t^{-1}+J_{0}(u)+J_{1}(u)t^1+...\right)\left(...+J_{-1}(v)t^{-1}+J_{0}(v)+J_{1}(v)t^1+...\right)$$

but $$\displaystyle J_{-n}(u)=(-1)^{n}J_{n}(u)$$
so...
$$\displaystyle J_{0}(u+v)= J_{0}(u)J_{0}(v)+2J_{1}(u)J_{-1}(v)+2J_{2}(u)J_{2}(v)+...$$
$$\displaystyle J_{0}(u+v)= J_{0}(u)J_{0}(v)+2\sum_{s=1}^{\infty}J_{s}(u)J_{-s}(v)$$

Funions!

Joppy said:
Funions!

oh sorry I mean function n and c missing from word

## 1. What is the Bessel function expansion?

The Bessel function expansion is a mathematical technique used to represent a wide range of functions as a linear combination of Bessel functions. These functions were first introduced by the German mathematician Friedrich Bessel in the early 19th century and have since been widely used in various fields, including physics, engineering, and applied mathematics.

## 2. What are the applications of the Bessel function expansion?

The Bessel function expansion has a wide range of applications, including solving differential equations, analyzing physical systems such as heat conduction and vibrations, and modeling electromagnetic and acoustic fields. It is also used in signal processing, image reconstruction, and data analysis.

## 3. How does the Bessel function expansion work?

The Bessel function expansion works by expressing a given function as a sum of Bessel functions multiplied by coefficients. These coefficients are determined by solving a system of equations, known as the Bessel equation, using various techniques such as power series, continued fractions, or numerical methods. The resulting expansion can then be used to approximate the original function with a desired level of accuracy.

## 4. What are the advantages of using Bessel function expansion?

The Bessel function expansion has several advantages, including its ability to represent a wide range of functions, its simplicity in form, and its convergence properties, which allow for accurate approximations with a relatively small number of terms. It also provides a systematic way to analyze and solve differential equations and physical problems, making it a valuable tool in scientific research and applications.

## 5. Are there any limitations to the Bessel function expansion?

While the Bessel function expansion has many applications and advantages, it also has some limitations. It may not be suitable for functions that do not have a certain level of smoothness or for functions with singularities. In addition, the coefficients in the expansion may be difficult to calculate for some functions, requiring the use of numerical methods. However, these limitations can often be overcome by using alternative techniques or approximations.

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