# How to add two bessel functions

1. Jan 4, 2012

### vvthuy

Hi,

I need to solve one problem like this:

(a+b)*J_{1}[x(a+b)]-(a-b)*J_{1}[x(a-b)]=c

J_{1} denotes the first order Bessel function. Do you think that it is possible to solve this function in an analytical way?

Thanks,

Viet.

2. Jan 4, 2012

### JJacquelin

Unfortunately not, I think.

3. Jan 5, 2012

4. Jan 5, 2012

### vvthuy

I found an way to modify the above bessel function using Multiplication theorem but I was stuck again at the following step
(b-1)^n-(b+1)^n

Do you know whether I can simplify this using series?

5. Jan 6, 2012

### JJacquelin

Generally, analytical approximate solutions are formulas which depend on the range of the parameters values. If nothing is known about the range of the numerical values of the parameters (a, b), probably it is impossible to say if such a formula can be derived.

6. Jan 6, 2012

### JJacquelin

The finite series development is :

File size:
3.2 KB
Views:
76
7. Jan 6, 2012

### vvthuy

Thank JJacquelin's comments and the equation. Do you know any other series which allow us to simplify (b-1)^n-(b+1)^n to the form of ()^n?

8. Jan 6, 2012

### JJacquelin

What do you mean (?)^n ?

9. Jan 6, 2012

### vvthuy

I meant, I tried to get the following form by expansion the left hand side of equation and then combine terms to get the right hand side

(b-1)^n-(b+1)^n =(f)^n

where f depends on b.

for example

(x-1)^2-(x+1)^2 =-[4sqrt(x)]^2

10. Jan 6, 2012

### JJacquelin

You can only do that in case of n<3 because, in this case, the number of terms of the series development (as shown in my preceeding post) is reduced to one.
If n>2, it is impossible to find a function f which do not depends of n.
(b-1)^n-(b+1)^n =(f(b,n))^n