# Recurrence relation for Bessel Functions

1. Aug 12, 2016

### Pual Black

1. The problem statement, all variables and given/known data
I want to prove this relation
$J_{n-1}(x) + J_{n+1}(x)=\frac{2n}{x}J_{n}(x))$
My problem is the part with comparing the coefficient. I don't understand how to compare them. They all have different powers. All pages that i visited have the same solution. Just said "compare the coefficient." They don't explain it.
Do you have a homepage that contains a detailed solution?

2. Aug 12, 2016

### Staff: Mentor

If two polynomials $x^2+2x+3$ and $a_3x^3+a_2x^2+a_1x+a_0$ are equal, what can you say about the $a_i$?

3. Aug 12, 2016

### Ray Vickson

If you have two power series in $t$ that are supposed to be equal for all $t$, then for each $n$ the coefficients of $t^n$ must be the same on both sides. So, look at the terms in $t^0$ on both sides; what will that give you? Then look at the terms in $t^1$ on both sides, and see what you get. Now try to extract the terms in $t^n$ on both sides.

Note that if you have something like $\sum_n a_n t^n + \sum b_n t^{n-1} + \sum_n c_n t^{n+1}$ then the terms in $t^n$ are $(a_n + b_{n+1} + c_{n-1}) t^n$. Do you see why?

4. Aug 13, 2016

### Pual Black

I think i got it.
Also
$a_0=3$
$a_1=2$
$a_2=1$
$a_3=0$

Sorry but i never had to use this method. I dont know why i can change the index "n". Is it because of the summation?

5. Aug 13, 2016

### Ray Vickson

For the summation I presented, write out ALL the terms for n = 1, 2 and 3; there are 9 of them. Now pick out all the terms in $t^2$.

6. Aug 13, 2016

### Pual Black

Thank you. I understand it now.

I got this
$(a_1t^1+a_2t^2+a_3t^3+b_1t^0+ b_2t^1+b3t^2+c_1t^2+c_2t^3+c_3t^4$.
The terms for $t^2$ are
$a_2+b_3+c_1$
Which is the same as

$(a_n + b_{n+1} + c_{n-1}) t^n$.