Recurrence relation for Bessel Functions

Pual Black
Messages
91
Reaction score
1

Homework Statement


I want to prove this relation
##J_{n-1}(x) + J_{n+1}(x)=\frac{2n}{x}J_{n}(x))##
from the generating function. The same question was asked in this page with solution.
http://www.edaboard.com/thread47250.html
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?
 
Physics news on Phys.org
Pual Black said:

Homework Statement


I want to prove this relation
##J_{n-1}(x) + J_{n+1}(x)=\frac{2n}{x}J_{n}(x))##
from the generating function. The same question was asked in this page with solution.
http://www.edaboard.com/thread47250.html
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?
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##?
 
Pual Black said:

Homework Statement


I want to prove this relation
##J_{n-1}(x) + J_{n+1}(x)=\frac{2n}{x}J_{n}(x))##
from the generating function. The same question was asked in this page with solution.
http://www.edaboard.com/thread47250.html
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?

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?
 
fresh_42 said:
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##?
I think i got it.
Also
##a_0=3##
##a_1=2##
##a_2=1##
##a_3=0##
Ray Vickson said:
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?

Sorry but i never had to use this method. I don't know why i can change the index "n". Is it because of the summation?
 
Pual Black said:
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 don't know why i can change the index "n". Is it because of the summation?

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##.
 
  • Like
Likes Pual Black
Ray Vickson said:
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##.

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##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top