Proving Recursion relations for Bessel Functions

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SUMMARY

This discussion focuses on solving recursion relations for Bessel functions, specifically J_{p+1}(x) and J_{p-1}(x), using the equations provided. The key equations are: 1) \(\frac{d}{dx}[x^{p}J_{p}(x)] = x^{p}J_{p-1}(x)\) and 2) \(\frac{d}{dx}[x^{-p}J_{p}(x)] = -x^{-p}J_{p+1}(x)\). By manipulating these equations through addition and subtraction, the results are derived as 3) \(J_{p-1}(x) + J_{p+1}(x) = \frac{2p}{x}J_{p}(x)\) and 4) \(J_{p-1}(x) - J_{p+1}(x) = 2J^{'}_{p}(x)\). The discussion highlights the importance of applying basic calculus rules rather than relying solely on infinite series expansions.

PREREQUISITES
  • Understanding of Bessel functions, specifically J_{p}(x)
  • Familiarity with differentiation and the product rule in calculus
  • Knowledge of recursion relations in mathematical functions
  • Ability to manipulate equations and perform algebraic operations
NEXT STEPS
  • Study the properties and applications of Bessel functions in mathematical physics
  • Learn about the product rule and its applications in differentiation
  • Explore infinite series expansions of Bessel functions
  • Investigate other recursion relations for special functions
USEFUL FOR

Mathematics students, physicists, and engineers who are working with Bessel functions and recursion relations in their studies or research.

gametheory
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Homework Statement


Solve equations 1) and 2) for J_{p+1}(x) and J_{p-1}(x). Add and subtract these two equations to get 3) and 4).


Homework Equations


1) \frac{d}{dx}[x^{p}J_{p}(x)] = x^{p}J_{p-1}(x)
2) \frac{d}{dx}[x^{-p}J_{p}(x)] = -x^{-p}J_{p+1}(x)
3) J_{p-1}(x) + J_{p+1}(x) = \frac{2p}{x}J_{p}(x)
4) J_{p-1}(x) - J_{p+1}(x) = 2J^{'}_{p}(x)


The Attempt at a Solution


My main problem is I'm not really sure what the question is asking me to do in the first part. Am I supposed to plug p+1 and p-1 into J_{p} on the left of each equation or am I supposed to simply solve equation 1) as J_{p-1}(x) = x^{-p}\frac{d}{dx}[x^{p}J_{p}(x)] and equation 2) as J_{p+1}(x) = -x^{p}\frac{d}{dx}[x^{-p}J_{p}(x)]? I tried this way and then differentiated the series and got two infinite series I didn't know what to do with.

Next, I tried to substitute J_{p+1} into J_{p} and I integrated on both sides and just got J_{p+1} = J_{p+1} after rearranging everything.

I feel like this isn't an overly difficult problem, but I just have no idea what direction to take with it.
 
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gametheory said:
My main problem is I'm not really sure what the question is asking me to do in the first part. Am I supposed to plug p+1 and p-1 into J_{p} on the left of each equation or am I supposed to simply solve equation 1) as J_{p-1}(x) = x^{-p}\frac{d}{dx}[x^{p}J_{p}(x)] and equation 2) as J_{p+1}(x) = -x^{p}\frac{d}{dx}[x^{-p}J_{p}(x)]? I tried this way and then differentiated the series and got two infinite series I didn't know what to do with.

Yes, it's saying the latter, but I don't know what you mean about getting infinite series; if you expand the derivative on the RHS in each case, you can just use the product rule to get one term with the Bessel function and one term with its derivative (and adding these or subtracting should give the two results).
 
Nevermind, you're right, I got it. All the other problems had us manipulate the bessel function as an infinite series and I was just used to doing this. Forgot to try basic rules from calculus...haha thanks
 

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