---
title: "Bessel recurrence J(n+3)+J(n+5)"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:bessel-recurrence-j-n-3"
source: https://www.bctengineer.com/q/semester_2/mathematics/bessel-recurrence-j-n-3
site: BCT Engineering
---

# Bessel recurrence J(n+3)+J(n+5)

Mathematics · Chapter 4: Solution of Differential Equations in Series and Special Functions

- Asked in: 2068 Magh, 5 marks
- Video: [13. Recurrence Formulae#3 and #4 | Bessel Function | Complete Concept](https://www.youtube.com/watch?v=HpPbf2w8b6Q) — Manoj Sir TUTES
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/bessel-recurrence-j-n-3
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Prove the following Bessel's function: J_(n+3)(x) + J_(n+5)(x) = (2/x)(n+4) J_(n+4)(x).

## Short answer

Apply the standard recurrence J_{p−1}(x)+J_{p+1}(x)=(2p/x)J_p(x) with p=n+4 to obtain the identity.

## Answer

The standard recurrence relation of Bessel functions is

J_(p−1)(x) + J_(p+1)(x) = (2p/x) J_p(x).

#### Proof of the recurrence
From the two derivative results

```text
x Jₚ′ = p Jₚ − x J_(p+1)       ... (a)   [from d/dx(x⁻ᵖ Jₚ) = −x⁻ᵖ J_(p+1)]
x Jₚ′ = −p Jₚ + x J_(p−1)      ... (b)   [from d/dx(xᵖ Jₚ) = xᵖ J_(p−1)]
```
Subtract (a) from (b): 0 = −2pJₚ + x(J_(p−1) + J_(p+1)), so

x[J_(p−1) + J_(p+1)] = 2p Jₚ.

#### Apply with p = n + 4
J_(n+3) + J_(n+5) = (2(n+4)/x) J_(n+4)

**Hence J_(n+3)(x) + J_(n+5)(x) = (2/x)(n + 4) J_(n+4)(x).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which standard recurrence is used?
   A. J(p−1) − J(p+1) = (2p/x) Jₚ
   B. J(p−1) + J(p+1) = (p/x) Jₚ
   C. J(p+1) = (p/x) Jₚ
   D. J(p−1) + J(p+1) = (2p/x) Jₚ
   Answer: D. It is the basic Bessel recurrence. (Key point: Starting recurrence)

2. How is the recurrence derived from the two derivative results?
   A. Integrate both expressions
   B. Add the two expressions and divide by 2
   C. Subtract the two expressions for x Jₚ'
   D. Differentiate again
   Answer: C. Subtracting gives 0 = −2pJₚ + x(J(p−1) + J(p+1)). (Key point: How derived)

3. To get the required identity we set:
   A. p = n + 3
   B. p = n + 4
   C. p = n + 5
   D. p = n
   Answer: B. With p = n + 4: J(n+3) + J(n+5) = (2(n+4)/x) J(n+4). (Key point: Choice of p)

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