---
title: "Express J₄(x) in terms of J₁ and J₀"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:express-j-x-terms-j"
source: https://www.bctengineer.com/q/semester_2/mathematics/express-j-x-terms-j
site: BCT Engineering
---

# Express J₄(x) in terms of J₁ and J₀

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

- Asked in: 2067 Mangsir, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/express-j-x-terms-j
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Show that J₄(x) = (48/x³ − 3/x) J₁(x) + (1 − 24/x²) J₀(x). [Coefficient '3/x' as printed; scan unclear]

## Short answer

Using J_{p+1}=(2p/x)J_p−J_{p−1} repeatedly yields J₄(x)=(48/x³−8/x)J₁(x)+(1−24/x²)J₀(x); the printed coefficient 3/x is a misprint.

## Answer

[Printed coefficient: 3/x of J₁. Working with the recurrence relation gives 8/x, so the 3/x appears to be a misprint; the identity is proved below with 8/x, i.e. J₄ = (48/x³ − 8/x)J₁ + (1 − 24/x²)J₀. Checked numerically at x = 2.3, where both sides equal 0.05560, while the 3/x version gives 1.229.]

Use the recurrence relation Jₚ₊₁ = (2p/x)Jₚ − Jₚ₋₁.

#### Step by step
```text
J₂ = (2/x)J₁ − J₀                                  (p = 1)
J₃ = (4/x)J₂ − J₁                                  (p = 2)
   = (4/x)[(2/x)J₁ − J₀] − J₁
   = (8/x² − 1)J₁ − (4/x)J₀
J₄ = (6/x)J₃ − J₂                                  (p = 3)
   = (6/x)[(8/x² − 1)J₁ − (4/x)J₀] − [(2/x)J₁ − J₀]
   = (48/x³ − 6/x − 2/x)J₁ + (1 − 24/x²)J₀
```

**Result: J₄(x) = (48/x³ − 8/x) J₁(x) + (1 − 24/x²) J₀(x).**

The J₀ coefficient (1 − 24/x²) matches the printed form; only the J₁ coefficient differs (8/x, not 3/x).

## Check yourself

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

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

2. To reach J₄ from J₀ and J₁ we find:
   A. J₄ directly with p = 4
   B. J₃ first with p = 3
   C. J₂ with p = 2
   D. J₂ (p=1), then J₃ (p=2), then J₄ (p=3)
   Answer: D. Each step uses the previous two. (Key point: Order of steps)

3. J₃ in terms of J₁ and J₀ is:
   A. (8/x − 1)J₁ − (4/x^2)J₀
   B. (4/x^2 − 1)J₁ − (8/x)J₀
   C. (8/x^2 − 1)J₁ − (4/x)J₀
   D. (8/x^2 + 1)J₁ + (4/x)J₀
   Answer: C. J₃ = (4/x)[(2/x)J₁ − J₀] − J₁. (Key point: J₃)

4. The coefficient of J₁ in J₄ is:
   A. 48/x^3 − 3/x
   B. 48/x^3 − 8/x
   C. 48/x^3 − 6/x
   D. 24/x^3 − 8/x
   Answer: B. −6/x − 2/x = −8/x, so the printed 3/x is a misprint. (Key point: Correct coefficient)

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