---
title: "Bessel function J₃/₂(x)"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:bessel-function-j-x-2"
source: https://www.bctengineer.com/q/semester_2/mathematics/bessel-function-j-x-2
site: BCT Engineering
---

# Bessel function J₃/₂(x)

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

- Asked in: 2075 Baisakh, 5 marks; 2073 Magh, 5 marks; 2072 Magh, 5 marks (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/bessel-function-j-x-2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Prove the Bessel's function J_{3/2}(x) = √(2/(πx)) (sin x / x − cos x).

## Short answer

Apply recurrence J_{-1/2} + J_{3/2} = (1/x) J_{1/2} with J_{1/2}=√(2/(πx)) sin x and J_{-1/2}=√(2/(πx)) cos x. This gives J_{3/2}(x) = √(2/(πx))[sin x / x - cos x].

## Answer

Bessel's function: J_p(x) = Σ (−1)^k / (k! Γ(k + p + 1)) (x/2)^(2k+p), and the recurrence relation is

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

#### J_(1/2) and J_(−1/2)
With Γ(k + 3/2) = (2k + 1)!√π / (2^(2k+1) k!):

```text
J_(1/2)(x) = Σ (−1)^k /(k! Γ(k+3/2)) (x/2)^(2k+1/2)
           = √(2/(πx)) Σ (−1)^k x^(2k+1)/(2k+1)!
           = √(2/(πx)) sin x
```
With Γ(k + 1/2) = (2k)!√π / (4^k k!):

```text
J_(−1/2)(x) = √(2/(πx)) Σ (−1)^k x^(2k)/(2k)! = √(2/(πx)) cos x
```

#### J_(3/2)
Put p = 1/2 in the recurrence:

J_(−1/2) + J_(3/2) = (1/x) J_(1/2)

```text
J_(3/2) = (1/x) J_(1/2) − J_(−1/2)
        = √(2/(πx)) [ (sin x)/x − cos x ]
```

**Hence J_(3/2)(x) = √(2/(πx)) ( sin x / x − cos x ).**

## Check yourself

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

1. Which recurrence is used to get J_(3/2)?
   A. p = 3/2
   B. p = -1/2
   C. p = 1/2 in J_(p-1) + J_(p+1) = (2p/x) J_p
   D. p = 0
   Answer: C. It links J_(-1/2), J_(1/2), J_(3/2). (Key point: Order used)

2. J_(1/2)(x) equals:
   A. sqrt(2/(πx)) sin x
   B. sqrt(2/(πx)) cos x
   C. sqrt(2x/π) sin x
   D. sin x / x
   Answer: A. From the series with Γ(k+3/2). (Key point: J_(1/2))

3. J_(-1/2)(x) equals:
   A. sqrt(2/(πx)) sin x
   B. sqrt(2/(πx)) cos x
   C. sqrt(2/(πx)) tan x
   D. -sqrt(2/(πx)) cos x
   Answer: B. From the series with Γ(k+1/2). (Key point: J_(-1/2))

4. J_(3/2) = (1/x) J_(1/2) - J_(-1/2) gives:
   A. sqrt(2/(πx)) (sin x / x + cos x)
   B. sqrt(2/(πx)) (cos x / x - sin x)
   C. sqrt(2/(πx)) (sin x / x - cos x)
   D. sqrt(2/(πx)) (cos x - sin x / x)
   Answer: C. (1/x) sin x - cos x with common factor. (Key point: Result)

## Close variants set in other exams

- [Bessel function J₋₅/₂(x)](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Abessel-function-j-x) — Mathematics, chapter 4, asked 7×

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