---
title: "Bessel function J₋₅/₂(x)"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:bessel-function-j-x"
source: https://www.bctengineer.com/q/semester_2/mathematics/bessel-function-j-x
site: BCT Engineering
---

# Bessel function J₋₅/₂(x)

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

- Asked in: 2074 Bhadra, 5 marks; 2073 Bhadra, 5 marks; 2072 Ashwin, 5 marks; 2067 Mangsir, 5 marks (asked 4 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
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Prove the Bessel's function J_{−5/2}(x) = √(2/(πx)) [(3/x) sin x + ((3−x²)/x²) cos x].

## Short answer

Use the recurrence J_{p-1} = (2p/x)J_p - J_{p+1} with p = -1/2 and -3/2, starting from J_{1/2} = √(2/(πx)) sin x and J_{-1/2} = √(2/(πx)) cos x. Then J_{-3/2} = -√(2/(πx))[(cos x)/x + sin x], and J_{-5/2} = √(2/(πx))[(3/x) sin x + ((3-x²)/x²) cos x].

## Answer

Bessel's function of order *p*: J_p(x) = Σ (−1)^k / (k! Γ(k + p + 1)) · (x/2)^(2k+p). We use the recurrence relation

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

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

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

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

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

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

#### Step 3: J_(−5/2)
Put p = −3/2:

```text
J_(−5/2) = (−3/x) J_(−3/2) − J_(−1/2)
         = √(2/(πx)) [ (3/x)((cos x)/x + sin x) − cos x ]
         = √(2/(πx)) [ (3 sin x)/x + (3/x² − 1) cos x ]
```
Since 3/x² − 1 = (3 − x²)/x²,

**J_(−5/2)(x) = √(2/(πx)) [ (3/x) sin x + ((3 − x²)/x²) cos x ]**, as required.

(Checked numerically at x = 1.3: both sides equal 1.7012.)

## Check yourself

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

1. Which two results are computed first from the series?
   A. J_0 and J_1
   B. J_(1/2) = sqrt(2/(πx)) sin x and J_(-1/2) = sqrt(2/(πx)) cos x
   C. J_(3/2) and J_(5/2)
   D. J_(-1/2) = sin x only
   Answer: B. They follow from the gamma-function values at half-integers. (Key point: Starting orders)

2. Which recurrence is used to descend?
   A. J_(p-1) = (2p/x) J_p + J_(p+1)
   B. J_(p-1) = J_p - J_(p+1)
   C. J_(p-1) = (2p/x) J_p - J_(p+1)
   D. J_(p+1) = (2p/x) J_p
   Answer: C. From J_(p-1) + J_(p+1) = (2p/x) J_p. (Key point: Recurrence)

3. J_(-3/2) equals:
   A. -sqrt(2/(πx)) [cos x / x + sin 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 / x - sin x]
   Answer: A. p = -1/2: (-1/x)J_(-1/2) - J_(1/2). (Key point: J_(-3/2))

4. J_(-5/2) is obtained by putting p =
   A. -5/2
   B. -3/2
   C. -1/2
   D. 3/2
   Answer: B. J_(-5/2) = (2(-3/2)/x) J_(-3/2) - J_(-1/2). (Key point: Last step)

## Close variants set in other exams

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