---
title: "Prove expansion of x⁵ in Legendre functions"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:prove-expansion-x-legendre-functions"
source: https://www.bctengineer.com/q/semester_2/mathematics/prove-expansion-x-legendre-functions
site: BCT Engineering
---

# Prove expansion of x⁵ in Legendre functions

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

- Asked in: 2072 Ashwin, 5 marks
- Video: [Express the f(x) interms of Legendre's polynomials example (PART-1)](https://www.youtube.com/watch?v=f86K0QerUug) — EASY MATHS EASY TRICKS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/prove-expansion-x-legendre-functions
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Prove that the Legendre's function x⁵ = (8/63)[P₅(x) + (7/2)P₃(x) + (27/8)P₁(x)].

## Short answer

Using P_5 = (63x^5 - 70x^3 + 15x)/8, P_3 = (5x^3 - 3x)/2, and P_1 = x, solve for x^5: 63x^5 = 8P_5 + 28P_3 + 27P_1, then divide by 63 to get x^5 = (8/63)[P_5 + (7/2)P_3 + (27/8)P_1].

## Answer

Legendre polynomials:

```text
P₁ = x
P₃ = (5x³ − 3x)/2
P₅ = (63x⁵ − 70x³ + 15x)/8
```

#### Solve for x⁵
From P₅: 63x⁵ = 8P₅ + 70x³ − 15x, so x⁵ = [8P₅ + 70x³ − 15x]/63.

From P₃: x³ = (2P₃ + 3P₁)/5, so 70x³ = 28P₃ + 42P₁.

Substitute (x = P₁):

```text
63x⁵ = 8P₅ + 28P₃ + 42P₁ − 15P₁
     = 8P₅ + 28P₃ + 27P₁
```
Divide by 63:

```text
x⁵ = (8/63)P₅ + (28/63)P₃ + (27/63)P₁
   = (8/63)[ P₅ + (28/8)P₃ + (27/8)P₁ ]
   = (8/63)[ P₅ + (7/2)P₃ + (27/8)P₁ ]
```

**Hence x⁵ = (8/63)[P₅(x) + (7/2)P₃(x) + (27/8)P₁(x)], as required.**

Check at x = 1: (8/63)(1 + 3.5 + 3.375) = (8/63)(7.875) = 1 ✓.

## Check yourself

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

1. P5(x) equals:
   A. (63x^5 + 70x^3 + 15x)/8
   B. (63x^5 - 70x^3 + 15x)/8
   C. (35x^4 - 30x^2 + 3)/8
   D. (63x^5 - 70x^3 - 15x)/8
   Answer: B. Standard Legendre polynomial. (Key point: P5)

2. 70x^3 in terms of P3 and P1 is:
   A. 14P3 + 42P1
   B. 28P3 - 42P1
   C. 28P3 + 42P1
   D. 70P3 + 42P1
   Answer: C. x^3 = (2P3+3P1)/5, times 70. (Key point: Eliminating x^3)

3. 63x^5 equals:
   A. 8P5 + 28P3 + 27P1
   B. 8P5 + 28P3 + 57P1
   C. 8P5 - 28P3 + 27P1
   D. 8P5 + 70P3 + 27P1
   Answer: A. 42P1 - 15P1 = 27P1. (Key point: Combined)

4. Factoring 8/63 gives the P3 coefficient:
   A. 28/63
   B. 7/2
   C. 7/4
   D. 3/2
   Answer: B. 28/8 = 7/2. (Key point: Factoring)

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