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Prove expansion of x⁵ in Legendre functions

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].

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

Legendre polynomials:

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):

63x⁵ = 8P₅ + 28P₃ + 42P₁ − 15P₁
     = 8P₅ + 28P₃ + 27P₁

Divide by 63:

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 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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