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Express x³−5x²+6x+1 in Legendre polynomials

Short answer

Express x³, x², x, 1 using P_0=1, P_1=x, P_2=(3x²-1)/2, P_3=(5x³-3x)/2. Substituting into x³-5x²+6x+1 gives (2/5)P_3(x) - (10/3)P_2(x) + (33/5)P_1(x) - (2/3)P_0(x).

Express in terms of Legendre's polynomials f(x) = x³ − 5x² + 6x + 1.

Legendre polynomials: P₀ = 1, P₁ = x, P₂ = (3x² − 1)/2, P₃ = (5x³ − 3x)/2.

Powers of x in terms of Pₙ

x³ = (2P₃ + 3P₁)/5        (from P₃ = (5x³ − 3x)/2)
x² = (2P₂ + P₀)/3         (from P₂ = (3x² − 1)/2)
x  = P₁,   1 = P₀

Substitute in f(x) = x³ − 5x² + 6x + 1

f = (2/5)P₃ + (3/5)P₁ − (5/3)(2P₂ + P₀) + 6P₁ + P₀
  = (2/5)P₃ − (10/3)P₂ + (3/5 + 6)P₁ + (1 − 5/3)P₀
  = (2/5)P₃ − (10/3)P₂ + (33/5)P₁ − (2/3)P₀

Answer: x³ − 5x² + 6x + 1 = (2/5)P₃(x) − (10/3)P₂(x) + (33/5)P₁(x) − (2/3)P₀(x).

Check at x = 1: 2/5 − 10/3 + 33/5 − 2/3 = 7 − 4 = 3, and f(1) = 1 − 5 + 6 + 1 = 3 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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