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Differential EquationsHardFrobenius & Special Functions

Frobenius & Special Functions · Hard problem

Expand f(x)=∣x∣f(x) = |x| on [−1,1][-1, 1] in Legendre polynomials, ∣x∣=∑n≥0cnPn(x)|x| = \sum_{n\ge0}c_nP_n(x), where P2(x)=12(3x2−1)P_2(x) = \tfrac12\left(3x^2 - 1\right). Find c2c_2. Express your answer as a common fraction.