- 2069 Bhadra · 5 marks
Derivative of xⁿJₙ(x)
Short answer
Multiply J_n series by x^n, differentiate termwise, simplify using Γ(n+k+1) = (n+k)Γ(n+k). The resulting series is x^n J_{n-1}, so d/dx[x^n J_n] = x^n J_{n-1}.
Prove the Bessel's function d/dx[xⁿ Jₙ(x)] = xⁿ J_(n−1)(x).
Jₙ(x) = Σₖ (−1)ᵏ / (k! Γ(n + k + 1)) · (x/2)^(2k+n).
Multiply by xⁿ
xⁿ Jₙ(x) = Σ (−1)ᵏ x^(2k+2n) / (2^(2k+n) k! Γ(n+k+1))
Differentiate term by term
d/dx [xⁿ Jₙ] = Σ (−1)ᵏ (2k+2n) x^(2k+2n−1) / (2^(2k+n) k! Γ(n+k+1))
Use Γ(n + k + 1) = (n + k)Γ(n + k) and 2k + 2n = 2(n + k):
= Σ (−1)ᵏ 2(n+k) x^(2k+2n−1) / (2^(2k+n) k! (n+k) Γ(n+k))
= Σ (−1)ᵏ x^(2k+2n−1) / (2^(2k+n−1) k! Γ(n+k))
Take xⁿ outside: x^(2k+2n−1) = xⁿ · x^(2k+n−1), and note n + k = (n − 1) + k + 1:
= xⁿ Σ (−1)ᵏ / (k! Γ((n−1)+k+1)) · (x/2)^(2k+(n−1))
= xⁿ J_(n−1)(x)
Hence d/dx[xⁿ Jₙ(x)] = xⁿ J_(n−1)(x).
✓ Answer checked by Claude Opus 5.5 (high effort)
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