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Derivative of x⁻ⁿJₙ(x)

Short answer

Start from the series for J_n, multiply by x^{-n}, differentiate term by term, then shift index. This yields the series for -x^{-n} J_{n+1}. Hence d/dx[x^{-n} J_n] = -x^{-n} J_{n+1}.

Prove Bessel's Function d/dx[x⁻ⁿ Jₙ(x)] = −x⁻ⁿ J_(n+1)(x).

Jₙ(x) = Σₖ (−1)ᵏ / (k! Γ(n + k + 1)) · (x/2)^(2k+n), k = 0, 1, 2, …

Multiply by x⁻ⁿ

x⁻ⁿ Jₙ(x) = Σ (−1)ᵏ x^(2k) / (2^(2k+n) k! Γ(n+k+1))

Differentiate term by term

d/dx [x⁻ⁿ Jₙ] = Σ (−1)ᵏ 2k x^(2k−1) / (2^(2k+n) k! Γ(n+k+1))

The k = 0 term vanishes. Put k = m + 1 (m = 0, 1, 2, …), and use k! = (m+1)! = (m+1)·m! and Γ(n+k+1) = Γ(n+m+2):

= Σ (−1)^(m+1) 2(m+1) x^(2m+1) / (2^(2m+n+2) (m+1) m! Γ(n+m+2))
= − Σ (−1)^m x^(2m+1) / (2^(2m+n+1) m! Γ(n+m+2))

Now take x⁻ⁿ outside: x^(2m+1) = x⁻ⁿ · x^(2m+n+1), so

= − x⁻ⁿ Σ (−1)^m / (m! Γ(n+1+m+1)) · (x/2)^(2m+n+1)
= − x⁻ⁿ J_(n+1)(x)

since the last sum is the series of J_(n+1).

Hence d/dx[x⁻ⁿ Jₙ(x)] = −x⁻ⁿ J_(n+1)(x).

✓ Answer checked by Claude Opus 5.5 (high effort)

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