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Leibnitz theorem: log y = tan⁻¹x
Short answer
Leibnitz theorem: (uv)_n = ∑ C(n,r) u_{n-r} v_r. For log y = tan⁻¹x, (1+x²)y₁ = y. Applying Leibnitz n+1 times gives (1+x²)y_{n+2} + (2nx+2x-1)y_{n+1} + (n²+n)y_n = 0.
State Leibnitz theorem. If log y = tan⁻¹x, then show that (1+x²)y_{n+2} + (2nx+2x−1)y_{n+1} + (n²+n)y_n = 0.
Leibnitz's theorem. If u and v are functions of x, the nth derivative of the product uv is
(uv)ₙ = uₙv + C(n,1)uₙ₋₁v₁ + C(n,2)uₙ₋₂v₂ + ... + u vₙ
where C(n,r) = n!/(r!(n−r)!).
Proof
log y = tan⁻¹x. Differentiate: y₁/y = 1/(1 + x²), so
(1 + x²) y₁ = y …(1)
Differentiate (1) (n+1) times by Leibnitz's theorem, with u = y₁ and v = 1 + x² (v₁ = 2x, v₂ = 2):
(1+x²)yₙ₊₂ + (n+1)(2x)yₙ₊₁ + [(n+1)n/2](2)yₙ = yₙ₊₁
(1+x²)yₙ₊₂ + (2nx + 2x − 1)yₙ₊₁ + (n² + n)yₙ = 0
Proved.
✓ Answer checked by Claude Opus 5.5 (high effort)
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