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Leibnitz theorem: y = e^(x²)

Short answer

Leibniz theorem states (uv)_n = sum_{k=0}^n C(n,k) u_{n-k} v_k. For y = e^(x^2), y_1 = 2x y; differentiating n times by the theorem gives the recurrence y_{n+1} − 2x y_n − 2n y_{n−1} = 0.

State Leibnitz theorem. If y = e^(x²), then show that y_{n+1} − 2xy_n − 2ny_{n−1} = 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

y = e^(x²). Then

y₁ = 2x e^(x²) = 2x y.

Differentiate y₁ = 2x y n times by Leibnitz's theorem, with u = y, v = 2x (v₁ = 2, v₂ = 0):

yₙ₊₁ = 2x yₙ + n(2) yₙ₋₁

yₙ₊₁ − 2x yₙ − 2n yₙ₋₁ = 0. Proved.

✓ Answer checked by Claude Opus 5.5 (high effort)

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