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Taylor series expansion

Short answer

Taylor series about 0: e^{-z} = ∑_{n=0}^∞ (-1)^n z^n/n! = 1 - z + z^2/2! - …, convergent for all z (radius ∞).

Expand f(z) = e^(−z) in a Taylor's series about z = 0.

Taylor series about z = 0: f(z) = Σ f⁽ⁿ⁾(0) zⁿ/n!. For f(z) = e^(−z), the nth derivative is f⁽ⁿ⁾(z) = (−1)ⁿ e^(−z), so f⁽ⁿ⁾(0) = (−1)ⁿ.

e^(−z) = Σ_{n=0}^∞ (−1)ⁿ zⁿ / n!
       = 1 − z + z²/2! − z³/3! + z⁴/4! − ...

Answer: e^(−z) = Σ (−z)ⁿ/n!, valid for all z (radius of convergence ∞, since e^(−z) is entire).

✓ Answer checked by Claude Opus 5.5 (high effort)

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