English
For differentiable f on an emetric ball, the Taylor expansion equals the function value at z exactly via HasSum.
Русский
Для дифференцируемой на эмитрическом шаре f разложение Тейлора совпадает с f(z) через HasSum.
LaTeX
$$$\text{HasSum}\left(\sum_{n=0}^{\infty} (n!)^{-1} (z-c)^n \operatorname{iteratedDeriv}_n f(c)\right) = f(z)$$$
Lean4
instance instTietzeExtensionTVS {𝕜 : Type v} [RCLike 𝕜] {E : Type w} [AddCommGroup E] [Module 𝕜 E] [TopologicalSpace E]
[IsTopologicalAddGroup E] [ContinuousSMul 𝕜 E] [T2Space E] [FiniteDimensional 𝕜 E] : TietzeExtension.{u, w} E :=
TietzeExtension.of_tvs 𝕜