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Project-declaredLean 4.32.0 · mathlib@81a5d257c8e4

Analytic On div Removable zero

AnalyticOn_divRemovable_zero

Plain-language statement

Let ff be analytic on an open set ss containing 00, and suppose f(0)=0f(0)=0. Define g(z)=f(z)/zg(z)=f(z)/z for z0z\ne0 and g(0)=f(0)g(0)=f'(0). Then the apparent singularity at 00 is removable and gg is analytic throughout ss.

Exact Lean statement

@[blueprint "AnalyticOn_divRemovable_zero"
  (title := "AnalyticOn-divRemovable-zero")
  (statement := /--
    Let $f$ be a complex function analytic on an open set $s$
    containing $0$ such that $f(0)=0$.
    Then, with $g$ defined as in
    Definition~\ref{divRemovable_zero}, $g$ is analytic on $s$.
  -/)
  (proof := /--
    We need to show that $g$ is complex differentiable at every
    point in $s$. For $z\neq 0$, this follows directly from the
    definition of $g$ and the fact that $f$ is analytic on $s$.
    For $z=0$, we use the definition of the derivative and the
    fact that $f(0)=0$:
    \[
    \lim_{z\to 0}\frac{g(z)-g(0)}{z-0}
    =\lim_{z\to 0}\frac{\frac{f(z)}{z}-f'(0)}{z}
    =\lim_{z\to 0}\frac{f(z)-f'(0)z}{z^2}
    =\lim_{z\to 0}\frac{f(z)-f(0)-f'(0)(z-0)}{(z-0)^2}=0,
    \]
    where the last equality follows from the definition of the
    derivative of $f$ at $0$. Thus, $g$ is complex differentiable
    at $0$ with derivative $0$, completing the proof.
  -/)
  (latexEnv := "lemma")]
lemma AnalyticOn_divRemovable_zero {f : ℂ → ℂ} {s : Set ℂ}
    (sInNhds0 : s ∈ nhds 0) (zero : f 0 = 0) (o : IsOpen s)
    (analytic : AnalyticOn ℂ f s) :
    AnalyticOn ℂ (divRemovable_zero f) s

Formal artifact

Lean source

Canonical source
Full Lean sourceLean 4
@[blueprint "AnalyticOn_divRemovable_zero"  (title := "AnalyticOn-divRemovable-zero")  (statement := /--    Let $f$ be a complex function analytic on an open set $s$    containing $0$ such that $f(0)=0$.    Then, with $g$ defined as in    Definition~\ref{divRemovable_zero}, $g$ is analytic on $s$.  -/)  (proof := /--    We need to show that $g$ is complex differentiable at every    point in $s$. For $z\neq 0$, this follows directly from the    definition of $g$ and the fact that $f$ is analytic on $s$.    For $z=0$, we use the definition of the derivative and the    fact that $f(0)=0$:    \[    \lim_{z\to 0}\frac{g(z)-g(0)}{z-0}    =\lim_{z\to 0}\frac{\frac{f(z)}{z}-f'(0)}{z}    =\lim_{z\to 0}\frac{f(z)-f'(0)z}{z^2}    =\lim_{z\to 0}\frac{f(z)-f(0)-f'(0)(z-0)}{(z-0)^2}=0,    \]    where the last equality follows from the definition of the    derivative of $f$ at $0$. Thus, $g$ is complex differentiable    at $0$ with derivative $0$, completing the proof.  -/)  (latexEnv := "lemma")]lemma AnalyticOn_divRemovable_zero {f : ℂ  ℂ} {s : Set ℂ}    (sInNhds0 : s  nhds 0) (zero : f 0 = 0) (o : IsOpen s)    (analytic : AnalyticOn ℂ f s) :    AnalyticOn ℂ (divRemovable_zero f) s := by  rw [Complex.analyticOn_iff_differentiableOn o,     Complex.differentiableOn_compl_singleton_and_continuousAt_iff sInNhds0]  constructor  · have (x) (hx : x  s \ {0}) : x  0 := by      rw [Set.mem_sdiff, Set.mem_singleton_iff] at hx      exact hx.right    rw [differentiableOn_congr (fun x hyp_x => Function.update_of_ne (this x hyp_x) _ _)]    exact      (analytic.mono Set.sdiff_subset).differentiableOn.fun_div      (differentiableOn_id.mono (Set.subset_univ (s \ {0})))      this  · have U :=      HasDerivAt.continuousAt_div (c := 0) (a := deriv f 0)        (f := f)        (DifferentiableOn.hasDerivAt          ((Complex.analyticOn_iff_differentiableOn o).mp analytic)          sInNhds0)    simpa [divRemovable_zero, zero] using U
Project
Prime Number Theorem and More
License
Apache-2.0
Commit
a93551347dce
Source
PrimeNumberTheoremAnd/BorelCaratheodory.lean:41-87

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Plain-language statement

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Source project: Prime Number Theorem and More

Person-level attribution pending.

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Project-declaredLean 4.32.0

Analytic On div Removable zero closed Ball

AnalyticOn_divRemovable_zero_closedBall

Plain-language statement

Suppose R>0R>0 and ff is analytic on the closed disc zR|z|\le R with f(0)=0f(0)=0. Define g(z)=f(z)/zg(z)=f(z)/z for z0z\ne0 and g(0)=f(0)g(0)=f'(0). Then gg is analytic on the entire closed disc, including at the removed singularity.

analytic number theoryprime numbersasymptotics

Source project: Prime Number Theorem and More

Person-level attribution pending.

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