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Project-declaredLean 4.32.0 · mathlib@81a5d257c8e4

Sum moebius pmul eq prod one sub

ArithmeticFunction.sum_moebius_pmul_eq_prod_one_sub

Plain-language statement

If g is a multiplicative arithmetic function, then for any n0n \neq 0, dnμ(d)g(d)=pn(1g(p))\sum_{d | n} \mu(d) \cdot g(d) = \prod_{p | n} (1 - g(p)).

Exact Lean statement

@[blueprint
  "sum_moebius_pmul_eq_prod_one_sub"
  (title := "sum moebius pmul eq prod one sub")
  (statement := /-- If $g$ is a multiplicative arithmetic function, then for any $n \neq 0$,
    $\sum_{d | n} \mu(d) \cdot g(d) = \prod_{p | n} (1 - g(p))$.
      \begin{verbatim}
  Upstream issue has been created #1240.
  \end{verbatim}

    -/)
  (proof := /--
  Multiply out and collect terms.
  -/)]
theorem sum_moebius_pmul_eq_prod_one_sub {R : Type*} [CommRing R]
    {g : ArithmeticFunction R} (hg : g.IsMultiplicative) (n : ℕ) : n ≠ 0 →
    ∑ d ∈ n.divisors, (moebius d : R) * g d = ∏ p ∈ n.primeFactors, (1 - g p)

Formal artifact

Lean source

Canonical source
Full Lean sourceLean 4
@[blueprint  "sum_moebius_pmul_eq_prod_one_sub"  (title := "sum moebius pmul eq prod one sub")  (statement := /-- If $g$ is a multiplicative arithmetic function, then for any $n \neq 0$,    $\sum_{d | n} \mu(d) \cdot g(d) = \prod_{p | n} (1 - g(p))$.      \begin{verbatim}  Upstream issue has been created #1240.  \end{verbatim}     -/)  (proof := /--  Multiply out and collect terms.  -/)]theorem sum_moebius_pmul_eq_prod_one_sub {R : Type*} [CommRing R]    {g : ArithmeticFunction R} (hg : g.IsMultiplicative) (n : ) : n  0     ∑ d  n.divisors, (moebius d : R) * g d = ∏ p  n.primeFactors, (1 - g p) := by  induction n using Nat.recOnPosPrimePosCoprime with  | zero => intro h; exact absurd rfl h  | one => exact fun _  by simp [hg.map_one]  | prime_pow p k hp hk =>    refine fun _  ?_    rw [Nat.primeFactors_prime_pow hk.ne' hp, Finset.prod_singleton, sum_divisors_prime_pow hp,      Finset.sum_range_succ']    simp only [pow_zero, moebius_apply_one, Int.cast_one, hg.map_one, mul_one]    rw [Finset.sum_eq_single_of_mem 0 (Finset.mem_range.mpr hk)]    · simp only [zero_add, pow_one, moebius_apply_prime hp, Int.reduceNeg, Int.cast_neg,        Int.cast_one, neg_mul, one_mul]; ring    · intro i _ hi      have hnsq : ¬Squarefree (p ^ (i + 1)) := by        rw [Nat.squarefree_pow_iff hp.ne_one (by omega : i + 1  0)]        omega      rw [moebius_eq_zero_of_not_squarefree hnsq]      simp  | coprime a b ha hb hab iha ihb =>    intro hn    rw [hab.primeFactors_mul, Finset.prod_union hab.disjoint_primeFactors,         iha (by omega),  ihb (by omega)]    let h : ArithmeticFunction R := fun n  ↑(moebius n) * g n, by simp    have h_mul : h.IsMultiplicative := by      refine ?_, ?_      · simp [h, ArithmeticFunction.coe_mk, hg.left]      · intro m n hmn        simp only [h, ArithmeticFunction.coe_mk]        rw [ArithmeticFunction.isMultiplicative_moebius.right hmn, hg.right hmn]        push_cast        ring    exact sum_divisors_mul_of_coprime h_mul hab (by omega) (by omega)
Project
Prime Number Theorem and More
License
Apache-2.0
Commit
a93551347dce
Source
PrimeNumberTheoremAnd/IwaniecKowalskiCh1.lean:99-145

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Plain-language statement

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Project-declaredLean 4.32.0

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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.

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Source project: Prime Number Theorem and More

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Project-declaredLean 4.32.0

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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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