AlexKontorovich/PrimeNumberTheoremAnd
Source indexedtheorem · leanprover/lean4:v4.32.0
ArithmeticFunction.LSeries_tau_eq_riemannZeta_sq
PrimeNumberTheoremAnd.IwaniecKowalskiCh1 · PrimeNumberTheoremAnd/IwaniecKowalskiCh1.lean:169 to 185
Source documentation
The L-series of equals the square of the Riemann zeta function for .
Exact Lean statement
@[blueprint
"LSeries_tau_eq_riemannZeta_sq"
(title := "LSeries tau eq riemannZeta sq")
(statement := /-- The L-series of $\tau$ equals the square of the Riemann zeta function for $\Re(s) > 1$. -/)
(proof := /--
From the previous theorem, we have that the Dirichlet convolution of $\zeta$ with itself is $\tau$.
Taking L-series on both sides, we get $L(\tau, s) = L(\zeta, s) \cdot L(\zeta, s)$.
Since $L(\zeta, s)$ is the Riemann zeta function $\zeta(s)$, we conclude that
$L(\tau, s) = \zeta(s)^2$ for $\Re(s) > 1$.
-/)]
theorem LSeries_tau_eq_riemannZeta_sq {s : ℂ} (hs : 1 < s.re) :
LSeries (↗τ) s = riemannZeta s ^ 2Complete declaration
Lean source
Full Lean sourceLean 4
@[blueprint "LSeries_tau_eq_riemannZeta_sq" (title := "LSeries tau eq riemannZeta sq") (statement := /-- The L-series of $\tau$ equals the square of the Riemann zeta function for $\Re(s) > 1$. -/) (proof := /-- From the previous theorem, we have that the Dirichlet convolution of $\zeta$ with itself is $\tau$. Taking L-series on both sides, we get $L(\tau, s) = L(\zeta, s) \cdot L(\zeta, s)$. Since $L(\zeta, s)$ is the Riemann zeta function $\zeta(s)$, we conclude that $L(\tau, s) = \zeta(s)^2$ for $\Re(s) > 1$. -/)]theorem LSeries_tau_eq_riemannZeta_sq {s : ℂ} (hs : 1 < s.re) : LSeries (↗τ) s = riemannZeta s ^ 2 := by have h1 : LSeries (↗(ζ * ζ)) s = LSeries (↗((ζ : ArithmeticFunction ℂ) * ζ)) s := by congr 1; ext n; simp only [← natCoe_mul, natCoe_apply] have h2 : LSeries (↗((ζ : ArithmeticFunction ℂ) * ζ)) s = LSeries (↗ζ) s * LSeries (↗ζ) s := LSeries_mul' (LSeriesSummable_zeta_iff.mpr hs) (LSeriesSummable_zeta_iff.mpr hs) rw [← zeta_mul_zeta, h1, h2, LSeries_zeta_eq_riemannZeta hs, pow_two]