Lesson 13 approached L(1, χ̄4) = 1.0864 − 0.5814i for p = 13 slowly, with an error still 0.004 after 390 terms. This lesson finds it exactly, twice. Expanding χ into waves (Lesson 14) turns the infinite series into p − 1 logarithms, and for an odd character almost everything cancels, leaving π, a Gauss sum and B1. For χ4 that B1 is β. A second route writes L(1, χ) with the digamma function. As a consequence, N(β) is p/π2 times |L(1, χ4)|2, which Lesson 16 turns into a class number.
Where this lesson is headed
In words: for an odd character the value at s = 1 is π times an algebraic number. For χ = χ4: L(1, χ̄4) = (πiτ(χ̄4)/p)β, so |L(1, χ4)|2 = π2N(β)/p. The digamma formula holds for every nontrivial χ and computes L(1, χ) to full precision from p − 1 terms.
The Fourier expansion χ(n) = τ(χ̄)−1∑χ̄(m)ζmn, |τ|2 = p and τ(χ)τ(χ̄) = χ(−1)p (Lesson 14); convergence of L(1, χ) and L(1, χ) ≠ 0 for non-real χ (Lesson 13); β = B1,χ4 (Lesson 08). From analysis (stated): ∑n≥1 zn/n = −log(1 − z) for |z| = 1, z ≠ 1, with the principal logarithm; and the series for the digamma function ψ.
p: a prime in P*; χ: a nontrivial character mod p
odd: χ(−1) = −1 (χ4, χ̄4); even: χ(−1) = 1 (the Legendre symbol here)
ψ(x) = −γ + ∑k≥0 [1/(k + 1) − 1/(k + x)]: the digamma function, γ ≈ 0.5772
log(1 − eiθ) = log(2 sin(θ/2)) + i(θ − π)/2 for 0 < θ < 2π
Start with p = 13 and χ̄4. Route 1 is the series of Lesson 13; route 2 sums p − 1 digamma values; route 3 is the closed form (πiτ(χ̄4)/p)·β. Routes 2 and 3 agree to every displayed digit, while the series is still settling. The chart below shows why route 3 works: the logarithmic parts cancel for an odd character.
| route | what is computed | value | distance from route 2 |
|---|
For each m, the wave expansion contributes χ̄(m)·log(1 − ζm) = χ̄(m)[log(2 sin(πm/p)) + i(πm/p − π/2)]. Bars show log(2 sin(πm/p)), which is symmetric in m ↔ p − m; each bar is labeled and colored by the weight χ̄(m).
Do these in order. "Set this up" chooses p and the character.
sin(πm/p) = sin(π(p − m)/p), so the log sine terms at m and p − m are equal. An odd character gives them opposite signs, so they cancel in pairs, and what remains is the linear part iπm/p, weighted by χ̄(m): a multiple of B1. That is why L(1, χ) is π times an algebraic number for odd χ.
For an even character the reverse happens: B1 = 0 (Lesson 08) and the log sine terms survive, so L(1, χ) is a combination of logarithms. For the Legendre symbol mod p ≡ 1 (mod 4) it is related to the logarithm of a unit in ℚ(√p). This course only needs the odd case.
Each step has the general argument on the left and your p and character from Stage 01 on the right.
With χ(n) = τ(χ̄)−1∑mχ̄(m)ζmn (Lesson 14), and exchanging the finite sum over m with the convergent sum over n:
For z = ζm on the unit circle, z ≠ 1, the series ∑zn/n converges (bounded partial sums of zn, as in Lesson 13) to −log(1 − z), principal branch (stated). Writing z = eiθ with θ = 2πm/p ∈ (0, 2π): 1 − eiθ = −2i sin(θ/2) eiθ/2 = 2 sin(θ/2) ei(θ−π)/2, so
So L(1, χ) = −τ(χ̄)−1∑χ̄(m)[log(2 sin(πm/p)) + iπm/p − iπ/2]. For odd χ:
the log sine terms at m and p − m are equal but χ̄(p − m) = −χ̄(m): they cancel. ∑χ̄(m)·(iπ/2) = 0 (orthogonality). What is left:
For odd χ, τ(χ)τ(χ̄) = −p, so 1/τ(χ̄) = −τ(χ)/p and L(1, χ) = (iπτ(χ)/p)B1,χ̄. Replacing χ by χ̄:
For χ = χ4: L(1, χ̄4) = (πiτ(χ̄4)/p)·β. The transcendental part is π; the rest is algebraic.
Taking absolute values with |τ|2 = p:
(|L(1, χ4)| = |L(1, χ̄4)| since the series are conjugate.) And since L(1, χ̄) ≠ 0 for non-real χ (Lesson 13), B1,χ ≠ 0 for every odd non-real χ: the promise made in Lesson 08.
Group n by its residue a mod p: n = kp + a. Since ∑aχ(a) = 0 we may subtract ∑aχ(a)/(p(k + 1)) inside each k-block without changing the sum, which makes each block converge absolutely:
and the γ terms vanish by orthogonality: L(1, χ) = −(1/p)∑χ(a)ψ(a/p), for every nontrivial χ, odd or even. With ψ computed by its recurrence and asymptotic series, this gives L(1, χ) to machine precision.
For even χ the roles reverse: the log sine terms at m and p − m have the same sign and survive, while the linear part is B1,χ̄ = 0. So
For the Legendre symbol this is a logarithm of a unit of ℚ(√p) divided by √p (stated); the course does not need it.
The same computation works for any primitive odd character χ mod m (|τ|2 = m for primitive χ), giving L(1, χ̄) = (πiτ(χ̄)/m)B1,χ. For the real character χ−4p = χ−4·(·/p) mod 4p this is L(1, χ−4p) = π|B1|/√(4p). Combined with Dirichlet's class number formula (stated) h(−4p) = √(4p)·L(1, χ−4p)/π, it gives h(−4p) = |B1,χ−4p|, and B1 < 0 here: the formula of Lesson 11, step 7.
Assemble the argument, then test it on every prime of P* below 1000.
Complete proof
Waves. By Lesson 14, L(1, χ) = τ(χ̄)−1∑mχ̄(m)∑nζmn/n = −τ(χ̄)−1∑mχ̄(m) log(1 − ζm), and log(1 − eiθ) = log(2 sin(θ/2)) + i(θ − π)/2 for 0 < θ < 2π.
Odd χ. The log sine part cancels under m ↦ p − m, the constant part vanishes by orthogonality, and the linear part is iπB1,χ̄. So L(1, χ) = −iπB1,χ̄/τ(χ̄) = (iπτ(χ)/p)B1,χ̄, using τ(χ)τ(χ̄) = −p. Swapping χ and χ̄ gives the stated form; taking absolute values gives |L(1, χ̄)|² = π²|B1,χ|²/p, and L(1, χ̄) ≠ 0 forces B1,χ ≠ 0.
Digamma. Split n = kp + a, subtract the zero quantity ∑aχ(a)/(p(k + 1)) from each block, and use ψ(x) + γ = ∑k≥0[1/(k + 1) − 1/(k + x)]; the γ terms vanish by orthogonality.
Even χ. Now the linear part is B1,χ̄ = 0 and the constant part vanishes, leaving the log sine sum. ∎
For all 42 primes of P* below 1000: L(1, χ̄4) by digamma versus (πiτ(χ̄4)/p)β to 9 decimals; N(β) = p|L|²/π²; the even formula for the Legendre symbol; and, for p ≤ 400, h(−4p) from reduced forms versus √(4p)·L(1, χ−4p)/π.
Use p = 29, where β = 1 − i.