Zeros and explicit formulas
Relate prime-power counts to zeta zeros with precise endpoint and truncation conventions.
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The question for today
What you’ll understand
- Use half-weighted endpoints in the explicit formula.
- Interpret conjugate-zero contributions as real oscillations.
- State a truncated formula together with its error range.
The logarithmic derivative of zeta packages prime-power weights into a meromorphic function. Its pole at one produces a main term in a cumulative count, while its poles at zeros of zeta produce oscillating corrections. The resulting explicit formula is an equality only after the summation convention and endpoint weights are specified. A short list of zeros is an approximation whose omitted contribution requires an error estimate.
A step function with a symmetric endpoint
Let for real . It jumps at prime powers, including composite prime powers, by the logarithm of their underlying prime. Define to use all terms strictly below the endpoint and half the endpoint weight when the endpoint is an integer. Away from prime powers, the two counts agree. This symmetric value is natural for inverse contour integrals, just as symmetric Fourier inversion returns the average of the two one-sided limits at a jump.
For positive real , define using the real logarithm, so no branch ambiguity enters. The nontrivial zeros are those in the open critical strip, each repeated according to multiplicity. Sum them with a symmetric height cutoff and let the cutoff tend to infinity through values avoiding zero ordinates. The exact formula includes a constant from the origin and a convergent contribution from the trivial zeros. Omitting these terms would change an exact identity even if they are small relative to a large main term.
From residues to a controlled finite sum
On a vertical line with real part greater than one, the logarithmic derivative has the absolutely convergent series with coefficients . Integrating it against and shifting a finite contour left crosses its pole at one and its poles at zeta zeros. Their residues have opposite signs: the main pole contributes , and a zero of multiplicity contributes . Horizontal contour pieces cannot be discarded merely because the desired residues have already appeared.
A convenient clean truncated statement restricts to half-integers with and , with the height not equal to a zero ordinate. In that range the displayed error absorbs the bounded origin and trivial-zero terms. Keeping the endpoint away from every integer controls its distance from prime powers, and the usual contour argument controls the horizontal integrals by choosing nearby safe heights. The full theorem for general endpoints includes a term depending on distance to the nearest distinct prime power; the cleaner formula should not be extended silently beyond its stated range.
Oscillations and conditional consequences
Conjugate zeros occur in pairs. Writing one as , its pair contributes twice the real part of . Its envelope grows like divided by the modulus of the zero, while its phase is proportional to . A larger real part allows a larger correction; a larger height produces faster oscillation in logarithmic position. Because the zero sum is not absolutely convergent, an arbitrary rearrangement is not interchangeable with the prescribed symmetric cutoff.
The prime number theorem is and follows from appropriate analytic information, including nonvanishing on the line with real part one. Under the additional Riemann-hypothesis assumption, the explicit-formula estimates yield . This is a conditional implication, not a deduction from reflection symmetry alone. In computation, approximating a handful of zeros does not certify either bound: one must also control the remaining zeros, their numerical uncertainty, and the chosen height and endpoint errors.
Work it through
Check the arithmetic side at a jump
Compute the weighted count at ten and the symmetrically weighted count at nine directly from prime powers, before using any zero information.
The contributing integers through ten are . The three powers of two contribute , the two powers of three contribute , and five and seven contribute their own logarithms.
Thus . Ten contributes no weight, so the full count at nine is the same. Counting only primes would miss the terms from four, eight, and nine.
At nine the endpoint jump is . Replacing its full weight by half gives . At ten, which is not a prime power, symmetric weighting makes no change.
The count at ten is log 2520; symmetric weighting at nine subtracts half of log three from that same full count.
The living notebook
Explore the model
Interactive lab · waves
Pairing oscillatory contributions
Pause/Reset and set phase 90 degrees and amplitude ratio 1. Predict resultant amplitude sqrt(2) and inspect the summed wave. For a fresh comparison Pause/Reset, set both phase 180 and ratio 1; predict complete cancellation. Reset restores both parameters and the clock; a control change alone retains time.
- Sum at x = 0
- 0.866
- Resultant amplitude
- 1.732
- Phase difference
- 60°
Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forPairing terms can convert phase information into a real oscillation or cancellation. These equal-frequency waves are not computed zeta zeros and do not produce a prime-counting error bound.
Pause & reason
Check your understanding
Your turn
Try it yourself
For a conjugate pair and , derive a real formula for when . Track the sign of the sine term.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Multiply the reciprocal of beta plus i gamma by its conjugate, and take the real part after multiplying by the complex exponential.
Compare with the solution
Put . Then . Its real part is , with a positive sine term because the product of the two imaginary factors contributes positively. The conjugate pair doubles this expression. The explicit formula then subtracts the pair; that outer minus sign is separate from the positive sine sign inside the pair.
Keep the idea
Take it with you
- Prime-power jumps and half-weighted endpoints are essential parts of the exact formula.
- Zero sums require multiplicity, ordering, and a specified complex-power convention.
- A truncated zero sum needs a theorem controlling its error in the stated parameter range.
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