Riemann zeta function/Appendix: Difference between revisions
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=== 1a. Dirichlet series for the von Mangoldt function === | === 1a. Dirichlet series for the von Mangoldt function === | ||
{{ | Start with the Euler Product for \(\zeta(s)\). | ||
For \(\mathrm{Re}(s) > 1\), the Riemann zeta function can be expressed as: | |||
<math>\displaystyle | |||
\zeta(s) = \prod_{p \text{ prime}} \left(1 - p^{-s}\right)^{-1} | |||
</math> | |||
Take the natural logarithm to convert the product into a sum: | |||
<math>\displaystyle | |||
\ln \zeta(s) = \sum_{p \text{ prime}} - \ln \left( 1 - p^{-s} \right) | |||
</math> | |||
=== Logarithmic Series Expansion === | |||
For \( |x| < 1 \), we have the well-known series: | |||
<math>\displaystyle | |||
-\ln(1 - x) = \sum_{k=1}^{\infty} \frac{x^k}{k} | |||
</math> | |||
Substitute \( x = p^{-s} \). | |||
When \( \mathrm{Re}(s) > 0 \), we have \( |p^{-s}| = p^{-\mathrm{Re}(s)} < 1 \) for any \( p \geq 2 \), ensuring convergence: | |||
<math>\displaystyle | |||
-\ln\left(1 - p^{-s}\right) = \sum_{k=1}^{\infty} \frac{\left(p^{-s}\right)^k}{k} | |||
</math> | |||
Simplify the expression. | |||
Noting that \( \left(p^{-s}\right)^k = \left(p^k\right)^{-s} \): | |||
<math>\displaystyle | |||
\sum_{k=1}^{\infty} \frac{\left(p^{-s}\right)^k}{k} | |||
= \sum_{k=1}^\infty \frac{1}{k} \left(p^k\right)^{-s} | |||
</math> | |||
We thus have: | |||
<math>\displaystyle | |||
- \ln\left(1 - p^{-s}\right) | |||
= \sum_{k=1}^\infty \frac{1}{k} \left(p^k\right)^{-s} | |||
</math> | |||
for all \( p \geq 2 \) and \( \mathrm{Re}(s) > 0 \). | |||
We can now substitute this back in our earlier sum: | |||
<math>\displaystyle | |||
\ln \zeta(s) | |||
= \sum_{p \text{ prime}} - \ln \left( 1 - p^{-s} \right) | |||
= \sum_{p \text{ prime}} \sum_{k=1}^\infty \frac{1}{k} \left(p^k\right)^{-s} | |||
</math> | |||
=== von Mangoldt function === | |||
The von Mangoldt function \(\Lambda(n)\) identifies prime powers: | |||
<math>\displaystyle | |||
\Lambda(n) = | |||
\begin{cases} | |||
\ln p & \text{if } n = p^k \text{ for prime } p \text{ and } k \geq 1 \\ | |||
0 & \text{otherwise} | |||
\end{cases} | |||
</math> | |||
Reindex the double sum. | |||
Each term in the double sum corresponds to a prime power \(n = p^k\). For any integer \(n \geq 2\), \(n\) contributes to the sum if and only if it's a prime power. | |||
For \(n = p^k\): | |||
<math>\displaystyle | |||
\frac{\Lambda(n)}{\ln n} = \frac{\ln p}{k \ln p} = \frac{1}{k} | |||
</math> | |||
Therefore: | |||
<math>\displaystyle | |||
\ln \zeta(s) = \sum_{n=2}^\infty \frac{\Lambda(n)}{\ln n} \cdot \frac{1}{n^s} | |||
</math> | |||
which is valid for \(\mathrm{Re}(s) > 1\), with the sum starting at \(n=2\) because \(\Lambda(1) = 0\). | |||
=== 1b. Conversion factor for removing primes === | === 1b. Conversion factor for removing primes === | ||