The Riemann zeta function and tuning/Appendix: Difference between revisions
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== 1. Euler product expression for the zeta function == | == 1. Euler product expression for the zeta function == | ||
{{ | Starting with the definition of the zeta function: | ||
<math>\displaystyle | |||
\zeta(s) = \sum_{n = 1}^\infty \frac{1}{n^s} | |||
</math>, | |||
by the Fundamental Theorem of Arithmetic, and for <math>\mathrm{Re}(s) > 1</math>, where the series is absolutely convergent, it can be rearranged into a sum over prime factorizations: | |||
<math>\displaystyle | |||
\zeta(s) = \sum_{r_k = 0}^\infty \frac{1}{\left(\prod_{k = 1}^\infty {p_k}^{r_k} \right)^s} = \sum_{r_k = 0}^\infty \prod_{k = 1}^\infty \frac{1}{{p_k}^{s r_k}} | |||
</math>, | |||
where <math>p_k</math> is the ''k''-th prime. | |||
The sum on the outside is really an infinite product of sum operators for each ''k'', and so: | |||
<math>\displaystyle | |||
\sum_{r_k = 0}^\infty \prod_{k = 1}^\infty \frac{1}{{p_k}^{s r_k}} = \prod_{k = 1}^\infty \sum_{r = 0}^\infty \frac{1}{\left({p_k}^s \right)^r} | |||
</math>, | |||
which is a geometric series, and given that <math>{p_k}^s</math> is greater than 1 for all ''k'' when <math>\mathrm{Re}(s) > 1</math>, | |||
<math>\displaystyle | |||
\zeta(s) = \prod_{k = 1}^\infty \frac{1}{1 - {p_k}^{-s}}, | |||
</math> | |||
which is the Euler product formula we desired. | |||
=== 1a. Dirichlet series for the von Mangoldt function === | === 1a. Dirichlet series for the von Mangoldt function === | ||
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<math>\displaystyle | <math>\displaystyle | ||
\zeta(s) = \prod_{p \text{ prime}} \left(1 - p^{-s}\right)^{-1} | \zeta(s) = \prod_{p \text{ prime}} \left(1 - p^{-s}\right)^{-1} | ||
</math> | </math>. | ||
Take the natural logarithm to convert the product into a sum: | Take the natural logarithm to convert the product into a sum: | ||
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<math>\displaystyle | <math>\displaystyle | ||
\ln \zeta(s) = \sum_{p \text{ prime}} - \ln \left( 1 - p^{-s} \right) | \ln \zeta(s) = \sum_{p \text{ prime}} - \ln \left( 1 - p^{-s} \right) | ||
</math> | </math>. | ||
==== Logarithmic Series Expansion ==== | ==== Logarithmic Series Expansion ==== | ||
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<math>\displaystyle | <math>\displaystyle | ||
-\ln(1 - x) = \sum_{k=1}^{\infty} \frac{x^k}{k} | -\ln(1 - x) = \sum_{k=1}^{\infty} \frac{x^k}{k} | ||
</math> | </math>. | ||
Substitute \( x = p^{-s} \). | Substitute \( x = p^{-s} \). | ||
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<math>\displaystyle | <math>\displaystyle | ||
-\ln\left(1 - p^{-s}\right) = \sum_{k=1}^{\infty} \frac{\left(p^{-s}\right)^k}{k} | -\ln\left(1 - p^{-s}\right) = \sum_{k=1}^{\infty} \frac{\left(p^{-s}\right)^k}{k} | ||
</math> | </math>. | ||
Simplify the expression. | Simplify the expression. | ||
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\sum_{k=1}^{\infty} \frac{\left(p^{-s}\right)^k}{k} | \sum_{k=1}^{\infty} \frac{\left(p^{-s}\right)^k}{k} | ||
= \sum_{k=1}^\infty \frac{1}{k} \left(p^k\right)^{-s} | = \sum_{k=1}^\infty \frac{1}{k} \left(p^k\right)^{-s} | ||
</math> | </math>. | ||
We thus have: | We thus have: | ||
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= \sum_{p \text{ prime}} - \ln \left( 1 - p^{-s} \right) | = \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} | = \sum_{p \text{ prime}} \sum_{k=1}^\infty \frac{1}{k} \left(p^k\right)^{-s} | ||
</math> | </math>. | ||
==== von Mangoldt function ==== | ==== von Mangoldt function ==== | ||
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0 & \text{otherwise} | 0 & \text{otherwise} | ||
\end{cases} | \end{cases} | ||
</math> | </math>. | ||
Reindex the double sum. | Reindex the double sum. | ||
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<math>\displaystyle | <math>\displaystyle | ||
\frac{\Lambda(n)}{\ln n} = \frac{\ln p}{k \ln p} = \frac{1}{k} | \frac{\Lambda(n)}{\ln n} = \frac{\ln p}{k \ln p} = \frac{1}{k} | ||
</math> | </math>. | ||
Therefore: | Therefore: | ||
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<math>\displaystyle | <math>\displaystyle | ||
\ln \zeta(s) = \sum_{n=2}^\infty \frac{\Lambda(n)}{\ln n} \cdot \frac{1}{n^s} | \ln \zeta(s) = \sum_{n=2}^\infty \frac{\Lambda(n)}{\ln n} \cdot \frac{1}{n^s} | ||
</math> | </math>, | ||
which is valid for \(\mathrm{Re}(s) > 1\), with the sum starting at \(n=2\) because \(\Lambda(1) = 0\). | which is valid for \(\mathrm{Re}(s) > 1\), with the sum starting at \(n=2\) because \(\Lambda(1) = 0\). | ||