===== Riemann zeta function ===== ==== Function ==== | @#FF9944: definition | @#FF9944: $ \zeta: \mathbb{C}\setminus\{1\} \to \mathbb C$ | | @#FF9944: definition | @#FF9944: $ \zeta(s) := \begin{cases} \sum_{n=1}^\infty\, n^{-s} &\hspace{.5cm} \mathrm{if}\hspace{.5cm} \mathfrak{R}(s)>1 \\\\ \text{analytic continuation}\hspace{.5cm} &\hspace{.5cm} \mathrm{else} \end{cases}$ | "$\text{analytic continuation}$" ----- === Theorems === Euler product: $\zeta(s)=\prod_{p\in\text{primes}}\frac{1}{1-p^{-s}}$ == Representations == ^ $\zeta(s) = \dfrac{\int_0^\infty\frac{x^{s}}{e^x-1}\frac{{\mathrm d}x}{x}}{\int_0^\infty \frac{x^{s}}{e^x-0}\frac{{\mathrm d}x}{x}}=\frac{1}{\Gamma(s)}\int_0^\infty\frac{x^{s}}{e^x-1}\frac{{\mathrm d}x}{x}$ ^ ^ $\log \zeta(s) = s \int_0^\infty \frac{\pi(x)}{x^s-1}\frac{{\mathrm d}x}{x}$ ^ where $\pi$ is the [[prime-counting function]]. == Functional equation == Tells you most values: ^ $ \zeta(s) = 2\,(2\pi)^{s-1}\sin{\left(\pi\,s/2\right)}\,\Gamma(1-s)\,\zeta(1-s)$ ^ == Specific values == ^ $\zeta(-2m)=0$ ^ ^ $\zeta(2m)=(-1)^{m+1}\frac{(2\pi)^{2m}}{2(2m)!} B_{2m}$ ^ so that $\zeta(1-2m) = \dfrac{2(2m-1)!}{(4\pi^2)^m}\cos(m\pi)\zeta(2m)$ $\zeta(1-2m)=(-1)^{m+1}\frac{1}{2m}B_{2m}$ E.g. ^ $\zeta(2)=\pi^2/6$ ^ ^ $\zeta(4)=\pi^4/90$ ^ Also ^ $\zeta(-1)=-1/12$ ^ == Expansions == Fix an $N$ and split the infinite sum into a finite part ($\sum_{n=1}^N\, n^{-s}$ is also the subject of Faulhaber's formula) and $\sum_{n=N}^\infty\, n^{-s}$. We can learn about its divergence via the following approximation $\sum_{n=N}^\infty\, n^{-s}\approx \lim_{\varepsilon\to 0}\int_N^{N/\varepsilon}n^{-s}$ $=\dfrac{N^{-s+1}}{-s+1}\left(\lim_{\varepsilon\to 0}\dfrac{1}{\varepsilon^{-s+1}}-1\right)$ $=\dfrac{1}{s-1}\dfrac{N}{N^s}\left(1-\lim_{\varepsilon\to 0}\varepsilon^{{\mathrm{Re}}(s)-1}{\mathrm e}^{i\,{\mathrm{Im}}(s)\log(\varepsilon)}\right)$ Due to the term $\frac{1}{s-1}$, the value $s=1$ is a proper singularity and this is also true for the analytical continuation of the sum, i.e. the zeta function. On the other hand, the term $\lim_{\varepsilon\to 0}\varepsilon^{{\mathrm{Re}}(s)-1}$ is responsible for the sum being undefined for $\mathrm{Re}(s)\le 1$. == 1+2+3+... == (maybe delete this) http://i.imgur.com/HCfGOYp.gif Sum[k, {k, 0, n}] Binomial[n + 1, 2] Plot[Binomial[n, 2], {n, -1, 4}, ImageSize -> 200] Integrate[Binomial[n, 2], {n, s, s + d}]/d // Simplify Solve[% == -1/12, d] Table[{s, %}, {s, 0, 1, 1/2}] // TableForm == On Riemanns paper == In [[https://upload.wikimedia.org/wikipedia/commons/d/d2/RiemannPrim1859.djvu|Riemanns original paper]], Riemann observed that a integration variable substitution x->n·x in the definition of the Gamma function $\Gamma(s) := \int_0^\infty x^{s-1} {\mathrm e}^{-x}\,{\mathrm d}x $ let's you write ${\frac {1} {n^s}} = {\frac {1} {\Gamma(s)}} \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x $ and thus $\sum_{n=1}^\infty \frac{1}{n^s} = \frac{1} { \Gamma(s) } \sum_{n=1}^\infty \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x $ Now consider the geometric series $\sum_{n=1}^\infty ({\mathrm e}^{-x})^n = \frac{1} { {\mathrm e}^{-x}-1} = \frac{1} {x} - \frac{1}{2} + \frac{1}{12}x+O(x^2)$ (check signs) The above sum over the integral is convergent for $s>1$, while the expression $\frac{1} { \Gamma(s) } \int_0^\infty \frac{x^{s-1}} { {\mathrm e}^x-1} \, {\mathrm d}x $ (check signs) works for all complex $s\ne 1$. We thus found the analytical continuation. The integrand $ \frac{1} { {\mathrm e}^x-1}$ diverges periodically in steps of $2\pi\,i$. He discovers that the function obeys a reflection formula https://en.wikipedia.org/wiki/Riemann_zeta_function#The_functional_equation http://boards.4chan.org/sci/thread/7746199#p7747084 === Reference === Wikipedia: [[http://en.wikipedia.org/wiki/Riemann_Zeta_function|Riemann zeta function]] [[https://upload.wikimedia.org/wikipedia/commons/d/d2/RiemannPrim1859.djvu|Riemanns original paper]] ----- === Subset of === [[Polylogarithm]] === Related === [[Prime number]], [[Bernoulli number]], [[prime-counting function]]