The LaTeX file used in the -1/12 video.

minus_one_over_twelve.tex

% This is the LaTeX file with all the formulas shown in the video on the 
% infamous -1/12 value in analytic number theory and other fields:
%
% https://youtu.be/az2WOnxsLhc

\newpage

$\lim_{q \to 1} n\, q^n = n$
\hspace{.5cm}

Proposition:

$\lim_{q \to 1} \left( \sum_{n=0}^\infty n\, q^n - \int_0^\infty n\, q^n\, {\mathrm d}n \right) = - \frac{1}{12} $

%Bild
\begin{figure}[h] % [h] steht dafür, dass das bild an dieser Stelle platziert wird. Alternativ gibt es noch [t], [b], und [p]
\centering
\includegraphics[width=14cm]{./pictures/graph.png}
\caption{$\sum_{n=0}^\infty n \, q^n - \int_0^\infty n\, q^n\, {\rm d}n=  - \frac{1}{12} + {\mathcal O}((q-1))$}
\end{figure}

We may make a guess about its value via Euler-Maclaurin.

\newpage

\section{Euler and Maclaurin}

How to infinite discrete sums compare to unbounded integrals?

Euler-Maclaurin formula:

\url{https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula}

\url{https://en.wikipedia.org/wiki/Bernoulli_number}

\hspace{.5cm}

$\sum _{n=m+1}^{N} f(n)=\int_m^{N} f(n)\, {\rm d}n+ {\mathcal R},$

\hspace{.5cm}

${\mathcal R} = -\frac{1}{2}\left( f(m)-f(N) \right) -\frac{1}{12}\left( f'(m)-f'(N)\right) + \frac 1{720}\left( f'''(m)-f'''(N)\right) + \cdots$

\hspace{1cm}

Examples: 

$\bullet$ Use $m=2$ and $N=4$ with $f(n)=n^2$, where $\int f(n)\, {\rm d}n = \tfrac{1}{3}n^3 + C$ 

$\implies 3^2+4^2 = \tfrac{1}{3}4^3-\tfrac{1}{3}2^3 - \frac{1}{2}(2^2-4^2) - \frac{1}{12}\,(2\cdot 2^1-2\cdot 4^1)$

\hspace{1cm}

$\bullet$ Use $m=2$ and $N=7$ with $f(n)=n$, where $\int f(n)\, {\rm d}n = \tfrac{1}{2}n^2 + C$ 

$\implies 3^1 + 4^1 + 5^1 + 6^1 + 7^1 = \tfrac{1}{2}7^2-\tfrac{1}{2}2^2 - \frac{1}{2}(2-7)$
\hspace{1cm}

Sidenote: $\sum_{n=0}^N n = \frac{1}{2}(N^2+N)$

\hspace{1cm}

Question to ask: \\ What aspects of the formula is preserved when the upper bound $N$ blows up and $f^{(k)}(N)$ becomes undefined?

Sidenote: Peano arithmetic is not a theory of infinite sums.

$\sum_{n=0}^\infty a_n = y$

means

$ \forall (\varepsilon\in{\mathbb R}_{>0}).\,\exists (m\in{\mathbb N}).\,\forall (N\ge_{\mathbb N} m).\,| \sum_{n=0}^N a_n - y \, |<\varepsilon $

\newpage

\section{Proof of the result of the expansion at $q=1$}

For $|q|<1$,

$\bullet \,(q-1)\cdot\sum_{n=0}^N q^n = (q-1) \cdot (q^N +  q^{N-1}+ \cdots  + q + 1)= q^{N+1}-1$

$ \implies \sum_{n=0}^\infty q^n =- \frac{1}{1-q}= \frac{1}{1-q}$

$\bullet \log(q)\cdot\int_0^N q^n \, {\rm d}n = \log(q)\cdot\int_0^N {\rm e}^{n \log(q)}  \, {\rm d}n= {\rm e}^{N \log(q)}-1$

$ \implies \int_0^\infty q^n \, {\rm d}n = -\frac{1}{\log(q)} = \frac{1}{\log(1/q)} $

\hspace{1cm}

$\bullet \,y=\frac{1}{2}x - \frac{1}{3}x^2 \implies y+y^2 = \frac{1}{2}x + \left(-\frac{1}{3}+(\frac{1}{2})^2\right)x^2+{\mathcal O}(x^3) = \frac{1}{2}x - \frac{1}{12} x^2+{\mathcal O}(x^3)$

$\bullet \,\log(1+x) = \int_1^{1+x}\frac{1}{s}\,{\rm d}s = \int_0^x\frac{1}{1-(-t)}\,{\rm d}t = \int_0^x\sum_{n=0}^\infty (-t)^n\,{\rm d}t = x\sum_{n=0}^\infty \frac{1}{n+1}(-x)^{n}$

$\bullet \,\frac {1} { \log(1+x)} = \frac {1} {x} \frac {1} {1 - \left(\frac{1}{2}x - \frac{1}{3}x^2 + {\mathcal O}(x^3)\right) } = \frac {1} {x} + \frac{1}{2} - \frac{1}{12} x + {\mathcal O}(x^2)$

\hspace{1cm}

$\bullet\, \sum_{n=0}^\infty q^n - \int_0^\infty q^n {\rm d}n = -\frac{1}{q-1}+\frac{1}{\log(q)} = \frac{1}{2} - \frac{1}{12} (q-1) + {\mathcal O}((q-1)^2)$

$\bullet \,\sum_{n=0}^\infty n \, q^n - \int_0^\infty n\, q^n\, {\rm d}n = q\frac{{\rm d}}{{\rm dq}} \left(\sum_{n=0}^\infty q^n - \int_0^\infty q^n {\rm d}n\right) = - \frac{1}{12} + {\mathcal O}((q-1)^2)$

\hspace{1cm}

qed.

\newpage

\section{Bernoulli and Euler}

How or why do the Bernoulli numbers ($\pm\tfrac{1}{2}, \pm\tfrac{1}{12}, \cdot$) tie to exponentiation?

$f(q)=\sum_{k=0}^\infty a_k q^k, \  \  \  \ f(0)=a_0$

$g(q):=\frac{f(0)}{f(f(0)\,q)}, \  \  \  \  g(0)=1$

\hspace{1cm}

$g(q)=\frac{a_0}{\sum_{k=0}^\infty a_k\left(a_0\,q\right)^k}=\frac{1}{1+\sum_{k=1}^\infty\frac{a_k}{a_0}\left(a_0\,q\right)^k}=$

$=1-a_1\,q+\left(a_1\,a_1-a_0\,a_2\right)\,q^2-\left(a_1\,a_1\,a_1-2\,a_0\,a_1\,a_2+a_0\,a_0\,a_3\right)\,q^3+{\mathcal O}(q^3)$

\hspace{1cm}

$f(q)=\sum_{k=0}^\infty \frac{1}{(k+1)!} q^k = \frac { {\mathrm{e}^q}-1 } { q }$

$\dfrac{q}{{\mathrm{e}^q}-1} = 1 + \left(-\frac{1}{2}\right) q + \left(\frac{1}{12} \right)   q^2+{\mathcal O}(q^3)$

\hspace{1cm}

Sidenote:

$\dfrac{q}{{\mathrm{e}^q}-1}\,{\mathrm{e}}^{x\,q}=1+\sum_{k=1}^\infty \frac{1}{k!}B_k(x) \,q^k=1+\frac{1}{1!}\left(x-\frac{1}{2}\right) q+\frac{1}{2!}\left(x^2-x+\frac{1}{6} \right)   q^2+{\mathcal O}(q^3)$

Sidenote:

\url{https://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%E2%8B%AF}

\url{https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF}

\section{Euler and Lagrange}

How does exponentiation tie to infinitesimal vs. finite movement?

$D_hf(x):=h\frac{{\rm d}}{{\rm d}x} f(x)$

$f(x+h)=\sum_{k=0}^\infty\frac{1}{k!}f^{k}(x)\,h^k = \sum_{k=0}^\infty\frac{1}{k!}D_h^kf(x) $ 

$\Delta_h f(x) = f(x+h)-f(x)=D_hf(x)+\sum_{k=2}^\infty\frac{1}{k!}D_h^kf(x)$ 

\hspace{1cm}

Sidenote: $\Delta_h = \exp(D_h) - 1$

$ \dfrac{\Delta_h f(x)}{D_hf(x)}=1+\frac{1}{D_hf(x)}\sum_{k=2}^\infty\frac{1}{k!}D_h^kf(x)$ 

$\frac{D_hf(x)}{\Delta_h f(x)}=1-\frac{f''(x)}{2!}\left(\frac{h}{f'(x)}\right)+\left(\frac{f''(x)\,f''(x)}{2!\,2!}-\frac{f'(x)\,f'''(x)}{1!\,3!}\right)\left(\frac{h}{f'(x)}\right)^2+{\mathcal O}(h^3).$

\section{Riemann and Bernoulli}

$\Gamma(s) := \int_0^\infty t^{s-1} {\mathrm e}^{-t}\,{\mathrm d}t = \int_0^\infty (n\,x)^{s-1} {\mathrm e}^{-(n\,x)}\,{\mathrm d}(n\,x)$

$\zeta(s) = \sum_{n=1}^\infty n^{-s} = \dfrac{1}{\Gamma(s)} \int_0^\infty \dfrac{x ^ {s-1}}{{\rm e} ^ x - 1} \, \mathrm{d}x$

$\zeta(2) = 1+\frac{1}{4} +\frac{1}{9} +\frac{1}{16} +\cdots = \int_0^\infty \dfrac{x}{{\rm e} ^ x - 1} \, \mathrm{d}x$

``'Where have I seen this before?''

$\zeta(2n) = (-1)^{n+1} \dfrac{(2\pi)^{2n}}{2(2n)!}\cdot B_{2n}$

$\zeta(2) = \pi^2\cdot \frac{1}{6}$

\hspace{1cm}

$\zeta(s) = 2\,\frac{1}{(2\pi)^{1-s}}\sin{\left(\pi\,s/2\right)}\,\Gamma(1-s)\,\zeta(1-s)$

$\zeta(-1) = 2\,\frac{1}{4\pi^2}(-1)\,\zeta(2) = -\frac{1}{2}\cdot\frac{1}{6}$

\section{Baker, Campell, Hausdorff, Dykin, etc.}

The adjoint algebra element representation makes for a derivation.

\url{https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula}

\url{https://en.wikipedia.org/wiki/Matrix_exponential#The_Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula}

\section{Todd}

\url{https://en.wikipedia.org/wiki/Todd_class}