Despite growing up in three dimensions, as a kid I did not recognize one of 3D’s deep and subtle properties: full rotations tangle, but double rotations untangle!

Following physicist Paul Dirac, twist a belt one full turn about its length. The 360° single twist cannot be undone without changing the belt buckles’ orientations, although the twist can be changed from clockwise to counterclockwise. Now twist the belt two full turns about its length. Amazingly, the 720° double twist can be undone without changing the belt buckles’ orientations. The double twist is the true identity.

The animations below show belts twisting concretely in 3D space and abstractly in a 3D projection of the 4D quaternion unit sphere. Points on the sphere represent all 3D rotations (twice). Each blue dot represents a belt cross section rotation and is located by a radius vector. Suppressing rotations about the 1‑direction, the radius vector’s projection onto the 2-3-plane is the section’s rotation axis and twice its co-latitude is the section’s rotation angle.

Elastic belt with a 360° twist. Blue dots on quaternion sphere projection represent belt cross section rotations. Blue curve connecting dots can not be smoothly contracted to the untwisted state represented by the north pole, but without changing the orientation of the belt’s ends, the twist can be changed from clockwise to counterclockwise as indicated.

The sphere’s north and south poles represent the same orientation, 0° ≡ 720° and 360°, but different orientation-entanglements. Identifying the north and south poles as the same orientation allows closed loops on the quaternion sphere to represent both 360° and 720° belt twists, but only the latter can smoothly contract to the north pole identity rotation. This multiple connectivity is reminiscent of a torus (or donut with hole), where toroidal loops (around-the-hole) are contractible but poloidal (through-the-hole) loops are not, rather than the simple connectivity of a sphere, where all loops are contractible.

Elastic belt with a 720° twist. Blue dots on quaternion sphere projection represent belt cross section rotations. Blue curve connecting dots can be smoothly contracted to the untwisted state represented by the north pole, so without changing the orientation of the belt’s ends, the twist can be undone as indicated.

In quantum electrodynamics, the bare charge of an electron is infinite, but the renormalized dressed charge is finite. The bare electron shields itself by polarizing the virtual electron-positron pairs of the nearby quantum vacuum to reduce its coupling at large distances to

\infty – \infty \stackrel{I}{=} \sqrt{\frac{1}{137}}

in natural units, where the “I” decorating the equals sign denotes an informal relationship. Renormalization techniques help physicists make sense of divergent sums and integrals used to calculate quantities like running coupling “constants” that vary with interaction energy and set the strengths of the fundamental forces.

For a simple example, the sum of natural numbers

\sum_{n=0}^N n = 1+2+3+\cdots+N= \frac{N}{2}(N+1)

diverges as N\rightarrow \infty. Similarly, the analogous integral of positive real numbers

\int_{0}^X \hspace{-0.8em}dx\ x =\frac{X^2}{2}

diverges as X\rightarrow \infty. To cancel these corresponding discrete and continuous divergences, introduce an exponential convergence factor and formally subtract the integral from the sum

\sum_{n=0}^\infty n -\int_{0}^\infty \hspace{-0.8em} dx\, x =\color{red}\lim_{\lambda \rightarrow 0}\color{black}\left( \sum_{n=0}^\infty n\, \color{red}e^{-\lambda n}\color{black}-\int_{0}^\infty \hspace{-0.8em} dx\, x\, \color{red}e^{-\lambda x} \color{black}\right)=\color{red}\lim_{\lambda \rightarrow 0}\color{black}\left(\mathcal{S}-\mathcal{I}\right).

To evaluate the cancellation, integrate the integral with respect to the convergence parameter \lambda>0 to get

\mathcal{I}=\int_{0}^\infty \hspace{-0.8em}dx\ x\, \color{red}e^{-\lambda x}\color{black} =-\frac{d}{d\lambda}\int_{0}^\infty \hspace{-0.8em}dx\ e^{-\lambda x} =-\frac{d}{d\lambda}\left(\frac{e^{-\lambda x}}{-\lambda}\right)\bigg|_{0}^\infty=-\frac{d}{d\lambda}\left(\frac{1}{\lambda}\right)=\frac{1}{\lambda^2}.

Next integrate the sum with respect to \lambda

\mathcal{S}=\sum_{n=0}^\infty n\,\color{red} e^{-\lambda n}\color{black} = -\frac{d}{d\lambda} \sum_{n=0}^\infty e^{-\lambda n} = -\frac{d}{d\lambda} \sum_{n=0}^\infty \left(e^{-\lambda}\right)^n

and sum the resulting geometric series to find

\mathcal{S}= -\frac{d}{d\lambda} \left( \frac{1}{1-e^{-\lambda}} \right)= \frac{e^{-\lambda}}{\left(1-e^{-\lambda}\right)^2}\color{red}\frac{e^{2\lambda}} {e^{2\lambda}}\color{black}= \frac{e^{\lambda}}{\left(e^{\lambda}-1\right)^2}\color{red}= \frac{e^{\lambda}}{\left(1-e^{\lambda}\right)^2}\color{black},

which is thus an even function of \lambda. Replace the exponentials by their power series expansions

\mathcal{S}=\frac{1+\lambda+\lambda^2/2!+\lambda^3/3!+\cdots}{\left(1+\lambda+\lambda^2/2!+\lambda^3/3!+\cdots – 1 \right)^2}=\frac{1+\lambda+\lambda^2/2+\cdots}{\lambda^2\left(1+\lambda/2+\lambda^2/6+\cdots \right)^2},

expand the denominator square

\begin{array}{ccc}\left(1+\lambda/2+\lambda^2/6+\cdots\right)^2=&+1\hspace{1.2em}+\lambda/2\hspace{0.7em}+\lambda^2/6\hspace{0.2em}\phantom{+\cdots}\\ \rule{0pt}{1.3em}&+\lambda/2\hspace{0.43em}+\lambda^2/4\hspace{0.37em}\color{red}+\lambda^3/12\color{black}\phantom{+\cdots}\\ \rule{0pt}{1.3em}&+\lambda^2/6\color{red}+\lambda^3/12+\lambda^4/36+\cdots&\color{black} = 1+\lambda+7\lambda^2/12+\cdots\end{array}

and use long division

\begin{array}{r}1-\lambda^2/12+\cdots\\ 1+\lambda+7\lambda^2/12+\cdots \,{\overline{\smash{\big)}\,1+\lambda+\phantom{9}\lambda^2/2\phantom{9} +\cdots}}\\ \underline{1+\lambda+7\lambda^2/12+\cdots }\\0-\lambda^2/12+\cdots\\ \underline{-\lambda^2/12+\cdots}\\0+\cdots \end{array}

to show

\mathcal{S} = \frac{1}{\lambda^2} \left(1-\frac{\lambda^2}{12}+\mathcal{O}[\lambda^4]\right) =\frac{1}{\lambda^2}-\frac{1}{12}+\mathcal{O}[\lambda^2].

For finite \lambda, the difference

\mathcal{S}-\mathcal{I}=\sum_{n=0}^\infty n\, e^{-\lambda n}-\int_{0}^\infty \hspace{-0.8em} dx\, x\,e^{-\lambda x} = \frac{1}{\lambda^2}-\frac{1}{12}+\mathcal{O}[\lambda^2]-\frac{1}{\lambda^2}= -\frac{1}{12}+\mathcal{O}[\lambda^2],

and for vanishing \lambda

\color{red}\lim_{\lambda \rightarrow 0}\color{black}\left( \sum_{n=0}^\infty n\, \color{red}e^{-\lambda n}\color{black}-\int_{0}^\infty \hspace{-0.8em} dx\, x\, \color{red}e^{-\lambda x} \color{black}\right) = -\frac{1}{12}.

Thus, removing nonphysical infinity in a controlled way exposes a finite component

\infty – \infty \stackrel{I}{=} -\frac{1}{12}

of the divergent natural numbers sum. To celebrate this hidden “golden nugget”, write

1+2+3+\cdots \stackrel{R}{=} -\frac{1}{12},

where”R” denotes regularized, renormalized, and remainder. “R” also denotes Ramanujan, who discovered this association without any formal mathematical training, and Riemann, whose famous zeta function \zeta[s] provides an alternate path to it.

In brief, use the convergence factor to tame infinity by isolating the diverging, remaining, and vanishing parts of the natural numbers sum

\sum_{n=1}^\infty n\stackrel{\color{red}\lambda \downarrow 0}{=}\color{black}\sum_{n=0}^\infty n\color{red}\, e^{-\lambda n}\color{black}\stackrel{\color{red}\lambda \downarrow 0}{=}\color{red}\underbrace{\ +\frac{1}{\lambda^2}\ \ }_\text{Diverge}\color{black}\underbrace{\ -\frac{1\vphantom{\lambda^2}}{12}\ \ }_\text{Remain}\color{red} \underbrace{\vphantom{-\frac{1\vphantom{\lambda^2}}{12}}+\mathcal{O}[\lambda^2]}_\text{Vanish}\color{black}\ \stackrel{R}{=}-\frac{1}{12}.

Discard the first, retain the second, and let go the third.

I just bought a new calculator. New to me, that is, but older than me.

Inspired by the 1600s Gottfried Leibniz stepped cylinder and the 1800s Thomas de Colmar arithmometer, the Curta mechanical calculator design was developed by Curt Herzstark while imprisoned in the Buchenwald concentration camp during World War II. Curta calculators were manufactured between 1947 to 1972 in Liechtenstein until electronic calculators became widely available.

A true number cruncher, the Curta has been nicknamed the pepper grinder because of its crank — or the math grenade due to its resemblance to some hand grenades. My Curta II has about 719 metal parts yet fits comfortably in my hand. I remember seeing advertisements for Curta calculators in Scientific American magazine. Now collector’s items, the one I just bought costs an order of magnitude more.

“The world’s first, last, and only hand-held mechanical calculator”, the Curta is an elegant marvel of mechanical engineering, the culmination of generations of work to mechanize arithmetic. And as I can now attest, it is a joy to hold and use.

A Liebniz wheel at the heart of the Curta computes 5 + 3 = 8 with 2 slides and 2 cranks.

Curta ad from my childhood.