@ColinTheMathmo Hard to be certain, but in school we were taught to ... make sanity checks.
In a similar vein, there's a story I've seen attributed to Vladimir Arnol'd, stating a problem about triangles that "American students can solve and Russian students can't". The upshot is, a triangle with stated properties doesn't exist, but a formula -- perhaps the Pythagorean theorem -- can be applied mindlessly.
Let (n \geq 2) be an integer. The regular (n)-gon inscribed in the complex unit circle and having (1) as a vertex, a.k.a., the convex hull of the (n)th roots of unity, is closed under complex multiplication.
The set of (n)th roots of unity is closed under multiplication, and
The product of two convex linear combinations of the vertices is itself a convex linear combination.
It's crucial to take the "standard" (n)-gon whose vertices are the (n)th roots of unity, i.e., not to take an arbitrary regular (n)-gon inscribed in the unit circle. The animations show the situation for (n = 7), with roots of unity ("standard") on the left, and the polygon rotated by one-tenth of a radian ("non-standard") on the right.
@christianp My oops ... how (\sigma) turned into an s. :)
<>
As for "semi-serious," it's idle speculation on my part, but I've commonly seen people use two ways of tracing a (\sigma), one like an s from bottom to top with an exaggerated loop at the start, and one similar to a counterclockwise spiral starting at the stem of the (\sigma), resembling a 6, or a (\mathfrak{S}) closed up.
(If this is more than coincidence, it's maybe noteworthy that I've never seen anyone write (\sigma) like an s that closes at the bottom, though that seems the likeliest path to the character s!)
Anyway, a fraktur S does have a certain resemblance to (\sigma), which in turn resembles s.
In an oddly touching anecdote, someone asked a purchaser of the image below, "Where is this?"
The questioner also found the caption on the back of the card [incomprehensible]:
"Students of real analysis are often surprised to learn that a `typical' continuous function is differentiable nowhere. Here, a sequence of trigonometric polynomials converges to a Weierstrass nowhere-differentiable function. Successive terms, shifted horizontally, look like ranges of ever-more-distant hills."
A core aim at the Differential Geometry math art shop is mathematical outreach, to invite toward mathematics, and to welcome in, as many people as possible as succinctly as possible. Thus, I imagined, an interested recipient of the card might read about "real analysis," "continuous function," "differentiable," and/or "Weierstrass nowhere-differentiable function," while an uninterested reader could simply smile and nod.
Three-sheet Monty, branched over the origin: The complex cubing map (top to bottom) and multi-valued cube root (bottom to top) branched along the negative real axis.
Complex analysis books generally describe the Riemann surface of the cube root as something like "three copies of the slit complex plane, with the lower edge of each cut joined cyclically to the upper edge of the next cut." This description is correct, but (for me, at least) hides the simple global picture: The Riemann surface of the multi-valued cube root function is itself a complex plane.
The discontinuity of the principal cube root across the branch cut is depicted geometrically in the top plane by the jump in position of the larger dot. The continuity of the multi-valued function is similarly depicted as a rotating equilateral triangle of cube roots.
Comparable pictures hold for square roots, fourth roots, etc.
@narain One natural approach comes from quotienting the domain of a mapping into level sets.
Formally, if (f:X \to Y) is a continuous surjection, we define an equivalence relation on (X) by (x \sim x') if and only if (f(x) = f(x')), whose equivalence classes are the levels of (f). The induced mapping (\bar{f}:X/R \to Y) is continuous and bijective, its inverse "is" the multi-valued inverse of (f), and we can ask if the inverse is continuous, which happens if (f) is an open mapping.
Here with (X = Y) the complex numbers and (f) the cubing map, the quotient is particularly nice; my claim about the multi-valued cube root can be interpreted in these terms.
Generally, non-constant holomorphic maps in one variable are open, and non-constant entire functions are either surjective or omit precisely one value by the big Picard theorem, so this framework works well for Riemann surfaces of inverse functions. :)
Implicit in Henning Makholm's comment just below the question is a wrinkle if we're working over the reals, which are not algebraically closed: The zero matrix, for example, is "close to" matrices with pairs of "nearly zero" real roots and to matrices with no real roots. Still, in the sense the real square root multi-function is continuous at (0), it looks to me that we get continuity of eigenvalues in the entries for real matrices by restricting the eigenvalue behavior of complex matrices (and ignoring non-real eigenvalues).
If you're asking about operators on infinite-dimensional spaces, however, we need a functional analyst....
Fix two points in the plane, say (S) and (N). For each positive real number (k), the locus of points (p) such that (|p - N| = k|p - S|) is readily checked to be a circle. (This circle has infinite radius -- i.e., is a line -- if (k = 1).) These circles turn out to form the family of curves orthogonal to the family of circles through (S) and (N). Collectively, the two families are sometimes known as "Apollonian circles" after Apollonius of Perga.
If the points are images of the poles of a sphere under stereographic projection, the circles through both points are images of longitudes, and the circles of the second family are images of latitudes. Physically, the circles are field lines and equipotentials of a dipole in the plane.
Thomae's "popcorn" function is defined by (f(x) = \frac{1}{q}) if (x = \frac{p}{q}) is rational in lowest terms, and (f(x) = 0) if (x) is irrational.
In pondering how to approach this function with real analysis students, it seemed helpful to write the rational numbers as a union over positive integers (q) of scaled copies (\frac{1}{q}\mathbf{Z}) of the integers.
Each finite union
[
\mathbf{Q}{N} := \bigcup{q=1}^{N} \tfrac{1}{q}\mathbf{Z} \subset \mathbf{Q}
]
consists entirely of isolated points. Consequently, for every real (x) and every positive integer (N), there exists a deleted neighborhood of (x) containing no rationals with denominator (N) or less, i.e., containing only rationals with denominator greater than (N). On such a neighborhood, (0 \leq f < \frac{1}{N}).
Unpacking the definition of limits now shows (\lim(f, x) = 0) for every real (x). (!!) Particularly, (f) is continuous at (x) if and only if (f(x) = 0), if and only if (x) is irrational. Riemann integrability of (f) is also not difficult to check.
Graph the parabola (y = x^{2}) on a floor with an infinite square grid. If we stand on the (y)-axis below the origin where we can see the entire parabola, it looks like an ellipse tangent to the horizon or -- from just the right vantage point -- a perfect circle.
This is likely either surprising or obvious, or perhaps one followed by the other.
Let (\varphi) denote the golden ratio. Fix a Euclidean cube and one of the regular tetrahedra obtained by taking "alternate vertices," introduce Cartesian coordinates with origin at the center and axes parallel to the cube's sides, and rotate about the axis ((0, 1, \varphi)) by multiples of one-fifth of a turn. The union of the five tetrahedra (and of the five cubes) has icosahedral symmetry. This image had a wintry palette. Other color schemes available for readers in the southern hemisphere, or at other times of the year.
The image below is an experimental crossed-eyes stereogram for a poster of the chaotic attractor for the ODE system
\begin{align*}
\frac{dx}{dt} &= 10(y - x), \
\frac{dy}{dt} &= x(28 - z) - y, \
\frac{dz}{dt} &= xy - \tfrac{8}{3}z.
\end{align*}
If you're willing to comment, it would be helpful to know whether the stereo effect is difficult to achieve and/or maintain.
@tokensane Thank you for having a look, and for your feedback!
When you say the frame is inconsistent with the image, does that refer to the relative thickness of the center white band separating the two "panels," something about how the respective panels are cropped, a general sensitivity of the stereo effect to surroundings, or something else...?
Due to very long habit that is proving hard to break, I use dollar signs to enter and leave inline math mode in LaTeX and MathJax. It would be convenient to have emacs convert opening and closing dollar signs to ( and ) automatically on the fly.
A bit of web searching on how to accomplish this hasn't turned up anything except man pages on writing emacs lisp functions (which I can presumably figure out/learn eventually) and other composition software (which is less convenient than continued habit-breaking).
I'd be grateful if you, dear reader, could point to an existing snippet of emacs lisp to do this intelligently: Able from context to distinguish an "opening $" and a "closing $", and both from an "intentional $". TIA!
A plane vector field assigns a vector to each point of a plane region. Near an isolated zero, a vector field is characterized qualitatively by an integer, the index, which counts how many times the value rotates upon tracing a small circle about the zero.
The formula (e^{i\theta} = \cos\theta + i\sin\theta) gives a pleasant explanation of the shapes in the poster. If (k) is an integer and (r) a positive real number, the field whose value is (re^{ikt}) at the unit complex number (e^{it}) models a field of index (k) along the unit circle. The tips of the arrows are the vector sum,
[
e^{it} + re^{ikt} = e^{it}(1 + re^{i(k-1)t}),
]
a path having (|k - 1|)-fold rotational symmetry about the origin.
