## Partitions of unity and monotonicity-preserving approximation

There are many ways to approximate a given continuous function ${f\colon [a, b]\to \mathbb R}$ (I will consider the interval ${[a, b]=[0, 1]}$ for convenience.) For example, one can use piecewise linear interpolation through the points ${(k/n, f(k/n))}$, where ${k=0, 1, \dots, n}$. The resulting piecewise linear function ${g}$ has some nice properties: for example, it is increasing if ${f}$ is increasing. But it is not smooth.

A convenient way to represent piecewise linear interpolation is the sum ${g(x) = \sum_{k=0}^n f(k/n) \varphi_k(x)}$ where the functions ${\varphi_k}$ are the triangles shown below: ${\varphi_k(x) = \max(0, 1 - |nx-k|)}$.

The functions ${{\varphi_k}}$ form a partition of unity, meaning that ${\sum_k \varphi_k \equiv 1}$ and all ${\varphi_k}$ are nonnegative. This property leads to the estimate

${\displaystyle |f(x) - g(x)| = \left| \sum_{k=0}^n (f(x) - f(k/n)) \varphi_k(x)\right| \le \sum_{k=0}^n |f(x) - f(k/n)| \varphi_k(x) }$

The latter sum is small because when ${x}$ is close to ${k/n}$, the first factor ${|f(x) - f(k/n)|}$ is small by virtue of continuity, while the second factor ${\varphi_k(x)}$ is bounded by ${1}$. When ${x}$ is far from ${k/n}$, the second factor ${\varphi_k(x)}$ is zero, so the first one is irrelevant. The upshot is that ${f-g}$ is uniformly small.

But if we want a smooth approximation ${g}$, we need a smooth partition of unity ${{\varphi_k}}$. But not just any set of smooth nonnegative functions that add up to ${0}$ is equally good. One desirable property is preserving monotonicity: if ${f}$ is increasing, then ${g}$ should be increasing, just as this works for piecewise linear interpolation. What does this condition require of our partition of unity?

An increasing function can be expressed as a limit of sums of the form ${\sum_{j} c_j [x \ge t_j]}$ where ${c_j>0}$ and ${[\cdots ]}$ is the Iverson bracket: 1 if true, 0 if false. By linearity, it suffices to have increasing ${g}$ for the case ${f(x) = [x \ge t]}$. In this case ${g}$ is simply ${s_m := \sum_{k=m}^n \varphi_k}$ for some ${m}$, ${0\le m\le n}$. So we want all ${s_m}$ to be increasing functions. Which is the case for the triangular partition of unity, when each ${s_m}$ looks like this:

One smooth choice is Bernstein basis polynomials: ${\displaystyle \varphi_k(x) = \binom{n}{k} x^k (1-x)^{n-k}}$. These are nonnegative on ${[0, 1]}$, and the binomial formula shows ${\displaystyle \sum_{k=0}^n \varphi_k(x) = (x + 1-x)^n \equiv 1}$. Are the sums ${\displaystyle s_m(x) = \sum_{k=m}^n \binom{n}{k} x^k (1-x)^{n-k}}$ increasing with ${x}$? Let’s find out. By the product rule,

${\displaystyle s_m'(x) = \sum_{k=m}^n \binom{n}{k} k x^{k-1} (1-x)^{n-k} - \sum_{k=m}^n \binom{n}{k} (n-k) x^{k} (1-x)^{n-k - 1}}$

In the second sum the term with ${k=n}$ vanishes, and the terms with ${k can be rewritten as ${\displaystyle \frac{n!}{k! (n-k)!} (n-k) x^{k} (1-x)^{n-k - 1}}$, which is ${\frac{n!}{(k+1)! (n-k-1)!} (k+1) x^{k} (1-x)^{n-k - 1}}$, which is ${\binom{n}{k+1} (k+1) x^{k} (1-x)^{n-k - 1} }$. After the index shift ${k+1\mapsto k}$ this becomes identical to the terms of the first sum and cancels them out (except for the first one). Thus,

${\displaystyle s_m'(x) = \binom{n}{m} m x^{m-1} (1-x)^{n-m} \ge 0 }$

To summarize: the Bernstein polynomials ${\displaystyle B_n(x) = \sum_{k=0}^n f(k/n) \binom{n}{k} x^k (1-x)^{n-k}}$ are monotone whenever ${f}$ is. On the other hand, the proof that ${B_n\to f}$ uniformly is somewhat complicated by the fact that the polynomial basis functions ${\varphi_k}$ are not localized the way that the triangle basis functions are: the factors ${\varphi_k(x)}$ do not vanish when ${x}$ is far from ${k/n}$. I refer to Wikipedia for a proof of convergence (which, by the way, is quite slow).

Is there some middle ground between non-smooth triangles and non-localized polynomials? Yes, of course: piecewise polynomials, splines. More specifically, B-splines which can be defined as follows: B-splines of degree ${1}$ are the triangle basis functions shown above; a B-spline of degree ${d+1}$ is the moving averages of a ${B}$-spline of degree ${d}$ with a window of length ${h = 1/n}$. The moving average of ${F}$ can be written as ${\frac{1}{h} \int_{x-h/2}^{x+h/2} f}$. We get a partition of unity because the sum of moving averages is the moving average of a sum, and averaging a constant function does not change it.

The splines of even degrees are awkward to work with… they are obtained from the triangles by taking those integrals with ${h/2}$ an odd number of times, which makes their knots fall in the midpoints of the uniform grid instead of the grid points themselves. But I will use ${d=2}$ anyway, because this degree is enough for ${C^1}$-smooth approximation.

Recall that a triangular basis function ${\varphi_k}$ has slope ${\pm n}$ and is supported on an interval ${[(k-1)h, (k+1)h]}$ where ${h=1/n}$. Accordingly, its moving average ${\psi_k}$ will be supported on ${[(k-3/2)h, (k+3/2)h]}$. Since ${\psi_k'(x) = n(\phi_k(x+h/2) - \phi_k(x-h/2))}$, the second derivative ${\psi_k''}$ is ${n^2}$ when ${-3/2< nx-k< -1/2}$, is ${-2n^2}$ when ${|nx-k| < 1/2}$, and is ${n^2}$ again when ${1/2 < nx-k < 3/2}$. This is enough to figure out the formula for ${\psi_k}$:

${\displaystyle \psi_k(n) = \begin{cases} (nx-k+3/2)^2 / 2, & -3/2\le nx -k\le -1/2 \\ 3/4 -(nx-k)^2, & -1/2\le nx-k \le 1/2 \\ (nx-k-3/2)^2 / 2, & 1/2\le nx -k \le 3/2 \\ \end{cases} }$

These look like:

Nice! But wait a moment, the sum near the endpoints is not constant: it is less than 1 because we do not get a contributions of two splines to the left and right of the interval. To correct for this boundary effect, replace ${\psi_0}$ with ${\psi_0 + \psi_{-1}}$ and ${\psi_n}$ with ${\psi_n + \psi_{n+1}}$, using “ghost” elements of the basis that lie outside of the actual grid. Now the quadratic B-spline basis is correct:

Does this partition of unity preserve monotinicity? Yes, it does: ${\displaystyle \sum_{k\ge m}\psi_k'(x) = n\sum_{k\ge m} (\phi_k(x+h/2) - \phi_k(x-h/2)) = n(s(x+h/2) - s(x-h/2))}$ which is nonnegative because the sum ${s := \sum_{k\ge m} \phi_k}$ is an increasing piecewise linear function, as noted previously. Same logic works for B-splines of higher degree.

In conclusion, here is a quadratic B-spline approximation (orange) to a tricky increasing function (blue).

One may wonder why the orange curve deviates from the line at the end – did we miss some boundary effect there? Yes, in a way… the spline actually approximates the continuous extension of our original function by constant values on the left and right. Imagine the blue graph continuing to the right as a horizontal line: this creates a corner at ${x=1}$ and the spline is smoothing that corner. To avoid this effect, one may want to extend ${f}$ in a better way and then work with the extended function, not folding the ghosts ${\psi_{-1}, \psi_{n+1}}$ into ${\psi_0, \psi_n}$.

But even so, B-spline achieves a better approximation than the Bernstein polynomial with the same number of basis functions (eight):

The reason is the non-local nature of the polynomial basis ${\varphi_k}$, which was noted above. Bernstein polynomials do match the function perfectly at the endpoints, but this is small consolation.

## The closure of periodic functions and sums of waves with incommensurable periods

Consider the space ${C(\mathbb R)}$ of all bounded continuous functions ${f\colon \mathbb R\to\mathbb R}$, with the uniform norm ${\|f\| = \sup |f|}$. Let ${P}$ be its subset that consists of all periodic continuous functions: recall that ${f}$ is periodic if there exists ${T>0}$ such that ${f(x+T)=f(x)}$ for all ${x\in \mathbb R}$.

The set ${P}$ is not closed in the topology of ${C(\mathbb R)}$. Indeed, let ${d(x) = \mathrm{dist}\,(x, \mathbb Z)}$ be the distance from ${x}$ to nearest integer. The function ${d}$ is periodic with ${T=1}$. Therefore, each sum of the form ${\displaystyle \sum_{k=0}^n 2^{-k} d(2^{-k} x)}$ is periodic with ${T=2^n}$. Hence the sum of the infinite series ${\displaystyle f(x) = \sum_{k=0}^\infty 2^{-k} d(2^{-k} x) }$ is a uniform limit of periodic functions. Yet, ${f}$ is not periodic, because ${f(0)=0}$ and ${f(x)>0 }$ for ${x\ne 0}$ (for every ${x\ne 0}$ there exists ${k}$ such that ${2^{-k}x}$ is not an integer).

The above example (which was suggested to me by Yantao Wu) is somewhat similar to the Takagi function, which differs from it by the minus sign in the exponent: ${\displaystyle T(x) = \sum_{k=0}^\infty 2^{-k} d(2^{k} x) }$. Of course, the Takagi function is periodic with period ${1}$.

Do we really need an infinite series to get such an example? In other words, does the set ${\overline{P}\setminus P}$ contain an elementary function?

A natural candidate is the sum of trigonometric waves with incommensurable periods (that is, the ratio of periods must be irrational). For example, consider the function ${g(x) = \cos (x) + \cos (\sqrt{2}x)}$ whose graph is shown below.

Since ${g(0)=2}$ and ${g(x) < 2}$ for all ${x\ne 0}$, the function ${g}$ is not periodic. Its graph looks vaguely similar to the graph of ${f}$. Is ${g}$ a uniform limit of periodic functions?

Suppose ${h\colon \mathbb R\to\mathbb R}$ is a ${T}$-periodic function such that ${\|h-g\|<\epsilon}$. Then ${h(0) > 2-\epsilon}$, hence ${h(nT)>2-\epsilon}$ for all ${n\in \mathbb Z}$, hence ${g(nT) > 2- 2\epsilon }$. By the definition of ${g}$ this implies ${\cos (nT) > 1-2\epsilon}$ and ${\cos (\sqrt{2}nT) > 1-2\epsilon}$ for all ${n\in \mathbb Z}$. The following lemma shows a contradiction between these properties.

Lemma. If a real number ${t}$ satisfies ${\cos nt > -1/2}$ for all ${n\in \mathbb Z}$, then ${t}$ is an integer multiple of ${2\pi}$.

Proof. Suppose ${t}$ is not an integer multiple of ${2\pi}$. We can assume ${0 < t < \pi}$ without loss of generality, because ${t}$ can be replaced by ${t - 2\pi k}$ to get it in the interval ${(0, 2\pi)}$ and then by ${2\pi - t}$ to get it in ${(0, \pi)}$. Since ${\cos t > -1/2}$, we have ${t\in (0, 2\pi/3)}$. Let ${k}$ be the smallest positive integer such that ${2^k t \ge 2\pi/3}$. The minimality of ${k}$ implies ${2^{k-1} t < 2\pi/3}$, hence ${2^k t \in [2\pi/3, 4\pi/3)}$. But then ${\cos (2^k t) \le -1/2}$, a contradiction. ${\quad \Box}$

The constant ${-1/2}$ in the lemma is best possible, since ${\cos (2n\pi/3)\ge -1/2}$ for all ${n\in \mathbb Z}$.

Returning to the paragraph before the lemma, choose ${\epsilon=3/4}$ so that ${1-2\epsilon = -1/2}$. The lemma says that both ${T}$ and ${\sqrt{2} T}$ must be integer multiples of ${2\pi}$, which is impossible since they are incommensurable. This contradiction shows that ${\|g-h\|\ge 3/4}$ for any periodic function ${h}$, hence ${g}$ is not a uniform limit of periodic functions.

The above result can be stated as ${\mathrm{dist}(g, P) \ge 3/4}$. I guess ${\mathrm{dist}(g, P)}$ is actually ${1}$. It cannot be greater than ${1}$ since ${|g(x)-\cos x|\le 1}$ for all ${x}$. (Update: Yantao pointed out that the density of irrational rotations implies the distance is indeed equal to 1.)

Note: the set ${\overline{P}}$ is a proper subset of the set of (Bohr / Bochner / uniform) almost periodic functions (as Yemon Choi pointed out in a comment). The latter is a linear space while ${\overline{P}}$ is not. I was confused by the sentence “Bohr defined the uniformly almost-periodic functions as the closure of the trigonometric polynomials with respect to the uniform norm” on Wikipedia. To me, a trigonometric polynomial is a periodic function of particular kind. What Bohr called Exponentialpolynom is a finite sum of the form ${\sum a_n e^{\lambda_n x}}$ where ${\lambda_n}$ can be any real numbers. To summarize: the set considered above is the closure of ${P}$ while the set of almost periodic functions is the closed linear span of ${P}$. The function ${g(x)=\cos (x) + \cos(\sqrt{2} x)}$ is an example of the latter, not of the former.

## Points of maximal curvature

In Calculus I students are taught how to find the points at which the graph of a function has zero curvature (that is, the points of inflection). The points of maximal curvature are usually not discussed. This post attempts to rectify this omission.

The (signed) curvature of ${y=f(x)}$ is

${\displaystyle \kappa = \frac{y''}{(1+(y')^2)^{3/2}}}$

We want to maximize the absolute value of ${\kappa}$, whether the function is positive or negative there (so both maxima and minima of ${\kappa}$ can be of interest). The critical points of ${\kappa}$ are the zeros of

${\displaystyle \kappa' = \frac{y''' (1+(y')^2) - 3 y' (y'')^2 }{(1+(y')^2)^{5/2}}}$

So we are lead to consider a polynomial of the first three derivatives of ${y}$, namely ${p := y''' (1+(y')^2) - 3 y' (y'')^2 }$.

Begin with some simple examples:

${y = x^2}$ has ${p = -24x}$ so the curvature of a parabola is maximal at its vertex. No surprise there.

${y = x^3}$ has ${p = 6(1 - 45x^4)}$, indicating two symmetric points of maximal curvature, ${x\approx \pm 0.386}$, pretty close to the point of inflection.

${y=x^4}$ has ${p = 24 x (1 - 56 x^6)}$. This has three real roots, but ${x=0}$ actually minimizes curvature (it vanishes there).

More generally, ${y=x^n}$ with positive integer ${n}$ yields ${\displaystyle p = n(n-1)x^{n-3} (n - 2 - (2n^3-n^2) x^{2n-2})}$ indicating two points ${\displaystyle x = \pm \left(\frac{n-2}{2n^3-n^2} \right)^{1/(2n-2)}}$ which tend to ${\pm 1}$ as ${n}$ grows.

The graph of a polynomial of degree ${n}$ can have at most ${n-2}$ points of zero curvature, because the second derivative vanishes at those. How many points of maximal curvature can it have? The degree of expression ${p}$ above is ${3n-5}$ but it is not obvious whether all of its roots can be real and distinct, and also be the maxima of ${|\kappa|}$ (unlike ${x=0}$ for ${y=x^4}$). For ${n=2}$ we do get ${3n-5 = 1}$ point of maximal curvature. But for ${n=3}$, there can be at most ${2}$ such points, not ${3n-5 = 4}$. Edwards and Gordon (Extreme curvature of polynomials, 2004) conjectured that the graph of a polynomial of degree ${n}$ has at most ${n-1}$ points of maximal curvature. This remains open despite several partial results: see the recent paper Extreme curvature of polynomials and level sets (2017).

A few more elementary functions:

${y = e^x}$ has ${p = e^x(1-2e^{2x})}$, so the curvature is maximal at ${x=-\log(2)/2}$. Did I expect the maximum of curvature to occur for a negative ${x}$? Not really.

${y=\sin x}$ has ${p = (\cos 2x - 3)\cos x}$. The first factor is irrelevant: the points of maximum curvature of a sine wave are at its extrema, as one would guess.

${y = \tan x}$ has ${p = -6\tan^8(x) - 16\tan^6(x) - 6\tan^4(x) + 8\tan(x)^2 + 4}$ which is zero at… ahem. The expression factors as

${\displaystyle p = -2(\tan^2 x+1)(3\tan^6(x) + 5\tan^4(x) - 2\tan^2(x) - 2)}$

Writing ${u = \tan^2(x)}$ we can get a cubic equation in ${u}$, but it is not a nice one. Or we could do some trigonometry and reduce ${p=0}$ to the equation ${8 - 41\cos 2x + \cos 6x =0}$. Either way, a numerical solution is called for: ${x \approx \pm 0.6937}$ (and ${+\pi n}$ for other periods).

## Branching

Multiple kinds of branching here. First, the motorsport content has been moved to formula7.blog. Two blogs? Well, it became clear that my Stack Exchange activity, already on hiatus since 2018, is not going to resume (context: January 14, January 15, January 17). But typing words in boxes is still a hobby of mine.

There may be yet more branching in the knowledge market space, with Codidact and TopAnswers attempting to rise from the ashes of Stack Exchange. (I do not expect either project to have much success.)

Also, examples of branching in complex analysis are often limited to the situations where any two branches differ either by an additive constant like ${\log z}$ or by a multiplicative constant like ${z^p}$. But different branches can even have different branch sets. Consider the dilogarithm, which has a very nice power series in the unit disk:

${\displaystyle f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2} = z + \frac{z^2}{4} + \frac{z^3}{9} + \frac{z^4}{16} + \cdots}$

The series even converges on the unit circle ${|z|=1}$, providing a continuous extension there. But this circle is also the boundary of the disk of convergence, so some singularity has to appear. And it does, at ${z=1}$. Going around this singularity and coming back to the unit disk, we suddenly see a function with a branch point at ${z=0}$, where there was no branching previously.

What gives? Consider the derivative:

${\displaystyle f'(z) = \sum_{n=1}^\infty \frac{z^{n-1}}{n} = -\frac{\log (1-z)}{z}}$

As long as the principal branch of logarithm is considered, there is no singularity at ${z}$ since ${\log(1-0) = 0}$ cancels the denominator. But once we move around ${z=1}$, the logarithm acquires a multiple of ${2\pi i }$, and so ${f'}$ gets an additional term ${cz^{-1}}$, and integrating that results in logarithmic branching at ${z=0}$.

Of course, this does not even begin the story of the dilogarithm, so I refer to Zagier’s expanded survey which has a few branch points itself.

Thus the dilogarithm is one of the simplest non-elementary functions one can imagine. It is also one of the strangest. It occurs not quite often enough, and in not quite an important enough way, to be included in the Valhalla of the great transcendental functions—the gamma function, Bessel and Legendre- functions, hypergeometric series, or Riemann’s zeta function. And yet it occurs too often, and in far too varied contexts, to be dismissed as a mere curiosity. First defined by Euler, it has been studied by some of the great mathematicians of the past—Abel, Lobachevsky, Kummer, and Ramanujan, to name just a few—and there is a whole book devoted to it. Almost all of its appearances in mathematics, and almost all the formulas relating to it, have something of the fantastical in them, as if this function alone among all others possessed a sense of humor.

## Fundamental limits as differences of reciprocals

The three fundamental limits (as they were taught to me) are

${\displaystyle \lim_{x\to 0} \frac{e^x-1}{x} = \lim_{x\to 0} \frac{\log(1+x)}{x} = \lim_{x\to 0} \frac{\sin x}{x} = 1}$

(exponential, logarithmic, trigonometric). Usually they come with pictures like

which show that the graph of the numerator in each limit, say ${y=f(x)}$, is indeed close to the line ${y=x}$.

But there are so many different degrees of “close”: for example, ${0.0103\approx 0.01}$ but the reciprocals of these numbers differ by almost ${3}$. As a stress-test of the approximation ${f(x)\approx x}$, let us consider the behavior of ${\displaystyle \frac{1}{f(x)} - \frac{1}{x}}$ for each of the fundamental limits.

## Exponential limit

Expressed as a difference: ${\displaystyle \frac{1}{e^x - 1} - \frac{1}{x}}$

Always negative, which is obvious once we recall that the graph of ${e^x-1}$ lies above its tangent line. The graph has central symmetry about ${(0, -1/2)}$ which is not as obvious but is an easy algebra exercise. Since the asymptote on the right is ${y=0}$, the other asymptote is ${y=-1}$. This looks a lot like the logistic function ${1/(1+\exp(-x))}$ … but it is not, because the logistic function approaches its asymptotes exponentially fast, while our function does so at the rate ${1/x}$.
For comparison, here is the logistic curve ${\displaystyle y = \frac{1}{1+\exp(-x/3)} - 1}$ (in green) scaled to match the behavior at ${0}$.

This could be potentially useful if one needs a logistic-type function that behaves like a rational function at infinity. Simply using a rational function for this purpose would not do: it cannot have two distinct horizontal asymptotes.

## Logarithmic limit

${\displaystyle \frac{1}{\log(x+1)} - \frac{1}{x}}$

This one is always positive, and has a vertical asymptote at ${x=-1}$ in addition to the horizontal asymptote at ${0}$. At a glance it may look like shifted/scaled hyperbola ${y=1/x}$. Indeed, ${y = (1+x/3)/(2+x)}$ is a decent approximation to it, shown in green below.

## Trigonometric limit

${\displaystyle \frac{1}{\sin x} - \frac{1}{x}}$

Unlike in the previous cases, the difference of reciprocals vanished at ${0}$ due to the approximation ${\sin x \approx x}$ being of higher order: the error term is cubic rather than quadratic. The graph looks like the tangent function but it cannot be exactly that since the nearest vertical asymptotes are ${x=\pm \pi}$ rather than ${x=\pm \pi/2}$. (Not to mention other reasons such as non-periodicity.) A stretched and rescaled tangent, namely ${\displaystyle \frac{1}{3}\tan \frac{x}{2}}$, sort of fits:

## Bonus limit

${\tan x \approx x}$, with the reciprocal difference being ${\displaystyle \frac{1}{\tan x} - \frac{1}{x}}$.

The limit ${\displaystyle \lim_{x\to 0} \frac{\tan x}{x} = 1}$ adds nothing new compared to the previous one, since ${\cos 0=1}$. But the difference of reciprocals is another story. For one thing, the principal error term for it is twice as large as for the sine limit: ${-x/3}$ versis ${x/6}$. Accordingly, the graph looks more like ${\displaystyle -\frac{2}{3}\tan \frac{x}{2}}$.

## Taylor series

All of the above differences have derivatives of all orders at the origin, which is not easy to prove with the standard calculus tools. Complex analysis makes the situation clear: the reciprocal of a holomorphic function with a zero at ${z=0}$ can be expanded into a Laurent series, and subtracting ${1/z}$ eliminates the principal part of that series, leaving a convergent power series, i.e., another holomorphic function. Let us take a look at these series:

${\displaystyle \frac{1}{e^x - 1} - \frac{1}{x} = - \frac{1}{2} + \frac{x}{12} - \frac{x^{3}}{720} + \frac{x^{5}}{30240} - \frac{x^{7}}{1209600} + \frac{x^{9}}{47900160} -\cdots }$

Nice, the coefficients have alternating signs and all of them have numerator ${1}$… oh no, next term is ${\displaystyle - \frac{691 x^{11}}{1307674368000}}$. The signs do continue to alternate. Apart from the constant term, only odd powers of ${x}$ appear, according to the central symmetry noted above. The coefficient of ${x^{n-1}}$ in this series is ${B_n/n!}$ where ${B_n}$ is the ${n}$th Bernoulli number. These are the “modern” Bernoulli numbers, with ${B_1 = -1/2}$ rather than ${1/2}$.

${\displaystyle \frac{1}{\log(x+1)} - \frac{1}{x} = \frac{1}{2} - \frac{x}{12} + \frac{x^{2}}{24} - \frac{19 x^{3}}{720} + \frac{3 x^{4}}{160} - \frac{863 x^{5}}{60480} + \cdots }$

Also alternating but not omitting even powers, and not decaying nearly as fast as the coefficients of the previous series. (Of course: this function has a singularity at ${-1}$ while the nearest singularities of the exponential thing are at ${\pm 2\pi i}$.)  These are Gregory coefficients, which according to Wikipedia “often appear in works of modern authors who do not recognize them”. I would not recognize them either without OEIS.

${\displaystyle \frac{1}{\sin x} - \frac{1}{x} = \frac{x}{6} + \frac{7 x^{3}}{360} + \frac{31 x^{5}}{15120} + \frac{127 x^{7}}{604800} + \frac{73 x^{9}}{3421440} + \cdots }$

These are all positive for some reason.

${\displaystyle \frac{1}{\tan x} - \frac{1}{x} = - \frac{x}{3} - \frac{x^{3}}{45} - \frac{2 x^{5}}{945} - \frac{x^{7}}{4725} - \frac{2 x^{9}}{93555} - \cdots }$

These are all negative for some reason.

## Discrete maximum principle for polynomials

The polynomially convex hull of a compact set ${K\subset \mathbb C}$ is defined as the set of all points ${z\in \mathbb C}$ such that the inequality ${|p(z)|\le \sup_K |p|}$ (a form of the maximum principle) holds for every polynomial ${p}$. For example, the polynomially convex hull of a simple closed curve is the union of that curve with its interior region. In general, this process fills up the holes in the set ${K}$, resulting in the complement of the unbounded connected component of ${\mathbb C\setminus K}$.

We can recover the usual convex hull from this construction by restricting ${p}$ to the polynomials of first degree. Indeed, when ${p}$ is linear, the set ${|p|\le M}$ is a closed disk, and we end up with the intersection of all closed disks that contain ${K}$. This is precisely the convex hull of ${K}$.

What if we restrict ${p}$ to the polynomials of degree at most ${n}$? Let’s call the resulting set the degree-${n}$ convex hull of ${K}$, denoted ${K_n}$. By definition, it is contained in the convex hull and contains the polynomially convex hull. To exactly compute ${K_n}$ for general ${K}$ appears to be difficult even when ${n=2}$.

Consider finite sets. When ${K}$ has at most ${n}$ points, we have ${K_n=K}$ because there is a polynomial of degree ${n}$ whose zero set is precisely ${K}$. So, the first nontrivial case is of ${K}$ having ${n+1}$ points. Let us write ${K=\{z_0, \dots, z_n\}}$.

Depending on the location of the points, ${K_n}$ may be strictly larger than ${K}$. For example, if ${K}$ consists of the vertices of a regular ${(n+1)}$-gon, then ${K_n}$ also contains its center. Here is why. By a linear transformation, we can make sure that ${K=\{\zeta^k\colon k=0, \dots, n\}}$ where ${\zeta = \exp(2\pi i/(n+1))}$. For ${j=1, \dots, n}$ we have ${\sum_{k=0}^n \zeta^{kj} = (\zeta^{(n+1)j}-1)/(\zeta^j - 1) = 0}$. Hence, for any polynomial ${p}$ of degree at most ${n}$, the sum ${\sum_{k=0}^n p(\zeta^k)}$ is equal to ${(n+1)p(0)}$. This implies ${|p(0)|\le \max_{0\le k\le n}|p(\zeta^k)|}$, a kind of a discrete maximum principle.

A more systematic approach is to use the Lagrange basis polynomials, that is ${L_j(z) = \prod_{k\ne j} (z-z_k)/(z_j-z_k)}$, which satisfy ${L_j(z_k) = \delta_{jk}}$. Since ${p = \sum_j p(z_j)L_j}$ for any polynomial of degree at most ${n}$, it follows that ${z\in K_n}$ if and only if ${\left|\sum_j c_j L_j(z)\right|\le \max |c_j| }$ holds for every choice of scalars ${c_0, \dots, c_n}$. The latter is equivalent to the inequality ${\sum_j |L_j(z)|\le 1}$.

This leads us to consider the function ${S=\sum_j |L_j|}$, the sum of the absolute values of the Lagrange basis polynomials. (Remark: S is called a Lebesgue function for this interpolation problem.) Since ${\sum_j L_j\equiv 1}$, it follows that ${S\ge 1}$ everywhere. By construction, ${S=1}$ on ${K}$. At a point ${z\notin K}$, the equality ${S(z)=1}$ holds if and only if ${\arg L_j(z)}$ is the same for all ${j}$.

In the trivial case ${K=\{z_0, z_1\}}$, the function ${S(z) = (|z-z_0|+|z-z_1|)/|z_0-z_1|}$ is equal to ${1}$ precisely on the linear segment with endpoints ${z_0, z_1}$. Of course, this only repeats what we already knew: the degree-1 convex hull is the ordinary convex hull.

If ${K=\{x_0, \dots, x_n\}}$ with ${x_0<\cdots real and ${n\ge 2}$, then ${K_n=K}$. Indeed, if ${x\in K_n\setminus K}$, then ${x}$ lies in the convex hull of ${K}$, and therefore ${x_{j-1} for some ${j}$. The basis polynomial ${L_j}$ is positive at ${x}$, since it is equal to ${1}$ at ${x_j}$ and does not vanish outside of ${K}$. Since a polynomial changes its sign at every simple zero, it follows that ${L_{j+1}(x) < 0}$. Well, there is no ${L_{j+1}}$ if ${j=n}$, but in that case, the same reasoning applies to ${L_{j-2}(x)<0}$. In any case, the conclusion is that ${\arg L_k(x)}$ cannot be the same for all ${k}$.

At this point one might guess that the vertex set of a regular polygon is the only kind of finite sets that admit a nontrivial discrete maximum principle for polynomials. But this is not so: the vertices of a rhombus work as well. Indeed, if ${K=\{a, -a, ib, -ib\}}$ with ${a, b>0}$, then ${L_j(0)>0}$ for all ${j}$, hence ${S(0)=1}$.

The vertices of a non-square rectangle do not work: if ${K}$ is the set of these vertices, the associated function ${S}$ is strictly greater than 1 on the complement of ${K}$.

Are there any other finite sets that support a discrete maximum principle for polynomials?

## Institutions ranked by the number of AMS Fellows, 2020 edition

Only those with the count of 4 or greater are included. Not counting the deceased. Considering CUNY as a single institution. Source of data.

Top 10 changes compared to 2019 ranking: MIT takes sole possession of the 3rd place with Berkeley dropping into 4th, UIUC rises from 8th to 6th, Princeton drops from 6th to 8th, Stanford rises from 11th to 9th, Wisconsin-Madison drops from 8th to 10th, Illinois-Chicago rises from 13th to 10th. Due to ties, the “top 10” are actually top 12.

Honorable mention: Texas A&M rises from 19th to 16th. Having once set “top 20” as the goal of their “Vision 2020” campaign, they achieved it at least by this measure.

1. Rutgers The State University of New Jersey New Brunswick : 44
2. University of California, Los Angeles : 39
3. Massachusetts Institute of Technology : 35
4. University of California, Berkeley : 32
5. University of Michigan : 31
6. Cornell University : 26
7. University of Illinois, Urbana-Champaign : 26
8. Princeton University : 25
9. Stanford University : 24
10. Brown University : 23
11. University of Illinois at Chicago : 23
12. University of Wisconsin, Madison : 23
13. New York University, Courant Institute : 22
14. University of California, San Diego : 22
15. University of Texas at Austin : 22
16. Texas A&M University : 21
17. University of Chicago : 21
18. The City University of New York : 20
19. University of Washington : 20
20. Stony Brook University : 19
21. University of Minnesota-Twin Cities : 19
22. Purdue University : 17
23. University of California, Santa Barbara : 17
24. University of Pennsylvania : 17
25. Duke University : 16
26. Indiana University, Bloomington : 16
27. Ohio State University, Columbus : 16
28. University of Maryland : 16
29. Georgia Institute of Technology : 15
30. Northwestern University : 15
31. Pennsylvania State University : 15
32. University of California, Irvine : 14
33. University of Utah : 14
34. Johns Hopkins University, Baltimore : 13
35. University of British Columbia : 12
36. Boston University : 11
37. Harvard University : 11
38. University of California, Davis : 11
39. University of Notre Dame : 11
40. University of Toronto : 11
41. Eidgenössische Technische Hochschule Zürich (ETH Zürich) : 10
42. University of North Carolina at Chapel Hill : 10
43. University of Virginia : 10
44. Vanderbilt University : 10
45. Brandeis University : 9
46. Columbia University : 9
47. Institute for Advanced Study : 9
48. University of Oregon : 9
49. Michigan State University : 8
50. Rice University : 8
51. Tel Aviv University : 8
52. University of Georgia : 8
53. University of Nebraska-Lincoln : 8
54. University of Southern California : 8
55. California Institute of Technology : 7
56. Ecole Polytechnique Fédérale de Lausanne (EPFL) : 7
57. Microsoft Research : 7
58. North Carolina State University : 7
59. University of Oxford : 7
60. University of Rochester : 7
61. Williams College : 7
62. Carnegie Mellon University : 6
63. Imperial College : 6
64. Louisiana State University, Baton Rouge : 6
65. The Hebrew University of Jerusalem : 6
66. Université Pierre et Marie Curie (Paris VI) : 6
67. University of Arizona : 6
68. Harvey Mudd College : 5
69. Northeastern University : 5
70. Temple University : 5
71. Université Paris-Diderot : 5
72. University of California, Riverside : 5
73. University of Colorado, Boulder : 5
74. Virginia Polytechnic Institute and State University : 5
75. Boston College : 4
76. Florida State University : 4
77. NYU Polytechnic School of Engineering : 4
78. Norwegian University of Science and Technology : 4
79. Rutgers The State University of New Jersey Newark : 4
80. Université Paris-Sud (Paris XI) : 4
81. University of California, Santa Cruz : 4
82. University of Cambridge : 4
83. University of Connecticut, Storrs : 4
84. University of Missouri-Columbia : 4
85. University of Tennessee, Knoxville : 4
86. University of Warwick : 4
87. Washington University : 4
88. Weizmann Institute of Science : 4
89. Yale University : 4
90.

## Zeros of Taylor polynomials of (1+z)^p

This is post is related to Extremal Taylor polynomials where it was important to observe that the Taylor polynomials of the function ${(1+z)^{-1/2}}$ do not have zeros in the unit disk. Let’s see how far this generalizes.

The function ${f(z)=(1+z)^{-1}}$ has the rare property that all zeros of its Taylor polynomial have unit modulus. This is clear from

${\displaystyle T_n(z) = \sum_{k=0}^n (-z)^k = (1-(-z)^{n+1})/(1+z)}$.

In this and subsequent illustrations, the zeros of the first 50 Taylor polynomials are shown as blue dots, with the unit circle in red for reference.

When the exponent is less than -1, the zeros move inside the unit disk and begin forming nice patterns in there.

When the exponent is strictly between -1 and 1, the zeros are all outside of the unit disk. Some of them get quite large, forcing a change of scale in the image.

Why does this happen when the exponent approaches 1? The function ${1+z}$ is its own Taylor polynomial, and has the only zero at -1.  So, when ${p\approx 1}$, the Taylor polynomials are small perturbations of ${1+z}$. These perturbations of coefficients have to create additional zeros, but being small, they require a large value of ${z}$ to help them.

For a specific example, the quadratic Taylor polynomial of ${(1+z)^p}$ is ${1 + pz + p(p-1)z^2/2}$, with roots ${(1\pm \sqrt{(2-p)/p})/(1-p) }$. When ${p\approx 1}$, one of these roots is near ${-1}$ (as it has to be) and the other is large.

Finally, when ${p>1}$ and is not an integer, we get zeros on both sides of the unit circle. The majority of them are still outside. A prominent example of an interior zero is ${-1/p}$ produced by the first-degree polynomial ${1 + pz}$.

Another related post: Real zeros of sine Taylor polynomials.

## Measuring the regularity of a graph by its Laplacian eigenvalues

Let ${G}$ be a graph with vertices ${1, 2, \dots, n}$. The degree of vertex ${i}$ is denoted ${d_i}$. Let ${L}$ be the Laplacian matrix of ${G}$, so that ${L_{ii}=d_i}$, ${L_{ij}}$ is ${-1}$ when the vertices ${i, j}$ are adjacent, and is ${0}$ otherwise. The eigenvalues of ${L}$ are written as ${\lambda_1\le \dots \le \lambda_n}$.

The graph is regular if all vertices have the same degree: ${d_1=\cdots = d_n}$. How can this property be seen from its Laplacian eigenvalues ${\lambda_1, \dots, \lambda_n}$?

Since the sum of eigenvalues is equal to the trace, we have ${\sum \lambda_i = \sum d_i}$. Moreover, ${\sum \lambda_i^2}$ is the trace of ${L^2}$, which is equal to the sum of the squares of all entries of ${L}$. This sum is ${\sum d_i^2 + \sum d_i}$ because the ${i}$th row of ${L}$ contains one entry equal to ${d_i}$ and ${d_i}$ entries equal to ${-1}$. In conclusion, ${\sum d_i^2 = \sum \lambda_i^2 - \sum\lambda_i}$.

The Cauchy-Schwarz inequality says that ${n\sum d_i^2 \ge \left(\sum d_i \right)^2}$ with equality if and only if all numbers ${d_i}$ are equal, i.e., the graph is regular. In terms of eigenvalues, this means that the difference
${\displaystyle D =n\sum d_i^2 - \left(\sum d_i \right)^2 = n\sum (\lambda_i^2 - \lambda_i) - \left( \sum\lambda_i \right)^2 }$
is always nonnegative, and is equal to zero precisely when the graph is regular. This is how one can see the regularity of a graph from its Laplacian spectrum.

As an aside, ${D }$ is an even integer. Indeed, the sum ${\sum d_i}$ is even because it double-counts the edges. Hence the number of vertices of odd degree is even, which implies that ${\sum d_i^k }$ is even for every positive integer  ${k }$.

Up to a constant factor, ${D}$ is simply the degree variance: the variance of the sequence ${d_1, \dots, d_n}$. What graph maximizes it for a given ${n}$? We want to have some very large degrees and some very small ones.

Let ${G_{m, n}}$ be the union of the complete graph ${K_m}$ on ${m}$ vertices and ${(n-m)}$ isolated vertices. The sum of degrees is ${m(m-1)}$ and the sum of squares of degrees is ${m(m-1)^2}$. Hence,

${D = nm(m-1)^2 - (m(m-1))^2 = m(m-1)^2(n-m)}$

For ${n=3, 4, 5, 6}$ the maximum is attained by ${m=n-1}$, that is there is one isolated vertex. For ${n=7, 8, 9, 10}$ the maximum is ${m=n-2}$. In general it is attained by ${m^*=\lfloor (3n+2)/4 \rfloor}$.

The graph ${G_{m, n}}$ is disconnected. But any graph has the same degree variance as its complement. And the complement ${G^c(m, n)}$ is always connected: it consists of a “center”, a complete graph on ${n-m}$ vertices, and “periphery”, a set of ${m}$ vertices that are connected to each central vertex. Put another way, ${G^c(m, n)}$ is obtained from the complete bipartite graph ${K_{m, n-m}}$ by connecting all vertices of the ${n-m}$ group together.

Tom A. B. Snijders (1981) proved that ${G(m^*, n)}$ and ${G^c(m^*, n)}$ are the only graphs maximizing the degree variance; in particular, ${G^c(m^*, n)}$ is the unique maximizer among the connected graphs. It is pictured below for ${n=4, \dots, 9}$.