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It’s not just Taylor series

Posted on 1 June 2026 by John

There is still active discussion on X about the approximation

exp(−x²) ≈ (1 + cos(sin(x) + x))/2

and some are saying this can just be explained by Taylor series: the series for the two sides differ for the first time at the x⁶ term, so that’s why you get a good approximation. As I wrote yesterday, that’s only part of it.

If it were just about Taylor series you could use

exp(−x²) ≈ 1 − x² + x⁴/2

which also has error O(x⁶). But this approximation is only good for fairly small x, say in [−0.5, 0.5], whereas the approximation at the top of the post is good over [−4, 4]. When x = 4, the error in the cosine approximation is 0.002579 but the error in the Taylor approximation is 113, five orders of magnitude larger.

If the accuracy of the cosine approximation were due to Taylor series, then we’d expect the function

exp(−x²) − (1 + cos(sin(x) + x))/2

to be small not just over the interval [−4, 4] but over a disk of radius 4 in the complex plane. But it’s not. When x = 4i the function is on the order of 10¹³.

Both the cosine approximation and the Taylor approximation are good for small disks, say of radius 0.5. They’re both bad for much larger disks, and in fact the cosine approximation is worse.

Subscribe by email

Posted on 1 June 2026 by John

Readers have subscribed to this blog via email almost from its beginning in 2008, but how they have subscribed has changed several times. I’ve used several services to provide email subscription that have come and gone.

For the past two years I’ve been using Substack to send out emails announcing new blog posts. That has worked out well. Substack delivers email reliably, and that’s all I wanted. I’m not active on Substack other than using it to an email. I give a brief introduction to the latest two or three blog posts in each email, and sometimes I include additional ideas that occurred to me later.

If you’d like to get blog post announcements and a little extra commentary via email, sign up for my free Substack newsletter here.

If you’d like to learn about new posts sooner, you can subscribe via RSS or follow me on X or Mastodon.

Another Gaussian approximation

Posted on 31 May 2026 by John

The function

(1 + cos(x))/2

gives a fair approximation to the Gaussian density

exp(−x²)

You can make the approximation much better by raising it to a power. The function

((1 + cos(x))/2)⁴

gives a good lower bound and

((1 + cos(x))/2)^3.5597

gives a good upper bound. More on that here.

There are other ways of improving the cosine approximation to the Gaussian. Yesterday I came across one I hadn’t seen before, adding a sin(x) term to x.

(1 + cos(sin(x) + x))/2

This function matches the first few terms of the power series for exp(−x²) and has an error on the order of x⁶/240. You can’t see the difference between the two functions in a plot for −4 ≤ x ≤ 4.

***

There’s a tension between the previous two statements. If the error in on the order of x⁶/240 then we’d expect the error to be huge at x = 4. We have

4⁶/240 = 17.07

and yet

exp(−4²) − ((1 + cos(4 + sin(4)))/2) = −0.002579,

i.e. the error is between 3 and 4 orders of magnitude smaller than we might expect.

We have an alternating series, so the truncation error should be roughly equal to the first term after the truncation, right? No, the alternating series theorem doesn’t apply because the absolute values of the terms in the series are not decreasing yet for x = 4. The terms have to decrease eventually because the series has infinite radius of convergence, but they’re not decreasing at the 6th term; the terms will get much larger in absolute value before they get smaller.

The basic alternating series theorem gives only an upper bound on truncation error, but there are extensions that also give a lower bound. I wrote about these extensions a few weeks ago. But they don’t apply here because the terms have not started decreasing in absolute value.

Update: See further discussion in the post It’s not just Taylor series.

Spot checking polynomial identities

Posted on 30 May 2026 by John

If a polynomial identity holds at a few random points, it’s very like true. We’ll make this statement more precise, but first let’s look at some applications.

You may want to test an identity that naturally presents itself as a statement that two polynomials are equal. Or you might use something like the binomial coefficient trick to reframe a problem that isn’t obviously an identity about polynomials. And with algebraic circuits, you can reformulate a wide range of computations as polynomial identities; this is widely used in zero-knowledge proofs.

The theorem alluded to at the top of the post is the Schwartz-Zippel lemma. It is formulated in terms of the probability of a non-zero polynomial P evaluating to zero. To prove that two polynomials Q₁ and Q₂ are equal, you can show that

P = Q₁(x) − Q₂(x) = 0.

Schwartz-Zippel lemma

Let F be a (typically large) finite field and let P be a non-zero polynomial in n variables

P(x₁, x₂, x₃, …, x_n)

of total degree d. If we choose the x‘s randomly from F then the probability that P evaluates to zero is no more than d/|F|. [1]

If the total degree d is small relative to the size of the field, then the probability of P evaluating to zero is small. As long as d is less than |F|, you can evaluate the polynomial k times to make

(d / |F|)^k

as small as you’d like. If d isn’t too large, and F is large, like the integers mod p = 2²⁵⁵ − 19 used in cryptography, one polynomial evaluation might be enough to give convincing evidence that the polynomial is zero.

[1] The Schwartz-Zippel lemma in its full generality applies to polynomials over an integral domain R with variables drawn from S, a finite subset of R. Here we’re setting R = S = F.

Online (one-pass) algorithms

Posted on 29 May 2026 by John

Canonical example

The sample variance of a set of numbers is defined in terms of the sum of the squared distances from each point to the mean.

$s^2 = \frac{1}{n-1}\sum_{i=1}^n (x_i -\bar{x})^2$

So it would seem that you first need to calculate the mean, then go back and compute the squared differences from the mean. And yet sample variance can be computed in one pass through the data.

You’ll find two equivalent equations in statistics books: the one described above and another based on the sum of the data points and the sum of the data points squared.

$s^2 = \frac{1}{n(n-1)}\left(n\sum_{i=1}^n x_i^2 -\left(\sum_{i=1}^n x_i\right)^2\right)$

While this equation is theoretically correct, it is numerically unstable. Code that directly implements this equation can return a negative value for a quantity that is theoretically positive. I’ve seen this happen with real data, causing a program to crash when taking the square root of the variance to get the standard deviation.

However, there is an algorithm that computes mean and variance in one pass that is accurate and numerically stable. This algorithm was developed by B. P. Welford in 1962. I discuss Welford’s algorithm and give code for implementing it here.

Online algorithms

Welford’s algorithm is known in computer science as an “online” algorithm. This term was coined well before the Internet. For example, see the paper [1] from 1965.

But of course now “online” means something else, and so the technical and colloquial uses of “online algorithm” have split. Technical literature uses the phrase to describe the kinds of algorithms in this post. Most people would take “online algorithm” to mean code that runs on a remote server. You may see “streaming algorithm” as a contemporary technical term, but I’d still search on “online algorithm” to find papers.

Computing higher moments online

Welford’s algorithm computes the first two moments, mean and variance, of a data set online. It is also possible to compute skewness and kurtosis online, as well as higher moments.

Online regression

Simple linear regression is closely related to calculating mean and variance, and it is possible to compute simple regression coefficients online. I have some old notes on this here.

This post was motivated by an email asking me about multiple regression. It is also possible to compute multiple regression coefficients online, but I haven’t done this. I found a couple references, [2] and [3], but I have not read them. There is a simple procedure for two predictor variables but I believe things get a little more complicated with three or more predictors, requiring a recursive least squares algorithm.

The notion of online algorithms is closely related to the notion of a fold in functional programming. Here are several posts on computing things with folds.

[1] One-Tape, Off-Line Turing Machine Computations by F. C. Hennie. Information and Control. 8, 553-578 (1965). Available here. In this paper Hennie writes “In an on-line computation the input data are supplied to the machine, one symbol at a time, at a special input terminal. … In an off-line computation all of the input symbols are written on one of the machine’s tapes prior to the start of the computation.

[2] Arthur Albert and Robert W. Sittler, “A Method for Computing Least Squares Estimators that Keep Up with the Data,” Journal of the Society for Industrial and Applied Mathematics, Series A: Control, 3(3), 384–417, 1965. DOI: 10.1137/0303026.

[3] Petre Stoica and Per Ashgren. Exact initialization of the recursive least-squares algorithm. Int. J. Adapt. Control Signal Process. 2002; 16:219&ndashh;230.

Turning K-L divergence into a metric

Posted on 27 May 2026 by John

Kullback-Leibler divergence

Kullback-Leibler divergence is defined for two random variables X and Y by

$D_{KL}(X || Y) = -\int f_X(x) \log \frac{f_Y(x)}{f_X(x)} \, dx$

K-L divergence is non-negative, and it’s zero if and only if X and Y have the same distribution. But it is not a metric, for reasons explained here. For one thing, it’s not symmetric.

Jeffreys divergence

We can fix the symmetry problem by defining

$J(X, Y) = D_{KL}(X || Y) + D_{KL}(Y || X)$

The J above stands for Jeffreys, for Harold Jeffreys. J is called either the symmetrized K-L divergence or Jeffreys’ divergence. It’s still a divergence, not a distance.

A distance (metric) d has to have four properties:

d(x, x) = 0
d(x, y) > 0 if x ≠ y
d(x, y) = d(y, x)
d(x, z) ≤ d(x, y) + d(y, z)

K-L divergence satisfies the first two properties. Jeffreys’ divergence satisfies the first three, but not the last one, the triangle inequality.

To show that J doesn’t satisfy the triangle inequality, let X, Y, and Z be Bernoulli random variables with p equal to 0.1, 0.2, and 0.3 respectively. Then the following Python code shows that the divergence from X to Y, plus the divergence from Y to Z, is less than the divergence from X to Z. This would be like saying you could get from LA to NYC faster by having a layover in Denver rather than taking a direct flight.

from math import log

kl = lambda p, q: p*log(p/q) + (1-p)*log((1-p)/(1-q))
j  = lambda p, q: kl(p, q) + kl(q, p)

a = j(0.1, 0.2)
b = j(0.2, 0.3)
c = j(0.1, 0.3)
print(a + b, c)

This prints 0.135 and 0.270.

Jensen-Shannon distance

Jensen-Shannon distance turns K-L divergence into a metric as follows. First, define the random variable M to be the average of X and Y. Then average the K-L divergence from M to each of X and Y. This defines the Jensen-Shannon divergence. It’s still not a metric, but its square root is, which defines the Jensen-Shannon distance.

$\begin{align*} M &= (X + Y)/2 \\ \text{JSD}(X, Y) &= \tfrac{1}{2}D_{KL}(X || M) + \tfrac{1}{2}D_{KL}(Y || M) \\ d_{JS}(X, Y) &= \sqrt{\text{JSD}(X, Y)} \end{align*}$

The following code gives an example of Jensen-Shannon distance satisfying the triangle inequality.

def d(p, q):
    m = 0.5*(p + q)
    jsd = 0.5*kl(p, m) + 0.5*kl(q, m) 
    return jsd**0.5

a = d(0.1, 0.2)
b = d(0.2, 0.3)
c = d(0.1, 0.3)
print(a + b, c)

This prints 0.1817 and 0.1801. Now a layover makes the trip longer.

The Meta logo and fitting Besace curves

Posted on 27 May 2026 by John

I saw a post yesterday saying that the Meta logo is a Besace curve.

Meta logo

A Besace curve has the implicit form

and the parametric form

$\begin{align*} x &= a\cos(t) - b \sin(t) \\ y &= -\sin(t) x\end{align*}$

where t ranges over [0, 2π].

So given a Besace curve, such as the Meta logo, how do you find the parameters a and b to fit the curve?

We can rewrite the parametric expression for x as a sine with a phase shift (see notes here)

$x = A \sin(t + \phi)$

where

$\begin{align*} A &= \sqrt{a^2 + b^2} \\ \phi &= -\arctan(a/b)\end{align*}$

Also, we can rewrite the parametric expression for y as

$\begin{align*} y &= A \sin(t) \sin(t + \phi) \\ &= \frac{A}{2} \left(\cos(\phi) - \cos(2t + \phi)\right) \\ \end{align*}$

Now the extreme values of x and y are easier to see. The maximum value of x is A and the minimum value is −A. The maximum value of y is A(cos(φ) + 1)/2 and the minimum value is A(cos(φ) − 1)/2.

We can simplify the cosine of an arctangent (see here) to find the height, i.e. the difference between the maximum and minimum y value, in terms of a and b.

$\begin{align*} \cos(\phi) &= \cos(-\arctan(a/b)) \\ &= \frac{1}{\sqrt{1 + (a/b)^2}} \\ &= \frac{b}{\sqrt{a^2 + b^2}} \end{align*}$

Then the height is given by

$\begin{align*} h &= \frac{A}{2}(\cos(\phi) + 1) - \frac{A}{2}(\cos(\phi) - 1) \\ &= A \cos(\phi) + A \\ &= b + \sqrt{a^2 + b^2} \end{align*}$

The width is given by

$w = 2A = 2\sqrt{a^2 + b^2}$

and so

and

$a = \pm \sqrt{\frac{w^2}{4} - b^2}$

Now the Meta logo is drawn with a thick line, and the line width isn’t constant. It’s a little fuzzy what the height and width of the middle of the curve are, but I estimated h = 120 and w = 200 from one image. This leads to b = 20 and a = 97.98.

The Mathematica code

ParametricPlot[{a Cos[t] + 
   b Sin[t], -Sin[t] ( a Cos[t] + b Sin[t])}, {t, 0, 2 Pi}, 
 PlotStyle -> Thickness[0.05]]

produces the following image.

Mathematica approximation of Meta logo

This is reminiscent of the Meta logo, but not a great match. I suspect the logo is not exactly a Besace curve. You could tinker with the a and b parameters and the aspect ratio to get a closer match. The logo may have been inspired by a Besace curve and then drawn by hand.

Calculating the expected range of normal samples

Posted on 26 May 2026 by John

The previous post looked at the expected IQ range in a jury of 12. This post will look more generally at computing the expected range of n samples from a N(0, 1) random variable. This will give the expected range in units of σ, i.e. multiply the results by σ if your σ isn’t 1.

As mentioned in the previous post, the expected range is given by

$d_n = 2n \int_{-\infty}^\infty \Phi(x)^{n-1} \, x\,\phi(x) \, dx$

where φ and Φ are the PDF and CDF of a standard normal. The integral can be calculated in closed form for n ≤ 5, but in general it requires numerical integration [1].

The following Python code can compute d_n.

from scipy.stats import norm
from scipy.integrate import quad
import numpy as np

def d(n):
    integrand = lambda x: x*norm.pdf(x)*norm.cdf(x)**(n-1)
    res, info = quad(integrand, -np.inf, np.inf)
    return 2*n*res

For large n we have the asymptotic approximation

$d_n = 2 \Phi^{-1}\left( \frac{n \,–\, 0.375}{ n + 0.25} \right)$

which we could implement in Python by

def approx(n):
    return 2*norm.ppf((n - 0.375)/(n + 0.25))

For very large n the asymptotic expression may be more accurate than the integral due to numerical integration error.

Here are a few example values.

|-----+-------|
|   n |   d_n |
|-----+-------|
|   2 | 1.128 |
|   3 | 1.693 |
|   5 | 2.326 |
|  10 | 3.078 |
|  12 | 3.258 |
|  23 | 3.858 |
|  50 | 4.498 |
| 100 | 5.015 |
|-----+-------|

[1] Order Statistics by H. A. David. John Wiley & Sons. 1970.

Expected IQ spread on a jury

Posted on 26 May 2026 by John

There’s been some discussion online lately about how a large difference in IQ makes it difficult for two people to communicate. There have been studies that confirm this effect. The difficulty is not insurmountable, but it takes deliberate effort to overcome.

Someone dismissed this communication difficulty by pointing out that the expected difference in IQ between two individuals is around 17, suggesting that most communication is between people who differ by more than one standard deviation in IQ. But this calculation assumes people are chosen at random, which they usually are not. People tend to live around and work around others of similar intelligence.

However, a jury is a random sample. It’s not a perfect random sample. For one thing, it starts with a random sample of people who are registered to vote, or who have a driver’s license, not all individuals. Furthermore, the pool of potential jurors is reduced to a jury through the process of voir dire, which is not random.

For this post I will make the simplifying assumption that a jury is a random sample from a population with normally distributed IQ with standard deviation σ = 15. The mean doesn’t matter here, but you could assume it’s 100 if you’d like.

By symmetry, the expected range of n samples from a normal random variable is twice the maximum. For n = 12 the range is about 3.26σ, which corresponds to nearly 50 IQ points.

This suggests there’s usually a big spread of IQ on a jury. Even if IQ doesn’t measure intelligence, it measures something, and that something varies a lot over 12 people chosen at random [1].

[1] In case you’re interested in the technical details, the expected range of n samples from a standard normal random variable is given by

$d_n = 2n \int_{-\infty}^\infty \Phi(x)^{n-1} \, x\,\phi(x) \, dx$

where φ and Φ are the PDF and CDF of a standard normal. Multiply this by σ to get the range of a normal random variable with standard deviation σ. As for how to calculate d_n, see the next post.

Hilbert transform as an infinite matrix

Posted on 23 May 2026 by John

The previous post linked to a post I wrote a few years ago about the Hilbert transform and Fourier series. That post says that if the Fourier series of a function is

$f(t) = \sum_{n=1}^\infty \left\{ a_n \sin(nt) + b_n\cos(nt) \right\}$

then the Fourier series of its Hilbert transform is

$f_H(x) = \sum_{n=1}^\infty \left\{ -b_n \sin(nx) + a_n\cos(nx) \right\}$

When I looked back at that post I thought about how if you thought of the Fourier coefficients as elements of an infinite vector then the Hilbert transform can be represented as multiplying by an infinite block matrix.

$\left[ \begin{array}{cc|cc|cc|c} 0 & -1 & 0 & 0 & 0 & 0 & \cdots \\ 1 & 0 & 0 & 0 & 0 & 0 & \cdots \\ \hline 0 & 0 & 0 & -1 & 0 & 0 & \cdots \\ 0 & 0 & 1 & 0 & 0 & 0 & \cdots \\ \hline 0 & 0 & 0 & 0 & 0 & -1 & \cdots \\ 0 & 0 & 0 & 0 & 1 & 0 & \cdots \\ \hline \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right] \left[ \begin{array}{c} a_1 \\ b_1 \\ \hline a_2 \\ b_2 \\ \hline a_3 \\ b_3 \\ \hline \vdots \end{array} \right]$

I rarely see infinite matrices except in older math books. Apparently they were more fashionable a few decades ago than they are now. I suppose the notation falls between two stools, too concrete for some tastes and not concrete enough for others. The former folks would prefer something like H and the latter would prefer the sum above.