2026-05-23 23:09:34
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
then the Fourier series of its Hilbert transform is
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]](https://www.johndcook.com/hilbert_transform_matrix.svg)
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.
The post Hilbert transform as an infinite matrix first appeared on John D. Cook.2026-05-23 21:57:21
The previous post announced some notes I wrote up based on an article by Henry Baker implementing functions of a complex variable in terms of functions of a real variable. That is, it finds functions u(x, y) and v(x, y) such that
f(x + iy) = u(x, y) + i v(x, y)
where x, y, u, and v are all real-valued. Not only that, but if f is an elementary function, so are u and v. Here “elementary” has a technical meaning, but essentially it means functions that you could evaluate on a scientific calculator. A couple of the functions might be unfamiliar, such as sgn and atan2, but there are no functions like the gamma function that are defined in terms of integrals.
One application of Baker’s equations would be to bootstrap a math library that doesn’t support complex numbers into one that does. But the equations could be useful in pure math when you’d like to have a convenient expression for the real or imaginary part of a function.
The real and imaginary parts of a complex analytic function are harmonic functions. So the functions on the right hand side of Baker’s equations satisfy Laplace’s equation:
uxx + uyy = 0
and
vxx + vyy = 0.
Furthermore, the functions u and v form harmonic conjugate pairs, meaning each is the Hilbert transform of the other.
The post Real and imaginary parts first appeared on John D. Cook.2026-05-23 11:24:10
A couple months ago I wrote about how to compute the sine and cosine of a complex number using only real functions of real variables using the equations
You can do something analogous for all the elementary functions, though some of the equations are quite a bit more complicated than the ones above. See the equations here.
The equations come from a paper by Henry G. Baker, cited in the linked page. I wrote up Baker’s equations in LaTeX, then used ChatGPT to generate Python code from the LaTeX to numerically verify the equations and my typesetting of them. This caught a few typos on my part.
The test code evaluated the equations at points from each quadrant. All matched NumPy, implying that Baker and NumPy use the same branch cuts on inverse functions.
***
This post is part of a thread that has gone on for a few days. Maybe it’s the last post in the thread; we’ll see.
It all started with a post on Markov’s equation
x² + y² + z² = 3 xyz
and an approximation to the equation that has a closed-form solution. That led to the identity
cosh( arccosh(a) + arccosh(b) ) = ab + √(a² − 1) √(b² − 1).
The approximation to Markov’s equation only needed the identity to be valid for real a and b greater than 1. But when I looked closer at the identity I found several complications with branch cuts. The identity doesn’t hold everywhere using the principle branch of the square root function. But if you define √(z² − 1) to have a branch cut along [−1, 1] then the equation holds everywhere in the complex plane. And that led to my writing up some notes on how to define all the elementary inverse functions in terms of log.
Someone reading these posts suggested I look at a paper that mentioned “couth” and “uncouth” function pairs, which led to this post and its warm up.
I find all this interesting because it’s an advanced perspective on a questions that are latent in an intro calculus class. What exactly do functions like arccos mean and why where they defined as they were? These are fairly deep and interesting questions that are swept under the rug, and swept there for good reason. A calculus class has to cover an enormous amount of material and there’s no time to dwell on fine points. Some of my favorite posts look back leisurely on things that go by in a blur when you’re a student.
The post Building complex functions out of real parts first appeared on John D. Cook.2026-05-22 01:31:13
“You can’t always get what you want. But sometimes you get what you need.” — The Rolling Stones
Circular functions and hyperbolic functions aren’t invertible, but we invert them anyway. These functions map many points in the domain to each point in the range, and we invert them by mapping a point in the range back to some point in the domain. Often this works as expected, but sometimes it doesn’t.
In the previous post I said
You can relate each trig function “foo” with its hyperbolic counterpart “fooh” by applying one of the functions to iz then multiplying by a constant c that depends on foo: c = i for sin and tan, c = 1 for cos and sec, and c = −i for csc and cot.
In symbols,
c foo(z) = fooh(iz).
Let’s suppose foo and fooh are invertible, ignoring any complications, and solve foo(z) = w for z. We get
i foo−1(w) = fooh−1(cw)
or using “arc” nomenclature for inverse functions
i arcfoo(w) = arcfooh(cw).
When the naive calculation above holds, except possibly at a finite number of points, we say the pair (foo, fooh) is couth. Otherwise we say the pair is uncouth. These term were coined by Robert Corless and his coauthors in their paper [1].
Whether the pair (foo, fooh) is couth depends not only on foo and fooh, but also on the details of how arcfoo and arcfooh are defined.
In Python’s NumPy library, the pairs (sin, sinh) and (tan, tanh) are couth, but the pair (cos, cosh) is uncouth.
Numpy doesn’t define the reciprocal functions sec, sech, csc, csch, cot, and coth. I used to find that annoying, but I’m beginning to think that was wise. These functions cause problems. For example, there may be two reasonable ways to define these functions, one of which forms a couth pair and one of which forms an uncouth pair.
For example, how should you define cot and coth? There would be no disagreement over the definition
cot = lambda x: 1/tan(x)
but there are at least two definitions of inverse coth that you’ll find in practice:
arccot = lambda z: 0.5*pi - arctan(z) arccot = lambda z: arctan(1/z).
Both definitions have their advantages, but the former is uncouth while the latter is couth. You can verify that both definitions are the same at z = 1 but not at z = −1.
With the following definitions, the pairs (cos, cosh) and (sec, sech) are uncouth and the rest are couth.
from numpy import * csc = lambda x: 1/sin(x) sec = lambda x: 1/cos(x) cot = lambda x: 1/tan(x) csch = lambda x: 1/sinh(x) sech = lambda x: 1/cosh(x) coth = lambda x: 1/tanh(x) arccot = lambda z: arctan(1/z) arcsec = lambda z: arccos(1/z) arccsc = lambda z: arcsin(1/z) arccoth = lambda z: arctanh(1/z) arcsech = lambda z: arccosh(1/z) arccsch = lambda z: arcsinh(1/z)
[1] “According to Abramowitz and Stegun” or arccoth needn’t be uncouth. Robert M. Corless et al. ACM SIGSAM Bulletin, Volume 34, Issue 2, pp 58 – 65 https://doi.org/10.1145/362001.362023
The post Couth and uncouth function pairs first appeared on John D. Cook.2026-05-21 23:53:56
The difference between a circular function and a hyperbolic function is a rotation or two.
For example, cosh(z) = cos(iz). You can read that as saying that to find the hyperbolic cosine of z, first you rotate z a quarter turn to the left (i.e. multiply by i) and then take the cosine.
For another example, sinh(z) = −i sin(iz). This says that you can calculate the hyperbolic sine of z by rotating z to the left, taking the sine, and then rotating to the right.
You can relate each trig function “foo” with its hyperbolic counterpart “fooh” by applying one of the functions to iz then multiplying by a constant c that depends on foo: c = i for sin and tan, c = 1 for cos and sec, and c = −i for csc and cot.
Note that if the constant for foo is c, the constant for 1/foo is 1/c. So, for example, the constant for tan is i and the constant for cot is 1/i = −i.
We have four groups of equations, depending on whether the left side of the equation is foo(iz), fooh(iz), foo(z), or fooh(z).
This post was written as a warm-up for the next post on couth and uncouth function pairs.
2026-05-20 08:49:50
How should we define √(z² − 1)? Well, you could square z, subtract 1, and take the square root. What else would you do?!
The question turns out to be more subtle than it looks.
When x is a non-negative real number, √x is defined to be the non-negative real number whose square is x. When x is a complex number √x is defined to be a function that extends √x from the real line to the complex plane by analytic continuation. But we can’t extend √x as an analytic function to the entire complex plane ℂ. We have to choose to make a “cut” somewhere, and the conventional choice is to make a cut along the negative real axis.
The “principal branch” of the square root function is defined to be the unique function that analytically extends √x from the positive reals to ℂ \ (−∞, 0].
Assume for now that by √x we mean the principal branch of the square root function. Now what does √(z² − 1) mean? It could mean just what we said at the top of the post: we square z, subtract 1, and apply the (principal branch of the) square root function. If we do that, we must exclude those values of z such that (z² − 1) is negative. This means we have to cut out the imaginary axis and the interval [−1, 1].
This is what Mathematica will do when asked to evaluate Sqrt[z^2 - 1]. The command
ComplexPlot[Sqrt[z^2 - 1], {z, -2 - 2 I, 2 + 2 I}]
makes the branch cuts clear by abrupt changes in color.
Now let’s take a different approach. Consider the function √(z² − 1) as a whole. Do not think of it procedurally as above, first squaring z etc. Instead, think of a it as a black box that takes in z and returns a complex number whose square is z² − 1.
This function has an obvious definition for z > 1. And we can extend this function, via analytic continuation, to more of the complex plane. We can do this directly, not by extending the square root function. But as before, we cannot extend the function analytically to all of ℂ. We have to cut something out. A common choice is to cut out [−1, 1]. This eliminates the need for a branch cut along the imaginary axis.
The function
extends √(z² − 1) the way we want [1].
The Mathematica code
ComplexPlot[Exp[(1/2) (Log[z - 1 ] + Log[z + 1])], {z, -2 - 2 I, 2 + 2 I}]
shows that our function is now continuous across the imaginary axis, though there’s still a discontinuity as you cross [−1, 1].
We used this analytic extension of √(z² − 1) in the previous post to eliminate branch cuts in an identity.
[1] The principal branch of the logarithm has a cut along the negative real axis. Why does our square root function, defined using log, not have a branch cut along the negative axis?
First of all, the log function, and Mathematica’s implementation of it Log[], isn’t undefined on (−∞, 1), it just isn’t continuous there. The function still has a value. By convention, the value is taken to be the limit of log(z) approaching z from above, i.e. from the 2nd quadrant.
Second, the value of (log(z – 1) + log(z + 1))/2 differs by a factor of 2πi when approaching a value z < −1 from above versus from below. This factor goes away when taking the exponential. So our function f(z) is analytic across (−∞, 1) even though the log functions in its definition are not.
The post Square root of x² − 1 first appeared on John D. Cook.
