2025-04-25 02:10:26
I’ve heard that barbershop quartets often sign the 7th in a dominant 7th a little flat in order to bring the note closer in tune with the overtone series. This post will quantify that assertion.
The overtones of a frequency f are 2f, 3f, 4f, 5f, etc. The overtone series is a Fourier series.
Here’s a rendering of the C below middle C and its overtones.
We perceive sound on a logarithmic scale. So although the overtone frequencies are evenly spaced, they sound like they’re getting closer together.
Let’s look at the notes in the chord above and compare the frequencies between the overtone series and equal temperament tuning.
Let f be the frequency of the lowest note. The top four notes in this overtone series have frequencies 4f, 5f, 6f, and 7f. They form a C7 chord [1].
In equal temperament, these four notes have frequencies 224/12f, 228/12f, 231/12f, and 234/12f. This works out to 4, 5.0397, 5.9932, and 7.1272 times the fundamental frequency f
The the highest note, the B♭, is the furthest from its overtone counterpart. The frequency is higher than that of the corresponding overtone, so you’d need to perform it a little flatter to bring it in line with the overtone series. This is consistent with the claim at the top of the post.
How far apart are 7f and 7.1272f in terms of cents, 100ths of a semitone?
The difference between two frequencies, f1 and f2, measured in cents is
1200 log2(f1 / f2).
To verify this, note that this says an octave equals 1200 cents, because log2 2 = 1.
So the difference between the B♭ in equal temperament and in the 7th note of the overtone series is 17.66 cents.
The difference between the E and the 5th overtone is 13.69 cents, and the difference between the G and the 6th overtone is only 1.96 cents.
[1] The dominant 7th chord gets its name from two thing. First, it’s called “dominant” because it’s often found on the dominant (i.e. V) chord of a scale, and it’s made from the 1st, 3rd, 5th, and 7th notes of the scale. It’s a coincidence in the example above that the 7th of the chord is also the 7th overtone. These are two different uses of the number 7 that happen to coincide.
The post Overtones and Barbershop Quartets first appeared on John D. Cook.2025-04-24 21:49:49
I ran across a video that played around with the equipentatonic scale [1]. Instead of dividing the octave into 12 equal parts, as is most common in Western music, you divide the octave into 5 equal parts. Each note in the equipentatonic scale has a frequency 21/5 times its predecessor.
The equipentatonic scale is used in several non-Western music systems. For example, the Javanese slendro tuning system is equipentatonic, as is Babanda music from Uganda.
In the key of C, the nearest approximants of the notes in the equipentatonic scale are C, D, F, G, A#. 
In the image above [2], the D is denoted as half sharp, 50 cents higher than D. (A cent is 1/100 of a half step.) The actual pitch is a D plus 40 cents, so the half sharp is closer, but still not exactly right.
The F should be 20 cents lower, the G should be 20 cents higher, and the A# should be 10 cents higher.
The conventional chromatic scale is denoted 12-TET, an abbreviation for 12 tone equal temperament. In general n-TET denotes the scale that results from dividing the octave into n parts. The previous discussion was looking at how 5-TET aligns with 12-TET.
A step in 5-TET corresponds to 2.4 steps in 12-TET. This is approximately 5 steps in 24-TET, the scale we’d get by adding quarter tones between all the notes in the chromatic scale.
When we talk of dividing the octave evenly into n parts, we mean evenly on a logarithmic scale because we perceive music on a log scale.
The notation will be a little cleaner if we start counting from 0. Then the kth note the n-TET scale is proportional to 2k/n.
The proportionality constant is the starting pitch. So if we start on middle C, the frequencies are 262 × 2k/n Hz. The nth frequency is twice the starting frequency, i.e. an octave higher.
We can measure how well an m-TET scale can be approximated by notes from an n-TET scale with the following function:
Note that this function is asymmetric: d(m, n) does not necessarily equal d(n, m). For example, d(12, 24) = 0 because a quarter tone scale contains exact matches for every note in a semitone scale. But d(24, 12) = 1/24 because the quarter tone scale contains notes in between the notes of the semitone scale.
The equipentatonic scale lines up better with the standard chromatic scale than would a 7-note scale or an 11-note scale. That is, d(5, 12) is smaller than d(7, 12) or d(11, 12). Something similar holds if we replace 12 with 24: d(5, 24) is smaller than d(m, 24) for any m > 1 that is relatively prime to 24.
[1] The video first presents 10-TET and then defines 5-TET as taking every other note from this scale.
[2] The image was created with the following Lilypond code.
\score {
\new Staff {
\relative c' {
c1 dih f g aih c \bar "|."
}
}
\layout {
\context {
\Staff
\remove "Bar_engraver"
\remove "Time_signature_engraver"
}
}
}
The post Equipentatonic scale first appeared on John D. Cook.
2025-04-24 06:18:22
I’ve written about the Coupon Collector Problem and variations several times, most recently here.
Brian Beckman left a comment linking to an article he wrote, which in turn links to a paper on the Double Dixie Cup Problem [1].
The idea behind the Coupon Collector Problem is to estimate how long it will take to obtain at least one of each of n coupons when drawing coupons at random with replacement from a set of n.
The Double Dixie Cup Problem replaces coupons with dixie cups, and it asks how many couples you’d expect to need to collect until you have two of each cup.
The authors in [1] answer the more general question of how long to collect m of each cup, so m = 1 reduces to the Coupon Collector Problem. The punchline is Theorem 2 from the paper: the expected number of cups (or coupons) is
for fixed m as n → ∞.
Here Cm is a constant that depends on m (but not on n) and o(1) is a term that goes to zero as n → ∞.
Since log log n is much smaller than log n when n is large, this says that collecting multiples of each coupon doesn’t take much longer, on average, than collecting one of each coupon. It takes substantially less time than playing the original coupon collector game m times.
[1] Donald J. Newman and Lawrence Shepp. The Double Dixie Cup Problem. The American Mathematical Monthly, Vol. 67, No. 1 (Jan., 1960), pp. 58–61.
The post The multiple coupon collector problem first appeared on John D. Cook.2025-04-23 22:36:57
Musician Adam Neely made a video asking What is the fastest music humanly possible? He emphasizes that he means the fastest music possible to hear, not the fastest to perform.
The video cites a psychology article [1] from 1894 that found that most people can reliably distinguish an inter-onset interval (time between notes) of 100 ms [2]. It also gives examples of music faster than this, such as a performance of Vivaldi with an inter-onset interval of 83 ms [3]. The limit seem to be greater than 50 ms because a pulse train with an inter-onset interval of 50 ms starts to sound like a 20 Hz pitch.
People are able to receive Morse code faster than this implies is possible. We will explain how this works, but first we need to convert words per minute to inter-onset interval length.
Morse code is made of dots and dashes, but it is also made of spaces of three different lengths: the space between the dots and dashes representing a single letter, the space between letters, and the space between words.
According to an International Telecommunication Union standard
The same timing is referred to as standard in a US Army manual from 1957,
Notice that all the numbers above are odd. Since a dot or dash is always followed by a space, the duration of a dot or dash and its trailing space is always an even multiple of the duration of a dot.
If we think of a dot as a sixteenth note, Morse code is made of notes that are either sixteenth notes or three sixteenth notes tied together. Rests are equal to one, three, or seven sixteenths, and notes and rests must alternate. All notes start on an eighth note boundary, i.e. either on a down beat or an up beat but not in between.
Morse code speed is measured in words per minute. But what exactly is a “word”? Words have a variable number of letters, and even words with the same number of letters can have very different durations in Morse code.
The most common standard is to use PARIS as the prototypical word. Ten words per minute, for example, means that dots and dashes are coming at you as fast as if someone were sending the word PARIS ten times per minute. Here’s a visualization of the code for PARIS with each square representing the duration of a dot.
This has the duration of 50 dots.
How does this relate to inter-onset interval? If each duration of a dot is an interval, then n words per minute corresponds to 50n intervals per minute, or 60/50n = 1.2/n seconds per interval.
This would imply that 12 wpm would correspond to an inter-onset interval of around 100 ms, pushing the limit of perception. But 12 wpm is a beginner speed. It’s not uncommon for people to receive Morse code at 30 wpm. The world record, set by Theodore Roosevelt McElroy in 1939, is 75.2 wpm.
In the preceding section I assumed a dot is an interval when calculating inter-onset interval length. In musical terms, this is saying a sixteenth note is an interval. But maybe we should count eighth notes as intervals. As noted before, every “note” (dot or dash) starts on a down beat or up beat. Still, that would say 20 wpm is pushing the limit of perception, and we know people can listen faster than that.
You don’t have to hear with the precision of the diagram above in order to recognize letters. And you have context clues. Maybe you can’t hear the difference between “E E E” and “O”, but in ordinary prose the latter is far more likely.
At some skill level people quit hearing individual letters and start hearing words, much like experienced readers see words rather than letters. I’ve heard that this transition happens somewhere between 20 wpm and 30 wpm. That would be consistent with the estimate above that 20 wpm is pushing the limit of perception letter by letter. But 30 words per minute is doable. It’s impressive, but not unheard of.
What I find hard to believe is that there were intelligence officers, such as Terry Jackson, who could copy encrypted text at 50 wpm. Here a “word” is a five-letter code group. There are millions of possible code groups [4], all equally likely, and so it would be impossible to become familiar with particular code groups the way one can become familiar with common words. Maybe they learned to hear pairs or triples of letters.
[1] Thaddeus L. Bolton. Rhythm. The American Journal of Psychology. Vol. VI. No. 2. January, 1894. Available here.
[2] Interonset is not commonly hyphenated, but I’ve hyphenated it here for clarity.
[3] The movement Summer from Vivaldi’s The Four Seasons performed at 180 bpm which corresponds to 720 sixteenth notes per minute, each 83 ms long.
[4] If a code group consisted entirely of English letters, there are 265 = 11,881,376 possible groups. If a code group can contain digits as well, there would be 365 = 60,466,176 possible groups.
The post Morse code and the limits of human perception first appeared on John D. Cook.2025-04-21 22:26:23
Toward the end of last year I wrote several blog posts about calendars. The blog post about the Gregorian calendar began with this paragraph.
The time it takes for the earth to orbit the sun is not an integer multiple of the time it takes for the earth to rotate on its axis, nor is it a rational number with a small denominator. Why should it be? Much of the complexity of our calendar can be explained by rational approximations to an irrational number.
The post went on to say why the Gregorian calendar was designed as it was. In a nutshell, the average length of a year in the Gregorian calendar is 365 97/400 days, which matches the astronomical year much better than the Julian calendar, which has an average year length of 365 ¼ days.
In the Julian calendar, every year divisible by 4 is a leap year. The Gregorian calendar makes an exception: centuries are not leap years unless they are divisible by 400. So the year 2000 was a leap year, but 1700, 1800, and 1900 were not. Instead of having 100 leap years every 400 years, the Gregorian calendar has 97 leap years every 400 years.
Why does it matter whether the calendar year matches the astronomical year? In the short run it makes little difference, but in the long run it matters more. Under the Julian calendar, the Spring equinox occurred around March 21. If the world had remained on the Julian calendar, the date of the Spring equinox would drift later and later, moving into the summer and eventually moving through the entire year. Plants that bloom in March would eventually bloom in what we’d call July. And instead of the dog days of summer being in July, eventually they’d be in what we’d call November.
The Gregorian calendar wanted to do two things: stop the drift of the seasons. and restore the Spring equinox to March 21. The former could have been accomplished with little disruption by simply using the Gregorian calendar moving forward. The latter was more disruptive since it required removing days from the calendar.
The Julian year was too long, gaining 3 days every 400 years. So between the time of Jesus and the time of Pope Gregory, the calendar had drifted by about 12 days. In order to correct for this, the calendar would have to jump forward about a dozen years. If you think moving clocks forward an hour is disruptive, imagine moving the calendar forward a dozen days.
The Gregorian calendar didn’t remove 12 days; it removed 10. In the first countries to adopt the new calendar in 1582, Thursday, October 4th in 1582 was followed by Friday, October 15th. Note that Thursday was followed by Friday as usual. The seven-day cycle of days of the week was not disrupted. That’ll be important later on.
Why did the Gregorian calendar remove 10 days and not 12?
We can think of the 10 days that were removed as corresponding to previous years that the Julian calendar considered a leap year but that the Gregorian calendar would not have: 1500, 1400, 1300, 1100, 1000, 900, 700, 600, 500, and 300. Removing 10 days put the calendar in sync astronomically with the 300’s. This is significant because the Council of Nicaea met in 325 and made decisions regarding the calendar. Removing 10 days in 1582 put the calendar in sync with the calendar at the time of the council.
Now let’s push the calendar back further. Most scholars say Jesus was crucified on Friday, April 3, 33 AD. What exactly does “April 3, 33” mean? Was that a Julian date or a Gregorian date? There’s a possible difference of two days, corresponding to whether or not the years 100 and 200 were considered leap years.
If we were to push the Gregorian calendar back to the first century, the calendar for April in 33 AD would be the same as the calendar in 2033 AD (five cycles of 400 years later). April 3, 2033 is on a Sunday. (You could look that up, or use the algorithm given here.) April 3, 33 in the Julian calendar corresponds to April 1, 33 in the Gregorian calendar. So April 3, 33 was a Friday in the Julian calendar, the calendar in use at the time.
Some scholars date the crucifixion as Friday, April 7, 30 AD. That would also be a Julian date.
2025-04-21 10:09:54
The Fredholm alternative is so called because it is a theorem by the Swedish mathematician Erik Ivar Fredholm that has two alternative conclusions: either this is true or that is true. This post will state a couple forms of the Fredholm alternative.
Mr. Fredholm was interested in the solutions to linear integral equations, but his results can be framed more generally as statements about solutions to linear equations.
This is the third in a series of posts, starting with a post on kernels and cokernels, followed by a post on the Fredholm index.
Given an m×n real matrix A and a column vector b, either
Ax = b
has a solution or
ATy = 0 has a solution yTb ≠ 0.
This is essentially what I said in an earlier post on kernels and cokernels. From that post:
Suppose you have a linear transformation T: V → W and you want to solve the equation Tx = b. … If c is an element of W that is not in the image of T, then Tx = c has no solution, by definition. In order for Tx = b to have a solution, the vector b must not have any components in the subspace of W that is complementary to the image of T. This complementary space is the cokernel. The vector b must not have any component in the cokernel if Tx = b is to have a solution.
In this context you could say that the Fredholm alternative boils down to saying either b is in the image of A or it isn’t. If b isn’t in. the image of A, then it has some component in the complement of the image of A, i.e. it has a component in the cokernel, the kernel of AT.
I’ve seen the Fredholm alternative stated several ways, and the following from [1] is the clearest. The “alternative” nature of the theorem is a corollary rather than being explicit in the theorem.
As stated above, Fredholm’s interest was in integral equations. These equations can be cast as operators on Hilbert spaces.
Let K be a compact linear operator on a Hilbert space H. Let I be the identity operator and A = I − K. Let A* denote the adjoint of A.
The last point says that the kernel and cokernel have the same dimension, and the first point says these dimensions are finite. In other words, the Fredholm index of A is 0.
Where is the “alternative” in this theorem?
The theorem says that there are two possibilities regarding the inhomogeneous equation
Ax = f.
One possibility is that the homogeneous equation
Ax = 0
has only the solution x = 0, in which case the inhomogeneous equation has a unique solution for all f in H.
The other possibility is that homogeneous equation has non-zero solutions, and the inhomogeneous has solutions has a solution if and only if f is orthogonal to the kernel of A*, i.e. if f is orthogonal to the cokernel.
We said in the post on kernels and cokernels that kernels represent degrees of freedom and cokernels represent constraints. We can add elements of the kernel to a solution and still have a solution. Requiring f to be orthogonal to the cokernel is a set of constraints.
If the kernel of A has dimension n, then the Fredholm alternative says the cokernel of A also has dimension n.
If solutions x to Ax = f have n degrees of freedom, then right-hand sides f must satisfy n constraints. Each degree of freedom for x corresponds to a basis element for the kernel of A. Each constraint on f corresponds to a basis element for the cokernel that f must be orthogonal to.
[1] Lawrence C. Evans. Partial Differential Equations, 2nd edition
The post Fredholm Alternative first appeared on John D. Cook.
