Solving hard problems

We help companies solve hard problems in mathematics, statistics, and computing. Let’s explore how we might work together.

Hyperbolic versions of latest posts

Posted on 25 February 2026 by John

The post A curious trig identity contained the theorem that for real x and y,

$|\sin(x + iy)| = |\sin x + \sin iy|$

This theorem also holds when sine is replaced with hyperbolic sine.

The post Trig of inverse trig contained a table summarizing trig functions applied to inverse trig functions. You can make a very similar table for the hyperbolic counterparts.

$\renewcommand{\arraystretch}{2.2} \begin{array}{c|c|c|c} & \sinh^{-1} & \cosh^{-1} & \tanh^{-1} \\ \hline \sinh & x & \sqrt{x^{2}-1} & \dfrac{x}{\sqrt{1-x^2}} \\ \hline \cosh & \sqrt{x^{2} + 1} & x & \dfrac{1}{\sqrt{1 - x^2}} \\ \hline \tanh & \dfrac{x}{\sqrt{x^{2}+1}} & \dfrac{\sqrt{x^{2}-1}}{x} & x \\ \end{array}$

The following Python code doesn’t prove that the entries in the table are correct, but it likely would catch typos.

    from math import *

    def compare(x, y):
        print(abs(x - y) < 1e-12)

    for x in [2, 3]:
        compare(sinh(acosh(x)), sqrt(x**2 - 1))
        compare(cosh(asinh(x)), sqrt(x**2 + 1))
        compare(tanh(asinh(x)), x/sqrt(x**2 + 1))
        compare(tanh(acosh(x)), sqrt(x**2 - 1)/x)                
    for x in [0.1, -0.2]:
        compare(sinh(atanh(x)), x/sqrt(1 - x**2))
        compare(cosh(atanh(x)), 1/sqrt(1 - x**2))

Trig of inverse trig

Posted on 25 February 2026 by John

I ran across an old article [1] that gave a sort of multiplication table for trig functions and inverse trig functions. Here’s my version of the table.

$\renewcommand{\arraystretch}{2.2} \begin{array}{c|c|c|c} & \sin^{-1} & \cos^{-1} & \tan^{-1} \\ \hline \sin & x & \sqrt{1-x^{2}} & \dfrac{x}{\sqrt{1+x^2}} \\ \hline \cos & \sqrt{1-x^{2}} & x & \dfrac{1}{\sqrt{1 + x^2}} \\ \hline \tan & \dfrac{x}{\sqrt{1-x^{2}}} & \dfrac{\sqrt{1-x^{2}}}{x} & x \\ \end{array}$

I made a few changes from the original. First, I used LaTeX, which didn’t exist when the article was written in 1957. Second, I only include sin, cos, and tan; the original also included csc, sec, and cot. Third, I reversed the labels of the rows and columns. Each cell represents a trig function applied to an inverse trig function.

The third point requires a little elaboration. The table represents function composition, not multiplication, but is expressed in the format of a multiplication table. For the composition f( g(x) ), do you expect f to be on the side or top? It wouldn’t matter if the functions commuted under composition, but they don’t. I think it feels more conventional to put the outer function on the side; the author make the opposite choice.

The identities in the table are all easy to prove, so the results aren’t interesting so much as the arrangement. I’d never seen these identities arranged into a table before. The matrix of identities is not symmetric, but the 2 by 2 matrix in the upper left corner is because

sin(cos⁻¹(x)) = cos(sin⁻¹(x)).

The entries of the third row and third column are not symmetric, though they do have some similarities.

You can prove the identities in the sin, cos, and tan rows by focusing on the angles θ, φ, and ψ below respectively because θ = sin⁻¹(x), φ = cos⁻¹(x), and ψ = tan⁻¹(x). This shows that the square roots in the table above all fall out of the Pythagorean theorem.

See the next post for the hyperbolic analog of the table above.

[1] G. A. Baker. Multiplication Tables for Trigonometric Operators. The American Mathematical Monthly, Vol. 64, No. 7 (Aug. – Sep., 1957), pp. 502–503.

A curious trig identity

Posted on 24 February 2026 by John

Here is an identity that doesn’t look correct but it is. For real x and y,

$|\sin(x + iy)| = |\sin x + \sin iy|$

I found the identity in [1]. The author’s proof is short. First of all,

$\begin{align*} \sin(x + iy) &= \sin x \cos iy + \cos x \sin iy \\ &= \sin x \cosh y + i \cos x \sinh y \end{align*}$

Then

$\begin{align*} |\sin(x + iy)|^2 &= \sin^2 x \cosh^2 y + \cos^2 x \sinh^2 y \\ &= \sin^2 x (1 + \sinh^2 y) + (1 -\sin^2x) \sinh^2 y \\ &= \sin^2 x + \sinh^2 y \\ &= |\sin x + i \sinh y|^2 \\ &= |\sin x + \sin iy|^2 \end{align*}$

Taking square roots completes the proof.

Now note that the statement at the top assumed x and y are real. You can see that this assumption is necessary by, for example, setting x = 2 and y = i.

Where does the proof use the assumption that x and y are real? Are there weaker assumptions on x and y that are sufficient?

[1] R. M. Robinson. A curious trigonometric identity. American Mathematical Monthly. Vol 64, No 2. (Feb. 1957). pp 83–85

Copy and paste law

Posted on 23 February 2026 by John

I was doing some research today and ran into a couple instances where part of one law was copied and pasted verbatim into another law. I suppose this is not uncommon, but I’m not a lawyer, so I don’t have that much experience comparing laws. I do, however, consult for lawyers and have to look up laws from time to time.

Here’s an example from California Health and Safety Code § 1385.10 and the California Insurance Code § 10181.10.

The former says

The health care service plan shall obtain a formal determination from a qualified statistician that the data provided pursuant to this subdivision have been deidentified so that the data do not identify or do not provide a reasonable basis from which to identify an individual. If the statistician is unable to determine that the data has been deidentified, the health care service plan shall not provide the data that cannot be deidentified to the large group purchaser. The statistician shall document the formal determination in writing and shall, upon request, provide the protocol used for deidentification to the department.

The latter says the same thing, replacing “health care service plan” with “health insurer.”

The health insurer shall obtain a formal determination … health insurer shall not provide the data … for deidentification to the department.

I saved the former in a file cal1.txt and the latter in cal2.txt and verified that the files were the same, with a search and replace, using the following shell one-liner:

diff <(sed 's/care service plan/insurer/g' cal1.txt) cal2.txt

I ran into this because I often provide statistical determination of deidentification, though usually in the context of HIPAA rather than California safety or insurance codes.

Giant Steps

Posted on 23 February 2026 by John

John Coltrane’s song Giant Steps is known for its unusual and difficult chord changes. Although the chord progressions are complicated, there aren’t that many unique chords, only nine. And there is a simple pattern to the chords; the difficulty comes from the giant steps between the chords.

Giant Steps chords

If you wrap the chromatic scale around a circle like a clock, there is a three-fold symmetry. There is only one type of chord for each root, and the three notes not represented are evenly spaced. And the pattern of the chord types going around the circle is

minor 7th, dominant 7th, major 7th, skip
minor 7th, dominant 7th, major 7th, skip
minor 7th, dominant 7th, major 7th, skip

To be clear, this is not the order of the chords in Giant Steps. It’s the order of the sorted list of chords.

For more details see the video The simplest song that nobody can play.

Tritone substitution

Posted on 23 February 2026 by John

Big moves in roots can correspond to small moves in chords.

Imagine the 12 notes of a chromatic scale arranged around the hours of a clock: C at 12:00, C♯ at 1:00, D at 2:00, etc. The furthest apart two notes can be is 6 half steps, just as the furthest apart two times can be is 6 hours.

Musical clock

An interval of 6 half steps is called a tritone. That’s a common term in jazz. In classical music you’d likely say augmented fourth or diminished fifth. Same thing.

The largest possible movement in roots corresponds to almost the smallest possible movement between chords. Specifically, to go from a dominant seventh chord to another dominant seventh chord whose roots are a tritone apart only requires moving two notes of the chord a half step each.

For example, C and F♯ are a tritone apart, but a C⁷ chord and a F♯⁷ chord are very close together. To move from the former to the latter you only need to move two notes a half step.

Musical clock

Replacing a dominant seventh chord with one a tritone away is called a tritone substitution, or just tritone sub. It’s called this for two reasons. The root moves a tritone, but also the tritone inside the chord does not move. In the example above, the third and the seventh of the C⁷ chord become the seventh and third of the F♯⁷ chord. On the diagram, the dots at 4:00 and 10:00 don’t move.

Tritone substitutions are a common technique for making basic chord progressions more sophisticated. A common tritone sub is to replace the V of a ii-V-I chord progression, giving a nice chromatic progression in the bass line. For example, in the key of C, a D min – G⁷– C progression becomes D min – D♭⁷ – C.

Bitcoin mining difficulty

Posted on 22 February 2026 by John

The previous post looked at the Bitcoin network hash rate, currently around one zettahash per second, i.e. 10²¹ hashes per second. The difficulty of mining a Bitcoin block adjusts over time to keep the rate of block production relatively constant, around one block every 10 minutes. The plot below shows this in action.

Bitcoin hash rate, difficulty, and ratio of the two

Notice the difficulty graph is more quantized than the hash rate graph. This is because the difficulty changes every 2,016 blocks, or about every two weeks. The number 2016 was chosen to be the number of blocks that would be produced in two weeks if every block took exactly 10 minutes to create.

The ratio of the hash rate to difficulty is basically constant with noise. The noticeable dip in mid 2021 was due to China cracking down on Bitcoin mining. This caused the hash rate to drop suddenly, and it took a while for the difficulty level to be adjusted accordingly.

Mining difficulty

At the current difficulty level, how many hashes would it take to mine a Bitcoin block if there were no competition? How does this compare to the number of hashes the network computes during this time?

To answer these questions, we have to back up a bit. The current mining difficulty is around 10¹⁴, but what does that mean?

The original Bitcoin mining task was to produce a hash [1] with 32 leading zeros. On average, this would take 2³² attempts. Mining difficulty is defined so that the original mining difficult was 1 and current mining difficulty is proportional to the expected number of hashes needed. So a difficulty of around 10¹⁴ means that the expected number of hashes is around

10¹⁴ × 2³² = 4.3 × 10²³.

At one zetahash per second, the number of hashes computed by the entire network over a 10 minute interval would be

10²¹ × 60 × 10 = 6 × 10²³.

So the number of hashes computed by the entire network is only about 40% greater than what would be necessary to mine a block without competition.

[1] The hash function used in Bitcoin’s proof of work is double SHA256, i.e. the Bitcoin hash of x is SHA256( SHA256( x ) ). So a single Bitcoin hash consists of two applications of the SHA256 hash function.

Exahash, Zettahash, Yottahash

Posted on 22 February 2026 by John

When I first heard of cryptographic hash functions, they were called “one-way functions” and seemed like a mild curiosity. I had no idea that one day the world would compute a mind-boggling number of hashes every second.

Because Bitcoin mining requires computing hash functions to solve proof-of-work problems, the world currently computes around 1,000,000,000,000,000,000,000 hashes, one zettahash, per second. Other cryptocurrencies uses hash functions for proof-of-work as well, but they contribute a negligible amount of hashes per second compared to Bitcoin.

The hashrate varies over time because the difficulty of Bitcoin mining is continually adjusted to keep new blocks being produced at about one every ten minutes. As hardware has gotten faster, the difficulty level of mining has gotten higher. When the price of Bitcoin drops and mining becomes less profitable, the difficulty adjusts downward. There are other factors too, and hashrate is variable.

Bitcoin network hashes per second over time

The prefix giga- (10⁹) wasn’t widely known until computer memory and storage entered the gigabyte range. Then the prefix prefix tera- (10¹²) became familiar when disk drives entered terabyte territory. The prefixes for larger units such as peta- (10¹⁵) and exa- (10¹⁸) are still not widely known. The prefixes zetta- (10²¹) and yotta- (10²⁴) were adopted in 1991 and ronna- (10²⁷) and quetta (10³⁰) were adopted in 2022.

When the Bitcoin hashrate is relatively low, it’s in the range of hundreds of exahashes per second. At the time of writing the hashrate is 1.136 ZH/s, 1.136 × 10²¹ hashes per second. This puts the hashrate per day in the tens of yottahashes and the number of hashes per year in tens of ronnahashes.

10,000,000th Fibonacci number

Posted on 21 February 2026 by John

I’ve written a couple times about Fibonacci numbers and certificates. Here the certificate is auxiliary data that makes it faster to confirm that the original calculation was correct.

This post puts some timing numbers to this.

I calculated the 10 millionth Fibonacci number using code from this post.

n = 10_000_000    
F = fib_mpmath(n)

This took 150.3 seconds.

As I’ve discussed before, a number f is a Fibonacci number if and only if 5f² ± 4 is a square r². And for the nth Fibonacci number, we can take ± to be positive if n is even and negative if n is odd.

I computed the certificate r with

r = isqrt(5*F**2 + 4 - 8*(n%2))

and this took 65.2 seconds.

Verifying that F is correct with

n = abs(5*F**2 - r**2)
assert(n == 4)

took 3.3 seconds.

About certificates

So in total it took 68.5 seconds to verify that F was correct. But if someone had pre-computed r and handed me F and r I could use r to verify the correctness of F in 3.3 seconds, about 2% of the time it took to compute F.

Sometimes you can get a certificate of correctness for free because it is a byproduct of the problem you’re solving. Other times computing the certificate takes a substantial amount of work. Here computing F and r took about 40% more work than just computing F.

What’s not typical about this example is that the solution provider carries out exactly the same process to create the certificate that someone receiving the solution without a certificate would do. Typically, even if the certificate isn’t free, it does come at a “discount,” i.e. the problem solver could compute the certificate more efficiently than someone given only the solution.

Proving you have the right Fibonacci number

Now suppose I have you the number F above and claim that it is the 10,000,000th Fibonacci number. You carry out the calculations above and say “OK, you’ve convinced me that F is a Fibonacci number, but how do I know it’s the 10,000,000th Fibonacci number?

The 10,000,000th Fibonacci number has 2,089,877 digits. That’s almost enough information to verify that a Fibonacci number is indeed the 10,000,000th Fibonacci number. The log base 10 of the nth Fibonacci number is very nearly

n log₁₀ φ − 0.5 log₁₀ 5

based on Binet’s formula. From this we can determine that the nth Fibonacci number has 2,089,877 digits if n = 10,000,000 + k where k equals 0, 1, 2, or 3.

Because

log₁₀ F_10,000,000 = 2089876.053014785

and

10^0.053014785 = 1.129834377783962

we know that the first few digits of F_10,000,000 are 11298…. If I give you a 2,089,877 digits that you can prove to be a Fibonacci number, and its first digit is 1, then you know it’s the 10,000,000th Fibonacci number.

You could also verify the number by looking at the last digit. As I wrote about here, the last digits of Fibonacci number have a period of 60. That means the last digit of the 10,000,000th Fibonacci number is the same as the last digit of the 40th Fibonacci number, which is 5. That is enough to rule out the other three possible Fibonacci numbers with 2,089,877 digits.

Computing big, certified Fibonacci numbers

Posted on 21 February 2026 by John

I’ve written before about computing big Fibonacci numbers, and about creating a certificate to verify a Fibonacci number has been calculated correctly. This post will revisit both, giving a different approach to computing big Fibonacci numbers that produces a certificate along the way.

As I’ve said before, I’m not aware of any practical reason to compute large Fibonacci numbers. However, the process illustrates techniques that are practical for other problems.

The fastest way to compute the nth Fibonacci number for sufficiently large n is Binet’s formula:

F_n = round( φⁿ / √5 )

where φ is the golden ratio. The point where n is “sufficiently large” depends on your hardware and software, but in my experience I found the crossover to be somewhere 1,000 and 10,000.

The problem with Binet’s formula is that it requires extended precision floating point math. You need extra guard digits to make sure the integer part of your result is entirely correct. How many guard digits you’ll need isn’t obvious a priori. This post gave a way of detecting errors, and it could be turned into a method for correcting errors.

But how do we know an error didn’t slip by undetected? This question brings us back to the idea of certifying a Fibonacci number. A number f Fibonacci number if and only if one of 5f² ± 4 is a perfect square.

Binet’s formula, implemented in finite precision, takes a positive integer n and gives us a number f that is approximately the nth Fibonacci number. Even in low-precision arithmetic, the relative error in the computation is small. And because the ratio of consecutive Fibonacci numbers is approximately φ, an approximation to F_n is far from the Fibonacci numbers F_{n − 1} and F_{n + 1}. So the closest Fibonacci number to an approximation of F_n is exactly F_n.

Now if f is approximately F_n, then 5f² is approximately a square. Find the integer r minimizing |5f² − r²|. In Python you could do this with the isqrt function. Then either r² + 4 or r² − 4 is divisible by 5. You can know which one by looking at r² mod 5. Whichever one is, divide it by 5 and you have the square of F_n. You can take the square root exactly, in integer arithmetic, and you have F_n. Furthermore, the number r that you computed along the way is the certificate of the calculation of F_n.

Solving hard problems

Hyperbolic versions of latest posts

Trig of inverse trig

A curious trig identity

Copy and paste law

Related posts

Giant Steps

Related posts

Tritone substitution

Related posts

Bitcoin mining difficulty

Mining difficulty

Related posts

Exahash, Zettahash, Yottahash

Related posts

10,000,000th Fibonacci number

About certificates

Proving you have the right Fibonacci number

Computing big, certified Fibonacci numbers