When was Newton born?

Posted on 23 December 2025 by John

A young Isaac Newton unwrapping and apple as a Christmas present.

Newton’s birthday was on Christmas when he was born, but now his birthday is not.

When Newton was born, England was still using the Julian calendar, and would continue to use the Julian calendar until 25 years after his death.

On the day of Newton’s birth, his parents would have said the date was December 25, 1642. We would now describe the date as January 4, 1643.

You’ll sometimes see Newton’s birthday written as December 25, 1642 O.S. The “O.S.” stands for “Old Style,” i.e. Julian calendar. Of course the Newton family would not have written O.S. because there was no old style until the new style (i.e. Gregorian calendar) was adopted, just as nobody living in the years before Christ would have written a date as B.C.

In a nutshell, the Julian year was too long, which made it drift out of sync with the astronomical calendar. The Julian year was 365 ¹/₄ days, whereas the Gregorian calendar has 365 ⁹⁷/₄₀₀ days, which more closely matches the time it takes Earth to orbit the sun. Removing three Leap Days (in centuries not divisible by 400) put the calendar back in sync. When countries adopted the Gregorian calendar, they had to retroactively remove excess Leap Days. That’s why Newton’s birthday got moved up 10 days.

You can read more on the Julian and Gregorian calendars here.

The winter solstice in the northern hemisphere was two days ago: December 21, 2025. And in 1642, using the Gregorian calendar, the solstice was also on December 21. But in England, in 1642, people would have said the solstice occurred on December 11, because the civil calendar was 10 days ahead of the astronomical calendar.

Rolling n-sided dice to get at least n

Posted on 10 December 2025 by John

Dungeons and Dragons dice

Say you have a common 6-sided die and need to roll it until the sum of your rolls is at least 6. How many times would you need to roll?

If you had a 20-sided die and you need to roll for a sum of at least 20, would that take more rolls or fewer rolls on average?

According to [1], the expected number of rolls of an n-sided dice for the sum of the rolls to be n or more equals

$\left(1 + \frac{1}{n}\right)^{n-1}$

So for a 6-sided die, the expected number of rolls is (7/6)⁵ = 2.1614.

For a 20-sided die, the expected number of rolls is (21/20)¹⁹ = 2.5270.

The expected number of rolls is an increasing function of n, and it converges to e.

Here’s a little simulation script for the result above.

from numpy.random import randint

def game(n):
    s = 0
    i = 0
    while s < n:
        s += randint(1, n+1)
        i += 1
    return i

N = 1_000_000
s = 0
n = 20
for _ in range(N):
    s += game(n)
print(s / N)

This produced 2.5273.

[1] Enrique Treviño. Expected Number of Dice Rolls for the Sum to Reach n. American Mathematical Monthly, Vol 127, No. 3 (March 2020), p. 257.

TV tuned to a dead channel

Posted on 24 November 2025 by John

The opening line of William Gibson’s novel Neuromancer is famous:

The sky above the port was the color of a television, tuned to a dead channel.

When I read this line, I knew immediately what he meant, and thought it was a brilliant line. Later I learned that younger readers didn’t know what he was saying.

TV tuned to a dead channel circa 1960

My mind went to an old black-and-white television, one that received analog broadcasts, and that displayed “snow” when tuned to a channel that had no broadcast signal. Someone whose earliest memories of television are based on digital color broadcast might imagine the sky above the port was solid blue rather than crackly gray.

Gibson discusses how his book has aged in a preface to a recent edition. He says that science fiction that is too prescient would be received poorly.

Imagine a novel from the sixties whose author had somehow fully envisioned cellular telephony circa 2004, and had worked it, exactly as we know it today, into the fabric of her imaginary future. Such a book would have seemed highly peculiar in the sixties … in ways that would quickly overwhelm the narrative.

He then goes on to say

I suspect that Neuromancer owes much of its shelf life to my almost perfect ignorance of the technology I was extrapolating from. … Where I made things up from whole cloth, the colors remain bright.

I find it odd that many judge a work of science fiction by what it “got right.” I don’t read science fiction as a forecast; read it to enjoy a story. I don’t need a book to be prescient, but until reading Gibson’s remarks it hadn’t occurred to me that fiction that is too prescient might not be enjoyable fiction, at least for its first readers.

Four generalizations of the Pythagorean theorem

Posted on 13 November 2025 by John

Here are four theorems that generalize the Pythagorean theorem. Follow the links for more details regarding each equation.

1. Theorem by Apollonius for general triangles.

2. Edsgar Dijkstra’s extension of the Pythagorean theorem for general triangles.

$\text{sgn}(\alpha + \beta - \gamma) = \text{sgn}(a^2 + b^2 - c^2)$

3. A generalization of the Pythagorean theorem to tetrahedra.

$V_0^2 = \sum_{i=1}^n V_i^2$

4. A unified Pythagorean theorem that covers spherical, plane, and hyperbolic geometry.

$A(c) = A(a) + A(b) - \kappa \frac{A(a) \, A(b)}{2\pi}$

Weighting an average to minimize variance

Posted on 12 November 2025 by John

Suppose you have $100 to invest in two independent assets, A and B, and you want to minimize volatility. Suppose A is more volatile than B. Then putting all your money on A would be the worst thing to do, but putting all your money on B would not be the best thing to do.

The optimal allocation would be some mix of A and B, with more (but not all) going to B. We will formalize this problem and determine the optimal allocation, then generalize the problem to more assets.

Two variables

Let X and Y be two independent random variables with finite variance and assume at least one of X and Y is not constant. We want to find t that minimizes

$\text{Var}[tX + (1-t)Y]$

subject to the constraint 0 ≤ t ≤ 1. Because X and Y are independent,

$\text{Var}[tX + (1-t)Y] = t^2 \text{Var}[X] + (1-t)^2 \text{Var}[Y]$

Taking the derivative with respect to t and setting it to zero shows that

$t = \frac{\text{Var}[Y]}{\text{Var}[X] + \text{Var}[Y]}$

So the smaller the variance on Y, the less we allocate to X. If Y is constant, we allocate nothing to X and go all in on Y. If X and Y have equal variance, we allocate an equal amount to each. If X has twice the variance of Y, we allocate 1/3 to X and 2/3 to Y.

Multiple variables

Now suppose we have n independent random variables X_i for i running from 1 to n, and at least one of the variables is not constant. Then we want to minimize

$\text{Var}\left[ \sum_{i=1}^n t_i X_i \right] = \sum_{i=1}^n t_i^2 \text{Var}[X_i]$

subject to the constraint

$\sum_{i=1}^n t_i = 1$

and all t_i non-negative. We can solve this optimization problem with Lagrange multipliers and find that

$t_i \text{Var}[X_i] = t_j \text{Var}[X_j]$

for all 1 ≤ i, j ≤ n. These (n − 1) equations along with the constraint that all the t_i sum to 1 give us a system of equations whose solution is

$t_i = \frac{\prod_{j \ne i} \text{Var}[X_j]}{\sum_{i = 1}^n \prod_{j \ne i} \text{Var}[X_j]}$

Incidentally, the denominator has a name: the (n − 1)st elementary symmetric polynomial in n variables. More on this in the next post.

How much is a gigawatt?

Posted on 7 November 2025 by John

There’s increasing talk of gigawatt data centers. Currently the largest data center, Switch’s Citadel Campus in Nevada, uses 850 megawatts of power. OpenAI’s Stargate data center, under construction, is supposed to use 1.2 gigawatts.

Gigawatt

An average French nuclear reactor produces about a gigawatt of power. If the US were allowed build nuclear reactors, we could simply build one reactor for every gigantic data center. Unfortunately, the Nuclear Regulatory Commission essentially prohibits the construction of profitable nuclear reactors.

An American home uses about 1200 watts of power, so a gigawatt of electricity could power 800,000 homes. So roughly, a gigawatt is a megahome.

Gigawatt-year

A gigawatt is a unit of power, not energy. Energy is power over some time period.

A gigwatt-year is about 3 × 10¹⁶ joules, or 30 petajoules.

A SpaceX Starship launch releases 50 terajoules of energy, so a gigawatt-year is 60 Starship launches.

A couple months ago I wrote about illustrating cryptographic strength in terms of the amount of energy needed to break it, and how much water that much energy would boil. Let’s do something similar for a gigawatt-year.

It takes about 300 kilojoules of energy to boil a liter of water [1], so 30 petajoules would boil 100 billion liters of water. So a gigawatt-year of energy would be enough to boil Coniston Water, the third largest lake in England.

If you could convert a kilogram of matter to energy according to E = mc², this would release 90 petajoules. So a gigawatt-year is the energy in about 300 grams of matter.

[1] In detail, boiling a liter of water is defined as increases the temperature from 20° C to 100° C at sea level.

Impedance and Triangular Numbers

Posted on 2 November 2025 by John

A few days ago I wrote two posts about how to create a Smith chart, a graphical device used for impedance calculations. Then someone emailed me to point out the connection between the Smith chart and triangular numbers.

The Smith chart is the image of a rectangular grid in the right half-plane under the function

f(z) = (z − 1)/(z + 1).

If you subtract the values of f at consecutive integers, you get the reciprocal of a triangular number.

f(n) − f(n − 1) = 2/(n(n + 1)) = 1 / T_n

Or to put it another way,

f(n) − f(n − 1) = 1 / (1 + 2 + 3 + … + n).

In the first post on the Smith chart we showed that the function f maps vertical lines

in the z plane to circles in the w plane all touching at w = 1.

The circles are symmetric about the real axis and the diameter runs from f(n) to 1. The separation between the circles on the left side is thus

f(n) − f(n − 1) = 1 / T_n.

Number the circles starting from the outermost as 0, 1, 2, …. Then the maximum distance between circle n and circle n − 1 is 1 / T_n. You can see in the graph above that the distance between circle 0 and circle 1 is 1. It’s a little harder to see that the distance between circle 1 and circle 2 is 1/3. It looks like the distance between circles 2 and 3 is about half of that between circles 1 and 2, so it would be 1/6.

Cross ratio

Posted on 1 November 2025 by John

The cross ratio of four points A, B, C, D is defined by

$(A, B; C, D) = \frac{AC \cdot BD}{BC \cdot AD}$

where XY denotes the length of the line segment from X to Y.

The idea of a cross ratio goes back at least as far as Pappus of Alexandria (c. 290 – c. 350 AD). Numerous theorems from geometry are stated in terms of the cross ratio. For example, the cross ratio of four points is unchanged under a projective transformation.

Complex numbers

The cross ratio of four (extended [1]) complex numbers is defined by

$(z_1, z_2; z_3, z_4) = \frac{(z_3 - z_1)(z_4 - z_2)}{(z_3 - z_2)(z_4 - z_1)}$

The absolute value of the complex cross ratio is the cross ratio of the four numbers as points in a plane.

The cross ratio is invariant under Möbius transformations, i.e. if T is any Möbius transformation, then

This is connected to the invariance of the cross ratio in geometry: Möbius transformations are projective transformations on a complex projective line. (More on that here.)

If we fix the first three arguments but leave the last argument variable, then

$T(z) = (z_1, z_2; z_3, z) = \frac{(z_3 - z_1)(z - z_2)}{(z_3 - z_2)(z - z_1)}$

is the unique Möbius transformation mapping z₁, z₂, and z₃ to ∞, 0, and 1 respectively.

The anharmonic group

Suppose (a, b; c, d) = λ ≠ 1. Then there are 4! = 24 permutations of the arguments and 6 corresponding cross ratios:

$\lambda, \frac{1}{\lambda}, 1 - \lambda, \frac{1}{1 - \lambda}, \frac{\lambda - 1}{\lambda}, \frac{\lambda}{\lambda - 1}$

Viewed as functions of λ, these six functions form a group, generated by

$\begin{align*} f(\lambda) &= \frac{1}{\lambda} \\ g(\lambda) &= 1 - \lambda \end{align*}$

This group is called the anharmonic group. Four numbers are said to be in harmonic relation if their cross ratio is 1, so the requirement that λ ≠ 1 says that the four numbers are anharmonic.

The six elements of the group can be written as

$\begin{align*} f(\lambda) &= \frac{1}{\lambda} \\ g(\lambda) &= 1 - \lambda \\ f(f(\lambda)) &= g(g(\lambda) = z \\ f(g(\lambda)) &= \frac{1}{\lambda - 1} \\ g(f(\lambda)) &= \frac{\lambda - 1}{\lambda} \\ f(g(f(\lambda))) &= g(f(g(\lambda))) = \frac{\lambda}{\lambda - 1} \end{align*}$

Hypergeometric transformations

When I was looking at the six possible cross ratios for permutations of the arguments, I thought about where I’d seen them before: the linear transformation formulas for hypergeometric functions. These are, for example, equations 15.3.3 through 15.3.9 in A&S. They relate the hypergeometric function F(a, b; c; z) to similar functions where the argument z is replaced with one of the elements of the anharmonic group.

I’ve written about these transformations before here. For example,

$F(a, b; c; z) = (1-z)^{-a} F\left(a, c-b; c; \frac{z}{z-1} \right)$

There are deep relationships between hypergeometric functions and projective geometry, so I assume there’s an elegant explanation for the similarity between the transformation formulas and the anharmonic group, though I can’t say right now what it is.

[1] For completeness we need to include a point at infinity. If one of the z equals ∞ then the terms involving ∞ are dropped from the definition of the cross ratio.

How blocks are chained in a blockchain

Posted on 27 October 2025 by John

The high-level explanation of a blockchain says that each block contains a cryptographic hash of the previous block. That’s how the blocks are chained together.

That’s not exactly true, and it leaves out a lot of detail. This post will look in full detail at how Bitcoin blocks are chained together by inspecting the bits of two consecutive blocks. Looking at the low-level details reveals that some statements, like the paragraph above, are simplifications and half-truths.

For the purpose of this post, I downloaded two blocks that were added to the blockchain overnight, blocks 920993 and 920994, and saved the blocks in the binary files 920993.dat and 920994.dat.

Hashing headers

According to the simplistic description, the hash of block 920993 should be contained in block 920994. That’s not correct. We will see that the hash of the header of block 920993 is contained in block 920994 [1].

What exactly is the header?

You may hear that the header of a Bitcoin block is the first 80 bytes. That’s also not quite true. The first 4 bytes of a (production) Bitcoin block are the magic number 0xf9beb4d9. This is number chosen by Satoshi with no apparent significance, a random number unlikely to conflict with anything else. Blocks used in test versions of the blockchain begin with different magic numbers.

The next 4 bytes represent the size of the block as an unsigned integer in little endian layout. The magic number and the block size form a sort of pre-header 8 bytes long.

The API that I used to download the blocks does not include the pre-header, so the header is the first 80 bytes of the files I downloaded, though strictly speaking headers are bytes 9 through 89 of the full block.

We can see a hex dump of the header of block 920993 by running

xxd -l 80 920993.dat

which shows us the following.

00e0 ff3f e31d 6937 0e1c dba2 5321 5546
0ecc 00bf 678c a2e1 255e 0100 0000 0000
0000 0000 996f 2b91 fefc dc17 e530 6c70
9672 af27 4361 7608 1ded fde3 1157 10a5
200f 0f83 7022 ff68 21eb 0117 96f2 fbb6

How to hash?

OK, so we’re supposed to hash the header. What hash function should we apply? Bitcoin uses double SHA256,

SHA256²(header) = SHA256( (SHA256(header) )

We can compute this with openssl by running

head -c 80 920993.dat | openssl dgst -sha256 -binary | openssl dgst -sha256

Note that the first invocation of openssl dgst uses the option -binary, instructing the software to pass the raw bytes to the rest of the pipeline rather than display a text representation. The last part of the pipeline does not have that option because we want a human-readable representation at the end. The output is

c2dfb57ef58275c27a6433c5edfddfc1af6ab9ae708c00000000000000000000

Note that there are a lot of zeros on the end of the hash. That’s not a coincidence. More on that later.

Header of the next block

In the previous section we found the hash of block 920993, and we expect to see it inside the header of block 920994.

xxd -l 80 920994.dat

we can see that the header of block 920994 contains the following.

0000 002e c2df b57e f582 75c2 7a64 33c5
edfd dfc1 af6a b9ae 708c 0000 0000 0000
0000 0000 bb15 accc 8940 3811 0b9b e1cf
b125 c0fa a65e fb1b 2fa4 61ed 989f 8d6f
e41e 04cf 5522 ff68 21eb 0117 f2bc ab1a

The first 4 bytes (8 hex characters) are a version number, and following the version number we see the hash we were looking for.

Byte order

Why all the zeros in the hash of the block header? That’s a result of the proof of work problem that had to be solved in order to add block 920993 to the blockchain. Bitcoin miners tweak the details of the block [2] until they create a block whose hash begins with the required number of 0 bits [3].

But the hash value above doesn’t begin with zeros; it ends with zeros. What’s up with that?

The Bitcoin header and openssl dgst both display hashes in little endian order, i.e. the reverse of what you’d expect from a positional number system.

[1] But if you only hash the header, couldn’t someone change the rest of the block? No, because the header contains the Merkle tree root. If someone changed a bit in the body of the block, that would change the Merkle tree root, which would change the header, which would change its hash.

[2] They don’t change the substance of the transactions, but they can change the order in which the transactions are included. And there’s a nonce value that they can change too. The only known way to produce a given number of zeros is by trial and error, and so this takes a lot of work. Having a block with the right leading zeros in its hash proves that you’ve put in the work, hence proof of work.

[3] This is another simplified half-truth. You don’t need to produce a certain number of leading zeros per se; you need to find a hash value less than a target value, and the target value is not in general an exact power of 2.

Spacing the circles on the Smith chart

Posted on 25 October 2025 by John

The previous post looked at the basics of how to create a Smith chart. The Smith chart is the image of a Cartesian grid in the right half-plane under the function

f(z) = (z − 1)/(z + 1).

At the end of the post I noted that evenly distributed grid lines in the z plane result in very unevenly spaced circles on the Smith chart in the w plane. We can fill in the Smith chart how we please by working backward, starting from a desired spacing of circles in the chart and tracing them back to the z plane.

The inverse of the function f is

g(w) = (1 + w)/(1 − w).

Complete circles

The Smith chart contains two kinds of circles: circles that fit entirely inside the chart, which are the images of vertical lines in the z plane, and circles that extend outside the chart, which are the images of horizontal lines. In the image below, the former are blue and the latter are orange.

We can space out the complete (blue) circles how we wish and use the inverse transformation g(w) to determine the corresponding spacing of the vertical lines in the z plane. If we take a point w₀ along the horizontal diameter of the Smith chart, w₀ is a real number, and so is its inverse z₀ = g(w₀).

So if we’d like the intersections of the complete circles with the diameter to be uniformly spaced, we pick uniformly spaced points w_i and use as our vertical lines in the z plane the lines with real part g(w_i).

Here’s the image we’d get if we wanted circles passing through the diameter of the Smith chart at increments of 0.1.

And here’s the corresponding set of vertical lines in the z plane that would produce these circles.

Incomplete circles

Now onto the incomplete circles, the orange circles in the graph above that are perpendicular to the blue circles.

As we showed in the previous post, the outer rim of the Smith chart is the image of the imaginary axis in the z plane under the function f(z). So if we pick a point w_i on the rim that we’d like a circle to pass through, we use the horizontal line with imaginary part equal to g(w_i).

So if we wanted orange circles to cross the boundary of the Smith chart every 10°, we pick w‘s from the interval [−π, π] spaced π/18 apart.

We can create these lines on the Smith chart below

from these lines in the z plane.

Putting it all together

The image of this grid in the z plane

is the following in the w plane.

Compromise

The Smith charts printed in textbooks use some compromise between even spacing in the w plane and even spacing in the z plane.

Smith chart

Uncategorized

When was Newton born?

Rolling n-sided dice to get at least n

TV tuned to a dead channel

Four generalizations of the Pythagorean theorem

Weighting an average to minimize variance

Two variables

Multiple variables

Related posts

How much is a gigawatt?

Gigawatt

Gigawatt-year

Impedance and Triangular Numbers

Related posts

Cross ratio

Complex numbers

The anharmonic group

Hypergeometric transformations

Related posts

How blocks are chained in a blockchain

Hashing headers

What exactly is the header?

How to hash?

Header of the next block

Byte order

Related posts

Spacing the circles on the Smith chart

Complete circles

Incomplete circles

Putting it all together

Compromise