Author: John

Solar system means

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Yesterday I stumbled on the fact that the size of Jupiter is roughly the geometric mean between the sizes of Earth and the Sun. That’s not too surprising: in some sense (i.e. on a logarithmic scale) Jupiter is the middle sized object in our solar system.

What I find more surprising is that a systematic search finds mean relationships that are far more accurate. The radius of Jupiter is within 5% of the geometric mean of the radii of the Earth and Sun. But all the mean relations below have an error less than 1%.

\begin{eqnarray*} R_\Mercury &=& \mbox{GM}\left(R_\Moon, R_\Mars\right) \\ R_\Mars &=& \mbox{HM}\left(R_\Moon, R_\Jupiter\right) \\ R_\Uranus &=& \mbox{AGM}\left(R_\Earth, R_\Saturn\right) \\ \end{eqnarray*}

The radius of Mercury equals the geometric mean of the radii of the Moon and Mars, within 0.052%.

The radius of Mars equals the harmonic mean of the radii of the Moon and Jupiter, within 0.08%.

The radius of Uranus equals the arithmetic-geometric mean of the radii of Earth and Saturn, within 0.0018%.

See the links below for more on AM, GM, and AGM.

Now let’s look at masses.

\begin{eqnarray*} M_\Earth &=& \mbox{GM}\left(M_\Mercury, M_\Neptune\right) \\ M_\Pluto &=& \mbox{HM}\left(M_\Moon, M_\Mars\right) \\ M_\Uranus &=& \mbox{AGM}\left(M_\Moon, M_\Saturn\right) \\ \end{eqnarray*}

The mass of Earth is the geometric mean of the masses of Mercury and Neptune, within 2.75%. This is the least accurate approximation in this post.

The mass of Pluto is the harmonic mean of the masses of the Moon and Mars, within 0.7%.

The mass of Uranus is the arithmetic-geometric mean of the masses of of the Moon and Saturn, within 0.54%.

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Earth : Jupiter :: Jupiter : Sun

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The size of Jupiter is approximately the geometric mean of the sizes of Sun and Earth.

In terms of radii,

\frac{R_\Sun}{R_{\text{\Jupiter}}} \approx \frac{R_\Jupiter}{R_\Earth}

The ratio on the left equals 9.95 and the ratio on the left equals 10.98.

The subscripts are the astronomical symbols for the Sun (☉, U+2609), Jupiter (♃, U+2643), and Earth (, U+1F728). I produced them in LaTeX using the mathabx package and the commands \Sun, Jupiter, and Earth.

The the mathabx symbol for Jupiter is a little unusual. It looks italicized, but that’s not because the symbol is being used in math mode. Notice that the vertical bar in the symbol for Earth is vertical, i.e. not italicized.

 

Gravity on Jupiter

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NASA image of Jupiter

I was listening to the latest episode of the Space Rocket History podcast. The show includes some audio from a documentary on Pioneer 11 that mentioned that a man would weigh 500 pounds on Jupiter.

My immediate thought was “Is that all?! Is this ‘man’ a 100 pound boy?”

The documentary was correct and my intuition was wrong. And the implied mass of the man in the documentary is 190 pounds.

Jupiter has more than 300 times more mass than the earth. Why is its surface gravity only 2.6 times that of the earth?

Although Jupiter is very massive, it is also very large. Gravitational attraction is proportional to mass, but inversely proportional to the square of distance.

A satellite in orbit 100,000 km from the center of Jupiter would feel 300 times as much gravity as one in orbit the same distance from the center of Earth. But the surface of Jupiter is further from its center of mass than the surface of Earth is from its center of mass.

The mass of Jupiter is 318 times that of Earth, and the its mean radius is 11 times that of Earth. So the ratio of gravity on the surface of Jupiter to gravity on the Earth’s surface is

318 / 11² = 2.63

Now suppose a planet had the same density as Earth but a radius of r Earth radii. Then its mass would be r³ times greater, but its surface gravity would only be r times greater since gravity follows an inverse square law. So if Jupiter were made of the same stuff as Earth, its surface gravity would be 11 times greater. But Jupiter is a gas giant, so its surface gravity is only 2.6 times greater.

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Are guidance documents laws?

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Are guidance documents laws? No, but they can have legal significance.

The people who generate regulatory guidance documents are not legislators. Legislators delegate to agencies to make rules, and agencies delegate to other organizations to make guidelines. For example [1],

Even HHS, which has express cybersecurity rulemaking authority under the Health Insurance Portability and Accountability Act (HIPAA), has put a lot of the details of what it considers adequate cybersecurity into non-binding guidelines.

I’m not a lawyer, so nothing I can should be considered legal advice. However, the authors of [1] are lawyers.

The legal status of guidance documents is contested. According to [2], Executive Order 13892 said that agencies

may not treat noncompliance with a standard of conduct announced solely in a guidance document as itself a violation of applicable statutes or regulations.

Makes sense to me, but EO 13992 revoked EO 13892.

Again according to [3],

Under the common law, it used to be that government advisories, guidelines, and other non-binding statements were non-binding hearsay [in private litigation]. However, in 1975, the Fifth Circuit held that advisory materials … are an exception to the hearsay rule … It’s not clear if this is now the majority rule.

In short, it’s fuzzy.

 

[1] Jim Dempsey and John P. Carlin. Cybersecurity Law Fundamentals, Second Edition, page 245.

[2] Ibid., page 199.

[3] Ibid., page 200.

 

More Laguerre images

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A week or two ago I wrote about Laguerre’s root-finding method and made some associated images. This post gives a couple more examples.

Laguerre’s method is very robust in the sense that it is likely to converge to a root, regardless of the starting point. However, it may be difficult to predict which root the method will end up at. To visualize this, we color points according to which root they converge to.

First, let’s look at the polynomial

(x − 2)(x − 4)(x − 24)

which clearly has roots at 2, 4, and 24. We’ll generate random starting points and color them blue, orange, or green depending on whether they converge to 2, 4, or 24. Here’s the result.

To make this easier to see, let’s split it into each color: blue, orange, and green.

Now let’s change our polynomial by moving the root at 4 to 4i.

(x − 2)(x − 4i)(x − 24)

Here’s the combined result.

And here is each color separately.

As we explained last time, the area taken up by the separate colors seems to exceed the total area. That is because the colors are so intermingled that many of the dots in the images cover some territory that belongs to another color, even though the dots are quite small.

A Bayesian approach to proving you’re human

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I set up a GitHub account for a new employee this morning and spent a ridiculous amount of time proving that I’m human.

The captcha was to listen to three audio clips at a time and say which one contains bird sounds. This is a really clever test, because humans can tell the difference between real bird sounds and synthesized bird-like sounds. And we’re generally good at recognizing bird sounds even against a background of competing sounds. But some of these were ambiguous, and I had real birds chirping outside my window while I was doing the captcha.

You have to do 20 of these tests, and apparently you have to get all 20 right. I didn’t. So I tried again. On the last test I accidentally clicked the start-over button rather than the submit button. I wasn’t willing to listen to another 20 triples of audio clips, so I switched over to the visual captcha tests.

These kinds of tests could be made less annoying and more secure by using a Bayesian approach.

Suppose someone solves 19 out of 20 puzzles correctly. You require 20 out of 20, so you have them start over. When you do, you’re throwing away information. You require 20 more puzzles, despite the fact that they only missed one. And if a bot had solved say 8 out of 20 puzzles, you’d let it pass if it passes the next 20.

If you wipe your memory after every round of 20 puzzles, and allow unlimited do-overs, then any sufficiently persistent entity will almost certainly pass eventually.

Bayesian statistics reflects common sense. After someone (or something) has correctly solved 19 out of 20 puzzles designed to be hard for machines to solve, your conviction that this entity is human is higher than if they/it had solved 8 out of 20 correctly. You don’t need as much additional evidence in the first case as in the latter to be sufficiently convinced.

Here’s how a Bayesian captcha could work. You start out with some distribution on your probability θ that an entity is human, say a uniform distribution. You present a sequence of puzzles, recalculating your posterior distribution after each puzzle, until the posterior probability that this entity is human crosses some upper threshold, say 0.95, or some lower threshold, say 0.50. If the upper threshold is crossed, you decide the entity is likely human. If the lower threshold is crossed, you decide the entity is likely not human.

If solving 20 out of 20 puzzles correctly crosses your threshold of human detection, then after solving 19 out 20 correctly your posterior probability of humanity is close to the upper threshold and would only require a few more puzzles. And if an entity solved 8 out of 20 puzzles correctly, that may cross your lower threshold. If not, maybe only a few more puzzles would be necessary to reject the entity as non-human.

When I worked at MD Anderson Cancer Center we applied this approach to adaptive clinical trials. A clinical trial might stop early because of particularly good results or particularly bad results. Clinical trials are expensive, both in terms of human costs and financial costs. Rejecting poor treatments quickly, and sending promising treatments on to the next stage quickly, is both humane and economical.

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Antenna length: Another rule of 72

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The famous Rule of 72 says that to find out how many years it takes an investment to double in value, divide 72 by the annual percentage rate. I’ll come back to that in a little bit.

This morning I read a really good article, Fifty Things you can do with a Software Defined Radio. The article includes a rule of thumb for how long an antenna needs to be.

My rule of thumb was to divide 72 by the frequency in MHz, and take that as the length of each side of the dipole in meters [1]. That’d make the whole antenna a bit shorter than half of the wavelength.

Ideally an antenna should be as long as half a wavelength of the signal you want to receive. Light travels 3 × 108 meters per second, so one wavelength of a 1 MHz signal is 300 m. A quarter wavelength, the length of one side of a dipole antenna, would be 75 m. Call it 72 m because 72 has lots of small factors, i.e. it’s usually mentally easier to divide things into 72 than 75. Rounding 75 down to 72 results in the antenna being a little shorter than ideal. But antennas are forgiving, especially for receiving.

Update: There’s more to replacing 75 with 72 than simplifying mental arithmetic. See Markus Meier’s comment below.

Just as the Rule of 72 for antennas rounds 75 down to 72, the Rule of 72 for interest rounds 69.3 up to 72, both for ease of mental calculation.

\begin{align*} \left(1 + \frac{n}{100}\right)^{72/n} &= \exp\left(\frac{72}{n} \log\left(1 + \frac{n}{100}\right)\right) \\ &\approx \exp\left(\frac{72}{n} \, \frac{n}{100} \right) \\ &= \exp(72/100) \\ &= 2.05 \end{align*}

The approximation step comes from the approximation log(1 + x) ≈ x for small x, a first order Taylor approximation.

The last line would be 2 rather than 2.05 if we replaced 72 with 100 log(2) = 69.3. That’s where the factor of 69.3 mentioned above comes from.

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[1] The post actually says centimeters, but the author meant to say meters.

Double super factorial

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I saw someone point out recently that

10! = 7! × 5! × 3! × 1!

Are there more examples like this?

What would you call the pattern on the right? I don’t think there’s a standard name, but here’s why I think it should be called double super factorial or super double factorial.

Super factorial

The factorial of a positive number n is the product of the positive numbers up to and including n. The super factorial of n is the product of the factorials of the positive numbers up to and including n. So, for example, 7 super factorial would be

7! × 6! × 5! × 4! × 3! × 2! × 1!

Double factorial

The double factorial of a positive number n is the product of all the positive numbers up to n with the same parity of n. So, for example, the double factorial of 7 would be

7!! = 7 × 5 × 3 × 1.

Double superfactorial

The pattern at the top of the post is like super factorial, but it only includes odd terms, so it’s like a cross between super factorial and double factorial, hence double super factorial.

Denote the double super factorial of n as dsf(n), the product of the factorials of all numbers up to n with the same parity as n. That is,

dsf(n) = n! × (n − 2)! × (n − 4)! × … × 1

where the 1 at the end is 1! if n is odd and 0! if n is even. In this notation, the observation at the top of the post is

10! = dsf(7).

Super double factorial

We can see by re-arranging terms that a double super factorial is also a super double factorial. For example, look at

dsf(7) = 7! × 5! × 3! × 1!

If we separate out the first term in each factorial we have

(7 × 5 × 3 × 1)(6! × 4! × 2!) = 7!! dsf(6)

We can keep going and show in general that

dsf(n) = n!! × (n − 1)!! × (n − 2)!! … × 1

We could call the right hand side super double factorial, sdf(n). Just as a super factorial is a product of factorials, a super double factorials is a product of double factorials. Therefore

dsf(n) = sdf(n).

Factorials that equal double super factorials

Are there more solutions to

n! = dsf(m).

besides n = 10 and m = 7? Yes, here are some.

0! = dsf(0)
1! = dsf(1)
2! = dsf(2)
3! = dsf(3)
6! = dsf(5)

There are no solutions to

n! = dsf(m)

if n > 10. Here’s a sketch of a proof.

Bertrand’s postulate says that for n > 1 there is always a prime p between n and 2n. Now p divides (2n)! but p cannot divide dsf(n) because dsf(n) only has factors less than or equal to n.

If we can show that for some N, n > N implies (2n)! < dsf(n) then there are no solutions to

n! = dsf(m)

for n > 2N because there is a prime p between N and 2N that divides the left side but not the right. In fact N = 12. We can show empirically there are no solutions for n = 11 up to 24, and the proof shows there are no solutions for n > 24.

Laguerre’s root finding method

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Edmond Laguerre (1834–1886) came up with a method for finding zeros of polynomials. Unlike Newton’s method for finding zeros of general functions, Laguerre’s method is specialized for polynomials. Laguerre’s method converges an order of magnitude faster than Newton’s method, i.e. the error is cubed on each step rather than squared.

The most interesting thing about Laguerre’s method is that it nearly always converges to a root, no matter where you start. Newton’s method, on the other hand, converges quickly if you start sufficiently close to a root. If you don’t start close enough to a root, the method might cycle or go off to infinity.

The first time I taught numerical analysis, the textbook we used began the section on Newton’s method with the following nursery rhyme:

There was a little girl,
Who had a little curl,
Right in the middle of her forehead.
When she was good,
She was very, very good,
But when she was bad, she was horrid.

When Newton’s method is good, it is very, very good, but when it is bad it is horrid.

Laguerre’s method is not well understood. Experiments show that it nearly always converges, but there’s not much theory to explain what’s going on.

The method is robust in that it is very likely to converge to some root, but it may not converge to the root you expected unless you start sufficiently close. The rest of the post illustrates this.

Let’s look at the polynomial

p(x) = 3 + 22x + 20x² + 24x³.

This polynomial has roots at 0.15392, and at -0.3397 ± 0.83468i.

We’ll generate 2,000 random starting points for Laguerre’s method and color its location according to which root it converges to. Points converging to the real root are colored blue, points converging to the root with positive imaginary part are colored orange, and points converging to the root with negative imaginary part are colored green.

Here’s what we get:

This is hard to see, but we can tell that there aren’t many blue dots, about an equal number of orange and green dots, and the latter are thoroughly mixed together. This means the method is unlikely to converge to the real root, and about equally likely to converge to either of the complex roots.

Let’s look at just the starting points that converge to the real root, its basin of attraction. To get more resolution, we’ll generate 100,000 starting points and make the dots smaller.

The convergence region is pinched near the root; you can start fairly near the root along the real axis but converge to one of the complex roots. Notice also that there are scattered points far from the real root that converge to that point.

Next let’s look at the points that converge to the complex root in the upper half plane.

Note that the basin of attraction appears to take up over half the area. But the corresponding basin of attraction for the root in the lower half plane also appears to take up over half the area.

They can’t both take up over half the area. In fact. both take up about 48%. But the two regions are very intertwined. Due to the width of the dots used in plotting, each green dot covers a tiny bit of area that belongs to orange, and vice versa. That is, the fact that both appear to take over half the area shows how commingled they are.

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Breach Safe Harbor

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In the context of medical data, Safe Harbor typically refers to the Safe Harbor provisions of the HIPAA Privacy Rule explained here. Breach Safe Harbor is a little different. It basically means you’re off the hook if you breach encrypted health data. (But not necessarily. More on that below.)

I’m not a lawyer, so this isn’t legal advice. Even the HHS, who coin the term “Breach Safe Harbor” in their guidance portal, weasels out of saying they’re giving legal guidance by saying “The contents of this database lack the force and effect of law, except as authorized by law …”

Quality of encryption

You can’t just say that data were encrypted before they were breached. Weak encryption won’t cut it. You have to use acceptable algorithms and procedures.

How can you know whether you’ve encrypted data well enough to be covered Breach Safe Harbor? HHS cites four NIST publications for further guidance. (Not that I’m giving legal advice. I’m merely citing the HHS, who also is not giving legal advice.)

Here are the four publications.

Maybe encryption isn’t enough

At one point Tennessee law said a breach of encrypted data was still a breach. According to Dempsey and Carlin [1]

In 2016, Tennessee repealed its encryption safe harbor, requiring notice of breach of even encrypted data, but then in 2017, after criticism, the state restored a safe harbor for “information that has been encrypted in accordance with the current version of the Federal Information Processing Standard (FIPS) 140-2 if the encryption key has not been acquired by an unauthorized person.”

This is interesting for a couple reasons. First, there is a precedent for requiring notification of encrypted data. Second, this underscores the point above that encryption in general is not sufficient to avoid having to give notice of a breach: standard-compliant encryption is sufficient.

Consulting help

If you would like technical or statistical advice on how to prevent or prepare for a data breach, or how to respond after a data breach after the fact, we can help.

 

[1] Jim Dempsey and John P. Carlin. Cybersecurity Law Fundamentals, Second Edition.