Body roundness index

Posted on by John

Body Roundness Index (BRI) is a proposed replacement for Body Mass Index (BMI) [1]. Some studies have found that BRI is a better measure of obesity and a more effective predictor of some of the things BMI is supposed to predict [2].

BMI is based on body mass and height, and so it cannot distinguish a body builder and an obese man if both have the same height and weight. BRI looks at body shape more than body mass.

The basic idea behind Body Roundness Index is to draw an ellipse based on a person’s body and report how close that ellipse is to being a circle. The more a person looks like a circle, higher his BRI. The formula for BRI is

BRI = 364.2 − 365.5 e

where e is the eccentricity of the ellipse.

Now what is this ellipse we’ve been talking about? It’s roughly an ellipse whose major axis runs from head to toe and whose minor axis runs across the person’s width.

There are a couple simplifications here.

  1. You don’t actually measure how wide someone is. You measure the circumference of their waist and find the diameter of a circle with that circumference.
  2. You don’t actually measure how high their waist is [3]. You assume their waist is at exactly half their height.

It’s conventional to describe an ellipse in terms of its semi-major axis a and semi-minor axis b. For a circle, a = b = radius. But in general an ellipse doesn’t have a single radius and a > b. You could think of a and b as being the maximum and minimum radii.

So to fit an ellipse to our idealized model of a person, the major axis, 2a, equals the person’s height.

a = h/2

The minor axis b is the radius of a circle of circumference c where c is the circumference of the person’s waist (or hips [3]).

b = c / 2π

As explained here, eccentricity is computed from a and b by

 

e = \sqrt{1 - \frac{b^2}{a^2}}

As an example, consider a man who is 6 foot (72 inches) tall and has a 34 inch waist. Then

a = 36
b = 17/π = 5.4112
e = √(1 − b²/a²) = 0.9886
BRI = 364.2 − 365.5 e = 2.8526

Note that the man’s weight doesn’t enter the calculation. He could be a slim guy weighing 180 pounds or a beefier guy weighing 250 pounds as long as he has a 34 inch waist. In the latter case, the extra mass is upper body muscle and not around his waist.

thin ellipse graph

Related posts

[1] Diana M. Thomas et al. Relationships Between Body Roundness with Body Fat and Visceral Adipose Tissue Emerging from a New Geometrical Model. Obesity (2013) 21, 2264–2271. doi:10.1002/oby.20408.

[2] Researchers argue over which number to reduce a person to: BMI, BRI, or some other measure. They implicitly agree that a person must be reduced to a number; they just disagree on which number.

[3] Or waist. There are two versions of BRI, one based on waist circumference and one based on hip circumference.

A couple more variations on an ancient theme

Posted on by John

I’ve written a couple posts on the approximation

\cos x \approx \frac{\pi^2 - 4x^2}{\pi^2 + x^2}

by the Indian astronomer Aryabhata (476–550). The approximation is accurate for x in [−π/2, π/2].

The first post collected a Twitter thread about the approximation into a post. The second looked at how far the coefficients in Aryabhata’s approximation are from the optimal approximation as a ratio of quadratics.

This post will answer a couple questions. First, what value of π did Aryabhata have and how would that effect the approximation error? Second, how bad would Aryabhata’s approximation be if we used the approximation π² ≈ 10?

Using Aryabhata’s value of π

Aryabhata knew the value 3.1416 for π. We know this because he said that a circle of diameter 20,000 would have circumference  62,832. We don’t know, but it’s plausible that he knew π to more accuracy and rounded it to the implied value.

Substituting 3.1416 for π changes the sixth decimal of the approximation, but the approximation is good to only three decimal places, so 3.1416 is as good as a more accurate approximation as far as the error in approximating cosine is concerned.

Using π² ≈ 10

Substituting 10 for π² in Aryabhata’s approximation gives an approximation that’s convenient to evaluate by hand.

\cos x \approx \frac{10 - 4x^2}{10 + x^2}

It’s very accurate for small values of x but the maximum error increases from 0.00163 to 0.01091. Here’s a plot of the error.

Finding pi in the alphabet

Posted on by John

Write the letters of the alphabet around a circle, then strike out the letters that are symmetrical about a vertical line. The remaining letters are grouped in clumps of 3, 1, 4, 1, and 6 letters.

I’ve heard that this observation is due to Martin Gardner, but I don’t have a specific reference.

In case you’re interested, here’s the Python script I wrote to make the image above.

    from numpy import *
    import matplotlib.pyplot as plt
    
    for i in range(26):
        letter = chr(ord('A') + i)
        if letter in "AHIMOTUVWXY":
            latex = r"$\equiv\!\!\!\!\!" + letter + "$"
        else:
            latex = f"${letter}$"
        theta = pi/2 - 2*pi*i/26
        pt = (0.5*cos(theta), 0.5*sin(theta))
        plt.plot(pt[0], pt[1], ' ')
        plt.annotate(latex, pt, fontsize="xx-large")
    plt.axis("off")
    plt.gca().set_aspect("equal")
    plt.savefig("alphabet_pi.png")

Optimal rational approximation

Posted on by John

A few days ago I wrote about the approximation

\cos x \approx \frac{\pi^2 - 4x^2}{\pi^2 + x^2}

for cosine due to the Indian astronomer Aryabhata (476–550) and gave this plot of the error.

I said that Aryabhata’s approximation is “not quite optimal since the ripples in the error function are not of equal height.” This was an allusion to the equioscillation theorem.

Chebyshev proved that an optimal polynomial approximation has an error function that has equally large positive and negative oscillations. Later this theorem was generalized to rational approximations through a sequence of results by de la Vallée Poussin, Walsh, and Achieser. Here’s the formal statement of the theorem from [1] in the context of real-valued rational approximations with numerators of degree m and denominators of degree n

Theorem 24.1. Equioscillation characterization of best approximants. A real function f in C[−1, 1] has a unique best approximation r*, and a function r is equal to r* if and only if f − r equioscillates between at least m + n + 2 − d extremes where d is the defect of r.

When the theorem says the error equioscillates, it means the error alternately takes on ± its maximum absolute value.

The defect is non-zero when both numerator and denominator have less than maximal degree, which doesn’t concern us here.

We want to find the optimal rational approximation for cosine over the interval [−π/2, π/2]. It doesn’t matter that the theorem is stated for continuous functions over [−1, 1] because we could just rescale cosine. We’re looking for approximations with (m, n) = (2, 2), i.e. ratios of quadratic polynomials, to see if we can improve on the approximation at the top of the post.

The equioscillation theorem says our error should oscillate at least 6 times, and so if we find an approximation whose error oscillates as required by the theorem, we know we’ve found the optimal approximation.

I first tried finding the optimal approximation using Mathematica’s MiniMaxApproximation function. But this function tries to optimize relative error and I’m trying to minimize absolute error. Minimizing relative error creates problems because cosine evaluates to zero at the ends of interval ±π/2. I tried several alternatives and eventually decided to take another approach.

Because the cosine function is even, the optimal approximation is even. Which means the optimal approximation has the form

(a + bx²) / (c + dx²)

and we can assume without loss of generality that a = 1. I then wrote some Python code to minimize the error as a function of the three remaining variables. The results were b = −4.00487004, c = 9.86024544, and d = 1.00198695, very close to Aryabhata’s approximation that corresponds to b = −4, c = π², and d = 1.

Here’s a plot of the error, the difference between cosine and the rational approximation.

The absolute error takes on its maximum value seven times, alternating between positive and negative values, and so we know the approximation is optimal. However sketchy my approach to finding the optimal approximation may have been, the plot shows that the result is correct.

Aryabhata’s approximation had maximum error 0.00163176 and the optimal approximation has maximum error 0.00097466. We were able to shave about 1/3 off the maximum error, but at a cost of using coefficients that would be harder to use by hand. This wouldn’t matter to a modern computer, but it would matter a great deal to an ancient astronomer.

Related posts

[1] Approximation Theory and Approximation Practice by Lloyd N. Trefethen

Pell is to silver as Fibonacci is to gold

Posted on by John

As mentioned in the previous post, the ratio of consecutive Fibonacci numbers converges to the golden ratio. Is there a sequence whose ratios converge to the silver ratio the way ratios of Fibonacci numbers converge to the golden ratio?

(If you’re not familiar with the silver ratio, you can read more about it here.)

The Pell numbers Pn start out just like the Fibonacci numbers:

P0 = 0
P1 = 1.

But the recurrence relationship is slightly different:

Pn+2 = 2Pn+1 + Pn.

So the Pell numbers are 0, 1, 2, 5, 12, 29, ….

The ratios of consecutive Pell numbers converge to the silver ratio.

Metallic ratios

There are more analogs of the golden ratio, such as the bronze ratio and more that do not have names. In general the kth metallic ratio is the larger root of

x² − kx − 1 = 0.

The cases n = 1, 2, and 3 correspond to the gold, silver, and bronze ratios respectively.

The quadratic equation above is the characteristic equation of the recurrence relation

Pn+2 = kPn+1 + Pn.

which suggests how we construct a sequence of integers such that consecutive ratios converge to the nth metallic constant.

So if we use k = 3 in the recurrence relation, we should get a sequence whose ratios converge to the bronze ratio. The results are

0, 1, 3, 10, 33, 109, 360, 1189, 3927, 12970, …

The following code will print the ratios.

    def bronze(n):
        if n == 0: return 0
        if n == 1: return 1
        return 3*bronze(n-1) + bronze(n-2)

    for n in range(2, 12):
        print( bronze(n)/bronze(n-1) )

Here’s the output.

    3.0
    3.3333333333333335
    3.3
    3.303030303030303
    3.302752293577982
    3.3027777777777776
    3.302775441547519
    3.3027756557168324
    3.302775636083269
    3.3027756378831383

The results are converging to the bronze ratio

(3 + √13)/2 = 3.302775637731995.

Plastic ratio

The plastic ratio is the real root of x³ − x − 1 = 0. Following the approach above, we can construct a sequence of integers whose consecutive ratios converge to the plastic ratio with the recurrence relation

Sn+3 = Sn+1 + Sn

Let’s try this out with a little code.

    def plastic(n):
        if n < 3: return n
        return plastic(n-2) + plastic(n-3)

    for n in range(10, 20):
        print( plastic(n)/plastic(n-1) )

This prints

    1.3
    1.3076923076923077
    1.3529411764705883
    1.3043478260869565
    1.3333333333333333
    1.325
    1.320754716981132
    1.3285714285714285
    1.3225806451612903

which shows the ratios are approaching the plastic constant 1.324717957244746.

Related posts

Miles to kilometers

Posted on by John

The number of kilometers in a mile is k = 1.609344 which is close to the golden ratio φ = 1.6180334.

The ratio of consecutive Fibonacci numbers converges to φ, and so you can approximately convert miles to kilometers by multiplying by a Fibonacci number and dividing by the previous Fibonacci number. For example, you could multiply by 8 and divide by 5, or you could multiply by 13 and divide by 8.

As you start going down the Fibonacci sequence, consecutive ratios get closer to k and closer to φ. But since the ratios converge to φ, at some point the ratios get closer to φ and further from k. That means there’s an optimal Fibonacci ratio for converting miles to kilometers.

I was curious what this optimal ratio is, and it turns out to be 21/13. There we have

|k − 21/13| = 0.0060406

and so the error in the approximation is 0.375%. The error is about a third smaller than using φ as the conversion factor.

The Lucas numbers satisfy the same recurrence relation as the Fibonacci numbers, but start with L0 = 2 and L1 = 1. The ratio of consecutive Lucas numbers also converges to φ, and so you could also use Lucas numbers to convert miles to kilometers.

There is an optimal Lucas ratio for converting miles to kilometers for the same reasons there is an optimal Fibonacci ratio. That ratio turns out to be 29/18, and

|k − 29/18| = 0.001767

which is about 4 times more accurate than the best Fibonacci ratio.

Ancient accurate approximation for sine

Posted on by John

This post started out as a Twitter thread. The text below is the same as that of the thread after correcting an error in the first part of the thread. I also added a footnote on a theorem the thread alluded to.

***

The following approximation for sin(x) is remarkably accurate for 0 < x < π.

\sin(x) \approx \frac{16x(\pi - x)}{5\pi^2 - 4x(\pi - x)}

The approximation is so good that you can’t see the difference between the exact value and the approximation until you get outside the range of the approximation.

Here’s a plot of just the error.

This is a very old approximation, dating back to Aryabhata I, around 500 AD.

In modern terms, it is a rational approximation, quadratic in the numerator and denominator. It’s not quite optimal since the ripples in the error function are not of equal height [1], but the coefficients are nice numbers.

***

As pointed out in the comments, replacing x with π/2 − x in order to get an approximation for cosine gives a much nicer equation.

\cos x \approx \frac{\pi^2 - 4x^2}{\pi^2 + x^2}

***

[1] The equioscillation theorem says that the optimal approximation will have ripples of equal positive and negative amplitude. This post explores the equioscillation theorem further and finds how far Aryabhata’s is from optimal.

Mentally multiply by π

Posted on by John

This post will give three ways to multiply by π taken from [1].

Simplest approach

Here’s a very simple observation about π :

π ≈ 3 + 0.14 + 0.0014.

So if you need to multiply by π, you need to multiply by 3 and by 14. Once you’ve multiplied by 14 once, you can reuse your work.

For example, to compute 4π, you’d compute 4 × 3 = 12 and 4 × 14 = 56. Then

4π ≈ 12 + 0.56 + 0.0056 = 12.5656.

The correct value is 12.56637… and so the error is .00077.

First refinement

Now of course π = 3.14159… and so the approximation above is wrong in the fourth decimal place. But you can squeeze out a little more accuracy with the observation

π ≈ 3 + 0.14 + 0.0014 + 0.00014 = 3.14154.

Now if we redo our calculation of 4π we get

4π ≈ 12 + 0.56 + 0.0056 + 0.00056 = 12.56616.

Now our error is .00021, which is 3.6 times smaller.

Second refinement

The approximation above is based on an underestimate of π. We can improve it a bit by adding half of our last term, based on

π ≈ 3 + 0.14 + 0.0014 + 0.00014 + 0.00014/2 = 3.14157

So in our running example,

4π ≈ 12 + 0.56 + 0.0056 + 0.00056 + 00028 = 12.5656 = 12.56654.

which has an error of 0.00007, which is three times smaller than above.

Related posts

[1] Trevor Lipscombe. Mental mathematics for multiples of π. The Mathematical Gazette, Vol. 97, No. 538 (March 2013), pp. 167–169

A better integral for the normal distribution

Posted on by John

For a standard normal random variable Z, the probability that Z exceeds some cutoff z is given by

\mbox{Prob}(Z \geq z) = Q(z) = \frac{1}{\sqrt{2\pi}} \int_z^\infty \exp(-x^2/2)\, dx

If you wanted to compute this probability numerically, you could obviously evaluate its defining integral numerically. But as is often the case in numerical analysis, the most obvious approach is not the best approach. The range of integration is unbounded and it varies with the argument.

J. W. Craig [1] came up with a better integral representation, better from the perspective of numerical integration. The integration is always over the same finite interval, with the argument appearing inside the integrand. The integrand is smooth and bounded, well suited to numerical integration.

For positive z, Craig’s integer representation is

Q(z) = \frac{1}{\pi} \int_0^{\pi/2} \exp\left( -\frac{z^2}{2\sin^2 \theta} \right) \, d\theta

Illustration

To show that the Craig’s integral is easy to integrate numerically, we’ll evaluate it using Gaussian quadrature with only 10 integration points.

    from numpy import sin, exp, pi
    from scipy import integrate
    from scipy.stats import norm

    for x in [0.5, 2, 5]:
        q, _ = integrate.fixed_quad(
            lambda t: exp(-x**2 / (2*sin(t)**2))/pi,
            0.0, pi/2, n=10)
        print(q, norm.sf(x))

(SciPy uses sf (“survival function”) for the CCDF. More on that here.)

The code above produces the following.

    0.30858301 0.30853754
    0.02274966 0.02275013
    2.86638437e-07 2.86651572e-07

So with 10 integration points, we get four correct figures. And the accuracy seems to be consistent for small, medium, and large values of x. (Five standard deviations is pretty far out in the tail of a normal distribution, as evidenced by the small value of the integral.)

Related posts

[1] J. W. Craig, A new, simple and exact result for calculating the probability of error for two-dimensional signal constellations, in TEEE MILCOM’91 Conf. Rec., Boston, MA (1991) рр. 25.2.1-25.5.5.

Drawing with a compass on a globe

Posted on by John

Take a compass and draw a circle on a globe. Then take the same compass, opened to the same width, and draw a circle on a flat piece of paper. Which circle has more area?

If the circle is small compared to the radius of the globe, then the two circles will be approximately equal because a small area on a globe is approximately flat.

To get an idea what happens for larger circles, let’s a circle on the globe as large as possible, i.e. the equator. If the globe has radius r, then to draw the equator we need our compass to be opened a width of √2 r, the distance from the north pole to the equator along a straight line cutting through the globe.

The area of a hemisphere is 2πr². If we take our compass and draw a circle of radius √2 r on a flat surface we also get an area of 2πr². And by continuity we should expect that if we draw a circle that is nearly as big as the equator then the corresponding circle on a flat surface should have approximately the same area.

Interesting. This says that our compass will draw a circle with the same area whether on a globe or on a flat surface, at least approximately, if the width of the compass sufficiently small or sufficiently large. In fact, we get exactly the same area, regardless of how wide the compass is opened up. We haven’t proven this, only given a plausibility argument, but you can find a proof in [1].

Note that the width w of the compass is the radius of the circle drawn on a flat surface, but it is not the radius of the circle drawn on the globe. The width w is greater than the radius of the circle, but less than the distance along the sphere from the center of the circle. In the case of the equator, the radius of the circle is r, the width of the compass is √2 r , and the distance along the sphere from the north pole to the equator is πr/2.

Related posts

[1] Nick Lord. On an alternative formula for the area of a spherical cap. The Mathematical Gazette, Vol. 102, No. 554 (July 2018), pp. 314–316