Clearing up the confusion around Jacobi functions

Posted on 12 October 2018 by John

The Jacobi elliptic functions sn and cn are analogous to the trigonometric functions sine and cosine. The come up in applications such as nonlinear oscillations and conformal mapping. Unfortunately there are multiple conventions for defining these functions. The purpose of this post is to clear up the confusion around these different conventions.

Plot of Jacobi sn

The image above is a plot of the function sn [1].

Modulus, parameter, and modular angle

Jacobi functions take two inputs. We typically think of a Jacobi function as being a function of its first input, the second input being fixed. This second input is a “dial” you can turn that changes their behavior.

There are several ways to specify this dial. I started with the word “dial” rather than “parameter” because in this context parameter takes on a technical meaning, one way of describing the dial. In addition to the “parameter,” you could describe a Jacobi function in terms of its modulus or modular angle. This post will be a Rosetta stone of sorts, showing how each of these ways of describing a Jacobi elliptic function are related.

The parameter m is a real number in [0, 1]. The complementary parameter is m‘ = 1 – m. Abramowitz and Stegun, for example, write the Jacobi functions sn and cn as sn(u | m) and cn(u | m). They also use m₁ = rather than m‘ to denote the complementary parameter.

The modulus k is the square root of m. It would be easier to remember if m stood for modulus, but that’s not conventional. Instead, m is for parameter and k is for modulus. The complementary modulus k‘ is the square root of the complementary parameter.

The modular angle α is defined by m = sin² α.

Note that as the parameter m goes to zero, so does the modulus k and the modular angle α. As any one of these three goes to zero, the Jacobi functions sn and cn converge to their counterparts sine and cosine. So whether your dial is the parameter, modulus, or modular angle, sn converges to sine and cn converges to cosine as you turn the dial toward zero.

As the parameter m goes to 1, so does the modulus k, but the modular angle α goes to π/2. So if your dial is the parameter or the modulus, it goes to 1. But if you think of your dial as modular angle, it goes to π/2. In either case, as you turn the dial to the right as far as it will go, sn converges to hyperbolic secant, and cn converges to the constant function 1.

Quarter periods

In addition to parameter, modulus, and modular angle, you’ll also see Jacobi function described in terms of K and K‘. These are called the quarter periods for good reason. The functions sn and cn have period 4K as you move along the real axis, or move horizontally anywhere in the complex plane. They also have period 4iK‘. That is, the functions repeat when you move a distance 4K‘ vertically [2].

The quarter periods are a function of the modulus. The quarter period K along the real axis is

$K(m) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-m\sin^2\theta}}$

and the quarter period K‘ along the imaginary axis is given by K‘(m) = K(m‘) = K(1 – m).

The function K(m) is known as “the complete elliptic integral of the first kind.”

Amplitude

So far we’ve focused on the second input to the Jacobi functions, and three conventions for specifying it.

There are two conventions for specifying the first argument, written either as φ or as u. These are related by

$u = \int_0^{\varphi} \frac{d\theta}{\sqrt{1-m\sin^2\theta}}$

The angle φ is called the amplitude. (Yes, it’s an angle, but it’s called an amplitude.)

When we said above that the Jacobi functions had period 4K, this was in terms of the variable u. Note that when φ = π/2, u = K.

Jacobi elliptic functions in Mathematica

Mathematica uses the u convention for the first argument and the parameter convention for the second argument.

The Mathematica function JacobiSN[u, m] computes the function sn with argument u and parameter m. In the notation of A&S, sn(u | m).

Similarly, JacobiCN[u, m] computes the function cn with argument u and parameter m. In the notation of A&S, cn(u | m).

We haven’t talked about the Jacobi function dn up to this point, but it is implemented in Mathematica as JacobiDN[u, m].

The function that solves for the amplitude φ as a function of u is JacobiAmplitude[um m].

The function that computes the quarter period K from the parameter m is EllipticK[m].

Jacobi elliptic functions in Python

The SciPy library has one Python function that computes four mathematical functions at once. The function scipy.special.ellipj takes two arguments, u and m, just like Mathematica, and returns sn(u | m), cn(u | m), dn(u | m), and the amplitude φ(u, m).

The function K(m) is implemented in Python as scipy.special.ellipk.

[1] The plot was made using JacobiSN[0.5, z] and the function ComplexPlot described here.

[2] Strictly speaking, 4iK‘ is a period. It’s the smallest vertical period for cn, but 2iK‘ is the smallest vertical period for sn.

Projecting the globe onto regular solids

Posted on 22 August 2018 by John

I was playing around with some geographic features of Mathematica this morning and ran across an interesting example in the documentation for the PolyhedronProjection function given here. Here’s what you get when you project a map of the earth onto each of the five regular (Platonic) solids.

dodecahedron

icosahedron
octahedron
cube
tetrahedron

How the images were made

At first I right-clicked on each image and saved as graphic files. This produced low quality images, even when I saved as SVG. I got better resolution by using the Export command and specifying the ImageSize and ImageResolution options.

The default view of the tetrahedron was face-on, so it looked like a flat triangle. By changing the ViewPoint I was able to get something that’s more clearly three dimensional.

By the way, you can use PolyhedronProjection to project onto more exotic polyhedra than the Platonic solids. For example,

    Export["rhomb.png", 
        PolyhedronProjection["ParagyrateDiminishedRhombicosidodecahedron"], 
        ImageResolution -> 72, 
        ImageSize -> Large]

produces this:

ParagyrateDiminishedRhombicosidodecahedron

Sine of five degrees

Posted on 1 August 2018 by John

Today’s the first day of a new month, which means the exponential sum of the day will be simpler than usual. The exponential sum of the day plots the partial sums of

$\sum_{n=0}^N \exp\left( 2\pi i \left( \frac{n}{m} + \frac{n^2}{d} + \frac{n^3}{y} \right ) \right )$

where m, d, and y are the month, day, and (two-digit) year. The n/d term is simply n, and integer, when d = 1 and so it has no effect because exp(2πn) = 1. Here’s today’s sum, the plot formed by the partial sums above.

exponential sum for August 1, 2018

It’s possible in principle to work out exactly each of the points are on the curve. In order to do so, you’d need to compute

$\exp\left(\frac{2\pi i}{72}\right) = \cos\left(\frac{2\pi}{72}\right) + i \sin\left(\frac{2\pi}{72}\right) = \cos(5^\circ) + i \sin(5^\circ)$

Since we can use the Pythagorean theorem to compute sine from cosine and vice versa, we only need to know how to compute the sine of 5°.

OK, so how do we do that? The sine and cosine of 45° and of 30° are well known, and we can use the trig identity for sin(A – B) to find that the sine of 15° is (√3 – 1)/ 2√2. So how do we get from the sine of 15° to the sine of 5°? We can borrow a trick from medieval astronomers. They did something similar to our work above, then solved the trig identity

$\sin 3\theta = 3 \sin\theta - 4\sin^3\theta$

as a cubic equation in sin θ to find the sine of 1/3 of an angle with known sine.

If we ask Mathematica to solve this equation for us

    Solve[3 x -4 x^3 == Sin[15 Degree], x]

we get three roots, and it’s the middle one that we’re after. (We can evaluate the roots numerically and tell that the middle one has the right magnitude.) So Mathematica says that the sine of 5° is

$\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{-\sqrt{2}+\sqrt{6}+i \sqrt{8+4 \sqrt{3}}}}{4\ 2^{2/3}}+\frac{1-i \sqrt{3}}{2 \sqrt[3]{2 \left(-\sqrt{2}+\sqrt{6}+i \sqrt{8+4 \sqrt{3}}\right)}}$

Unfortunately, this is the end of the line for Mathematica. We know that the expression above must be real, but Mathematica is unable to find its real part. I tried running FullSimplify on this expression, and aborted evaluation after it ran for 20 minutes or so. If anybody wants to pick up where Mathematica left off, let me know your solution. It may be possible to continue with Mathematica by giving the software some help, or it may be easier to continue with pencil and paper.

Update: Based on the comments below, it seems this is about as far as we can go. This was a surprise to me. We’re solving a cubic equation, less than 5th degree, so the roots are expressible in terms of radicals. But that doesn’t mean we can eliminate the appearance of the i‘s, even though we know the imaginary part is zero.

System of Jacobi elliptic functions

Posted on 22 June 2018 by John

Jacobi’s elliptic functions are sorta like trig functions. His functions sn and cn have names that reminiscent of sine and cosine for good reason. These functions come up in applications such as the nonlinear pendulum (i.e. when θ is too
large to assume θ is a good enough approximation to sin θ) and in conformal mapping.

I ran across an article [1] yesterday that shows how Jacobi’s three elliptic functions—sn, cn, and dn—could be defined by one dynamical system

$\begin{eqnarray*} x' &=& yz \\ y' &=& -zx \\ z' &=& -k^2 xy \end{eqnarray*}$

with initial conditions x(0) = 0, y(0) = 1, and z(0) = 1.

The parameter k is the modulus. (In Mathematica’s notation below, k² is the parameter. See this post on parameterization conventions.) As k decreases to 0, sn converges to sine, cn to cosine, and dn to 1. As k increases to 1, sn converges tanh, and cn and dn converge to sech. So you could think of k as a knob you turn to go from being more like circular functions (ordinary trig functions) to more like hyperbolic functions.

Since we have a dynamical system, let’s plot the solution, varying the modulus each time.The Jacobi functions are periodic (in fact they’re doubly periodic) and so the plots will be closed loops.

f[t_, m_] = {JacobiSN[t, m], JacobiCN[t, m], JacobiDN[t, m]}
ParametricPlot3D[f[t, 0.707], {t, 0, 10}, AspectRatio -> 1]

phase portrait k = 1/2

ParametricPlot3D[f[t, 0.99], {t, 0, 20}, AspectRatio -> 1]

phase portrait k = 1/2

ParametricPlot3D[f[t, 0.01], {t, 0, 20}, AspectRatio -> 1]

phase portrait k = 1/2

Note that this last plot is nearly flat because the modulus is small and so z is nearly constant. The small modulus also makes the phase portrait nearly circular because x is approximately sine and y is approximately cosine.

[1] Kenneth R. Meyer. Jacobi Elliptic Functions from a Dynamical Systems Point of View. The American Mathematical Monthly, Vol. 108, No. 8 (Oct., 2001), pp. 729-737

Almost prime generators and almost integers

Posted on 13 June 2018 by John

Here are two apparently unrelated things you may have seen before. The first is an observation going back to Euler that the polynomial

produces a long sequence of primes. Namely, the values are prime for n = 1, 2, 3, …, 40.

The second is that the number

$e^{\pi \sqrt{163}}$

is extraordinarily close to an integer. This number is known as Ramanujan’s constant. It differs from its nearest integer by 3 parts in 10³⁰. Ramanujan’s constant equals

262537412640768743.99999999999925…

There is a connection between these two facts: The polynomial

returns primes for n = 1, 2, 3, …, k-1 primes if 4k – 1 is a Heegner number, and

$e^{\pi \sqrt{d}}$

is almost an integer if d is a (large) Heegner number.

Source: The Book of Numbers by Conway and Guy.

Heegner numbers

So what’s a Heegner number and how many are there? An integer d is a Heegner number if the ring generated by appending √-d to the integers has unique factorization. There are nine such numbers:

1, 2, 3, 7, 11, 19, 43, 67, 163.

There’s deeper stuff going on here than I understand—modular forms, the j-function, etc.—so this post won’t explain everything. There’s something unsatisfying about saying something is “almost” an integer without quantifying. There’s a way to be more precise, but we won’t go there. Instead, we’ll just play with the results.

Mathematica computation

First we look at the claim that n² – n + k produces primes for n = 1 through k – 1 if 4k – 1 is a Heegner number. The values of k such that 4k-1 is a Heegner number are 2, 3, 5, 11, and 17. The following code shows that the claim is true for these values of k.

k = {2, 3, 5, 11, 17}
claim[x_] := AllTrue[
  Table[n^2 - n + x, {n, x - 1}], 
  PrimeQ
]
AllTrue[k, claim]

This returns True, so the claim is true.

As for exp(π √d) being close to an integer, this apparently only true for the last three Heegner numbers.

h = {1, 2, 3, 7, 11, 19, 43, 67, 163}
For[i = 1, i < 10, i++, 
  Print[
    AccountingForm[
      N[
        Exp[ Pi Sqrt[ h[[i]] ] ], 
        31
      ]
    ]
  ]
]

(The function AccountingForm suppresses scientific notation, making it easier to see where the decimal portion of the number starts.)

Here are the results:

                23.1406926327792
                85.0196952232072
               230.7645883191458
              4071.9320952252610
             33506.1430655924387
            885479.7776801543194
         884736743.9997774660349
      147197952743.9999986624548
262537412640768743.9999999999993

I manually edited the output to align the decimal points and truncate the decimal places beyond that needed to show that the last number is not an integer.

Partition numbers and Ramanujan’s approximation

Posted on 12 June 2018 by John

The partition function p(n) counts the number of ways n unlabeled things can be partitioned into non-empty sets. (Contrast with Bell numbers that count partitions of labeled things.)

There’s no simple expression for p(n), but Ramanujan discovered a fairly simple asymptotic approximation:

$p(n) \sim \frac{1}{4n\sqrt{3}} \exp(\pi \sqrt{2n/3})$

How accurate is this approximation? Here’s a little Matheamtica code to see.

    p[n_] := PartitionsP[n]
    approx[n_] := Exp[ Pi Sqrt[2 n/3]] / (4 n Sqrt[3])
    relativeError[n_] := Abs[p[n] - approx[n]]/p[n]
    ListLinePlot[Table[relativeError[n], {n, 100}]]

So for values of n around 100, the approximation is off by about 5%.

Since it’s an asymptotic approximation, the relative error decreases (albeit slowly, apparently) as n increases. The relative error for n = 1,000 is about 1.4% and the relative error for n = 1,000,000 is about 0.044%.

Update: After John Baez commented on the oscillation in the relative error I decided to go back and look at it more carefully. Do the oscillations end or do they just become too small to see?

To answer this, let’s plot the difference in consecutive terms.

    relerr[a_, b_] := Abs[a - b]/b
    t = Table[relerr[p[n], approx[n]], {n, 300}];
    ListLinePlot[Table[ t[[n + 1]] - t[[n]], {n, 60}]]

first differences of relative error

The plot crosses back and forth across the zero line, indicating that the relative error alternately increases and decreases, but only up to a point. Past n = 25 the plot stays below the zero line; the sign changes in the first differences stop.

But now we see that the first differences themselves alternate! We can investigate the alternation in first differences by plotting second differences [1].

    ListLinePlot[
        Table[ t[[n + 2]] - 2 t[[n + 1]] + t[[n]], 
        {n, 25, 120}]
    ]

first differences of relative error

So it appears that the second differences keep crossing the zero line for a lot longer, so far out that it’s hard to see. In fact the second differences become positive and stay positive after n = 120. But the second differences keep alternating, so you could look at third differences …

See also: Special numbers

[1] The code does a small algebraic simplification that might make some people scratch their heads. All it does is simplify

(t_n+2 – t_n+1) – (t_n+1 – t_n).

Computing SVD and pseudoinverse

Posted on 5 May 2018 by John

In a nutshell, given the singular decomposition of a matrix A,

$A = U \Sigma V^*$

the Moore-Penrose pseudoinverse is given by

$A^+ = V \Sigma^+ U^*.$

This post will explain what the terms above mean, and how to compute them in Python and in Mathematica.

Singular Value Decomposition (SVD)

The singular value decomposition of a matrix is a sort of change of coordinates that makes the matrix simple, a generalization of diagonalization.

Matrix diagonalization

If a square matrix A is diagonalizable, then there is a matrix P such that

$A = P D P^{-1}$

where the matrix D is diagonal. You could think of P as a change of coordinates that makes the action of A as simple as possible. The elements on the diagonal of D are the eigenvalues of A and the columns of P are the corresponding eigenvectors.

Unfortunately not all matrices can be diagonalized. Singular value decomposition is a way to do something like diagonalization for any matrix, even non-square matrices.

Generalization to SVD

Singular value decomposition generalizes diagonalization. The matrix Σ in SVD is analogous to D in diagonalization. Σ is diagonal, though it may not be square. The matrices on either side of Σ are analogous to the matrix P in diagonalization, though now there are two different matrices, and they are not necessarily inverses of each other. The matrices U and V are square, but not necessarily of the same dimension.

The elements along the diagonal of Σ are not necessarily eigenvalues but singular values, which are a generalization of eigenvalues. Similarly the columns of U and V are not necessarily eigenvectors but left singular vectors and right singular vectors respectively.

The star superscript indicates conjugate transpose. If a matrix has all real components, then the conjugate transpose is just the transpose. But if the matrix has complex entries, you take the conjugate and transpose each entry.

The matrices U and V are unitary. A matrix M is unitary if its inverse is its conjugate transpose, i.e. M^* M = MM^* = I.

Pseudoinverse and SVD

The (Moore-Penrose) pseudoinverse of a matrix generalizes the notion of an inverse, somewhat like the way SVD generalized diagonalization. Not every matrix has an inverse, but every matrix has a pseudoinverse, even non-square matrices.

Computing the pseudoinverse from the SVD is simple.

$A = U \Sigma V^*$

then

$A^+ = V \Sigma^+ U^*$

where Σ⁺ is formed from Σ by taking the reciprocal of all the non-zero elements, leaving all the zeros alone, and making the matrix the right shape: if Σ is an m by n matrix, then Σ⁺ must be an n by m matrix.

We’ll give examples below in Mathematica and Python.

Computing SVD in Mathematica

Let’s start with the matrix A below.

$A = \begin{bmatrix} 2 & -1 & 0 \\ 4 & 3 & -2 \end{bmatrix}$

We can find the SVD of A with the following Mathematica commands.

    A = {{2, -1, 0}, {4, 3, -2}}
    {U, S, V} = SingularValueDecomposition[A]

From this we learn that the singular value decomposition of A is

$\begin{bmatrix} \frac{1}{\sqrt{26}} & -\frac{5}{\sqrt{26}} \\ \frac{5}{\sqrt{26}} & \frac{1}{\sqrt{26}} \\ \end{bmatrix} \begin{bmatrix} \sqrt{30} & 0 & 0 \\ 0 & 2 & 0 \end{bmatrix} \begin{bmatrix} \frac{11}{\sqrt{195}} & \frac{7}{\sqrt{195}} & -\sqrt{\frac{5}{39}} \\ -\frac{3}{\sqrt{26}} & 2 \sqrt{\frac{2}{13}} & -\frac{1}{\sqrt{26}} \\ \frac{1}{\sqrt{30}} & \sqrt{\frac{2}{15}} & \sqrt{\frac{5}{6}} \end{bmatrix}$

Note that the last matrix is not V but the transpose of V. Mathematica returns V itself, not its transpose.

If we multiply the matrices back together we can verify that we get A back.

    U . S. Transpose[V]

This returns

    {{2, -1, 0}, {4, 3, -2}}

as we’d expect.

Computing pseudoinverse in Mathematica

The Mathematica command for computing the pseudoinverse is simply PseudoInverse. (The best thing about Mathematica is it’s consistent, predictable naming.)

    PseudoInverse[A]

This returns

    {{19/60, 1/12}, {-(11/30), 1/6}, {1/12, -(1/12)}}

And we can confirm that computing the pseudoinverse via the SVD

    Sp = {{1/Sqrt[30], 0}, {0, 1/2}, {0, 0}}
    V . Sp . Transpose[U]

gives the same result.

Computing SVD in Python

Next we compute the singular value decomposition in Python (NumPy).

    >>> a = np.matrix([[2, -1, 0],[4,3,-2]])
    >>> u, s, vt = np.linalg.svd(a, full_matrices=True)

Note that np.linalg.svd returns the transpose of V, not the V in the definition of singular value decomposition.

Also, the object s is not the diagonal matrix Σ but a vector containing only the diagonal elements, i.e. just the singular values. This can save a lot of space if the matrix is large. The NumPy method svd has other efficiency-related options that I won’t go into here.

We can verify that the SVD is correct by turning s back into a matrix and multiply the components together.

    >>> ss = np.matrix([[s[0], 0, 0], [0, s[1], 0]])
    >>> u*ss*vt

This returns the matrix A, within floating point accuracy. Since Python is doing floating point computations, not symbolic calculation like Mathematica, the zero in A turns into -3.8e-16.

Note that the singular value decompositions as computed by Mathematica and Python differ in a few signs here and there; the SVD is not unique.

Computing pseudoinverse in Python

The pseudoinverse can be computed in NumPy with np.linalg.pinv.

    >>> np.linalg.pinv(a)
    matrix([[ 0.31666667,  0.08333333],
            [-0.36666667,  0.16666667],
            [ 0.08333333, -0.08333333]])

This returns the same result as Mathematica above, up to floating point precision.

Narcissus prime

Posted on 5 March 2012 by John

This morning Futility Closet posted the following.

Repeat the string 1808010808 1560 times, and tack on a 1 the end. The resulting 15601-digit number is prime, and because it’s a palindrome made up of the digits 1, 8, and 0, it remains prime when read backward, upside down, or in a mirror.

I used Mathematica to verify that the number described above is indeed prime.

PrimeQ[ 10*Sum[1808010808*10^(10 i), {i, 0, 1559}] + 1 ]

After a little over two minutes, the function returned True.

For daily tweets on algebra and other math, follow @AlgebraFact on Twitter.

Words that are primes base 36

Posted on 9 April 2011 by John

This morning on Twitter, Alexander Bogomolny posted a link to his article that gives examples of words that are prime numbers when interpreted as numbers in base 36. Some examples are “Brooklyn”, “paleontologist”, and “deodorant.” (Numbers in base 36 are written using 0, 1, 2, …, 9, A, B, C, …, Z as “digits.” )

Tim Hopper replied with a snippet of Mathematica code that lists all words with up to four letters that correspond to base 36 primes.

Rest[ Flatten[ Union[
    DictionaryLookup /@ IntegerString[
        Table[Prime[n], {n, 1, 300000}], 36]]]]

That made me wonder whether you could estimate how many such words there are without doing an exhaustive search.

The Prime Number Theorem says that the probability of a number less than N being prime is approximately 1/log(N). If we knew how many English words there were of a certain length, then we could guess that 1/log(N) of that those words would be prime when interpreted as base 36 numbers. This assumes that forming an English word and being prime have independent probabilities, which may be approximately true.

How well would our guess have worked on Tim’s example? He prints out all the words corresponding to the first 300,000 primes. The last of these primes is 4,256,233. The exact probability that a number less than that upper limit is prime is then

300,000 / 4,256,233 ≈ 0.07.

There are about 4200 English words with four or fewer letters. (I found this out by running

grep -ciE '^[a-z]{1,4}$'

on the words file on a Linux box. See similar tricks here.) If we estimate that 7% of these are prime, we’d expect 294 words from Tim’s program. His program produces 275 words, so our prediction is pretty good.

If we didn’t know the exact probability of a number in our range being prime, we could have estimated the probability at

1/log(4,256,233) ≈ 0.0655

using the Prime Number Theorem. Using this approximation we’d estimate 4200*0.0655 = 275.1 words; our estimate would be exactly correct! There’s good reason to believe our estimate would be reasonably close, but we got lucky to get this close.

Mathematica

Clearing up the confusion around Jacobi functions

Modulus, parameter, and modular angle

Quarter periods

Amplitude

Jacobi elliptic functions in Mathematica

Jacobi elliptic functions in Python

Related posts

Projecting the globe onto regular solids

How the images were made

More regular solid posts

Sine of five degrees

System of Jacobi elliptic functions

Almost prime generators and almost integers

Heegner numbers

Mathematica computation

Partition numbers and Ramanujan’s approximation

Computing SVD and pseudoinverse

Singular Value Decomposition (SVD)

Matrix diagonalization

Generalization to SVD

Pseudoinverse and SVD

Computing SVD in Mathematica

Computing pseudoinverse in Mathematica

Computing SVD in Python

Computing pseudoinverse in Python

More linear algebra posts

Narcissus prime

Related posts

Words that are primes base 36

More prime number posts

Modulus, parameter, and modular angle

Quarter periods

Amplitude

Jacobi elliptic functions in Mathematica

Jacobi elliptic functions in Python

Related posts

More group theory posts

How the images were made

More regular solid posts

Heegner numbers

Mathematica computation

Singular Value Decomposition (SVD)

Matrix diagonalization

Generalization to SVD

Pseudoinverse and SVD

Computing SVD in Mathematica

Computing pseudoinverse in Mathematica

Computing SVD in Python

Computing pseudoinverse in Python

More linear algebra posts

Related posts

More prime number posts