Relating Airy and Bessel functions

The Airy functions Ai(x) and Bi(x) are independent solutions to the differential equation

y'' - xy = 0

For negative x they act something like sin(x) and cos(x). For positive x they act something like exp(x) and exp(-x). This isn’t surprising if you look at the differential equation. If you replace x with a negative constant, you get sines and cosines, and if you replace it with a positive constant, you get positive and negative exponentials.

The Airy functions can be related to Bessel functions as follows:

\mathrm{Ai}(x) = \left\{ \begin{array}{ll} \frac{1}{3}\sqrt{\phantom{-}x} \left(I_{-1/3}(\hat{x}) - I_{1/3}(\hat{x})\right) & \mbox{if } x > 0 \\<br /><br /><br /><br /> \\<br /><br /><br /><br /> \frac{1}{3}\sqrt{-x} \left(J_{-1/3}(\hat{x}) + J_{1/3}(\hat{x})\right) & \mbox{if } x < 0 \end{array} \right.


\mathrm{Bi}(x) = \left\{ \begin{array}{ll} \sqrt{\phantom{-}x/3} \left(I_{-1/3}(\hat{x}) + I_{1/3}(\hat{x})\right) & \mbox{if } x > 0 \\<br /> \\<br /> \sqrt{-x/3} \left(J_{-1/3}(\hat{x}) - J_{1/3}(\hat{x})\right) & \mbox{if } x < 0 \end{array} \right.

Here J is a “Bessel function of the first kind” and I is a “modified Bessel function of the first kind.” Also

\hat{x} = \frac{2}{3} \left(\sqrt{|x|}\right)^3

To verify the equations above, and to show how to compute these functions in Python, here’s some code.

The SciPy function airy computes both functions, and their first derivatives, at once. I assume that’s because it doesn’t take much longer to compute all four functions than to compute one. The code for Ai2 and Bi2 below uses np.where instead of if … else so that it can operate on NumPy vectors all at once. You can plot Ai and Ai2 and see that the two curves lie on top of each other. The same holds for Bi and Bi2.


from scipy.special import airy, jv, iv
from numpy import sqrt, where

def Ai(x):
    (ai, ai_prime, bi, bi_prime) = airy(x)
    return ai

def Bi(x):
    (ai, ai_prime, bi, bi_prime) = airy(x)
    return bi

def Ai2(x):
    third = 1.0/3.0
    hatx = 2*third*(abs(x))**1.5
    return where(x > 0,
        third*sqrt( x)*(iv(-third, hatx) - iv(third, hatx)),
        third*sqrt(-x)*(jv(-third, hatx) + jv(third, hatx)))

def Bi2(x):
    third = 1.0/3.0
    hatx = 2*third*(abs(x))**1.5
    return where(x > 0,
        sqrt( x/3.0)*(iv(-third, hatx) + iv(third, hatx)),
        sqrt(-x/3.0)*(jv(-third, hatx) - jv(third, hatx)))


There is a problem with Ai2 and Bi2: they return nan at 0. A more careful implementation would avoid this problem, but that’s not necessary since these functions are only for illustration. In practice, you’d simply use airy and it does the right thing at 0.

Related links:

SymPy book

There’s a new book on SymPy, a Python library for symbolic math.

The book is Instant SymPy Starter by Ronan Lamy. As far as I know, this is the only book just on SymPy. It’s only about 50 pages, which is nice. It’s not a replacement for the online documentation but just a quick start guide.

The online SymPy documentation is good, but I think it would be easier to start with this book. And although I’ve been using SymPy off and on for a while, I learned a few things from the book.


Need a 12-digit prime?

You may have seen the joke “Enter any 12-digit prime number to continue.” I’ve seen it floating around as the punchline in several contexts.

So what do you do if you need a 12-digit prime? Here’s how to find the smallest one using Python.

>>> from sympy import nextprime
>>> nextprime(10**11)

The function nextprime gives the smallest prime larger than its argument. (Note that the smallest 12-digit number is 1011, not 1012. Great opportunity for an off-by-one mistake.)

Optionally you can provide an addition argument to nextprime to get primes further down the list. For example, this gives the second prime larger than 1011.

>>> nextprime(10**11, 2)

What if you wanted the largest 12-digit prime rather than the smallest?

>>> from sympy import prevprime
>>> prevprime(10**12)

Finally, suppose you want to know how many 12-digit primes there are. SymPy has a function primepi that returns the number of primes less than its argument. Unfortunately, it fails for large arguments. It works for arguments as big as 2**27 but throws a memory error for 2**28.

The number of primes less than n is approximately n / log n, so there are about 32 billion primes between 1011 and 1012. According to Wolfram Alpha, the exact number of 12-digit primes is 33,489,857,205. So if you try 12-digit numbers at random, your chances are about 1 in 30 of getting a prime. If you’re clever enough to just pick odd numbers, your chances go up to 1 in 15.

* * *

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Synchronizing cicadas with Python

Suppose you want to know when your great-grandmother was born. You can’t find the year recorded anywhere. But you did discover an undated letter from her father that mentions her birth and one curious detail:  the 13-year and 17-year cicadas were swarming.

You do a little research and find that the 13-year cicadas are supposed to come out next year, and that the 17-year cicadas came out last year. When was your great-grandmother born?

Since 13 and 17 are relatively prime, the 13-year and 17-year cicadas will synchronize their schedules every 13 × 17 = 221 years. Suppose your great-grandmother was born n years ago. The remainder when n is divided by 13 must be 12, and the remainder when n is divided by 17 must be 1. We have to solve the pair of congruences n = 12 mod 13 and n = 1 mod 17. The Chinese Remainder Theorem says that this pair of congruences has a solution, and the proof of the theorem suggests an algorithm for computing the solution.

The Python library SymPy has a function for solving linear congruences.

>>> from sympy.ntheory.modular import solve_congruence
>>> solve_congruence( (12, 13), (1, 17) )
(103, 221)

This says we’re 103 years into the joint cycle of the 13-year and 17-year cicadas. So your great-grandmother was born 103 years ago. (The solution 324 = 103 + 221 is also mathematically possible, but not biologically possible.)

You can use the same SymPy function to solve systems of more congruences. For example, when is the next year in which there will be summer Olympic games and the 13-year and and 17-year cicadas will swarm? Here are a couple ways to approach this. First, you could find the last time this happened, then find when it will happen next. You’d need to solve n = 1 mod 4 (since we had summer Olympics last year) and  n = 12 mod 13 and n = 1 mod 17.

>>> solve_congruence( (1, 4), (12, 13), (1, 17) )
(545, 884)

So the cicadas and the summer Olympics were in sync 545 years ago. (Well, they would have been if the modern Olympics had existed in the middle ages.) This says they’ll be in sync again in 885 – 545 = 339 years.

Here’s a more direct approach. We want to know when the summer Olympics will be 3 years ahead of where they are now in the cycle, when the 13-year cicadas will be 1 year ahead, and the 17-year cicadas will be 16 years ahead.

>>> solve_congruence( (3, 4), (1, 13), (16, 17) )
(339, 884)

By the way, you can use negative integers with the congruences, so you could have used (-1, 17) to say the 17-year cicadas will be 1 year back instead of 16 years ahead in their cycle.

Related: Applied number theory

Recognizing numbers

I was playing around with SymPy, a symbolic math package for Python, and ran across nsimplify. It takes a floating point number and tries to simplify it: as a fraction with a small denominator, square root of a small integer, an expression involving famous constants, etc.

For example, suppose some calculation returned 4.242640687119286 and you suspect there’s something special about that number. Here’s how you might test where it came from.

>>> from sympy import *
>>> nsimplify(4.242640687119286)

Maybe you do a calculation numerically, find a simple expression for the result, and that suggests an analytical solution.

I think a more common application of nsimplify might be to help you remember half-forgotten formulas. For example, maybe you’re rusty on your trig identities, but you remember that cos(π/6) is something special.

>>> nsimplify(cos(pi/6))

Or to take a more advanced example, suppose that you vaguely remember that the gamma function takes on recognizable values at half integer values, but you don’t quite remember how. Maybe something involving π or e. You can suggest that nsimplify include expressions with π and e in its search.

>>> nsimplify(gamma(3.5), constants=[pi, E])

You can also give nsimplify a tolerance, asking it to find a simple representation within a neighborhood of the number. For example, here’s a way to find approximations to π.

>>> nsimplify(pi, tolerance=1e-5)

With a wider tolerance, it will return a simpler approximation.

>>> nsimplify(pi, tolerance=1e-2)

Finally, here’s higher precision approximation to π that isn’t exactly simple:

>>> nsimplify(pi, tolerance=1e-7)
exp(141/895 + sqrt(780631)/895)

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Rolling dice for normal samples: Python version

A handful of dice can make a decent normal random number generator, good enough for classroom demonstrations. I wrote about this a while ago.

My original post included Mathematica code for calculating how close to normal the distribution of the sum of the dice is. Here I’d like to redo the code in Python to show how to do the same calculations using SymPy. [Update: I’ll also give a solution that does not use SymPy and that scales much better.]

If you roll five dice and add up the spots, the probability of getting a sum of k is the coefficient of xk in the expansion of

(x + x2 + x3 + x4 + x5 + x6)5 / 65.

Here’s code to find the probabilities by expanding the polynomial and taking coefficients.

from sympy import Symbol

sides = 6
dice = 5
rolls = range( dice*sides + 1 )

# Tell SymPy that we want to use x as a symbol, not a number
x = Symbol('x')

# p(x) = (x + x^2 + ... + x^m)^n
# where m = number of sides per die
# and n = number of dice
p = sum([x**i for i in range(1, sides + 1)])**dice

# Extract the coefficients of p(x) and divide by sides**dice
pmf = [sides**(-dice) * p.expand().coeff(x, i) for i in rolls]

If you’d like to compare the CDF of the dice sum to a normal CDF you could add this.

from scipy import array, sqrt
from scipy.stats import norm

cdf = array(pmf).cumsum()

# Normal CDF for comparison
mean = 0.5*(sides + 1)*dice
variance = dice*(sides**2 -1)/12.0
temp = [norm.cdf(i, mean, sqrt(variance)) for i in roles]
norm_cdf = array(temp)

diff = abs(cdf - norm_cdf)
# Print the maximum error and where it occurs
print diff.max(), diff.argmax()

Question: Now suppose you want a better approximation to a normal distribution. Would it be better to increase the number of dice or the number of sides per dice? For example, would you be better off with 10 six-sided dice or 5 twelve-sided dice? Think about it before reading the solution.

Update: The SymPy code does not scale well. When I tried the code with 50 six-sided dice, it ran out of memory. Based on Andre’s comment, I rewrote the code using polypow. SymPy offers much more symbolic calculation functionality than NumPy, but in this case NumPy contains all we need. It is much faster and it doesn’t run out of memory.

from numpy.polynomial.polynomial import polypow
from numpy import ones

sides = 6
dice = 100

# Create an array of polynomial coefficients for
# x + x^2 + ... + x^sides
p = ones(sides + 1)
p[0] = 0

# Extract the coefficients of p(x)**dice and divide by sides**dice
pmf = sides**(-dice) * polypow(p, dice)
cdf = pmf.cumsum()

That solution works for up to 398 dice. What’s up with that? With 399 dice, the largest polynomial coefficient overflows. If we divide by the number of dice before raising the polynomial to the power dice, the code becomes a little simpler and scales further.

p = ones(sides + 1)
p[0] = 0
p /= sides
pmf = polypow(p, dice)
cdf = pmf.cumsum()

I tried this last approach on 10,000 dice with no problem.

* * *

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Moments of mixtures

I needed to compute the higher moments of a mixture distribution for a project I’m working on. I’m writing up the code here in case anyone else finds this useful. (And in case I’ll find it useful in the future.) I’ll include the central moments first. From there it’s easy to compute skewness and kurtosis.

Suppose X is a mixture of n random variables Xi with weights wi, non-negative numbers adding to 1. Then the jth central moment of X is given by

E[(X - \mu)^j] = \sum_{i=1}^n \sum_{k=0}^j {j \choose k} (\mu_i - \mu)^{j-k} w_i E[(X_i- \mu_i)^k]

where μi is the mean of Xi.

In my particular application, I’m interested in a mixture of normals and so the code below computes the moments for a mixture of normals. It could easily be modified for other distributions.

from scipy.misc import factorialk, comb

def mixture_central_moment(mixture, moment):

    '''Compute the higher moments of a mixture of normal rvs.
    mixture is a list of (mu, sigma, weight) triples.
    moment is the central moment to compute.'''

    mix_mean = sum( [w*m for (m, s, w) in mixture] )

    mixture_moment = 0.0
    for triple in mixture:
        mu, sigma, weight = triple
        for k in range(moment+1):
            prod = comb(moment, k) * (mu-mix_mean)**(moment-k)
            prod *= weight*normal_central_moment(sigma, k)
            mixture_moment += prod

    return mixture_moment

def normal_central_moment(sigma, moment):

    '''Central moments of a normal distribution'''

    if moment % 2 == 1:
        return 0.0
        # If Z is a std normal and n is even, E(Z^n) == (n-1)!!
        # So E (sigma Z)^n = sigma^n (n-1)!!
        return sigma**moment * factorialk(moment-1, 2)

Once we have code for central moments, it’s simple to add code for computing skewness and kurtosis.

def mixture_skew(mixture):

    variance = mixture_central_moment(mixture, 2)
    third = mixture_central_moment(mixture, 3)
    return third / variance**(1.5)

def mixture_kurtosis(mixture):

    variance = mixture_central_moment(mixture, 2)
    fourth = mixture_central_moment(mixture, 4)
    return fourth / variance**2 - 3.0

Here’s an example of how the code might be used.

# Test on a mixture of 30% Normal(-2, 1) and 70% Normal(1, 3)
mixture = [(-2, 1, 0.3), (1, 3, 0.7)]

print "Skewness = ", mixture_skew(mixture)
print "Kurtosis = ", mixture_kurtosis(mixture)

Related post: General formula for normal moments

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New introduction to SciPy

The Python stack for scientific computing is more modular than say R or Mathematica. Python is a general-purpose programming language that has libraries for scientific computing. R and Mathematica are statistical and mathematical programming languages that have general-purpose features. The Python approach has its advantages — I’d rather do math in a general language than do general programming in a mathematical language — but it takes longer to learn. The components of the Python stack work well together, but someone new to Python has to discover what components they’ll need.

Several books have come out recently to help someone learn Python and the components for numerical computing. The latest is Learning SciPy for Numerical and Scientific Computing by Francisco J. Blanco-Silva.

This book covers the things you’d expect, including SciPy, NumPy, and Matplotlib. The only exception may be IPython. But no book can cover everything. And since IPython is an environment more than a library, it makes some sense to leave it out.

In addition to the usual topics, the book includes several important topics that are not as commonly covered. For example, it devotes a good amount of space to special functions and integration in a chapter on numerical analysis. I share the author’s assessment that this is “one of the most interesting chapters in the book.”

There are three chapters on more specific applications: signal processing, data mining, and computational geometry. These chapters give an introduction to their topics as well as how to carry out computations in SciPy.

The final chapter of the book is on integration with other programming languages.

Learning SciPy for Numerical and Scientific Computing covers the material you need to get going. It’s easy to read, and still fairly small: 150 pages total, about 130 pages of content. This is the right size for such a book in my opinion. There’s plenty of online documentation for more details, but it helps to have a good overview such as this book before diving into reference material.

* * *

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Suffix primes

MathUpdate tweeted this afternoon that

Any number made by removing the first n digits of 646216567629137 is still prime.

and links to sequence A012885 in the Online Encyclopedia of Integer Sequences (OEIS). The OEIS heading for the sequence is

Suffixes of 357686312646216567629137 (all primes)

which implies you can start with an even larger number, cutting off the first digit each time and producing a sequence of primes.

The following Python code verifies that this is indeed the case.

    from sympy.ntheory import isprime

    x = "357686312646216567629137"

    while x:
        print isprime(int(x))
        x = x[1:]

Update: lucio wrote a program to show that the prime given here is the longest one with the suffix property.

    def extend_prime(n, result):
        for i in range(10):
            nn = int(str(i) + str(n))
            if nn == n: continue
            if isprime(nn):
                extend_prime(nn, result)
        return result        

    print "Max Prefix Prime:", max(extend_prime("", []))

One minor suggestion: by using range(1, 10) rather than range(10) above, i.e. eliminating 0, the line if nn == n: continue could be eliminated.

Instead of calling max, you could call len to find that there are 4260 suffix primes.

Here’s a list of all suffix primes created by the code above and sorting the output.

Other novelty primes:

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Wallpaper and phase portraits

Suppose you want to create a background image that tiles well. You’d like it to be periodic horizontally and vertically so that there are no obvious jumps when the image repeats.

Functions like sine and cosine are period along the real line. But if you want to make a two-dimensional image by extending the sine function to the complex plane, the result is not periodic along the imaginary axis but exponential.

There are functions that are periodic horizontally and vertically. If you restrict your attention to functions that are analytic except at poles, these doubly-periodic functions are elliptic functions, a class of functions with remarkable properties. See this post if you’re interested in the details. Here we’re just after pretty wallpaper. I’ll give Python code for creating the wallpaper.

Here I’ll take a particular elliptic function sn(x). This is one of the Jacobi elliptic functions, somewhat analogous to the sine function, and use its phase portrait. Phase portraits use hue to encode the phase of a complex number, the θ value when a complex number is written in polar coordinates. The brightness of the color indicates the magnitude, the r value in polar coordinates.

Here’s the plot of sn(z, 0.2). (The sn function takes a parameter m that I arbitrarily chose as 0.2.) The plot shows two periods, horizontally and vertically. I included two periods so you could more easily see how it repeats. If you wanted to use this image as wallpaper, you could use 1/4 of the image, one period in each direction, to get by with a smaller image.

phase plot of sn(z, 0.2) - 0.2

Here’s the Python code that was used to create the image.

from mpmath import cplot, ellipfun, ellipk
sn = ellipfun('sn')
m = 0.2
x = 4*ellipk(m) # horizontal period
y = 2*ellipk(1-m) # vertical period
cplot(lambda z: sn(z, m) - 0.2, [0, 2*x], [0, 2*y], points = 100000)

I subtracted 0.2 from sn just to shift the color a little. Adding a positive number shifts the color toward red. Subtracting a positive number shifts the color toward blue. You could also multiply by some constant to increase or decrease the brightness.

You could also play around with other elliptic functions, described in the mpmath documentation here. And you can find more on cplot here. For example, you could supply your own function for how phase is mapped to color. The saturated colors used by default are good for mathematical visualization, but more subtle colors could be better for aesthetics.

If you make some interesting images, leave a comment with a link to your image and a description of how you made it.

Read more: Applied complex analysis

Exact chaos

Pick a number x between 0 and 1. Then repeatedly replace x with 4x(1-x). For almost all starting values of x, the result exhibits chaos. Two people could play this game with starting values very close together, and eventually their sequences will diverge.

It’s somewhat surprising that the iterative process described above can be written down in closed form. Starting from a value x0, the value after n iterations is

sin( 2n arcsin( √ x0 ) )2.

Now suppose two people start with the same initial value. One repeatedly applies 4x(1-x) and the other uses the formula above. If both carried out their calculations exactly, both would produce the same output at every step. But what if both used a computer?

The two approaches correspond to the Python functions f and g below. Because both functions are executed in finite precision arithmetic, both have errors, but they have different errors. Suppose we want to look at the difference between the two functions as we increase n.

from scipy import arcsin, sin, sqrt, linspace
from matplotlib import pyplot as plt

def f(x0, n):
    x = x0
    for _ in range(n):
        x = 4*x*(1-x)
    return x

def g(x0, n):
    return sin(2.0**n * arcsin(sqrt(x0)))**2

n = 40
x = linspace(0, 1, 100)
plt.plot(x, f(x, n) - g(x, n))
plt.ylim(-1, 1)

When we run the code, nothing exciting happens. The difference is a flat line.

Next we increase n to 45 and we start to see the methods diverge.

The divergence is large when n is 50.

And the two functions are nearly completely uncorrelated when n is 55.


So which function is more accurate, f or g? As noted in the comments, the two functions have different kinds of numerical errors. The former accumulates arithmetic precision error at each iteration. The latter shifts noisy bits into significance by multiplying by 2^n. Apparently both about the same overall error, though they have different distributions of error.

I recomputed g using 100-digit precision with mpmath and used the results as the standard to evaluate the output of f and g in ordinary precision. Here’s a plot of the errors when n = 45, first with f

and then with g.

The average absolute errors of f and g are 0.0024 and 0.0015 respectively.

Python animation for mechanical vibrations

Stéfan van der Walt wrote some Python code to animate the system described in yesterday’s post on mechanical vibrations.

Stéfan posted his code on github. It currently illustrates undamped free vibrations, but could be modified to work with damped or driven vibrations.

The code includes an option to save the output as an mp4 file. You must have ffmpeg installed to save a matplotlib animation as mp4.

Continued fractions with Sage

My previous post looked at continued fractions and rational approximations for e and gave a little Python code.  I found out later there’s a more direct way to do this in Python using Sage.

At its simplest, the function continued_fraction takes a real number and returns a truncated continued fraction representation. For example, continued_fraction(e) returns

[2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1]

Optionally, you can also specify the number of bits of precision in the real argument and the number of terms desired.

By calling the convergents method on the return value of continued_fraction(e) you can find a sequence of rational approximations based on the continued fraction. For example,

print continued_fraction(e).convergents()


[2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465,
1457/536, 2721/1001, 23225/8544, 25946/9545, 49171/18089,
517656/190435, 566827/208524, 1084483/398959,
13580623/4996032, 14665106/5394991, 28245729/10391023].

To get higher precision output, you need higher precision input. For example, you could pass in


rather than simply e to tell Sage that you’d like to use the 200-bit representation of e rather than the default precision.

Rational approximations to e

This morning Dave Richeson posted a humorous fake proof that depends on the famous approximation 22/7 for pi. It occurred to me that nearly everyone knows a decent rational approximation to pi. Some people may know more. But hardly anyone, myself included, knows a decent rational approximation for e.

Another approximation for pi is 355/113. I like this approximation because it’s easy to remember: take the sequence 113355, split it in the middle, and make it into a fraction. It’s accurate to six decimal places, which is sufficient for most practical applications.

The approximations 22/7 and 355/113 are part of the sequence of approximations coming from the continued fraction approximation for pi. So to come up with rational approximations for e, I turned to its continued fraction representation.

The best analog of the approximation 22/7 for pi may be the approximation 19/7 for e. Obviously the denominators are the same, and the accuracy of the two approximations is roughly comparable.

Here’s how you can make your own rational approximations for e. Find the coefficients in the continued fraction for e, for example here. You can turn this into a sequence of approximations by using the following Python code:

from __future__ import division
from math import e

e_frac = [2,1,2,1,1,4,1,1,6,1,1,8]

def display(n, d, exact):
    print n, d, n/d, n/d - exact

def approx(a, exact):
    # initialize the recurrence
    n0 = a[0]
    d0 = 1
    n1 = a[0]*a[1] + 1
    d1 = a[1]

    display(n0, d0, exact)
    display(n1, d1, exact)

    for x in a[2:]:
        n = x*n1 + n0 # numerator
        d = x*d1 + d0 # denominator
        display(n, d, exact)
        n1, n0 = n, n1
        d1, d0 = d, d1

approx(e_frac, e)

This will print the numerator, denominator, value, and error for each approximation. You could include more terms in the continued fraction for e if you’d like. Here are some of the results: 19/7, 87/32, 106/39, etc. Unfortunately it doesn’t look like there are any approximations as memorable as 355/113 for pi.

You could also use the code to create rational approximations to other numbers if you’d like. For example, you can find the continued fraction expansion for pi here and use the code above to find rational approximations for pi.

Update: There’s a more direct way to find continued fractions and rational approximations in Python using Sage. See the next post.

Footnote: The continued fraction coefficients for e have a nice pattern:
… 1, 1, 4, 1, 1, 6, 1, 1, 8, … 1, 1, 2k, 1, 1, 2k+2, …

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Python / Emacs setup

When I got a new computer a few days ago, I installed the latest version of Emacs, 24.2, and broke my Python environment. I decided to re-evaluate my environment and start over. I asked a question on the Python Google+ group and here’s a summary of the Emacs packages recommended.

  • python-mode
  • rainbow-mode, rainbow-delimiters-mode
  • flymake-mode (hooked up to flymake-pyflakes)
  • linum-on
  • jedi
  • pycheckers + pyflakes
  • rope
  • electric-pair
  • show-paren
  • auto-complete
  • yassnippets

What recommendations do you have for packages? Links to relevant articles?

It might be better to leave your comments on Google+ rather than here so the answers will be in one place.