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function counter(times, callback) {
  var ct = 0;
  return function() {
    ct++;
    if (ct === times)
      callback();
  };
};

// register a counter function

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var _ = require('underscore');

var freqVector = function(word, letters) {
  var freq = _.groupBy(word.split(''), function(l) {return l;});
  return _.map(letters, function(l) {
    return freq[l] ? freq[l].length : 0;
  }); 
};

var dot = function(v1, v2) {

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Proposition: If $f : A \to \mathbb{R}$ is continuous at $c \in A$, $kf$ is continuous at $c$ for $k \in \mathbb{R}$ Proof: Let $\epsilon > 0$ be given. Since $f$ is continuous at $c$, $\exists \delta > 0$ such that if $|x - c| < \delta$ then $|f(x) - f(c)| < \epsilon / (|k| + 1)$. Thus if $|x - c| < \delta$:

$$ |kf(x) - kf(c)| = |k||f(x) - f(c)| \leq \frac{|k|\epsilon}{|k| + 1} = \frac{|k|}{|k|+1}\epsilon < \epsilon $$

If we want to show that the product of two continuous functions is continuous, we need the following lemma:

Lemma: If $f : A \to \mathbb{R}$ is continuous at $c \in A$, $f$ is bounded in a neighborhood of $c$. Proof: Let $\epsilon = 1$. As $f$ is continuous at $c$ we know there exists some $\delta &gt; 0$ such that if $|x - c| &lt; \delta$, then $|f(x) _ f(c)| &lt; \epsilon = 1$. By the triangle inequality, we can go on to say that $|f(x)| - |f(c)| \leq |f(x) - f(c)| &lt; 1$, therefore $|f(x)| &lt; 1 + |f(c)| = M$. Thus if $|x - c| &lt; \delta$, then $|f(x)| &lt; M$, i.e. it is bounded.

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var quote = String.fromCharCode(39);
var source = [
  'var quote = String.fromCharCode(39);',
  'var source = [',
  '  ',
  '];',
  'console.log(source[0]);',
  'console.log(source[1]);',
  'source.forEach(function(line) {',
  '  console.log(source[2] + quote + line + quote + ",");',

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from random import randint

def run(goal, runs):
  total = 0
  for x in xrange(runs):
    sequence = []
    for i in xrange(3):
      sequence.append(randint(0,1))
    while sequence[-3:] != goal:
      sequence.append(randint(0,1))

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import requests
import os

twilio_sid = os.environ['TWILIO_SID']
twilio_auth = os.environ['TWILIO_AUTH']
twilio_url = 'https://api.twilio.com/2010-04-01/Accounts/%(twilio_sid)s/SMS/Messages.json' % locals()

weather_url = 'http://api.openweathermap.org/data/2.5/weather'
weather_params = {'q': 'montreal'}
r = requests.get(weather_url, params=weather_params)

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Javascript Review

Table of contents

1. Objects
2. Functions
3. Closures
4. The call and apply methods
5. Prototypes and constructors

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Fun problems (rendered at http://mathb.in/18258)

1. Find $\displaystyle\sum_{n=1}^\infty \frac{n}{2^n}$

2. Find $\displaystyle\sum_{n=1}^\infty \frac{n^2}{2^n}$

3. Let $a(k) = \displaystyle\sum_{n=1}^\infty \frac{n^k}{2^n}$. Find a recurrence relation for $a(k)$.

4. Let $\mathcal{N}$ be the set of natural numbers that do not contain a 6 in their decimal expansion (so, $\mathcal{N} = { 1, 2, 3, 4, 5, 7, \ldots, 14, 15, 17, \ldots }$). Prove that $$ \sum_{n\in\mathcal{N}} \frac{1}{n} < 80 $$

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1. Give an algorithm to find the longest common subsequence for two given strings using edit distance (the minimal number of additions, deletions and substitutions needed to transform one string into the other).

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\documentclass[11pt, oneside]{article}   	% use "amsart" instead of "article" for AMSLaTeX format
\usepackage{geometry}                		% See geometry.pdf to learn the layout options. There are lots.
\geometry{letterpaper}                   		% ... or a4paper or a5paper or ... 
%\geometry{landscape}                		% Activate for for rotated page geometry
%\usepackage[parfill]{parskip}    		% Activate to begin paragraphs with an empty line rather than an indent
\usepackage{graphicx}				% Use pdf, png, jpg, or eps with pdflatex; use eps in DVI mode
								% TeX will automatically convert eps --> pdf in pdflatex		
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{epigraph}