@fawnpnguyen whoa. Need to try and make this.

— Dan Anderson (@dandersod) June 4, 2013

I started off and used VPython to create the fractal, but it was slow and buggy. Here’s a movie of my first attempt. When you click you create a new “seed” for the fractal to start. 

So I rewrote it in Processing.org (2.0 just came out!), and the results are far more satisfying. Left click to start a seed, and right click to clear the scene. Check it out!

Processing source code.

VPython source code.

RAY: Tommy, Dougie and I are sitting around the office one day at Car Talk Plaza. We were noticing how dingy the place looked. We’d been there 15 years, and the place had never been painted. So, we decided to paint Car Talk Plaza.

We didn’t know which team of us was going to do it, so we sat down and decided to do a little math. We determined that Tommy and I together could paint the entire Car Talk plaza in 10 days. After all, we had a lot of painting experience as kids, having painted Dad’s car a couple of times with brushes.

Dougie and I could do it in 15 days. And, if Doug and Tom worked together, they could do it in 30 days.

The question is how long would it take each of us, painting by ourselves, to paint the whole of Car Talk Plaza?

Dr. Drang looks at the math behind it.  This needs some more investigation. Also: sphere icecubes.

So bigger is better right? On their original ad they claimed that the ASUS VivoTab Smart tablet had more area than the iPad.

But Elliot Temple claims the following:

How can the screen with a larger diagonal measurement be smaller? Because it’s a different shape. Long and thin gets you a bigger diagonal but a smaller screen, for the same diagonal inches.

Who’s right? Microsoft or Elliot? Great math involved here.

[edit - 5/28/13: This post links in very nicely with the Mathalicious lesson Viewmongous. Thanks Fawn Nguyen!]

So 50 billion apps downloaded on the iTunes store has come and gone, and I have nothing to show for it (except for a bunch of fun learning and some excitement for the next time).

Things that happened:

• We made a prediction in class to within 15 min of the actual stopping time.
• I found the javascript file that the counter is checking to keep up to date.

Reasons Why I’m Excited For The Next Countdown In The App or Music Store:

• I’ll make a script to check the javascript file and get the hourly rates for several days in a row.
• Find a function to model the app download rate.
• Algebra/Geometry/Pre-Calculus: Find the rate using the slope secant lines for two data points. Use the data to make a guess for when the store will hit #next# billion.
• Calculus: use Riemann sums and definite integrals to approximate the number of apps downloaded in a day, a week, a month, a year. Use the actual app download data for one day to see how find the accuracy of the area underneath the curve.
• Can you help me out? Am I missing anything? (answer: yup)

The truth is, most of us discover where we are headed when we arrive.

Must read advice from Bill Waterson on Brain Pickings, Maria Popova’s blog.

Your preparation for the real world is not in the answers you’ve learned, but in the questions you’ve learned how to ask yourself.

Old posts from when Apple was at 25 billion app downloads.
More questions:
When will it hit 100 billion? 1 trillion? How many apps will be sold by December 25th? Any more?

Edit 5/11/13
More info:

Allows you to paste in a roster and make random groups. No friction, no mess. Cool stuff.

Bailey–Borwein–Plouffe formula

Bellard’s formula

and
Chudnovsky algorithm

Holy smokes is Chudnovsky algorithm’s fast!

			 Plouff 		 Bellard 			 Chudnovsky
Iteration number  1   3.133333333333333333333333   3.141765873015873015873017   3.141592653589734207668453
Iteration number  2   3.141422466422466422466422   3.141592571868390306374053   3.141592653589793238462642
Iteration number  3   3.141587390346581523052111   3.141592653642050769944284   3.141592653589793238462642
Iteration number  4   3.141592457567435381837004   3.141592653589755368080514   3.141592653589793238462642
Iteration number  5   3.141592645460336319557021   3.141592653589793267843377   3.141592653589793238462642
Iteration number  6   3.141592653228087534734378   3.141592653589793238438852   3.141592653589793238462642
Iteration number  7   3.141592653572880827785241   3.141592653589793238462664   3.141592653589793238462642
Iteration number  8   3.141592653588972704940778   3.141592653589793238462644   3.141592653589793238462642
Iteration number  9   3.141592653589752275236178   3.141592653589793238462644   3.141592653589793238462642
Iteration number  10   3.141592653589791146388777   3.141592653589793238462644   3.141592653589793238462642
Iteration number  11   3.141592653589793129614171   3.141592653589793238462644   3.141592653589793238462642
Iteration number  12   3.141592653589793232711293   3.141592653589793238462644   3.141592653589793238462642
Iteration number  13   3.141592653589793238154767   3.141592653589793238462644   3.141592653589793238462642
Iteration number  14   3.141592653589793238445978   3.141592653589793238462644   3.141592653589793238462642
Iteration number  15   3.141592653589793238461733   3.141592653589793238462644   3.141592653589793238462642
Iteration number  16   3.141592653589793238462594   3.141592653589793238462644   3.141592653589793238462642
Iteration number  17   3.141592653589793238462641   3.141592653589793238462644   3.141592653589793238462642
Iteration number  18   3.141592653589793238462644   3.141592653589793238462644   3.141592653589793238462642
Iteration number  19   3.141592653589793238462644   3.141592653589793238462644   3.141592653589793238462642

Source code (Python)

from decimal import *

#Sets decimal to 25 digits of precision
getcontext().prec = 25

def factorial(n):
    if n<1:
        return 1
    else:
        return n * factorial(n-1)

def plouffBig(n): #http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula
    pi = Decimal(0)
    k = 0
    while k < n:
        pi += (Decimal(1)/(16**k))*((Decimal(4)/(8*k+1))-(Decimal(2)/(8*k+4))-(Decimal(1)/(8*k+5))-(Decimal(1)/(8*k+6)))
        k += 1
    return pi

def bellardBig(n): #http://en.wikipedia.org/wiki/Bellard%27s_formula
    pi = Decimal(0)
    k = 0
    while k < n:
        pi += (Decimal(-1)**k/(1024**k))*( Decimal(256)/(10*k+1) + Decimal(1)/(10*k+9) - Decimal(64)/(10*k+3) - Decimal(32)/(4*k+1) - Decimal(4)/(10*k+5) - Decimal(4)/(10*k+7) -Decimal(1)/(4*k+3))
        k += 1
    pi = pi * 1/(2**6)
    return pi

def chudnovskyBig(n): #http://en.wikipedia.org/wiki/Chudnovsky_algorithm
    pi = Decimal(0)
    k = 0
    while k < n:
        pi += (Decimal(-1)**k)*(Decimal(factorial(6*k))/((factorial(k)**3)*(factorial(3*k)))* (13591409+545140134*k)/(640320**(3*k)))
        k += 1
    pi = pi * Decimal(10005).sqrt()/4270934400
    pi = pi**(-1)
    return pi
print "\t\t\t Plouff \t\t Bellard \t\t\t Chudnovsky"
for i in xrange(1,20):
    print "Iteration number ",i, " ", plouffBig(i), " " , bellardBig(i)," ", chudnovskyBig(i)

2^86 is largest known pwr of 2 that does not contain 0.What about other digits?Does each non-zero digit occur inf often among the pwrs of 2?

— James Tanton (@jamestanton) March 6, 2013

Interesting. The Intro to Programming Students gave a crack at this one; and after 30 minutes all were close and half had it. Some worked in Python (easier large numbers and string manipulation) and some in Java (muuuuch faster). Here’s my Java (BlueJ) code. It checks all powers of 2 up to 2^10,000, and then it prints out the counts of all the digits 0-9. The digits are spread out pretty evenly.

import java.math.BigInteger;
public class Powersof2
{
    public int[] digits = new int[11];;
    public Powersof2()
    {
        BigInteger number = new BigInteger ("1");;
        int power = 1;
        
        for (int i = 0; i < digits.length; i++)
        {
            digits[i]=0;
        }
        
        while (power <= 10000)
        {
            number = number.multiply(new BigInteger ("2"));
            
            if (HasAZeroInIt(number))
            {
                System.out.println("Next power of 2 is " + number + " with an exponent of " + power);
            }
            
            
            power += 1;
        }
        
        for (int i = 0; i < digits.length-1; i++)
        {
            System.out.println("index " + i + " and number of digits is " + digits[i]);
        }
                   
       
    }
    
    public boolean HasAZeroInIt(BigInteger n)
    {
        String strn = String.valueOf(n);
        
        int index = 0;
        while (index < strn.length())
        {
            digits[Character.getNumericValue(strn.charAt(index))] += 1;
            index += 1;
        }
        index = 0;
        while (index < strn.length())
        {
            if (strn.charAt(index) == '0')
            {
                return false;
            }
            index += 1;
        }
        return true;
    }    
}