Math Quiz 9

[Aren Siekmeier]

Originally Posted by Ether

Really nice work Aren and Gus.

Here's how I did it.

Cool, I much prefer how you dropped to the double integral, I don't think about stats enough when doing these things.

Originally Posted by MechEng83

Now that this has been satisfactorily solved, I'll post a youtube video of this problem that I literally saw the week before Ether posted. I didn't feel right in just posting it, or claiming it as my own solution.

The problem gets to the 4ʃʃ(1-x)(1-y)√(x^2+y^2)dxdy GeeTwo derived, but then it does a polar coordinate substitution to make the integration "easier"
It's another approach which gives a closed form solution, demonstrating that there can be multiple ways to validly solve a problem.

While you can certainly say that there are multiple ways to solve a problem, I'll venture that the one presented in the video is considerably more elegant than mine The immediate trig sub works wonders on the rest of it.

Thanks for the link!

[GeeTwo]

Originally Posted by MechEng83

The problem gets to the 4ʃʃ(1-x)(1-y)√(x^2+y^2)dxdy GeeTwo derived, but then it does a polar coordinate substitution to make the integration "easier"
It's another approach which gives a closed form solution, demonstrating that there can be multiple ways to validly solve a problem.

This is the route I was following. (I have not looked at the video, so perhaps this is exactly what is there.) My original thought was to substitute r for √(x²+y²), rcosθ for x, and rsinθ for y. dxdy then becomes rdrdθ. While driving to mom's house for Sunday dinner, I realized that I could take advantage of the symmetry of x and y by rotating θ by π/4, so that y is √2(cosθ + sinθ)r/2 and x is √2(cosθ - sinθ)r/2. Then the integration is over an interval symmetric over θ=0, and any odd terms in θ can be tossed (I already figured out while driving that there aren't any, however), and then the integration can be done from 0 to π/4. The fun part is the limit of integration - for r<=1, there's no issue. For 1<r<√2, the limits of one integration need to be set so that the maximum of r is √2/(cosθ + sinθ) (ignoring the required absolute values because I've limited θ to the first quadrant). Alternately, the limits of θ can be set based on r.

Originally Posted by Caleb Sykes

There are uncountably infinite lines of length 0.01, just as there are an uncountably infinite number of lines of length 0.52.

It makes no logical sense to say that there are "more" or "less" of one uncountably infinite thing than another uncountably infinite thing.

It makes just as much sense as to ask "What fraction of integers are even?" There are uncountably many even integers and uncountably many integers, and it is possible to identify a different even integer for every integer (e=2i), so that the sets have the same number of elements. This does not change the equally meaningful statement that exactly half of the integers are even.

[Ether]

Originally Posted by GeeTwo

There are uncountably many even integers and uncountably many integers...

You are using the word "uncountably" incorrectly here.

The integers are countably infinite.

The reals are uncountably infinite.

This does not change the equally meaningful statement that exactly half of the integers are even.

It's meaningful only if you define what you mean

If you mean "the set of all even integers has an asymptotic density of ½", then yes, it's meaningful. Otherwise, not.

[Gregor]

These solutions are quite above my mathematics level, but nonetheless I can somewhat follow along, thanks for the cool thread.

[GeeTwo]

Originally Posted by Ether

You are using the word "uncountably" incorrectly here.

The integers are countably infinite.

The reals are uncountably infinite.

"Countably infinite" is still uncountable for us mortals. When you finish, please get back to me.

Originally Posted by Ether

... the concept of half of an infinite set is not well-defined. It’s possible to pair up the even integers with the odd integers with none left over in either set, and if we were talking about finite sets, that would be a demonstration that each was half of the whole set of integers. However, it’s also possible to pair up the multiples of 100, say, with all the rest of the integers with none left over in either set, and the multiples of 100 are obviously only part of the set of even numbers. Clearly, then, this kind of pairing argument cannot lead to any very useful notion of half of the set of integers.

There is a notion of asymptotic density of a set of positive integers that does a pretty good job of capturing many people’s intuitive sense of what half (or any other fraction) of the set of positive integers should mean.

excerpted from http://math.stackexchange.com/questi...ll-numbers-odd

I fully admit to having (yes, consciously and deliberately) blown through the difference between aleph null and aleph one in my earlier reply, to make a statement that might make sense to shose who found the original problem "meaningless". (That is, I was more worried about enlightenment than mathematical rigor.) I specifically addressed your concern that the even integers and all integers are of the same cardinality.

A far as I am aware, any argument that decides the statement "Half of the integers are even." as meaningless can also be used to decide the concept of "average length of a line segment located within the unit square" as meaningless -- or worse.

Edit:

Originally Posted by Ether, as edited after I prepared the response above

It's meaningful only if you define what you mean,

If you mean "the set of all even integers has an asymptotic density of ½", then yes, it's meaningful. Otherwise, not.

OK, I'll go with that -- and it's a remarkably fast asymptotic function, as of every pair of consecutive integers, exactly one is even.

Here's a bit more precise statement: For every set of consecutive integers with a non-zero, even number of members, exactly half are even.

[Ether]

FWIW

[GeeTwo]

Originally Posted by Ether

FWIW

It's worth enough to make me reconsider doing theta integration first so I can generate a closed form for this distribution.

[Ether]

Descriptive statistics for 10 million random samples:

mean      0.52151
median    0.51212
mode*     0.48 
meansq    0.33346
IQR       0.37633
std       0.24795
var       0.06148
skewness  0.18406
kurtosis -0.66077

*based on 100 bins, max slope of smoothed percentile vs length

[GeeTwo]

Originally Posted by Ether

 mode*     0.48 
*based on 100 bins, max slope of smoothed percentile vs length

I finished the polar integral up to r=1, found the mode to be (4+√(16-3π))/3 = .47859...

So thanks for the sanity check!

Other stats so far:

• The first part of the histogram curve (0..1) is 2πr - 8r² + 2r³.
• I have covered π/4 of the square (78.54%) by area
• I have integrated over π - 13/6 of the segments (97.49%).
• The numerator of the integral is up to 2π/3 - 8/5 = .494395..
• There were only three terms in the integral over Θ, the most complicated one being cos2Θ.
• I have only have four terms to integrate over r=[1..√2]. Two are powers of r, one looks familiar from the rectangular integration, and the last one looks strange, but probably no worse than anything in the rectangular problem.

Definitely an easier way to go unless that last one proves nastier than it looks.

Also, doing the integral this way explains the sudden change in behavior of the histogram at length=1.

[Greg Woelki]

Here is a generalized Monte Carlo simulation in python for the average line segment length in any regular polygon with one square unit of area:

%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
import time

def isInPoly(r,theta,cr,n):
    theta = theta%(2*np.pi/n) #Rotates point to first "sector" of the polygon for simplicity
    
    x = r*np.cos(theta)
    y = r*np.sin(theta)    
    
    x1 = cr #x coordinate of vertex at theta = 0
    y1 = 0.0 #y coordinate of vertex at theta = 0
    x2 = cr*np.cos(2*np.pi/n) #x coordinate of vertex at theta = 2pi/n
    y2 = cr*np.sin(2*np.pi/n) #y coordinate of vertex at theta = 2pi/n
    
    if y < ((y2 - y1)/(x2 - x1))*(x - x1) + y1: #Check using 2-point line formula
        return True
    else:
        return False


def genPolyPt(n, cr):
    '''Return tuple of random rectangular coordinate within an n-sided regular polygon centered
    at (0,0) with circumradius = cr and a vertex at (cr,0).'''

    r = cr*(np.random.random()**0.5) #Square root correction ensures even probability distribution over area
    theta = 2*np.pi*np.random.random()
    
    x = r*np.cos(theta)
    y = r*np.sin(theta)
    
    if r < cr*np.cos(np.pi/n): #If r < apothem of polygon then point is in the polygon's incircle
        return x,y
    if isInPoly(r,theta,cr,n): #Checks if point is between polygon and incircle
        return x,y
    else: #If point not in polygon
        return genPolyPt(n, cr)

t = time.time()

n = 4 #Number of sides of polygon
area = 1.0
trials = 10**7 #Number of random samples

cr = (area/(n*np.sin(2*np.pi/n)/2))**0.5 #Solves for circumradius that yields polygon with specified area

total = 0.0

#fig = plt.figure()
#ax1 = fig.gca()

#xlist = []
#ylist = []

for i in range(trials):
    x1,y1 = genPolyPt(n,cr)
    x2,y2 = genPolyPt(n,cr)

    #xlist.append(x1)
    #ylist.append(y1)
    
    total += ((x2-x1)**2 + (y2-y1)**2)**0.5

#plt.xlim(-.85, .85)
#plt.ylim(-.85, .85)
#ax1.set_aspect('equal')
#ax1.scatter(xlist,ylist,s=5)
    
print(total/trials)
print(time.time()-t)

The following are average lengths based on ten million samples:

Triangle: 0.554367342879
Square: 0.521395852676
Pentagon: 0.514797872593
Hexagon: 0.512588030594
Heptagon: 0.51176618103
Octagon: 0.511213933924
Nonagon: 0.511120940654
Decagon: 0.511039068646
25-gon: 0.510860752385
100-gon: 0.510794294963
1000-gon: 0.510750983962
1000000-gon: 0.510902061585

Can anyone think of another general way of generating a random point in an n-gon that is more efficient than mine for small values of n (without sacrificing significantly for larger values of n, of course)? When n=3, my program is discarding almost 60% of the points it generates.

[Richard Wallace]

Originally Posted by Ether

FWIW

So the distribution of lengths is Rayleigh. Neat example.

What would the distribution look like if the line segments were contained in a cube (3D) boundary? That one may be easier to visualize using a Monte Carlo simulation.

[Caleb Sykes]

Originally Posted by Richard Wallace

So the distribution of lengths is Rayleigh. Neat example.

Rayleigh distributions can't have upper bounds though, right? This distribution will clearly have an upper bound at sqrt(2).

[Ether]

Originally Posted by Richard Wallace

What would the distribution look like if the line segments were contained in a cube (3D) boundary?

...

[Aren Siekmeier]

Originally Posted by GeeTwo

... I specifically addressed your concern that the even integers and all integers are of the same cardinality.

FTFY. When discussing cardinality, distinction between aleph-0 and aleph-1 is key. The integers (aleph-0) have measure 0 in the real line (aleph-1), so any "proportion" there would be 0.

Originally Posted by GeeTwo

A far as I am aware, any argument that decides the statement "Half of the integers are even." as meaningless can also be used to decide the concept of "average length of a line segment located within the unit square" as meaningless -- or worse.

Rather, what composes half of an infinite set is not well-defined at all, since there is no well-defined limit of half of a finite set. The average of an infinite set, on the other hand, is well understood as the limit of the average of a finite set (an infinite series or an integral).

Originally Posted by GeeTwo

OK, I'll go with that -- and it's a remarkably fast asymptotic function, as of every pair of consecutive integers, exactly one is even.

The error bounding condition is still error <= 1/2n, so not quite that fast. You just happen to hit the limit exactly on every other term.

[Ether]

Originally Posted by Ether

What's the average length of all the line segments which can be drawn inside a 1 inch square?

(accurate to 5 decimal digits)

(show your work)

OK I'm going to drop a bombshell here.

The correct answer to the OP, as worded, is 0.33634

Aren was the first to articulate the key principle when he wrote:

Originally Posted by Aren Siekmeier

Randomly sample the set of all line segments, and find the mean length of only the segments in the sample (a finite number chosen to suit your computational resources, rather than the uncountably infinite number in the entire set). If the sampling matches the probability distribution/weighting of the set, then the law of large numbers says your mean will approach the true mean as sample size increases.

Sampling segments by choosing the segment endpoint coordinates from a random uniform distribution does not match the probability density of the segment lengths in the infinite set.