A FINITE REPRESENTATION OF AN INFINITELY SMALL
WORLDCHRIS RACKAUCKAS

Abstract. A common symbol of it's a small world is the cutout ofpeople holding hands.

These gures were transformed into functionsdened on the interval [0, 8] and thus a generalized interval fourier decomposition. The constants were then solved using a nite sum approximation. These values were then used to generate a function that repeats
indenitely that draws an approximation to this gure with a periodicity of 8. Since this method solved for a nite sum approximation, it is a
nitely approximated solution of an innite version of the it's a smallworld cutout, hence a nite representation of an innitely small world.

1. Introduction to the Problem

The picture above is a very common image. It shows unity among people,small the world is, and represents how we should all come together as one.
Although we think of the cutout like this, it is lacking in one signicantproperty: it is nite and the number of people is much less than the population. Thus I set out to construct an innite version of this image. Beingthat innite actual representations must either take an innite amount ofwhatever is used to construct it, or if the innite representation does take anite amount of paper then the size of the persons would be some geometric series where the people are getting smaller. Since I want to show thatall people are the same and actually be able to make this construction, anactual physical representation was not possible. Thus I decided to create a
representation in function space.
2. The Image as a Function
To solve this problem in function space, I rst had to describe what thefunction was.

A few abstractions were used to ensure that the function

1

2

CHRIS RACKAUCKAS

could be described in the function space of R . All of the people were to be2

described using what in the image above was the male body. Most cutouts
do not distinguish between male and female and so this does not seem tostray from the original (also, it is so passe to believe that all women must
be wearing a dress!). Also, the heads were changed to be a rectangle. Thisis because it would have been impossible for the function to get to the sideof the head while staying on the body which would have been necessary tosplit the circle into two function. However, this will not end up as a problem
since the fourier decomposed version will round out the edges and thus givea pretty circular looking head. Lastly, although many cutouts use a version
with the hands going down, I chose to have the hands going up like in theimage above. All of the same steps would apply to either image.
To transform this image into a function, I had to make sure it passed thevertical line test.

Since no one function would pass the vertical line test

(using non-parametric functions), I split the image into 4 functions.

One

describes the inside of the legs, the other the outside of the legs (and gives
a waist line), another describes the bottom of the arms (and gives the samewaist line), and the last gives the top of the arms and the head.

These

functions were labeled f1 through f4 respectively. By drawing this image ona graph of width 8 and height 8, I solved for these functions to be:0f1 (x) = 2x − 4
−2x + 12

if 0 ≤ x ≤ 2 or 6 ≤ x ≤ 8if 2 ≤ x ≤ 4if 4 ≤ x ≤ 6

0



4x − 4f2 (x) =

4



−4x + 28

if 0 ≤ x ≤ 1 or 7 ≤ x ≤ 8
if 1 ≤ x ≤ 2if 2 ≤ x ≤ 6
if 6 ≤ x ≤ 73− 2 x + 7
f3 (x) = 4
32x − 5

7




−x + 8f4 (x) = 6


8



x

if 0 ≤ x ≤ 2if 2 ≤ x ≤ 6if 6 ≤ x ≤ 8

if 0 ≤ x ≤ 1 or 7 ≤ x ≤ 8
if 1 ≤ x ≤ 2if 2 ≤ x ≤ 3 or 5 ≤ x ≤ 6if 3 ≤ x ≤ 5
if 6 ≤ x ≤ 7

The image below shows functions f1 through f4 respectively then plotsthem together using Mathematica.

A FINITE REPRESENTATION OF AN INFINITELY SMALL WORLD

3Figure 2.1. The plots of f1 through f4

Figure 2.2. A plot of f1 through f4 together

3. Fourier Decomposition
The next step was to decompose these functions to innite periodic functions using what is known as a Fourier decomposition. The Fourier Decoposition method is dened as writing the function gi (x) as an innite summationsine and cosine functions. The method is described as writing∞gi (x) =A0 X+(Ani cos(nx) + Bni sin(nx))
2n=1

4CHRIS RACKAUCKAS

where1Ani =πand

1Bni =πˆ πfi (x) cos(nx)dx−πˆ πfi (x) sin(nx)dx.
−πHowever, this Fourier decomposition requires the function to be dened on
the interval [−π, π].

Thus I instead had to use a dierent version of theFourier Decomposition called the Generalized Interval Fourier Decomposition[2]which denes the summed function as a summation of functions dened on
the interval [0, L] as∞

A0 X
2πnx
2πnx+(Ani cos() + Bni sin())
2LL

gi (x) =n=1

where2Ani =Land

2Bni =Lˆ L
fi (x) cos(2πnx)dx
Lfi (x) sin(2πnx)dx
L0ˆ L0

Notice that this simply resizes the periodicity of the functions to matchthe length of the interval and thus is an intuitive extension of the standard
Fourier Decomposition.
4. Infinite Sum Representation
In order for these to be calculated, I needed these integrals to be put into asummation form. First take note that the integral of a piecewise function is
simply the sum of the integrals on the dierent intervals. Since each intervalof every fi function can be written as some linear function ax + b, it suces
to solve for what these integrals would be.
Noting that the sine function can be representated as a summation bysin(x) =∞X(−1)k x2k+1k=0(2k + 1)!

by using a U-substituion and solving we see that2Lˆ s∞b cos(r2k2πnx
2b X (−1)k (s2k+1 − r2k+1 )( 2πnL ))dx =LL(2k + 1)!
k=0

and by using integration by parts we see that2Lˆ s∞ax cos(r2k2πnx
2a X (−1)k (s2k+2 − r2k+2 )( 2πnL ))dx =(2k+2)!(2k+1)!
LL
k=0(2k+2)!−(2k+1)!

The derivations are included in the notes. It is interesting to see that theax integral can be put into one summation.

This is due to the fact that

A FINITE REPRESENTATION OF AN INFINITELY SMALL WORLD

5the integration requires the one portion be integrated once and the otherintegrated twice. However, while the one that is integrated twice is a cosine
while the other is a sine, the u-substitution occurs on the integrated twice
version twice while the u-substition occurs on the integrated once version
just once.

Thus while the summation of the cosine contains only the 2kvalues and the sine contains the 2k + 1 values, after the u-substitions cancelout the n terms a bit, it turns out that both parts are dened as summations
with the values of n only being even (since the cosine term loses n twice,
one for each u-substition, and the sine term loses n once for its single usubstitution). Thus the value(2k+2)!(2k+1)!
(2k+2)!−(2k+1)! is simply nding the common

denominator to add together these two terms to make a single function.
Notice by putting this together we get the equation necessary for computation:2Lˆ s∞(ax + b) cos(r2k2πnx
2 X (−1)k (s2k+2 − r2k+2 )( 2πnL ))dx = [a(2k+2)!(2k+1)!
LL+bk=0∞
Xk=0(2k+2)!−(2k+1)!

2k(−1)k (s2k+1 − r2k+1 )( 2πnL )](2k + 1)!

Notice that for the functions we are interested in, functions dened on[0, 8], L = 8. Thus these sums are sucient for the computation.
5. A Note on the Accuracy
When I rst solved the equations, I solved the equations using just therst 3 terms in the summation for both a and b. At n = 1this solves theintegral of (2x−4) cos(x) from 2 to 4 (using L =12π ) as approximately 15.39.

However, using WolframAlpha I saw that the actual value is approximately-1.178585. Why such the large discrepency?
Upon investigation, I noticed that there were some signs that this error
gives that would end up being useful for computation. First of all, noticethat both of the summations are alternating series. Also notice that since
the factorial function grows quicker than the expoential function, this gives
us the fact that each term of the series is smaller than the next. Thus what
we see with the errors is that the estimation of the Ani starts as a large
value and then decreases k increases. Since in general for a Fourier series then+1th term is smaller than the nth term, we can easily see if we did not solve
the sum of the k values enough if we see this did not decrease. A simple ofdata not ran on enough k 's is included. What is shown is the sum for solving(2x − 4) cos( 2πnxL ) where n = 3 and L = 8. Notice how the total value of thesum starts high and decreases with each k . If k was large enough, this would
have approached its actual value, approximately 0.18012654869748941859,
but since the sum was not run long enough, its value is higher than the6CHRIS RACKAUCKAS

summed value for the n = 2 term which was 0.81056946913870223475, whichindicates that the max k used for the summation must be increased.

6. Computation
Since the accuracy required was far too many terms to be computed byhand, the summations were solved using a program in Python. The method
was to solve out the terms using a given maximum k and check to see if theterms were decreasing and if not increase the maximum k value by hand. I
did not resort to having the computation increment the k values since eachrun of the program already takes approximately an hour, and thus letting thecomputer nd a good k could have made the program run on a standard PC
for more than a day. Also, normal oats and integers were not able to be usedfor computation. This is because the integers from factorials of large numbers
quickly created integers larger than what could be represented by 32 bits andthe decimal precision required for much of the division was much more thana 32 bit oat could oer. Thus the computation had to be done using Longand Decimal formats. Python automatically converts int types to long types
which have innite accuracy (which was why I chose Python, I knew this from
Crypotography exercises). However, some conversions had to occur to havethe Decimal class used instead of oats since operations between decimalsand oats were undened (though operations between longs and Decimals aredened). In particular, a decimal accuracy must be set for the computation.
I noticed a decimal accuracy of 200 siginicant gures. Thus by using thedecimal class for the computation, I was guarenteed a computation that wascorrect to 200 siginicant gures (which, gives at least 50 signicant guresin the decimal place even for the largest of numbers seen in the computation).
A sample of how the computation was done for a simple function is included. That program was then extended to be able to compute the integrals
of an arbitrary number of functions dened piecewise by linear functions. By
trial and error using the facts mentioned in the Note on Accuracy, I foundthat a good maximum k value for the computation was 2000. Thus the nalcomputation consisted of using innite accuracy integers called longs, 200signicant gure decimals, used the rst 2000 terms of the summation, andsolved for the rst 10 Ani terms for each function f1 through f4 (notice thatthe rst 10 terms means solving for A0i through A9i ).

This computation

took approximately an hour for a single run (making debugging be a pretty
lengthly process!).

7. Final Results
The program was run and the nal results were obtained.
are included.

Notice that any value|Ani | ≤The results

10−10 is set equal to 0 to

account for errors in the computation. The transformed functions can thus

A FINITE REPRESENTATION OF AN INFINITELY SMALL WORLDbe approximated as:

g1 (x) = 1 + −1.62113893827740437022 cos(πx)
4πx− 0.18012654869748941859 cos(3 )4πx− 0.06484555753109615893 cos(5 )4πx+ 0.09006327434874469275 cos(6 )4πx− 0.03308446812811030948 cos(7 )4πx− 0.02001406096638769847 cos(9 )4πx)4+ 0.81056946913870223475 cos(2g2 (x) = 2.5 − 2.292636673003025218162 cos(πx)
4πx+ 0.254737408111447384170 cos(3 )4πx+ 0.405284734569351117373 cos(4 )4πx+ 0.091705466920120927387 cos(5 )4πx− 0.090063274348744742383 cos(6 )4πx− 0.046788503530673986010 cos(7 )4πx− 0.028304156456827470640 cos(9 )4πx)4− 0.810569469138702085847 cos(2g3 (x) = 4.75 + 1.21585420370805288082 cos(πx)
4πx+ 0.13509491152311646856 cos(3 )4πx+ 0.04863416814832160323 cos(5 )4πx+ 0.06754745576155797382 cos(6 )
4πx+ 0.02481335109608216509 cos(7 )4πx+ 0.01501054572479024465 cos(9 )4+ 0.60792710185402613032 cos(2πx)
478CHRIS RACKAUCKAS

g4 (x) = 6.875 − 0.32715714790635031829 cos(πx)
4πx− 0.36378979074689778757 cos(3 )4πx− 0.10132118364233828611 cos(4 )4πx+ 0.15713689650139041578 cos(5 )4πx− 0.18969077220200814945 cos(6 )
4πx+ 0.14031374247654027899 cos(7 )4πx− 0.09295910712547210659 cos(9 )4πx)4+ 0.83926213965225634361 cos(2The results were then put into Mathematica to plot the generated functions. The functions are plotted in order and on top of the function they areapproximating:Figure 7.1. Functions g1 through g4 plotted on top of f1

through f4 respectively

Notice that f4 , being the most complex function and including a discontinuity, is the least well approximated.

However, when these are all put

together, these form a picture that I would say is quite pretty:

A FINITE REPRESENTATION OF AN INFINITELY SMALL WORLD

9Figure 7.2. The nite sum representation of the innite

functions g1 through g4

8. Summary
The small world cutout is quite a lovely image, but as humans we can dobetter than just a nite representation. Thus I created a representation of
the cutout in function space using a Fourier Decomposition. After dealing
with many computational problems due to accuracy, nal equations wereproduced. With the nite approximation of the innitely dened function, I
draw out the picture to make what I would call a lovely picture that expresses
how innitely many people are one. This is complete unity. This is the nite
representation of the innitely small world.
References[1] Khamsi, M. A. "Fourier Series: Basic Results." S.O.S Math. MathMedics, LLC. Web.
16 May 2012. <http://www.sosmath.com/fourier/fourier1/fourier1.html>.
[2] Olver, Peter. "Chapter 12, Fourier Series." Peter Olver's Home Page. University ofMinnesota. Web. 16 May 2012. <http://www.math.umn.edu/~olver/am_/fs.pdf>.
[3] Weisstein, Eric W. "Fourier Series." Wolfram MathWorld. Wolfram Research, Inc.
Web. 16 May 2012. <http://mathworld.wolfram.com/FourierSeries.html>.