# Polyhedra: Conclusion

In this lab, we set out to explore the five regular polyhedra including their qualities as well as their exclusive nature. Through analyzing these shapes, it is clear that regular polyhedra are only created through three basic 2D shapes: the equilateral triangle, the square, and the pentagon. It is also clear that the vertices on a regular polyhedra require at least three of these shapes to obtain their 3D nature. Another requirement for their 3D nature is that each vertex can only be surrounded by interior angle equaling less than 360 degrees in sum. This is because 3D shapes must have a height and this comes from bending the shapes at their connected edges.

These findings helped to prove that there are only five regular polyhedra. It was found that there are only five regular polyhedra because the limits on the the total amount of degrees that can surround a vertex in the third dimension as well as the minimum amount of shapes that must share the vertices. Because equilateral triangles have small interior angles of 60 degrees, the triangle is capable of forming three regular polyhedra with 3, 4, and 5 triangles around a vertex. The 90 degree angles of the square allow for the cube to form, and the 108 degree angles allow for the dodecahedron to form. It is clear that no other regular polyhedra exist because the interior angles of regular shapes only grow larger the more edges they have. Therefore, no sum of three interior angles is less than 360 after the pentagon and never will be. Another experiment that helped to further show the exclusivity of regular polyhedra was the truncating of their vertices which resulted in the shapes to form into one-another. It is clear that the truncating of a shape creates a new shape with the number of faces equal to the number of vertices from the previous shape. These new faces have the same number of edges as the number of old faces shared at the cut vertex.

In further studying regular polyhedra, it would be interesting to analyze the incomplete truncation of these solids. When truncating, polygons are seen at the corners that eventually become the faces of the new shape when truncation is complete. When truncation is incomplete, each corner is inhabited by these miniature polygons while the old faces shrink in size and gain an extra edge at each of their vertices. However, are these shapes still symmetrical throughout their entire transformation? If so, how many of these solids can be created. For example, a cube with slight truncation would have triangles on its vertices and the squares would have four new edges, making them octagons. How many of these shapes exist? Is there a finite amount? These are the questions that would be explored in further experimentation.

# The Sieve of Eratosthenes

Because there was a question in our first post about what a sieve is, it would be good to give a quick example of what a sieve is supposed to do, and what it looks like.

For the sake of an example, we are going to sieve all of the primes from 2-30, in other words we are going to separate the primes from the non-prime numbers in the set.

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Let’s strikethrough all of the multiples of 2 greater than 2.

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Now, let’s strikethrough all of the multiples of 3 greater than 3.

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Now, you may be saying to me, but why didn’t you make 6, 12, 18…green too? Well, they’ve already been highlighted once in blue, so there’s no real reason to do more work than I have to.

So, let’s continue with 5.

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

I’m going to continue going up using the non-crossed out numbers as a guide, and highlighting the multiples of those non-highlighted numbers.

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Because I only chose to go up to 30, we are actually done. The non-crossed out numbers are our prime numbers:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

So, here are some closing thoughts about the sieve “method” of finding primes:

Pro: Easy algorithm, an easy way to loop through a set of numbers to find the primes, easy to program by hand

Con: Can take a long time with larger numbers, eats memory in larger number sets.

And just for a thing to brighten your day, here is a poem that I found about the sieve on Wikipedia. I take no credit because it’s not mine, but I thought it was a fun way to think about the sieve method. More information can be found at the wikipedia page on the topic, including a very cool animation of the sieve from 2-120.

Sift the Two’s, and Sift the Three’s,

The Sieve of Eratosthenes.

When the multiples sublime,

The numbers that remain are prime.

http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

Previous Post: Prime-Generating Functions

# Prime-Generating Functions

In the eighteenth century the supercomputers we have today which people can use to find prime numbers did not exist. Mathematicians have always sought to create order in a very chaotic world. Many people looked for a way to produce prime numbers. Is there a polynomial that when inputted with the set of all non-negative integers, would always produce distinct prime numbers?

In 1772 Euler discovered a polynomial whch produces primes from integers. This polynomial is x^2+x+41. After running this polynomial through Sage, we find that it does indeed produce distinct prime numbers but only for all integer values of x between 0 and 39. After that, the formula begins to degrade and not always produce prime numbers.

{41,43,47,53,61,71,83,97,113,131,151,173,197,223,251,281,313,347,383,421,461,503,547,593,641,691,743,797,853,911,971,1033,1097,1163,1231,1301,1373,1447,1523,1601}

If we then look at a second polynomial, x^2-79x+1601, and find its outputs with Sage, we find that it produces primes for all integer values between 0 and 79. Though, this polynomial simply outputs all primes that Euler’s original formula produced, twice each.

These quadratic polynomials both do pretty well. They both produce uninterrupted strings of primes of lengths 40 and 80, respectively. To show that there are polynomials that do a very poor job of this, we can look at x^2 + x + 2. The largest string of primes this produces for non negative integers is… 1, as it produces a 2 at x = 0, and nonprime numbers otherwise.

The fact is that there does not exist a polynomial with integer coefficients that always produces prime numbers. Their behavior is very difficult to predict. So although primes cannot be described with a polynomial, maybe there are other things we can learn about them by observing their distributions, another time.

# Absolute Values on Q (Part I)

Real numbers, Q, can be constructed from the rational numbers Q={a/b where a,b ϵ the integers Z}.  Construction of p-adic numbers, Qp, is done exactly the same way.

Let’s now consider absolute values.  An absolute value is a map from Q to [0,∞) that has the following properties for any x,y ϵ Q:

1. |x| ≥ 0, and |x| = 0 if and only if x = 0,
2. |x∙y| = |x|∙|y|,
3. |x+y| ≤ |x|+|y| (The triangle inequality).

Now that we have some properties, we can look at types of absolute values that can be put on the rationals.  Alexander Ostrowski showed in 1935 that there are only three types.

Definition 1:  Let x be a rational number.

The trivial absolute value of x, denoted |x|0, is defined by

$|x|_0 = \left\{ \begin{array}{l l} 1 & \quad \text{if x\ne0 }\\ 0 & \quad \text{if x=0} \end{array} \right.$

The usual absolute value of x, denoted |x|, is defined by

$|x|_{\infty} = \left\{ \begin{array}{l l} x & \quad \text{if x\ge0 }\\ -x & \quad \text{if x\le0} \end{array} \right.$

The p-adic absolute value of x, denoted |x|p, is defined for a given prime p by

$|x|_p = \left\{ \begin{array}{l l} 1/p^{ord_p(x)} & \quad \text{if x\ne0 }\\ 0 & \quad \text{if x=0} \end{array} \right.$

The quantity ordp(x) is called the order of x.  It is the highest power of p that divides x.  For example, let p=5 and let x=75.  Then ord5(75)=2.  Since 25=52 and 25 is the highest power of 5 that divides 75, then the order of x is 2.  Confused yet?  Here are some examples of p-adic absolute values:

|75|5 = |52 ∙ 3|5 = 5-2

|10|5 = |51 ∙ 2|5 = 5-1

|13|5 = |50 ∙ 13|5 = 1

$|\frac{2}{75}|5 = |5-2 ∙ \frac{2}{3} |5 = 52$

|-375|5 = |53 ∙ -3|5 = 5-3

In these examples, we were looking at 5-adic absolute values.  We broke the number up into the form 5n ∙ a.  The 5-adic absolute value of 5n ∙ a is 5-n.

Conjecture 1:  Given a rational number x, the p-adic absolute value of x is p-n, where pn divides x.

|x|p = |pn ∙ a|p = p-n

In the introduction post (http://math.boisestate.edu/m287/quick-intro-to-p-adics/), it was mentioned that large p-adic numbers are closer to 0 than smaller p-adic numbers.  If one was to think of absolute value as a measure of distance, notice in the above examples that the larger the exponent on p, the resulting absolute value (or distance from 0) gets smaller.

# Numerical Integration Analysis — Data!

We wanted to follow up with a post that contains a dump of our data. This includes, essentially, our percent error from the expected value of the given test functions:

• $\cos{x}$ over [0, $\pi{}$]
• $2x + 1$ over [0, 1]
• $4-x^2$ over [0, 2]
• $5x^3 – 6x^2 + 0.3x$ over [-1, 3]
• $x^3$ over [-1, 3]
• $x^3 -27x^2 + 8x$ over [0, 3]

We tested 5 different deltas (rectangle widths), $dx$, namely, $0.1$, $0.01$, $0.001$, $0.0001$, $0.00001$. But we are not going to put tables for each method and each delta; it’s just too much. However, we will do the first delta ($0.1$) and the last delta ($0.00001$).

### Summary of Methods for $\cos{x}$ over [0, $\pi{}$]

Method Delta Percent Error
Trapezoidal $0.100000$ -0.33364
Trapezoidal $0.000010$ -0.00000
Midpoint $0.100000$ -0.20893
Midpoint $0.000010$ -0.00000
Simpsons $0.100000$ 0.05475
Simpsons $0.000010$ 0.00000
Left Rectangle $0.100000$ -4.97995
Left Rectangle $0.000010$ -0.00050
Right Rectangle $0.100000$ 4.31267
Right Rectangle $0.000010$ 0.00050

### Summary of Methods for $2x + 1$ over [0, 1]

Method Delta Percent Error
Trapezoidal $0.100000$ -14.50000
Trapezoidal $0.000010$ -0.00150
Midpoint $0.100000$ -14.50000
Midpoint $0.000010$ -0.00150
Simpsons $0.100000$ 0.00000
Simpsons $0.000010$ 0.00000
Left Rectangle $0.100000$ -10.00000
Left Rectangle $0.000010$ -0.00100
Right Rectangle $0.100000$ -19.00000
Right Rectangle $0.000010$ -0.00200

### Summary of Methods for $4-x^2$ over [0, 2]

Method Delta Percent Error
Trapezoidal $0.100000$ -0.42813
Trapezoidal $0.000010$ -0.00000
Midpoint $0.100000$ -0.33906
Midpoint $0.000010$ -0.00000
Simpsons $0.100000$ 0.00000
Simpsons $0.000010$ -0.00000
Left Rectangle $0.100000$ -3.81250
Left Rectangle $0.000010$ -0.00038
Right Rectangle $0.100000$ 2.95625
Right Rectangle $0.000010$ 0.00037

### Summary of Methods for $5x^3 – 6x + 0.3x$ over [-1, 3]

Method Delta Percent Error
Trapezoidal $0.100000$ -16.93086
Trapezoidal $0.000010$ -0.00181
Midpoint $0.100000$ -17.10882
Midpoint $0.000010$ -0.00181
Simpsons $0.100000$ -0.00000
Simpsons $0.000010$ -0.00000
Left Rectangle $0.100000$ -7.67699
Left Rectangle $0.000010$ -0.00078
Right Rectangle $0.100000$ -26.18473
Right Rectangle $0.000010$ -0.00284

### Summary of Methods for $x^3$ over [-1, 3]

Method Delta Percent Error
Trapezoidal $0.100000$ -14.87537
Trapezoidal $0.000010$ -2.44034
Midpoint $0.100000$ -15.01091
Midpoint $0.000010$ -2.44034
Simpsons $0.100000$ -2.43902
Simpsons $0.000010$ -2.43902
Left Rectangle $0.100000$ -8.68293
Left Rectangle $0.000010$ -2.43966
Right Rectangle $0.100000$ -21.06780
Right Rectangle $0.000010$ -2.44102

### Summary of Methods for $x^3 – 27x^2 + 8x$ over [0, 3]

Method Delta Percent Error
Trapezoidal $0.100000$ -9.88570
Trapezoidal $0.000010$ -0.00103
Midpoint $0.100000$ -9.97363
Midpoint $0.000010$ -0.00103
Simpsons $0.100000$ 0.00000
Simpsons $0.000010$ -0.00000
Left Rectangle $0.100000$ -5.08032
Left Rectangle $0.000010$ -0.00051
Right Rectangle $0.100000$ -14.69108
Right Rectangle $0.000010$ -0.00154

Of course, we must mention that there is some rounding in the percent errors. Simpsons, Midpoint, and Trapezoidal methods are not perfect.

# Prime Numbers – Expanded Modivations and Definitions

The explanation of primes, and their importance to number theory by definition must be both explainable, and justifiable with words as mathematics is an extension of language. That extension is by way of the principle that mathematics consists of ideas that are not only consistent, they are also communicable. Our proof that shows that an transfinite number of primes exists, does so by method of substitution. The principle that an entire idea can be expressible in a single symbol. It is quite possible to reverse substitute all of the mathematical symbols for the language that defines each symbol. Though the proof would become exceedingly long.

Basis for all numbers, and by consequence the basis of Number Theory, stems from the principle that all numbers are products of primes. This is the definition of Prime Factorization. What prime factorization does for us is put the composites in terms of primes. Further, the Sieve method eliminates composite numbers by consequence of their not meeting the definition of prime. That definition being any number that has exactly two factors, one, and itself. By this definition all negative numbers are not prime because they contain the additional factor of negative one. A composite number would be any number that contains more than two factors. Such as four, for example, in that Four has the factors of one, two, and four.

For primes this is obvious by the given definition, but less so for composites. Each composite number essentially is an unfactored product of primes. Knowing this further helps us when we reach beyond the primes we know into the realm of higher primes. This is quintessential when we are working with things like the Euclid algorithm or the Sieve algorithm. At this point eliminating numbers that are not prime is the only real method we know to find primes, which returns us the the “basis for all numbers” hinted at earlier.

# Polyhedra: Proof Of The Five

While the perfect structure of the regular polyhedra are fascinating, there are only five that are currently known: the tetrahedron consisting of four triangular faces; the octahedron with eight triangular faces; the icosahedron with twenty triangular faces; the cube with six square faces, and the dodecahedron with twelve pentagonal faces. These five unique polyhedra have been recognized throughout the ages dating all the way back to the ancient Greeks. The Greeks believed that these five shapes were the only in existence and none others exist naturally. Through analyzing these five polyhedra, we have come to the conclusion that the Greeks were right in their theory.

Why is this, however? First, it helps to understand the differences between 2D and 3D. Think of the vertices of the faces on a 2D polygon. When connecting polygons at their vertices, each vertex has a 360 degree radius for which shapes can be placed. Once all of the degrees are occupied, the vertex can no longer hold any other shapes. The angles that center around a vertex are the interior angles of the polygons attached. Regular polygons have equal interior angles, and regular polygons make up regular polyhedra.

Now, what makes 2D different from 3D? In order for an object to shift from the second dimension to the third dimension, it must have a height. An object that has no height is simply flat, a 2D polygon. Therefore, in order to be a 3D object, the object must also have faces. As a result, a regular polygon cannot enter the third dimension on its own. There must be more polygons. In the case of the regular polyhedra, there must be more regular and equal polygons. These polygons are attached at their vertices and line up along their edges. Also notice that two regular polygons are not enough to create an enclosed object because the two polygons would simply fold onto one-another. A third polygon is needed, and as a result, a vertex must connect a minimum of three polygons. Notice that this is the case for all of the five regular polyhedra: the tetrahedron has three triangles connected at each vertex while the octahedron has four and the icosahedron has five. The cube has three squared at each vertex and the dodecagon has three pentagons at each vertex.

Remember that a vertice has 360 degrees to work with in two dimensions. Now, remember that a 3D object must have a height. In order for a polyhedra to obtain this height, its regular polygon faces must be at angles to one-another, or they must bend at their edges. When the faces bend, the vertices no longer have a 360 degree radius. In other words, the vertices can only connect a total number of polygons whose interior angles equal a total less that 360 degrees!

Notice that each for each of the five regular polyhedra, this is the case. The equilateral triangle has interior angles of 60 degrees. In a tetrahedron, each vertex needs only to hold 180 degrees. In an octagedron, each vertex needs only hold 240 degrees. In a icosahedron, each vertex needs only to hold 300 degrees of interior angles. However, notice that there exists no regular polyhedron where each vertex connects six triangles. This is because six triangles would require a vertex to hold 360 degrees, something possible in 2D but not in 3D. The same applies to the cube which has vertices connecting three squares with interior angles of 90. The sum of these three interior angles is 270 which is possible, but four squares would contain a sum of interior angles too high for a 3D vertex. The dodecagon has three pentagons at every vertex. Pentagons have interior angles of 108 degrees. Three pentagons would therefore take up 324 degrees on a vertex which is enough for the vertex to handle. Four, once again, is too much.

The next regular polygon, the hexagon, does not form a regular polyhedron. This is because the hexagon has interior angles of 120 degrees. Because at least three regular polygons are needed at each vertex in order to form a regular polyhedron, a vertex would need to be able to connect three hexagons. However, 120 multiplied by three is 360 which just barely exceeds the maximum angular capabilities of a vertex. Three are able to share a vertex in 2D but not 3D as shown in the figures below.

How do we know that no regular polygons besides triangles, squares, and pentagons can form regular polyhedra, through? Notice that, as sides are added to regular polygons, their interior angles are always increasing. Therefore, the interior angles of the hexagon are smaller that that of every other regular polygon with a larger number of sides. Therefore, their angles will never be small enough to form a regular polyhedra.

# Numerical Integration Analysis Part 2

In addition to looking at the percent errors of different methods for integrating, our group also wanted to explore the effectiveness of the various methods for different types of functions. We were guided to explore an example of trigonometric, linear, quadratic, and cubic functions, but wanted to see if patterns emerged within these different types of functions and ultimately if a particular method of integration gives more accurate estimations given a specific type of function.

The actual data we collected, or a detailed summary of the data can be seen here. From this data we found that Simpson’s method is by far the best method of the five methods we explored, particularly for cubic functions. For symmetric functions such as trigonometric functions as well as quadratic functions, the midpoint method is also very accurate. Trapezoidal method provides an accurate estimation especially for small deltas, but has trouble with cubic formulas. The other Reimann sum methods provide over or underestimates depending on the type of function but, especially for larger deltas, are not as accurate as the other methods.

# Numerical Integration Analysis

Now that we have a number of numerical methods implemented, we want to compare them to see which method is best and in what circumstances.

We have a few test functions we were trying these methods over. Namely the following:

• $\cos{x}$ over [0, $\pi{}$]
• $2x + 1$ over [0, 1]
• $4-x^2$ over [0, 2]
• $5x^3 – 6x^2 + 0.3x$ over [-1, 3]

### Summary of Methods for $\cos{x}$ over [0, $\pi{}$]

Method Delta Percent Error
Midpoint $0.1$ -0.208927
Simpsons $0.1$ 0.054748
Right Rectangle $0.1$ 4.31267
Riemann $0.1$ 5.020046
Trapezoidal $0.1$ -0.33364
Left Rectangle $0.1$ -4.979954

### Summary of Methods for $5x^3 – 6x^2 + 0.3x$ over [-1, 3] (*)

Method Delta Percent Error
Midpoint $0.00001$ -0.00181
Simpsons $0.00001$ 0.00000
Right Rectangle $0.00001$ -0.00284
Riemann $0.00001$ -0.00103
Trapezoidal $0.00001$ -0.00181
Left Rectangle $0.00001$ -0.00078

Between these two sets of data, Let’s take a closer look at the magnitudes of the percent errors to see which method is more correct and then rank them.

Looking back at the first table, we can fairly easily tell that Simpsons rule is the best and Riemman sums was the worst. A little more difficult to pull out an order, so we made the computer compute the order: Simpsons, Midpoint, Trapezoidal, Right Rectangle, Left Rectangle, Riemann.

And same for the second table, we can easily see Simpsons was the best and Right Rectangle was the worst. And the order: Simpsons, Left Rectangle, Riemann, Trapezoidal, Midpoint, Right_rectangle.

Remember, when looking at the order between these two functions, we cannot say anything about each of the methods because we are changing two things in the comparison (the width of the rectangles summed and the function).

Now let’s take a closer look at the overall most correct method for all functions for each delta. That is, we will be varying the delta and looking at which method was the best.

Looking at a delta of $0.1$, we see that the Simpsons method is the most accurate for all our test functions. Interestingly though, with a delta of $0.01$ and $0.001$, the Midpoint method is better than Simpsons for $\cos{x}$, but Simpsons is still better for the other three. Moving to a delta of $0.0001$, we see Simpsons method is best for all functions again and remains to be for $0.00001$ as well.

So far, we can see that Simpsons method is amazing at single variable integration. But we will want to know how good? What’s the relative rates of accuracy increase between the methods?

Look for a follow up post where we post more data about our analysis and try to answer the above questions.

* Simpsons method looks to be $0.000$ here; this is the result of some rounding for presentation. The actual value is really close to zero but not quite zero.

Anyone who has not touched upon the p-adic numbers might find the topic rather confusing. Truth be told, it is confusing. The arithmetic part of the process can be a bit challenging due to its abstraction. Assuming that the reader is not familiar with the p-adic numbers, we will consider an example to clarify the matter.

Let’s consider the equation $x+64=0$. Any one individual might tell us that the solution is $x=-64$ when we ask them to solve the equation and of course, they would be right. But we wish to know the 5-adic representation of $-64$. We follow the following procedure:

Solution to $x + 64 = 0$ $mod$ $5$ is $x = 1 = 1 * 5^0$

Solution to $x + 64 = 0$ $mod$ $5^2$ is $x = 11 = 1 * 5^0 + 2 * 5^1$

Solution to $x + 64 = 0$ $mod$ $5^3$ is $x = 61 = 1 * 50^0 + 2 * 5^1 + 2 * 5^2$

Solution to $x + 64 = 0$ $mod$ $5^4$ is $x = 561 = 1 * 50^0 + 2 * 5^1 + 2 * 5^2 + 4 * 5^3$

Solution to $x + 64 = 0$ $mod$ $5^5$ is $x = 3061 = 1 * 50^0 + 2 * 5^1 + 2 * 5^2 + 4 * 5^3 + 4 * 5^4$

.

.

Solution to $x + 64 = 0$ $mod$ $5^n$ is $x = 1 *5^0 + 2 * 5^1 . . . 4 * 5^{n-1}$

For  $x + 64 = 0$ $mod$ $5$, we can instantly see that the solution is $x = 1$ since $1 + 64 = 65$ is evenly divisible by $5$. Notice that $x = 1$ can be written as $x = 1 * 5^0$. Now for $x + 64 = 0$ $mod$ $5^2$, we know that when we add $11$ to $64$ which gives us $75$, it will be divisible by $25$. Once again, notice that we can write $x=11$ as $11 = 1 * 5^0 + 2 * 5^1$. The same procedure applies from $x + 64 = 0$ $mod$ $5^3$ all the way up to $x + 64 = 0$ $mod$ $5^n$. In general, we take our $x$ value for $x + 64 = 0$ $mod$ $5^n$ and write the linear combination for it in the form of $x = 1 *5^0 + 2 * 5^1 . . . 4 * 5^{n-1}$.

Following the same procedure, we represent $-64$ in its 5-adic representation as $(1,2,2,4,4,…)$. Although quite interesting, the arithmetic for p-adic representation can be a tedious process. Now that we have seen an example, next we will consider p-adic representation of rational numbers and absolute value on $Q$.