With a fresh game, I decided to explore and try to find the rate at which Attack Rating increases based upon leveling up the Strength or Dexterity stats.

Seen as most weapons get additional parameter bonuses using both Strength and Dexterity, I decided to use the Butcher Knife as it only requires the strength stat, but a minimum of 24 strength.

The equation to find the total attack rating goes as follows:

TotalAttack = TotalPhysical + TotalMagic + TotalFire + TotalDark + TotalLightning

Since the Butcher Knife only uses physical damage, the other damage types were irrelevant. The calculation for total physical rating goes as follows:

TotalPhysical = BasePhysical + BasePhysical*StrengthBonus*ParameterBonus + BasePhysical*DexterityBonus*ParameterBonus

Explanation here can be hard to follow and charts seem intimidating, but the power point should help clear things up.

https://docs.google.com/presentation/d/1GuLZJay4Cvi8VvyoGwdwp4gVy1q64VX6hh67zfwA3OQ/edit#slide=id.g1310044aa8_0_59

]]>Webster’s Dictionary defines a Yo Yo as a round toy that has two flat sides with a string attached to its center, that is held in your hand, and that is made to go up and down by unwinding and rewinding the string with a movement of your wrist…….. But is it really just that?

**That the X Position graph of each yoyo is the antiderivative of the yoyo’s X Velocity graph, and the X Velocity is the derivative of the X Position.**

**That the X Acceleration graph of each yoyo is the derivative of each yoyo’s velocity graph.**

- Each yoyo although similar, have very slight differences in their stats even when performing the same tricks at the same string lengths.

**We first took videos of Yo-Yo champion Andrew doing yo yo tricks.**

- We then downloaded software that allows us to track an object throughout the video (the program is “Tracker”), which then produced a graph of position, velocity, and acceleration for both X and Y values.

- Using the different graph settings we are able to prove that the position, velocity, and acceleration are all derivatives of each other.

- MINDS BLOWN!!

First Trick:

https://docs.google.com/presentation/d/1KTSQMhCj8E-XDuvS4YGY170TBDidMv73NyfL3qnvHsA/edit?usp=sharing

]]>

In order to make this pendulum wave we started off by trying to figure out the lengths of the strings for all 7 pendulums that are in our project. If done correctly with the perfect lengths, the machine is supposed to initially look like a sine or cosine wave, then move into a chaotic phase, and then return to the original wave. The more pendulums in the wave, the better it will look.

String Length:

The lengths of the string are what affect the Pendulum’s period. For the Pendulum Wave to work you need to find the correct lengths. To determine these lengths we used this formula.

l(n) = g(T/(2pi(N+n)))^2

l(n) = length of string, g = gravity, T = period, N = number of oscillations, n = pendulum index (starting at 0)

Our Pendulum Wave:

As I said before, we used 7 pendulums on a yard stick and it was held by binder clips. I feel like we could have constructed the base so it was sturdier and so it could hold more pendulums. A common problem is that the string would tangle but we were able to stop that by looping the string through the nut and allowing space by the two sides of the string.

Presentation:

https://docs.google.com/presentation/d/1ueVVNH53MzA8BDO7PY-6wmmRPOV9qHiiQJBsO-N_Hw0/edit#slide=id.g1350596cc1_1_1

Citation:

http://www.instructables.com/id/Wave-Pendulum/

Our project is a game styled after Family Feud. The class will be split up into 2 teams and a player from each team will come up to answer a question. There are 8 non-calculator questions and 7 calculator questions. The person who can answer the question correctly first will earn their team a point. If they answer incorrectly it will be sent to the other player who will have a chance to answer. There will be random bonus questions throughout the deck for the team who answers correctly. The whole team can collaborate to answer this question and if they are correct they will receive another point. The team with the most points at the end of the game wins.

Our objective with this game was to create a fun way to prepare for the AP test. We tried to find a variety of different problems as well as make people think a little bit with the silly calculus questions. They seem funny when you read them but they really make you think about the basics of calculus and mathematics.

Link to Presentation

]]>Hashing is taking data and assigning it a unique numerical value through an algorithm. Many different hashing algorithms exist today, such as MD5, SHA-1, and djb2. In this presentation I used *djb2*.

Hashing data, similar to encryption, is widely used today in all sorts of applications. A common use of hashing is file integrity.

Let’s say you have a simple text file that says something like this:

`Hello World!`

Using the djb2 algorithm, our output is this:

`bc517990f19c4304`

However changing it slightly can result in a catastrophic change.

`hello world` –> `bf169756f8c65345`

Using a program called Vim in my Mac Terminal, I created a small file to give us hash values for whatever we enter (as long as it is 256 characters or less). I wrote the program using the C Programming Language.

Not really.

Here’s a few links for people who would like to try this themselves.

Source Code: CMDkOeOI.c

Windows Executable: CMDkOeOI.exe

Mac OS X / Linux: GQ35laY4.out

Tessellations are created from a repeating shape that can infinitely cover a plane without gaps or overlapping.

One regular shape, a triangle, square, or hexagon, can be used to to create a regular tessellation. An equilateral triangle and a square are most commonly seen, but any triangle can be used to tessellate, and any quadrilateral.

A semi-regular tessellation is composed of two or more regular polygons. To determine if a tessellation works, you add up the interior angles at the vertex (the point where the shapes meet). All the vertex’s in a tessellation should be identical, and should add to 360 degrees.

To name a semi-regular tessellations, name the number of sides on the shapes surrounding the vertex, starting with the least number of sides.

Demi-regular tessellations are the most abstract. They can be any shape that will tessellate. These were most commonly seen when we were in grade school, and colored in tiled shapes or animals. M.C. Escher created the famous Regular Division of the Plane with Birds, shown below.

He was born in Leeuwarden, Holland in 1898. The basis of his tessellations all respect three, four, or six-fold symmetry. The birds on a plane have a triangle basis, and Development I (shown below) has a more obvious square basis.

In other examples of tessellations with abstract shapes, the basis shape was simply changed. For example, you can start with a square, and draw a line through the square, cut that piece off, and slide it to the opposite side, resulting in something like this.

Or, you could take a side of a square, draw a line to the midpoint, and then rotate that shape around the axis. The bottom half will then be a cut out, and the top half will have that shape added on. These new shapes may make it hard to prove the tessellation works, but you can determine if the vertex adds to 360 degrees by drawing tangent lines to the curves at the vertex.

Although it seems as if M.C. Escher simply drew birds that fit together, there is mathematical reasoning behind them to explain why they fit together, and the geometry makes it possible. As a kid, when you were coloring fish on a tile, it seemed like a simple art activity, but you were already practicing geometry.

]]>This is to help visualize how we turn a simple graph into a 3D model by rotating it around the x-axis.

We We used a 5 piece piece-wise function to match our graph.

We then used integrated the equations to find a volume of 393 mL, a 10% error from the measured 355 mL. The error is due to human error, the odd shape of the bottle making it hard to measure, and that our functions don’t 100% match the points.

If you want to play around with your own solids or revolution click here.

]]>

There is a more technical vocabulary for some of the things in string art, the tangent lines are called chords and the curved line that the chords form is called the intersection envelope. Above, you can see the chords in blue and the intersection envelope in red.

The simplest of the curves that can be made with string art only uses the positive ends of the x and y axis which puts you in the first quadrant. To make a curved line you must start in the first position on the x axis and the nth position on the y axis. From there you move to n-1 and the x+1 place and continue this process until you wish to stop and you will be left with a curve, as shown below. Depending on how far away the positions are on the x or y axis, the intersection envelope will be either smoother or more edged, in theory the smoothness of the curve is infinite. Also, it is possible to make the curve wider or thinner by changing the angle of the axis.

Further, derivatives can be used to form a circle using a series of straight lines.

String art itself is made up of many Bezier curves. Bezier curves are lines that start and end at two different points but are also attracted to other points although they never touch them. The points that the lines are attracted to but never touch are called “control points”. There are three main degrees of Bezier curves, as shown below, but the one most common in string art is degree 1.

String art is more than just a pretty project with some math to pass the time however. Suspension bridges, such as the Jerusalem Chords Bridge that is shown below, use the same math so that the weight of the bridge is distributed towards the top so that the bridge can hold massive amounts of weight.

The math behind string art can be found in many aspects of many day life, whether it is a fun pass time, holding up a bridge, the curves in letters or something else entirely.

https://docs.google.com/presentation/d/1uVnh1G6kVfmXJR6fSndaw8yq131rE96Bf5IY8RywlPE/edit?usp=sharing

]]>

**Blooms are based on a single number**: 137.5º is a very special angle, called the golden angle, based on the golden ratio. When that angle is used by nature as a growth strategy, it leads to the formation of spiral patterns.

The spiral patterns that are recreated in the bloom are based on mathematics. These spirals are seen all across the world in nature. That is something that peeked our interest. To us, mathematics seemed like something detached from the natural world. Something very rigid and man-made. Seeing these patterns in nature was something that came as a shock to us and was the cause of our further exploration into the topic.

To begin, we went back to the basics of these spiral patterns. We researched a couple of mathematical explanations: The Fibonacci Sequence & The Golden Ratio.

**The Fibonacci Sequence**

The Fibonacci sequence is a series of numbers where a number is found by adding up the two numbers before it. Starting with 0 and 1, the sequence goes 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, and so forth.

This sequence is essential to the spiral pattern because this is where the dimensions for the spiral come from. This can be seen below.

The numbers in the Fibonacci sequence are seen everywhere in nature from the way bunnies breed, to how many spirals there are on a pineapple, to how spiral galaxies are shaped!

The Rule for the sequence is **x _{n} = x_{n-1} + x_{n-2}**

Using the Fibonacci Sequence, we can actually move right into the next topic:

**The Golden Ratio.**

If we divide Fibonacci numbers (especially larger values), a peculiar thing occurs. There seems to be a limit. This can be seen below.

1.618…….The Greeks called this PHI. We call it THE GOLDEN RATIO!

To prove that the limit of dividing Fibonacci numbers is in fact the Golden Ratio. We completed a proof seen below.

This is all very important to the construction of a 3-D bloom. Remember… **Blooms are based on a single number**: 137.5º is a very special angle, called the golden angle, based on the golden ratio. When that angle is used by nature as a growth strategy, it leads to the formation of spiral patterns.

To construct a bloom, each petal must be strategically placed based on the golden angle, just like how they are made in nature!

When a strobe light is shone on the bloom as it spins, amazing affects can be seen as seen in the video below.

For our project, we 3-D printed a bloom in the innovations lab and created a turntable for it where speed could be adjusted. We then obtained a strobe light that had adjustable frequencies as an attempt to recreate the above effects. Below are some pictures of our project.

Bada Bing Bada Bloom.

Link to Presentation:

https://docs.google.com/presentation/d/1iDk9sS3IC450426fwezYpMhSPziRsuMaZxiqdvdJOWY/edit?usp=sharing

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The three problems we talked about and tested were:

- The rope around the world problem- How many feet of rope is required to wrap around the world, and how much more would we need if we raised the rope ONE foot off of the earth’s surface? Upon our calculations, the answer turned out to be a mere 2pi feet, or 6.28 feet, which is much less than people would expect.

2. You are on the Monty Hall game show as a contestant. Monty, the host, calls you down and you play this game. In the game, there are 3 doors. Behind one is the car of your dreams (you name it) and behind the other two are goats. You have to choose one of the three doors, whatever one you think has the car behind it. Once you choose, Monty then reveals one of the other two remaining doors that has a goat behind it, leaving you now with two doors instead. He then asks you if you wish to keep the door you chose. Or to switch your choice to the other door. So, what do you do?

The answer to this, believe it or not, is to switch doors EVERY TIME. The math is in the slideshow linked to this post.

3. Finally, we explored and proved the birthday paradox. Basically, the birthday paradox asks this:

Theoretically, how many people can you put in the same room to find a pair of people with the same birthday?

The answer is 23 people. (Well, sort of)

We used permutations and probability to prove that in a room of 23 people, the liklihood of 2 people sharing the same birthday is 50%, and that percentage only raises exponentially as more and more people enter the room. For example, we experimented with 40 people, and the percentage rose to around 80%.https://docs.google.com/presentation/d/1ve4i7-Mh5iK4d-mUPo6KPs8xoAP5t1NeBLqUmZA8h3A/edit#slide=id.g1447889c57_1_0

]]>**“A fractal is a never ending pattern and a natural phenomenon. Fractals are infinitely complex patterns that are self-similar across different scales. They are created by repeating a simple process over and over in an ongoing feedback loop. Driven by recursion, fractals are images of dynamic systems.”**

If the replication is the same on every scale, it is called a self similar pattern.

An example of this the** Menger Sponge**:

- Begin with a cube
- Divide every face of the cube into 9 squares (imagine a Rubik’s Cube). This will sub-divide the cube into 27 smaller cubes.
- Remove the smaller cube in the middle of each face, and remove the smaller cube in the very center of the larger cube. Thus leaving 20 smaller cubes.
- Repeat steps 2 and 3 for each of the remaining smaller cubes, and continue infinitely.

**Koch Snowflake **

- Begin with an equilateral triangle
- Replace the middle third (median) of every line segment with with a pair of lie segments that form an equilateral triangle.
- Erase the base of the triangle.

**The Mandelbrot Set**

On the complex plane, is the set of all points where:

How **WE** created this:

- Begin with a rectangle, labeling the length as X and the height as Y.
- Divide the large rectangle into four similar rectangles using the dimensions of 4/9X and 4/9Y. A cross-like shape should appear in the middle with the dimensions of 1/9X and 1/9Y.
- Repeat this process, dividing each triangle in the same manner an infinite amount of times.

At time I was writing code in processing to generate organic looking structures.

ArrayList myBugs = new ArrayList(); void setup() { size(500, 500); frameRate(60); background(0, 0, 0); colorMode(HSB, 255); color seedColor = color(200, 255, 255); Bug seedBug = new Bug(int(width/2),int(height/2),seedColor); myBugs.add(seedBug); } void draw() { for(int i = 0; i<myBugs.size();i++) { myBugs.get(i).update(); } if(frameCount%10==0) { saveFrame("Frame###.jpg"); } } class Bug { int xpos, ypos; color myColor; Bug(int x, int y, color c) { xpos = x; ypos = y; myColor = c; } void update() { int huemin = 0; int huemax = 255; int randChance = 45; set(xpos, ypos, myColor); if(get(xpos-1, ypos) == color(0, 0, 0)) { if(random(100)<randChance) { myBugs.add(new Bug(xpos-1, ypos,color(random(huemin, huemax), 255, 255))); } } if(get(xpos+1, ypos) == color(0, 0, 0)) { if(random(100)<randChance) { myBugs.add(new Bug(xpos+1, ypos,color(random(huemin, huemax), 255, 255))); } } if(get(xpos, ypos-1) == color(0, 0, 0)) { if(random(100)<randChance) { myBugs.add(new Bug(xpos, ypos-1,color(random(huemin, huemax), 255, 255))); } } if(get(xpos, ypos+1) == color(0, 0, 0)) { if(random(100)<randChance) { myBugs.add(new Bug(xpos, ypos+1,color(random(huemin, huemax), 255, 255))); } } println(myBugs.size()); if(myBugs.size()>10) { myBugs.remove(this); return; } return; } }

It starts out by generating a seed “cell” the cell then randomly generates neighboring cells. The seed cell then deletes it self. All the newly generated cells then pick a random color, and repeat the same processes as taken by the seed cell. Later I decided to base the color on the number of pixels colored in. It does not use every color once, but it does go through all the possible hues.

I can vary the general shape by changing the percent chance for a cell to create a neighbor.

Now for the fun part, getting every color to show up once, and only once. Computers represent colors as three distinct values red, green, and blue. Typically these values have a range of 0 to 255. 255 can also be represented as 2^{8}-1, the largest possible value represented in 8 binary bits. For 8-bit colors there are 255 x 255 x 255 or 2^{24} possible colors. To store these colors in memory I created a three dimensional array where each set of three coordinates represents that colors red, green, and blue values respectively. For example the data located at 100, 30, 65 is the data for the color red = 100, green = 30, blue = 65.

Here’s the color in case you were wondering what it looks like.

Every frame for each cell I calculate the average color of all the surrounding cells. With this average I search through the (three dimensional) array of available colors and select the closest available. I determine the distance using the Euclidean distance between the color values.

So I started off by rendering images with 6 bits per color channel, also know and 18 bit colors. The images generated contain 262,144 or 2^{18} or 2^{6} x 2^{6} x 2^{6} colors. These images are generated at a resolution of 512 x 512 pixels.

Here is the first image I generated.

Now we’re onto something. This image took about 20 minutes to generate. After some optimizations to my code, I can now generate an image this size in seconds.

Once I saw that my algorithm worked, I want it to run as fast as possible. With some careful reasoning I figured that I don’t need to search the entire space for available colors, only colors close by, then expand my search if none were found. With this change, images generated hundreds of time faster. So I decided to generated some larger images. I stepped up to 7 bits per channel or 21 bit colors. With 21 bits there are 2,097,152 or 2^{21} possible color combinations. The images are 2,048 by 1,024 pixels.

Here are a few pictures and videos of the images created each used a slightly different variation of the same algorithm.

And for the grand finale I present an image with every possible color with 8 bits per channel also know as 24 bit colors. These images contain 16,777,216 or 2^{24} different colors. Each image is 4,096 by 4,096 pixels. Click on the image to view a full size JPEG. Because of the compression required to put these image on the web they do not contain every RGB color, the original TIFs do, and are 48 MB each.

And my personal favorite.

]]>https://drive.google.com/file/d/0B4re6ufaa0KAeUEtenVCWnZKREU/view?usp=sharing

The simulation was changed several times, from 20 total possible Blairs, to 40 total possible Blairs, to 80 total possible Blairs, to almost 600 total possible Blairs and 80 total possible Blairs in a smaller space.

After doing 4 trials with all these versions (except the 600 Blair version) I put all the numbers on a spreadsheet.

https://docs.google.com/spreadsheets/d/1yqByTGBer7dZ7cZBNnpe139XnI8EuYxbFPdbn4q70as/edit?usp=sharing

Then, using the equation dy/dt = ky(M-y) and turning that equation into y/M-y=e^Mtk*e^c I can and did use these equations to find the k values at each point, the average k values for each trial, and the average k value for each version. These are my findings:

The larger the population, the faster the simulation ends, and the k value gets smaller.

In a smaller environment, the trial ends faster, but the k value gets bigger.

So now in the next Apoca-Blair, you know what changes the k value of his logistic growth.

]]>So…when I was first coming up for ideas about this project my first thought was that I wanted to do food because if I can remember anything about my BC Calc class second block it was that I was always hungry. The idea of rock candy somehow came up. So I started looking up the mathematics of rock candy which then led me to this whole universe of geometry, vectors, 3D structure and all in all CRYSTAL MATH…

The idea with the rock candy was to see what the trend of the rate of volume for the crystals would look like. My hypothesis was that the volume would increase exponentially since the growth of sugar crystals depends on the amount of surface area that’s available to stick to.

I took out the rock candy to measure about every twelve hours to get volumes by both integration and by water displacement.

Graphs

Day .5

https://www.desmos.com/calculator/ndsc4akgow

Day 1

https://www.desmos.com/calculator/qyjxuwezyj

Day 1.5

https://www.desmos.com/calculator/hhrtmg8wnn

Day 2

https://www.desmos.com/calculator/uv8decod43

Day | Volume by Integration | Volume by Water Disp. (cc) |

.5 | 3.498 | 3 |

1 | 7.35 | 6 |

1.5 | 24.15 | 17 |

2 | 27.35 | 22 |

The graph of this data is here:

https://www.desmos.com/calculator/uhhcn9g6qf

My hypothesis turned out to be for the most part correct!

As I was making rock candy I also started looking up the properties and structure of sugar…turns out there are 7 basic crystal structures…

There is actually a lot of math that relates to these structures such as vectors and the differences in bonding angles depending on which structure your crystal happens to be…

To see what these vectors look like, here’s the link…

http://www.materials.ac.uk/elearning/matter/crystallography/3dcrystallography/7crystalsystems.html

As it turns out, sugar is monoclinic that’s why its crystals form into cube-like shapes.

So if I were to go to one corner of the sugar cube and label three vectors a, b, and c these vectors would ideally stay equal to each other as the crystal grows.

Then I bumped into a few names as I was researching and this one name popped up everywhere…Dmitry Kondrashov.

Dmitry Kondrashov is a professor at the University of Chicago. He studies crystal structure and crystal lattices using x-ray technology. Kondrashov composed a geometric theory of crystal growth in 1998 and I decided to give it a read. The math involved in calculating errors and factors that impact the ideal crystal shape go far beyond my competence of math knowledge however, I still learned quite a bit. There were many components to his theory but adjusting to time and my knowledge of this subject I focused on what Kondrashov mentions to be a “Wulff shape”

The Wulff shape is related to the Wulff construction method and it’s used to determine the equilibrium shape of a crystal that has a fixed volume in a separate phase like vapor or saturated solution, like the sugar solution used to make rock candy. In order to obtain this equilibrium it must abide by the principles of Gibbs thermodynamics by minimizing the total free energy of the crystal’s surface. When two fluids touch the portion of total free energy is proportional to the area of the surface in which the fluids are in contact. Motion of the planes created by the two fluids can be calculated using several equations such as these…the only issue is figuring out where the corners will meet so a system of linear equations would need to be simultaneously calculated as well. Kondrashov’s theory takes into account many of the factors that impact crystal growth such as energy loss, spacing,phase changes, and surface tension among other things. Another interesting Kondrashov states is that in order to concentrate on the shape of the crystal rather than the size, energy is often replaced by surface tension in many of the equations that are part of his geometric algorithm.

Kondrashov briefly discusses the main search in the world of crystallography, the main objective is to find and predict the equilibrium shape of a crystal from its internal structure using math.

All in all, through this project I’ve continued my love for 3D structure, not only are crystals beautiful to look at the amount of calculations and math that go into them is pretty interesting as well.

Thank you for reading!

Recipe for Rock candy….

On a stick

1 cup of water

3.5 cups of sugar

lollipop sticks

glass jars

clothespins

flavoring (optional)

1. First dip the lollipop sticks in water and coat the area you’d like the crystals to grow in sugar, this gives the 2. crystals something to latch onto.

3. Heat water and sugar and bring to a boil, boil for about ten minutes or until you can’t see the sugar crystals anymore.

4. Let solution cool and add flavoring and food coloring if desired.

5. Pour solution into glass jars take clothespins to dip only the portion you’d like to grow crystals on, into the solution (make sure the lollipop sticks are dry!).

6. Wait for crystals to grow!

“Broken Glass”

half a cup of water

¾ cup of light corn syrup

2 tsp of flavoring

2 cups of sugar

food coloring (blue if you’re a fan of Breaking Bad)

1. Heat sugar, corn syrup, and water and bring to 285 degrees

2. Remove from burner, add food coloring and flavoring

3. Pour into a thin layer on a baking sheet that has been lined with foil and greased with cooking spray.

4. Let sit for at least 30 minutes until brittle

5. Break with a mallet and eat

Works Cited

http://gamediv1.weebly.com/uploads/9/8/4/6/9846625/399563333_orig.gif

http://www.materials.ac.uk/elearning/matter/crystallography/3dcrystallography/7crystalsystems.html

http://users-phys.au.dk/philip/pictures/surface_science/fig6_0.gif

I ended up reading only about half of the whole book. The parts about Edward Frenkel’s life I found very interesting. A little bit about Edward Frenkel’s life:

He’s a mathematician that was originally interested in quantum physics, but was converted to mathematics by Evgeny Evgenievich, who became one of his mentors at a young age. He wanted to study mathematics at Moscow State University, but was not accepted due to the extreme discrimination against Jewish people. Instead, he had to enroll in the applied mathematics program at the Moscow Institute of Oil and Gas. He snuck into lectures at MGU, attended Israel Gelfand’s seminar and worked with mathematicians like Dmitry Fuchs. Frenkel received his PhD from Harvard University in 1991, and now has been a professor at UC Berkeley since 1997.

The math parts of the book I found….not so interesting.

Frenkel talks about some pretty advanced stuff, and a lot of it was over my head. It took a while to even get through a chapter because I constantly had to keep re-reading to try to understand what he was talking about.

At one point in the book, Frenkel talks about Fibonacci numbers briefly, and I was like woah something I actually understand!

So, I decided to take this topic from the book and research it a little more.

We all know that the Fibonacci numbers are defined as each number being equal to the sum of the preceding two numbers. (1,1,2,3,5,8,13,21,34,…)

Using this rule, you can only find the nth Fibonacci number if you know the previous two, which isn’t very effective.

Is there a way to find the nth Fibonacci term without finding all the previous numbers?

In his book, Frenkel talks about how the Fibonacci numbers can be generated using the following series:

q+q(q+q^2)+q(q+q^2)2+q(q+q^2)3+q(q+q^2)4+…

When expanded out, the coefficient in front of q^n should be the nth Fibonacci number.

I did this for n=1 through n=4 and showed that the coefficients do come out to be the Fibonacci numbers.

I also was researching the Fibonacci sequence online and found another way to find the nth term of the sequence. This can be done using Phi (comes from golden ratio) and phi (the reciprocal of Phi).

Obviously by hand is not the most efficient way of going about that, and it would be much easier to do with a calculator. Using this, we can find essentially any Fibonacci number we want. For example, we can find the 50th term.

The Fibonacci sequence appears a lot in nature, and it also occurs in the sums of the diagonals of Pascal’s Triangle.

So this was me expanding on a topic I could fully comprehend from the book I read, Love and Math, by Edward Frenkel.

I’m sure this book is really good if you’re smarter than I am and can get more out of it.

Credits:

http://www.motherjones.com/files/Frenkel-LM.jpghttp://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.htmlhttp://www.goldennumber.net/wp-content/uploads/2012/05/pascals-triangle-fibonacci.gif

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The spider web is actually comprised of numerous radii, a logarithmic spiral (given by the polar equation r=ae^{bθ} ) and the arithmetic spiral (given by the polar equation r = a+bθ ). The constants a and b are adjustable.

Construction begins with the spider’s identification of two vertical, structural supports. After ascending one of the supports, the spider perfectly predicts the length of silk that it will need to reach the other vertical support. A gentle gust of wind is all the spider needs to carry it to the other side. Once the very first strand is constructed and strengthened by multiple strands, the spider will build three radii and a web-frame. More radii are constructed to the frame and then a temporary logarithmic spiral is built. Keith Devlin describes the logarithmic spiral that increases distance between successive turns by a constant growth factor (Devlin 89). View my graph of a logarithmic spiral below as well as further reading about the logarithmic spiral.

http://www.desmos.com/calculator/cis4o1sxc8

Logarithmic Spiral

http://mathworld.wolfram.com/LogarithmicSpiral.html

Next, the spider will turn around and spiral inward with an arithmetic spiral while simultaneously eating/ destructing the temporary spiral. According to Keith Devlin:

*In an arithmetic spiral, a line drawn outward from the center crosses each successive turn at the same distance, and the tangent angles increase at a constant rate, approaching 90 degrees as the number of turns increases (Devlin 89).*

Below is a picture of my graph of an arithmetic spiral as well as some further reading.

Arithmetic Spiral

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

At the completion of the web, the spider retreats to the the center of its web, suspends itself upside-down, motionless while awaiting its prey. The final product of the web, if undisturbed and created perfectly, would look something like this.

Pattern of Final Product created by Alicia Bautista

As part of my experiment, I decided to attempt a time logged construction of an orb web. I am certainly much slower than this guy.

My construction, which is incomparable due to my lack of engineering capability, took me about three hours. However, I bypassed the logistic spiral and went straight for observable end product of radii and arithmetic spiral. The average garden spider, an orb-weaver, takes between one and three hours to construct its web. Spidy Speed!

In order to stimulate the stronger, structural silk that spiders use, I used fishing line. In place of the arithmetic silk spiral, which is used to catch the prey, I constructed a spiral of hot glue. I got to “be the spider” so to speak. The properties of hot glue are actually pretty similar to the “sticky” silk that spiders use since both materials solidify almost instantly.

However, my web is very fragile and was only sticky for about 2 seconds. If I were a spider, I would die.

It just goes to show that spiders are true engineers by nature. I mean just look at them!

Video

https://www.youtube.com/watch?v=DwLn11KJS9Y

For further reading on the the spider and its construction methods:

http://ednieuw.home.xs4all.nl/Spiders/InfoNed/webthread.html

http://grandpacliff.com/Animals/Spiders/OrbWebCon.htm

Blog by: Alicia Bautista

]]>Everyone loves popcorn. People have been popping corn for 5000 years. Over the last century, there has been an expanding market for mass produced popcorn in the United States and around the world. There are multiple ways to pop popcorn. The original way would be to cook the kernels over a heat source such as a fire. This developed into cooking corn over a stove and eventually evolved into mass producing product such has Jiffy Pop. With the development of the microwave, making popcorn went from being a clumsy time consuming process, to clicking a button. But….

How do you know how long to click that button for?

To answer this question, we designed an experiment using a standard recipe for making your own popcorn at home. All you do is put kernels of popcorn in a paper bag, fold the top of the bag over, and put it in the microwave. The question we were looking to answer was “How long do you put the bag of popcorn in the microwave for?”. At first, we tried this experiment with a 1/4 cup of kernels in every bag. This didn’t quite work out as planned.

Original Trial and Errors

We were trying to make each time as realistic as possible by using an online recipe for homemade popcorn, which consisted of simple ¼ cup of plain kernels in a paper bag. We assumed that ¼ cup would have the same amount of kernels in each cup, which we counted to be 380. However, this was not true as we found in one specific trial upwards of 450 kernels. Because of inaccurate measurements and differences in kernel size, the real number varied from trial to trial. This cause the data to fluctuate in a non logistic pattern, so we decided we needed to start over.

Effects on Optimal Time

We determined that there are several factors which affect the amount of time it takes for the kernels to pop and burn. The number of kernels popped at one time is the main cause. We first tested ¼ cup where we found all popcorn was completely burnt at 4 minutes and ten seconds. We then changed the amount to 100 kernels per bag, but at 4 minutes 15 seconds only 7 kernels were burned in our first trial and only 2 in our second. Even in our largest trial, 5 minutes, only 11 of the 100 burned in the first trial and 9 in the second. We can conclude from this data that the fewer kernels there are in a bag, the longer you will need to set your timer.

Another factor that affects each individual kernel is the moisture content of the kernel. The amount of water in each kernel determines how long each kernel will take to pop because the microwave vaporizes the water causing the kernel to expand. If there is more water in a kernel it will take longer it to pop. However, the lower the water content, the less likely the kernel is to pop. While you can calculate the moisture content, you can’t control it. The only way we would be able to calculate it with the technology available to us would be to weigh each kernel before and after popping, subtracting the difference and finding the percentage of water. But because we still only would know the content after the trial is complete it is not a effect control. Therefore, time and technology forced us to ignore this variable and assure it was constant.

Using the function we graphed our data, and found the function that models our data the best.

- has variables; K-constant, Po-Initial popped kernels, r-rate, t-time (in seconds)
- Although there are no popped kernels when we start our trial, we cannot use zero for Po because all values on the line are zero when we do so. Because we could not use zero, we tried using other numbers close to zero but found that when we used .01 for Po, the line of best fit had the highest correlation value.
- For the first trial, the line fit our graph pretty well and had a correlation value of R^2=.904
- K= 40.922 r=.048932
- In the second trial, the line fit our graph the best out of any trials. We had a correlation value of R^2=.994
- K=43.342 r=.04891
- In our third and final trial, the again fit pretty well, but not quite as well as the second trial. We had a correlation value of R^2=.988
- K=44.202 r=.047704
- All of our data on the same graph shows that our trials were very similar, and for the most part the best fit lines were alike. Even though while recording it didn’t seem it, our data was fairly accurate.
- After looking at all of our data, and of course taste testing all of the popcorn to see which trials were really the best, we came to a conclusion. Although we never reached a point were burning the popcorn was a worry, the popcorn did eventually get crispier than preferred. The best tasting popcorn and most popped kernels occurred around three minutes. Although we did have a couple crispy pieces of popcorn, three minutes is the optimal time to pop your popcorn.

- Desmos- https://www.desmos.com/calculator/sv9rggmkop
- Data Sheets- https://docs.google.com/spreadsheets/d/1QpqPWmf05LWFISOYXU-HCI8CbWUb_rTvKRUoiI4K7R8/edit?usp=sharing
- All trials were done using a 1100 watt, over the stove microwave.

**The Friendship Paradox:**

Most people when asked about their relative number of friends, will predict that they have more friends than their friends have. However, a mathematical phenomenon termed the “Friendship Paradox” disproves this. First discovered in 1991, the paradox essentially asserts that on average, your friends have more friends than you.

To examine the paradox on a real world level, we decided to enter the Twitter world and record the number of followers within 100 “chains” of Twitter users. To begin, we took the 5th most recent follower of @POTUS, or President Obama. We used @POTUS because at the time of our study, this account was the fastest growing account in twitter history. In addition, the President is followed by the widest range of account types and does necessarily appeal to a specific aspect of society. From this 5th most recent follower, we recorded the amount of “followers” and the amount of “following.” Then, we took the 5th most recent person this user had followed, and did the same. Each “chain” of users consisted of 5 accounts. The hope was that on average, the amount of followers would increase as the chain progressed to the 5th person.

Before we reveal the trends that were discovered from our data, let’s examine the paradox on a simpler level.

**Situation:** There are four people, named ‘A,’ ‘B,’ ‘C,’ and ‘D.’ The following is a visual representation of their friend group, with the lines indicating a friendship.

A has just 1 friend, B has 3 friends, C and D each has 2 friends. A has 3 friends of friends, B has 5 friends friends, C has 5 friends of friends, and D has 5 friends of friends.

For persons A, C, and D, the number of friends of friends is greater than the number of friends. For Person B, this is not the case as he represents a “popular” person, who is likely to have more friends than his friends have.

There are 8 total “friendships” in this situation, meaning the average person has 2 friends.

There are 18 total “friends of friends” among the 8 friendships meaning the average friend has 2.25 friends. This is the friendship paradox.

**The math behind it:**

If person *i *has a number *xi* friends and there are *n* amount of people, we must figure out what the average amount of friends is for the entire graph. To do this, we divide the total number of friends by the total number of people. So,

We need to calculate the total sum for this average, and so this means we only need to count the friends of *i* if we count the friends of the friends of *i*. Each of the friends of *i* will give the term *xi* to the total sum, so the holistic summation has (*xi*)(*xi*) = *xi*2. The numerator will be the sum of the squares of each friend’s friend, and the denominator will be the sum of the total number of friends. This gives us the mean:

Using the variance formula,

We can rearrange it into the following equation.

If we substitute this into the original equation we get…

This means that the average amount of friends is less than or equal to the average amount of friend’s friends.

Benford’s Law

Benford’s Law is a mathematical phenomenom named after physicist Frank Benford, although it was first discovered by Canadian Simon Newcomb. While looking through his book of log tables, Newcomb realized that the earlier pages of the book, (the pages that had numbers with a leading digit of 1), were of much greater quantity than the other pages.

Benford’s Formula: the probability of the leading digit being of a certain value can be described by

Probability (D) = log10 (d+1) – log10 (d) or Probability (D) = log10(1+(1/d))

Connection to enhance understanding:

Think of a pencil of length 1x. In order for the leading digit of the length of this pencil to become 2, the length must change by a factor of 100%. From here, for the leading digit of the length of this pencil to become 3, the length must change by a factor of 50%. And so on.

Examples:

-Distribution of First digit (Meters)

-Daily volume of shares on NASDAQ

-Import/Export Volumes for sales of fish from USA

-Distribution of first digits of altitude in top 120,000 towns in the world

Fraud application:

The majority of people who attempt to commit fraud in either tax returns, sales reports, or voting records are not aware of this phenomenon. Therefore, they attempt to “unsuspiciously” provide an array of numbers with a “random” distribution of leading digits. Data that does not adhere to Benford’s Law is viewed as suspicious and often leads to speculation over the legitimacy of the data.

Because we had recorded data on 1000 twitter accounts, we decided to see if the phenomenon applied to Twitter followers as well. As is indicated in the table below, the data generally followed a trend typical for data that adheres to Benford’s Law.

Thank you for reading!

Sources:

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50 degree average velocity (inches/second): 279.5

54 degree average velocity (inches/second): 232.7

58 degree average velocity (inches/second): 220.7

50 degree average time (seconds): .98

54 degree average time (seconds): 1.26

58 degree average time (seconds): 1.06

**After integral:**

50 degree average distance (inches): 273.91

54 degree average distance (inches): 293.202

58 degree average distance (inches): 233.942

Image:http://smock.foreuphosting5.com/wp-content/uploads/2015/04/tight.jpg

]]>**Step 1: Calorie Intake**

In order to determine the population limit, we first had to do some basic calculations.

- Current total population: 7.2441 billion = 7,244,100,000 people

**Assumption: Every person needs the calorie intake of a moderately active average adult male (age 30, 5’10”, 180 pounds)

- Calorie intake needed to maintain body weight: 2,500 calories per day, 912,500 calories per year
- Total calorie intake for entire world population: 1.811025 x 10^13 calories per day, 6.61024125 x 10^15 calories per year.

**Step 2: Calories in Corn**

**Assumption: Everyone’s diet consists of only corn. Even though this would not be feasible in an actual situation, it provides a simplified model of food production.

- Calories: 62.8 calories per ear of corn
- Ears of corn needed to sustain everyone for a year: 1.05258618631 x 10^14 ears of corn
- One acre of land can yield 18,000 ears of corn per year
- Amount of land available to grow corn on: 7.68 billion acres on Earth = 7,680,000,000 acres
- Maximum amount of corn produced: 1.3824 x 10^14 ears of corn per year
- Maximum amount of calories produced from maximum amount of corn: 8.681472 x 10^15 calories per year

**Step 3: Population Limit**

Based on this data, we were able to calculate the maximum population on Earth based on the current maximum production of corn. We concluded that the carrying capacity of Earth is **9.513941917 billion people**.

**Step 4: Population Growth Over Time**

Using the population limit from the previous step and the last 200 years of population growth data, we were able to calculate the equation of the graph that displays the logistic growth of the population.

Using the equations from sheets 3 and 4, we graphed two logistic graphs to see the growth of the world population. (Red graph is from sheet 3 and black graph is from sheet 4)

__Step 5: Arable Land__

**Assumption: Everyone’s diet still consists of only corn and the yield per acre of corn is constant over time.

- Arable land is being lost at the rate of approximately 38,610 square miles (24.7 million acres) per year = 24,700,000 acres per year
- We used the amount of arable land available to find the amount of corn produced per year, the maximum amount of calories from corn, and the maximum sustainable population.
- In 2015, there is 7,680,000,000 acres of arable land.
- We used the same data as we did in Steps 1 and 2 to determine the new sustainable population given that the amount of arable land is decreasing.

- Based on this data, a linear relationship can be determined for the sustainable population. The equation of the linear data set is:
- (y – 9513941917.81) = -30598224.658(x – 2015) (green line on graph below)

- Using this graph and the black logistic growth graph, we determined that with decreasing arable land and constant corn production, the population of the Earth will just be sustained in the year 2055. After 2055, the amount of food produced will not be enough for the calories required by the entire world population.

**Step 6: Technological Innovation**

**Assumption: Due to continuous farming innovation, every five years the yield of corn per acre increases by 5% of the preceding five year’s yield of corn. The amount of arable land is still decreasing as it was in Step 5.

- We plotted the points as shown in the graph below (orange line).

- We would never run out of food in this case because the rate of technological innovation is increasing the yield of corn faster than the rate at which arable land (therefore amount of corn) is lost.

**Step 7:**

**Assumption: Relationships between oil-producing nations and the rest of the world slowly deteriorate with no hope of reconciliation. As a result, despite the need of corn for consumption, the corn-producing nations of the world must switch machinery and vehicles over to ethanol. Every 5 years the percent of ethanol needed increases by 3% as gasoline becomes more scarce. More corn must be collected to convert into ethanol by the government. (Also assume that the amount of corn is based on decreasing amount of arable land and increasing amount of technology).

- We plotted the points as shown in the graph below (purple line).

- In 2085, the amount of food produced will just barely be able to support a population of about 8.466 billion people.
- (Due to the scale of this graph, it is difficult to see the new curve produced by the loss of corn to ethanol production. To see the data more clearly, please click on the link below.)

**Link to Desmos Graph:** https://www.desmos.com/calculator/lbgqzbt6uj

Based on our results, the most accurate conclusion comes from Step 7 with a prediction that after 2085, the amount of food produced will not be able to sustain the world population growth. However, this calculation is based on many assumptions and the future is always subject to change, therefore our number is just an estimate and not a definite conclusion.

Katelyn Ripley and Elena Silverberg

Works Cited:

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