I’ll not go into the mathematics of the binomial theorem here.  Instead, I’ll introduce you to math tricks which can be used instead.  First, let me refresh your mind on why we were taught the binomial theorem.  Remember when you were asked to expand the equation:

(x + y)²

If you recall, this equation can be expanded to the equivalent equation:

x² + 2xy + y²

The binomial theorem will allow you to solve a higher order problem of the example above.  For instance, what is the expansion of the equation:

(x + y)⁵

Generally, an equation of this type can be expanded as:

where c₁, c₂, … are the binomial coefficients in the expansion.   So given any n, you can determine the expansion without the coefficients.  Expanding our example above:

(x + y)⁵ = c₁x⁵y⁰ + c₂x⁴y¹ + c₃x³y² + c₄x²y³ + c₅x¹y⁴ + c₆x⁰y⁵

So how do you determine the binomial coefficients?  You can determine the binomial coefficients individually using the equation:

for k=0 to k=n.  This works fine, but is a little bit cumbersome – especially for large values of n!  So what is the math trick to solve this quickly?

Before I can answer this, I have to introduce to you Pascal’s Triangle.  Pascal’s triangle is a mathematical progression which is determined by constructing a triangle with numbers using a very simple algorithm.  First, take a look at this example of Pascal’s triangle:

At the very top is row 0, which is simply a 1.  In row 1, there are two numbers, both 1s.  In row 2, there are three numbers: 1, 2, and 1.  Notice that the 2 in row two is the sum of the two numbers above it; this is how you determine the numbers in the triangle – simply add two side-by-side numbers to get the result below and between the numbers:

Construction of Pascal’s Triangle¹

So how can you use Pascal’s triangle to find the binomial coefficients when you expand the equation (x + y)⁵?  First, notice that the equation is raised to the 5^th power.  So now simply go to the 5^th row of Pascal’s triangle (remember, the top row is row 0), and those numbers are the required coefficients:

1 5 10 10 5 1

And so,

(x + y)⁵ = x⁵y⁰ + 5x⁴y¹ + 10x³y² + 10x²y³ + 5x¹y⁴ + x⁰y⁵

Quite a time saver!

¹File by Hersfold, en.wikipedia.org/wiki/User:Hersfold

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As a child, there was nothing like the first snowfall of the year.  It was a sign that the joyous upcoming holiday season was upon us.  Looking around at the growing accumulation, one would begin to anticipate snowball fights and sledding down white, glistening hills.  If you caught a snowflake on your tongue, the cold, crisp beauty of the event would precipitate within your soul.

Now consider the snowflakes themselves.  Their beautiful geometric patterns have fascinated us.  Taken individually, they seem to us lovely enough, but considering the innumerable variety of complex patterns, one is almost overwhelmed with this marvel of nature.

Snowflake patterns do form as a consequence of the physical properties of water, but this does not take away from our appreciation of the astonishing symmetry within them.  Indeed, one can come away with a feeling of the tight bond between mathematics and nature.  To many, a picture of a snowflake represents order within nature.   Simply by looking at a snowflake, you may be convinced that there is an underlying order in our universe.

Natural snowflakes can be imitated mathematically with fractal patterns.  Helge von Koch was a pioneer in fractal mathematics, and in 1904, he came up with a fractal pattern (the Koch snowflake) that resembled a snowflake:

The Koch Snowflake¹

Constructing this fractal is an easy process.  You begin with an equilateral triangle.  For each side of the triangle, divide the side into thirds.  For the middle third, draw an equilateral triangle.  When complete, repeat this process for each triangle of your new construct.  This process is illustrated in the figure below:

Construction of the Koch Snowflake²

When properly considered, there is more to the intrinsic beauty within a natural snowflake or one constructed using fractal mathematics.  If you consider the Koch snowflake carefully, you will realize that there is much more to it than what you can see.  Indeed, if you analyze the Koch snowflake mathematically, you will even be able to glimpse of the handiwork of God!

Let me show you two mathematical properties of the Koch snowflake that, taken together, are really very remarkable philosophically.  First, what is the area of the Koch snowflake?  I will not derive the formula for you – I will just give it to you here for iteration k to n:

where s is the length of each side of the original equilateral triangle.

Now, as n approaches infinity, the area of the Koch snowflake becomes a finite value, and is equal to:

So what about the length of the perimeter of the Koch snowflake?  The length of the perimeter at iteration n is given by:

Now, as n approaches infinity, can you see what happens to the length?  Amazingly, the length of the perimeter of the Koch snowflake becomes infinite!  So the Koch snowflake is a mathematical object with finite area bounded by an infinite boundary – how awesome is that?!

There are many things I like to think about when I philosophize about the nature of the universe, and the mathematical properties of the Koch snowflake is one of them.  I like to think of the Koch snowflake as a metaphor for existence outside of our realm of perception.  We are all bounded by the limits of our lives, but is there something beyond our existence that we sometimes fail to see?  Look at the Koch snowflake, and you may see, evidence of intelligence, and what may be.

¹Illustration by Wrtlprnft: commons.wikimedia.org/wiki/User:Wrtlprnft

²Illustration by Wxs: commons.wikimedia.org/wiki/User:Wxs

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Yes, it is the infamous 2=1 “proof”.  Here I present it to you as a graphic an also as a video set to Beethoven’s 5th (yea – it was raining yesterday so I had a lot of time on my hands!).

First, here is the graphic of the proof:

And now for the video proof:

So that is the proof.  Can you spot where the error is?

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So what’s the deal here??  Where did the hole come from?  Please post your answers  :)

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