Back in grade school, I was first introduced to the Binomial Theorem.  The title alone was quite enough to intimidate me, let alone the seemingly impossible to understand equations involved with it.

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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Tags: binomial expansion, binomial theorem, Math Patterns, Math Tricks, Pascal's Triangle

Many of you may have seen this one before – it is very old (I first saw this in grade school, which makes this example of math tricks ancient!).

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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I came across a very interesting problem – one that seems to involve math tricks of some sort.  Or is it magic?  Whatever it is, it is certainly a math brain teaser!  Anyway, here is the graphic that had me scratching my head:

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

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So you have seen already a few posts about the golden ratio, which is approximately equal to 1.618. Are there math tricks that will allow you to determine the value for the golden ratio?  Well, since the title of this post is “The Golden Ratio and the Fibonacci Sequence”, then you might have guessed that there is a way to determine the golden ratio with the Fibonacci Sequence.

What the heck is the Fibonacci Sequence?  True, I have not written a post on it yet – but you just wait because there is a LOT to say about it later on – a LOT!!  But for now, let me just give you the sequence:

0,1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, …

Do you see the pattern?  Each value, starting with the second “1″ in the sequence, is simply the sum of the preceding two values.  The general formuls for the Fibonacci sequence is:

F(n) = F(n – 1) + F(n – 2)

So how does this interesting sequence of numbers relate to the golden ratio?  Take any value in the sequence and divide it by the preceding value – what do you get?

For example:

34/21 = 1.619

Looks familiar, eh?

Try it again for a pair farther down the sequence:

233/144 = 1.61806

In fact, this manipulation of the Fibonacci series converges to the golden ratio.

Also, you can perform this manipulation using an “out of frame” Fibonacci series – that is, choose any two consecutive numbers, apply the general formula F(n) = F(n – 1) + F(n – 2) to get a new sequence, and then from the new sequence you will be able to determine an approximation for the golden ratio by following the same procedure as outlined above.  For example, starting with 887 and 888, we get the series:

887, 888, 1775, 2663, 4438, 7101, 11539, 18640, 30179 ,48819, 78998, 127817, 206815, 334632

Notice here that you do not get a very good approximation if you divide 888 by 887.  But as you move down the sequence, the value you obtain gets closer and closer to the golden ratio:

334632/206815 = 1.6180258

A superb example of math tricks in nature!

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Tags: 1.618, 1.61803399, fibonacci numbers, Fibonacci Sequence, fibonacci series, golden ratio, Math Tricks