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	<title>Math Tricks &#187; Math Patterns</title>
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		<title>The Binomial Theorem and Pascal’s Triangle</title>
		<link>http://mathtricks.org/math-tricks/the-binomial-theorem-and-pascal%e2%80%99s-triangle/</link>
		<comments>http://mathtricks.org/math-tricks/the-binomial-theorem-and-pascal%e2%80%99s-triangle/#comments</comments>
		<pubDate>Wed, 24 Feb 2010 23:52:48 +0000</pubDate>
		<dc:creator>Math Tricks</dc:creator>
				<category><![CDATA[Math Patterns]]></category>
		<category><![CDATA[Math Tricks]]></category>
		<category><![CDATA[Pascal's Triangle]]></category>
		<category><![CDATA[binomial expansion]]></category>
		<category><![CDATA[binomial theorem]]></category>

		<guid isPermaLink="false">http://mathtricks.org/?p=226</guid>
		<description><![CDATA[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.  [...]]]></description>
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<p>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.</p>
<p>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:</p>
<p>(x + y)<sup>2</sup></p>
<p>If you recall, this equation can be expanded to the equivalent equation:</p>
<p>x<sup>2</sup> + 2xy + y<sup>2</sup></p>
<p>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:</p>
<p>(x + y)<sup>5</sup></p>
<p>Generally, an equation of this type can be expanded as:</p>
<p style="text-align: center;"><a href="http://mathtricks.org/wp-content/uploads/2010/02/eq1.jpg"><img class="size-full wp-image-227 aligncenter" title="binomial expansion" src="http://mathtricks.org/wp-content/uploads/2010/02/eq1.jpg" alt="binomial expansion" width="638" height="76" /></a></p>
<p>where c<sub>1</sub>, c<sub>2</sub>, … are the binomial coefficients in the expansion.   So given any n, you can determine the expansion without the coefficients.  Expanding our example above:</p>
<p>(x + y)<sup>5</sup> = c<sub>1</sub>x<sup>5</sup>y<sup>0</sup> + c<sub>2</sub>x<sup>4</sup>y<sup>1</sup> + c<sub>3</sub>x<sup>3</sup>y<sup>2</sup> + c<sub>4</sub>x<sup>2</sup>y<sup>3</sup> + c<sub>5</sub>x<sup>1</sup>y<sup>4</sup> + c<sub>6</sub>x<sup>0</sup>y<sup>5</sup></p>
<p>So how do you determine the binomial coefficients?  You can determine the binomial coefficients individually using the equation:</p>
<p style="text-align: center;"><a href="http://mathtricks.org/wp-content/uploads/2010/02/eq-2.jpg"><img class="size-full wp-image-228 aligncenter" title="binomial coefficients" src="http://mathtricks.org/wp-content/uploads/2010/02/eq-2.jpg" alt="binomial coefficients" width="164" height="97" /></a></p>
<p>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?</p>
<p>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:</p>
<p style="text-align: center;"><a href="http://mathtricks.org/wp-content/uploads/2010/02/hexagonal-pascal-triangle.jpg"><img class="size-full wp-image-230 aligncenter" title="hexagonal pascal triangle" src="http://mathtricks.org/wp-content/uploads/2010/02/hexagonal-pascal-triangle.jpg" alt="hexagonal pascal triangle" width="500" height="488" /></a></p>
<p>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:</p>
<p style="text-align: center;"><a href="http://mathtricks.org/wp-content/uploads/2010/02/PascalTriangleAnimated2.gif"><img class="size-full wp-image-231 aligncenter" title="Animated Pascal Triangle" src="http://mathtricks.org/wp-content/uploads/2010/02/PascalTriangleAnimated2.gif" alt="Animated Pascal Triangle" width="260" height="240" /></a>Construction of Pascal’s Triangle<sup>1</sup></p>
<p>So how can you use Pascal’s triangle to find the binomial coefficients when you expand the equation (x + y)<sup>5</sup>?  First, notice that the equation is raised to the 5<sup>th</sup> power.  So now simply go to the 5<sup>th</sup> row of Pascal’s triangle (remember, the top row is row 0), and those numbers <em>are</em> the required coefficients:</p>
<p>1 5 10 10 5 1</p>
<p>And so,</p>
<p>(x + y)<sup>5</sup> = x<sup>5</sup>y<sup>0</sup> + 5x<sup>4</sup>y<sup>1</sup> + 10x<sup>3</sup>y<sup>2</sup> + 10x<sup>2</sup>y<sup>3</sup> + 5x<sup>1</sup>y<sup>4</sup> + x<sup>0</sup>y<sup>5</sup></p>
<p>Quite a time saver!</p>
<p><sup>1</sup>File by Hersfold, en.wikipedia.org/wiki/User:Hersfold</p>
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		<title>The Golden Ratio and the Fibonacci Sequence</title>
		<link>http://mathtricks.org/math-tricks/the-golden-ratio-and-the-fibonacci-sequence/</link>
		<comments>http://mathtricks.org/math-tricks/the-golden-ratio-and-the-fibonacci-sequence/#comments</comments>
		<pubDate>Sun, 27 Dec 2009 21:14:28 +0000</pubDate>
		<dc:creator>Math Tricks</dc:creator>
				<category><![CDATA[Fibonacci Sequence]]></category>
		<category><![CDATA[Golden Rectangle]]></category>
		<category><![CDATA[Math Tricks]]></category>
		<category><![CDATA[1.618]]></category>
		<category><![CDATA[1.61803399]]></category>
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		<description><![CDATA[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 &#8220;The Golden Ratio and the Fibonacci Sequence&#8221;, then you might have guessed that [...]]]></description>
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<p>So you have seen already a few posts about the golden ratio, which is approximately equal to 1.618<strong>.</strong> Are there <strong>math tricks</strong> that will allow you to determine the value for the golden ratio?  Well, since the title of this post is &#8220;The Golden Ratio and the Fibonacci Sequence&#8221;, then you might have guessed that there is a way to determine the golden ratio with the Fibonacci Sequence.</p>
<p>What the heck is the Fibonacci Sequence?  True, I have not written a post on it yet &#8211; but you just wait because there is a LOT to say about it later on &#8211; a LOT!!  But for now, let me just give you the sequence:</p>
<p>0,1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, &#8230;</p>
<p>Do you see the pattern?  Each value, starting with the second &#8220;1&#8243; in the sequence, is simply the sum of the preceding two values.  The general formuls for the Fibonacci sequence is:</p>
<p>F(<em>n</em>) = F(<em>n</em> – 1) + F(<em>n</em> – 2)</p>
<p>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 &#8211; what do you get?</p>
<p>For example:</p>
<p>34/21 = 1.619</p>
<p>Looks familiar, eh?</p>
<p>Try it again for a pair farther down the sequence:</p>
<p>233/144 = 1.61806</p>
<p>In fact, this manipulation of the Fibonacci series <em>converges </em>to the golden ratio.</p>
<p>Also, you can perform this manipulation using an &#8220;out of frame&#8221; Fibonacci series &#8211; that is, choose any two consecutive numbers, apply the general formula F(<em>n</em>) = F(<em>n</em> – 1) + F(<em>n</em> – 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:</p>
<p>887, 888, 1775, 2663, 4438, 7101, 11539, 18640, 30179 ,48819, 78998, 127817, 206815, 334632</p>
<p>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:</p>
<p>334632/206815 = 1.6180258</p>
<p>A superb example of math tricks in nature!</p>
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		<title>Math Patterns</title>
		<link>http://mathtricks.org/math-patterns/math-patterns/</link>
		<comments>http://mathtricks.org/math-patterns/math-patterns/#comments</comments>
		<pubDate>Tue, 13 Oct 2009 01:53:06 +0000</pubDate>
		<dc:creator>Math Tricks</dc:creator>
				<category><![CDATA[Math Patterns]]></category>
		<category><![CDATA[happy numbers]]></category>

		<guid isPermaLink="false">http://mathtricks.org/?p=128</guid>
		<description><![CDATA[Math Patterns This is the first post to the new category Math Patterns.  Here in this category, I will post many interesting math patterns; some of these are very well known, and some are obscure.  For many of these mathematical patterns, one can derive general formulas very easily by just carefully observing how the math [...]]]></description>
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<h2 style="text-align: left;"><strong>Math Patterns</strong></h2>
<p>This is the first post to the new category <strong>Math Patterns</strong>.  Here in this category, I will post many interesting <strong>math patterns</strong>; some of these are very well known, and some are obscure.  For many of these mathematical patterns, one can derive general formulas very easily by just carefully observing how the math pattern develops.  Other numeric patterns will have you pounding your head on your keyboard while trying to come up with the next number in the series.</p>
<p>So, I give you here the first in the series (pun intended!) of <strong>math patterns</strong>:</p>
<p>1, 7, 10, 13, 19, 23, 28, 31, 32, 44, 49, 68, 70, 79, 82, 86, 91, 94, 97, 100, 103, 109, 129, 130, 133, 139, 167, 176, 188, 190, 192, 193, 203, 208, 219, 226, 230, 236, 239, 262, 263, 280, 291, 293, 301, 302, 310, 313, 319, 320, 326, 329, 331, 338, 356, 362, 365, 367, 368, 376, 379 . . .</p>
<p>Can you see the next number in the pattern?</p>
<p>Here is a subset of this math pattern:</p>
<p>7, 13, 19, 23, 31, 79, 97, 103, 109, 139, 167, 193, 239, 263, 293, 313, 331, 367, 379 . . .</p>
<p>The subset above is simply all of the prime numbers in the first set.  So then, what is the pattern of the first set?  This is the set of <strong>happy numbers</strong> up to and including 379.</p>
<p>So what the heck is a <strong>happy number</strong>?  A happy number is happy if you take each digit in the number, square them, add the squares together, and then repeat the process with the result and then eventually get the value 1.</p>
<p>For example, take the number 338.  First square each digit and add them together:</p>
<p>9 + 9 + 64 = 82</p>
<p>Now repeat the process with the result (82):</p>
<p>64 + 4 = 68</p>
<p>Continue the process with 68:</p>
<p>36 + 64 = 100</p>
<p>And then with 100:</p>
<p>1 + 0 + 0 = 1</p>
<p>Here the result is 1, and thus 338 is a happy number!</p>
<p>So what if a number is not happy?  Interestingly, an unhappy number ends up in a cyclic loop with the pattern:</p>
<p>4, 16, 37, 58, 89, 145, 42, 20, 4 . . .</p>
<p>So is there a limit to the number of happy numbers?  Happily, there are an infinite number of happy numbers – and an infinite number of unhappy numbers as well.</p>
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