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	<title>StaticMethod &#187; Math</title>
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		<title>Summations / Sigma Notation</title>
		<link>http://www.staticmethod.net/2009/05/01/summations-sigma-notation/</link>
		<comments>http://www.staticmethod.net/2009/05/01/summations-sigma-notation/#comments</comments>
		<pubDate>Fri, 01 May 2009 18:10:38 +0000</pubDate>
		<dc:creator>Chris</dc:creator>
				<category><![CDATA[Learning]]></category>
		<category><![CDATA[Math]]></category>
		<category><![CDATA[summation]]></category>

		<guid isPermaLink="false">http://www.staticmethod.net/?p=17</guid>
		<description><![CDATA[A few years ago I had a job interview with a company that ended up being pretty math focussed.  I had been out of college for a while at the time, and on a day to day basis in doing software development I hadn&#8217;t had the opportunity to exercise a lot of the math [...]]]></description>
			<content:encoded><![CDATA[<p>A few years ago I had a job interview with a company that ended up being pretty math focussed.  I had been out of college for a while at the time, and on a day to day basis in doing software development I hadn&#8217;t had the opportunity to exercise a lot of the math that I had learned while in college.  At some point in my life I had taken: Calc, Calc II, Linear Algebra, Discrete Math, and some others&#8230;but because I haven&#8217;t been using it, my math knowledge had kind of gone down the tubes.</p>
<p>So when I was asked a question about summations, I was kind of stumped.  Now that I&#8217;ve started reading a algorithms text book, I&#8217;m running into summations all over again.  In my search to re-learn them, I&#8217;ve found the following sites useful:</p>
<ul>
<li> <a href="http://tutorial.math.lamar.edu/Classes/CalcI/SummationNotation.aspx">Paul&#8217;s Online Math Notes</a> &#8211; Prof. Paul Dawkins, of Lamar University, has a great math site, and has a good, short, re-introduction to summations here.</li>
<li><a href="http://planetmath.org/?method=l2h&amp;from=objects&amp;id=6361&amp;op=getobj">PlanetMath&#8217;s Summation Page</a> &#8211; PlanetMath has a good page on summations that goes a little more into detail on some summation properties and useful formulas</li>
<li><a href="http://polysum.tripod.com/">Summation Formulas</a> &#8211; Good outline of summation formulas</li>
</ul>
<p>What I haven&#8217;t been able to find is a page explaining how the solutions for common summations is found, such as the one for a simple summation:</p>
<p style="text-align: center;"><img src='http://s.wordpress.com/latex.php?latex=%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20x_%7Bi%7D%20%3D%20%5Cdfrac%7Bn%20%28n%20%2B%201%29%7D%20%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sum_{i=1}^{n} x_{i} = \dfrac{n (n + 1)} {2}' title='\sum_{i=1}^{n} x_{i} = \dfrac{n (n + 1)} {2}' class='latex' /></p>
<p><span style="text-decoration: line-through;">I&#8217;m not sure how that formula is reached, but I&#8217;m sure I&#8217;ll be trying to figure it out soon.</span></p>
<p><strong>Update:</strong> Finally found a good description of the steps in solving the summation above, here&#8217;s my attempt at explaining it:</p>
<p><span id="more-17"></span></p>
<p>The simple summation above can be thought of as the series:</p>
<p style="text-align: center;"><img src='http://s.wordpress.com/latex.php?latex=1%2B2%2B3%2B%5Ccdots%2Bn&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1+2+3+\cdots+n' title='1+2+3+\cdots+n' class='latex' /></p>
<p style="text-align: left;">It can also be thought of (in reverse) as:</p>
<p style="text-align: center;"><img src='http://s.wordpress.com/latex.php?latex=n%2B%28n-1%29%2B%28n-2%29%2B%5Ccdots%2B1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n+(n-1)+(n-2)+\cdots+1' title='n+(n-1)+(n-2)+\cdots+1' class='latex' /></p>
<p style="text-align: left;">Now, imagine writing each of the terms of those two series above eachother, then adding the ones that are above and below eachother.  You would end up with something like:</p>
<p style="text-align: center;"><img src='http://s.wordpress.com/latex.php?latex=%281%2Bn%29%2B%282%2Bn-1%29%2B%283%2Bn-2%29%2B%5Ccdots%2B%28n%2B1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(1+n)+(2+n-1)+(3+n-2)+\cdots+(n+1)' title='(1+n)+(2+n-1)+(3+n-2)+\cdots+(n+1)' class='latex' /></p>
<p style="text-align: left;">This can be simplified by doing the math inside the parens to:</p>
<p style="text-align: center;"><img src='http://s.wordpress.com/latex.php?latex=%28n%2B1%29%2B%28n%2B1%29%2B%28n%2B1%29%2B%5Ccdots%2B%28n%2B1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(n+1)+(n+1)+(n+1)+\cdots+(n+1)' title='(n+1)+(n+1)+(n+1)+\cdots+(n+1)' class='latex' /></p>
<p style="text-align: left;">Which is simplified even further to:</p>
<p style="text-align: center;"><img src='http://s.wordpress.com/latex.php?latex=n%28n%2B1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n(n+1)' title='n(n+1)' class='latex' /></p>
<p style="text-align: left;">At this point, you&#8217;ve basically described adding the summation twice, or</p>
<p style="text-align: center;"><img src='http://s.wordpress.com/latex.php?latex=2%28%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20x_%7Bi%7D%29%20%3D%20n%28n%2B1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='2(\sum_{i=1}^{n} x_{i}) = n(n+1)' title='2(\sum_{i=1}^{n} x_{i}) = n(n+1)' class='latex' /></p>
<p style="text-align: left;">Divide each side by two, and you&#8217;ve got the solution we were looking for:</p>
<p style="text-align: center;"><img src='http://s.wordpress.com/latex.php?latex=%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20x_%7Bi%7D%20%3D%20%5Cdfrac%7Bn%20%28n%20%2B%201%29%7D%20%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sum_{i=1}^{n} x_{i} = \dfrac{n (n + 1)} {2}' title='\sum_{i=1}^{n} x_{i} = \dfrac{n (n + 1)} {2}' class='latex' /></p>
<p style="text-align: left;">I finally found this explanation at the <a href="http://mathrefresher.blogspot.com/2006/09/summation-formula.html">Math Refresher</a> weblog.  The article I linked to goes into a lot more detail, so I would suggest reading it if you want to know more.  I tried, but my eyes are getting tired from math at ths point in time, and his equation representations are a little hard to read.</p>
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