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	<title>Comments on: I need an average</title>
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	<link>http://threesixty360.wordpress.com/2009/07/16/i-need-an-average/</link>
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		<title>By: Veky</title>
		<link>http://threesixty360.wordpress.com/2009/07/16/i-need-an-average/#comment-11250</link>
		<dc:creator><![CDATA[Veky]]></dc:creator>
		<pubDate>Wed, 08 Jun 2011 07:34:11 +0000</pubDate>
		<guid isPermaLink="false">http://threesixty360.wordpress.com/?p=2672#comment-11250</guid>
		<description><![CDATA[Well, of course. Look at it this way: there is only one mean, and it is usually called &quot;arithmetical mean&quot;. It can be calculated in the continuous case using integrals, as shown above (result 4/pi).

What about other &quot;means&quot;? It turns out these are also instances of the above mean, just in different codomains. To calculate f-mean (mean pulled by f), we map the values by function f into some other space, calculate the mean there, and pull back by f^{-1}.

For example, geometric mean is log-mean, and harmonic mean is reciprocal-mean. (Square mean is literally squaring-mean.) And these are useful in many applications, so they have special names.

Now I suppose it is clear what happened: you have calculated the angle-mean of all the chords, where angle is the function that maps the length of the chord to size of its central angle. You&#039;ve mapped chord lengths to angle sizes, calculated the mean (very easily, by symmetry) in the angle-space, and pulled that mean (=0) to chord-space, yielding sqrt2.

So you can get almost any value in domain space (at least in the interval spanned by actual values) as f-mean for suitably chosen f. You have just chosen f so that the codomain mean can be easily calculated, not as some intrinsic property of chord lengths. But you don&#039;t have to be too disappointed, since great mathematicians had almost the same problems in a slightly more difficult setting. See http://en.wikipedia.org/wiki/Bertrand_paradox_(probability)]]></description>
		<content:encoded><![CDATA[<p>Well, of course. Look at it this way: there is only one mean, and it is usually called &#8220;arithmetical mean&#8221;. It can be calculated in the continuous case using integrals, as shown above (result 4/pi).</p>
<p>What about other &#8220;means&#8221;? It turns out these are also instances of the above mean, just in different codomains. To calculate f-mean (mean pulled by f), we map the values by function f into some other space, calculate the mean there, and pull back by f^{-1}.</p>
<p>For example, geometric mean is log-mean, and harmonic mean is reciprocal-mean. (Square mean is literally squaring-mean.) And these are useful in many applications, so they have special names.</p>
<p>Now I suppose it is clear what happened: you have calculated the angle-mean of all the chords, where angle is the function that maps the length of the chord to size of its central angle. You&#8217;ve mapped chord lengths to angle sizes, calculated the mean (very easily, by symmetry) in the angle-space, and pulled that mean (=0) to chord-space, yielding sqrt2.</p>
<p>So you can get almost any value in domain space (at least in the interval spanned by actual values) as f-mean for suitably chosen f. You have just chosen f so that the codomain mean can be easily calculated, not as some intrinsic property of chord lengths. But you don&#8217;t have to be too disappointed, since great mathematicians had almost the same problems in a slightly more difficult setting. See <a href="http://en.wikipedia.org/wiki/Bertrand_paradox_(probability)" rel="nofollow">http://en.wikipedia.org/wiki/Bertrand_paradox_(probability)</a></p>
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		<title>By: TwoPi</title>
		<link>http://threesixty360.wordpress.com/2009/07/16/i-need-an-average/#comment-9928</link>
		<dc:creator><![CDATA[TwoPi]]></dc:creator>
		<pubDate>Wed, 22 Jul 2009 17:25:07 +0000</pubDate>
		<guid isPermaLink="false">http://threesixty360.wordpress.com/?p=2672#comment-9928</guid>
		<description><![CDATA[How does one define the median in the continuous context?  What is the median of an uncountable set of data?]]></description>
		<content:encoded><![CDATA[<p>How does one define the median in the continuous context?  What is the median of an uncountable set of data?</p>
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		<title>By: Ξ</title>
		<link>http://threesixty360.wordpress.com/2009/07/16/i-need-an-average/#comment-9923</link>
		<dc:creator><![CDATA[Ξ]]></dc:creator>
		<pubDate>Tue, 21 Jul 2009 05:38:24 +0000</pubDate>
		<guid isPermaLink="false">http://threesixty360.wordpress.com/?p=2672#comment-9923</guid>
		<description><![CDATA[Thanks for all the ways to think about this [and sorry for the delay in response -- our internet access was really limited over the past week, so I&#039;m just getting a chance to read them].

I think I was viewing the problem continuously, not discretely, but I&#039;m still not wrapping my brain around why it would be a median but not a mean -- I&#039;m seeing symmetry, so the two should be the same.  (But I think it&#039;s quite possible I&#039;m making some other assumptions, as jedwards points out.  I was certainly assuming Euclidean, but being in a plane follows from that and the fact that there are three points.  [Wait, is that right?  Maybe the mean would change if C varies over a sphere rather than a circle.  OK, I think I&#039;m seeing how the mean and median might be different.]]]></description>
		<content:encoded><![CDATA[<p>Thanks for all the ways to think about this [and sorry for the delay in response -- our internet access was really limited over the past week, so I'm just getting a chance to read them].</p>
<p>I think I was viewing the problem continuously, not discretely, but I&#8217;m still not wrapping my brain around why it would be a median but not a mean &#8212; I&#8217;m seeing symmetry, so the two should be the same.  (But I think it&#8217;s quite possible I&#8217;m making some other assumptions, as jedwards points out.  I was certainly assuming Euclidean, but being in a plane follows from that and the fact that there are three points.  [Wait, is that right?  Maybe the mean would change if C varies over a sphere rather than a circle.  OK, I think I'm seeing how the mean and median might be different.]</p>
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		<title>By: jedward706</title>
		<link>http://threesixty360.wordpress.com/2009/07/16/i-need-an-average/#comment-9922</link>
		<dc:creator><![CDATA[jedward706]]></dc:creator>
		<pubDate>Mon, 20 Jul 2009 19:53:47 +0000</pubDate>
		<guid isPermaLink="false">http://threesixty360.wordpress.com/?p=2672#comment-9922</guid>
		<description><![CDATA[perhaps the &quot;discrete&quot; view could be considered like &quot;taxi-cab&quot; geometry -- the only allowed positions for the points are on a Cartesian grid ?]]></description>
		<content:encoded><![CDATA[<p>perhaps the &#8220;discrete&#8221; view could be considered like &#8220;taxi-cab&#8221; geometry &#8212; the only allowed positions for the points are on a Cartesian grid ?</p>
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	<item>
		<title>By: Sue</title>
		<link>http://threesixty360.wordpress.com/2009/07/16/i-need-an-average/#comment-9920</link>
		<dc:creator><![CDATA[Sue]]></dc:creator>
		<pubDate>Fri, 17 Jul 2009 17:25:29 +0000</pubDate>
		<guid isPermaLink="false">http://threesixty360.wordpress.com/?p=2672#comment-9920</guid>
		<description><![CDATA[@Jonathan: If the points were colinear, the answer would be 2, not 1. I was guessing the joke was that the A and C in the snetence were one inch apart. Also, any 3 points are in a plane, so that wasn&#039;t an assumtion.

@Mitchell: Can you explain what you see as the difference between a discrete and a continuous perspective on this problem?

Interesting question.]]></description>
		<content:encoded><![CDATA[<p>@Jonathan: If the points were colinear, the answer would be 2, not 1. I was guessing the joke was that the A and C in the snetence were one inch apart. Also, any 3 points are in a plane, so that wasn&#8217;t an assumtion.</p>
<p>@Mitchell: Can you explain what you see as the difference between a discrete and a continuous perspective on this problem?</p>
<p>Interesting question.</p>
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		<title>By: Mitchell</title>
		<link>http://threesixty360.wordpress.com/2009/07/16/i-need-an-average/#comment-9919</link>
		<dc:creator><![CDATA[Mitchell]]></dc:creator>
		<pubDate>Fri, 17 Jul 2009 04:05:03 +0000</pubDate>
		<guid isPermaLink="false">http://threesixty360.wordpress.com/?p=2672#comment-9919</guid>
		<description><![CDATA[The problem with the reasoning in the article is that he seems to have considered the problem discretely, rather than continuously. Thinking continuously, you easily get Dave&#039;s answer above.]]></description>
		<content:encoded><![CDATA[<p>The problem with the reasoning in the article is that he seems to have considered the problem discretely, rather than continuously. Thinking continuously, you easily get Dave&#8217;s answer above.</p>
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		<title>By: Fadam</title>
		<link>http://threesixty360.wordpress.com/2009/07/16/i-need-an-average/#comment-9918</link>
		<dc:creator><![CDATA[Fadam]]></dc:creator>
		<pubDate>Thu, 16 Jul 2009 19:35:44 +0000</pubDate>
		<guid isPermaLink="false">http://threesixty360.wordpress.com/?p=2672#comment-9918</guid>
		<description><![CDATA[I agree with Dave, \pi/4 would be the mean.  What you calculated was the median length.  

A simpler way of asking the same question would be what is the average length of a chord in a circle with a radius of 1.]]></description>
		<content:encoded><![CDATA[<p>I agree with Dave, \pi/4 would be the mean.  What you calculated was the median length.  </p>
<p>A simpler way of asking the same question would be what is the average length of a chord in a circle with a radius of 1.</p>
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		<title>By: Dave Richeson</title>
		<link>http://threesixty360.wordpress.com/2009/07/16/i-need-an-average/#comment-9917</link>
		<dc:creator><![CDATA[Dave Richeson]]></dc:creator>
		<pubDate>Thu, 16 Jul 2009 19:27:06 +0000</pubDate>
		<guid isPermaLink="false">http://threesixty360.wordpress.com/?p=2672#comment-9917</guid>
		<description><![CDATA[Oops. It should have been $latex 4/\pi$, not $latex \pi/4$.]]></description>
		<content:encoded><![CDATA[<p>Oops. It should have been <img src='http://s0.wp.com/latex.php?latex=4%2F%5Cpi&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='4/&#92;pi' title='4/&#92;pi' class='latex' />, not <img src='http://s0.wp.com/latex.php?latex=%5Cpi%2F4&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;pi/4' title='&#92;pi/4' class='latex' />.</p>
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		<title>By: Dave Richeson</title>
		<link>http://threesixty360.wordpress.com/2009/07/16/i-need-an-average/#comment-9916</link>
		<dc:creator><![CDATA[Dave Richeson]]></dc:creator>
		<pubDate>Thu, 16 Jul 2009 19:21:48 +0000</pubDate>
		<guid isPermaLink="false">http://threesixty360.wordpress.com/?p=2672#comment-9916</guid>
		<description><![CDATA[How about $latex pi/4$ for an average (approx. 1.27324...)? 

Here&#039;s my reasoning. If the angle between the segments is $latex \theta$, then the distance between A and C is $latex 2\sin(\theta/2)$. Then the average value of this function on the interval $latex [0,\pi]$ is $latex \displaystyle\frac{1}{\pi}\int_0^{\pi}2\sin(\theta/2)d\theta=\frac{\pi}{4}$.]]></description>
		<content:encoded><![CDATA[<p>How about <img src='http://s0.wp.com/latex.php?latex=pi%2F4&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='pi/4' title='pi/4' class='latex' /> for an average (approx. 1.27324&#8230;)? </p>
<p>Here&#8217;s my reasoning. If the angle between the segments is <img src='http://s0.wp.com/latex.php?latex=%5Ctheta&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;theta' title='&#92;theta' class='latex' />, then the distance between A and C is <img src='http://s0.wp.com/latex.php?latex=2%5Csin%28%5Ctheta%2F2%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='2&#92;sin(&#92;theta/2)' title='2&#92;sin(&#92;theta/2)' class='latex' />. Then the average value of this function on the interval <img src='http://s0.wp.com/latex.php?latex=%5B0%2C%5Cpi%5D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='[0,&#92;pi]' title='[0,&#92;pi]' class='latex' /> is <img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7B1%7D%7B%5Cpi%7D%5Cint_0%5E%7B%5Cpi%7D2%5Csin%28%5Ctheta%2F2%29d%5Ctheta%3D%5Cfrac%7B%5Cpi%7D%7B4%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle&#92;frac{1}{&#92;pi}&#92;int_0^{&#92;pi}2&#92;sin(&#92;theta/2)d&#92;theta=&#92;frac{&#92;pi}{4}' title='&#92;displaystyle&#92;frac{1}{&#92;pi}&#92;int_0^{&#92;pi}2&#92;sin(&#92;theta/2)d&#92;theta=&#92;frac{&#92;pi}{4}' class='latex' />.</p>
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		<title>By: jonah</title>
		<link>http://threesixty360.wordpress.com/2009/07/16/i-need-an-average/#comment-9915</link>
		<dc:creator><![CDATA[jonah]]></dc:creator>
		<pubDate>Thu, 16 Jul 2009 19:18:39 +0000</pubDate>
		<guid isPermaLink="false">http://threesixty360.wordpress.com/?p=2672#comment-9915</guid>
		<description><![CDATA[Also, it&#039;s the RMS average.]]></description>
		<content:encoded><![CDATA[<p>Also, it&#8217;s the RMS average.</p>
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