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	<title>Comments on: Get out of here!</title>
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	<link>http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/</link>
	<description>seductive math problems for the modern mind</description>
	<pubDate>Thu, 11 Mar 2010 00:50:30 +0000</pubDate>
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		<title>By: Rich</title>
		<link>http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-5875</link>
		<dc:creator>Rich</dc:creator>
		<pubDate>Wed, 04 Jun 2008 20:34:18 +0000</pubDate>
		<guid isPermaLink="false">http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-5875</guid>
		<description>You know, I got curious about just this sort of thing a little while ago. I wrote a bash script which generated 10 random numbers (between 0 and 99) and computed the average, then I found the high and low values and averaged those. Here's what I got...

$ ./numbers

85 58 6 47 7 66 55 64 61 42

   Average = 49.10
   Min = 6, Max = 85, Total = 91
   Average of Min and Max = 45.50

$ ./numbers

75 21 85 19 41 73 66 88 53 81

   Average = 60.20
   Min = 19, Max = 88, Total = 107
   Average of Min and Max = 53.50

$ ./numbers

26 1 26 12 89 35 70 79 81 54

   Average = 47.30
   Min = 1, Max = 89, Total = 90
   Average of Min and Max = 45.00

-----------------------------

As you can see, they are close, but they ain't the same thing.
In these examples it looks like they can be up to 12% off. There
can be a significant difference.</description>
		<content:encoded><![CDATA[<p>You know, I got curious about just this sort of thing a little while ago. I wrote a bash script which generated 10 random numbers (between 0 and 99) and computed the average, then I found the high and low values and averaged those. Here&#8217;s what I got&#8230;</p>
<p>$ ./numbers</p>
<p>85 58 6 47 7 66 55 64 61 42</p>
<p>   Average = 49.10<br />
   Min = 6, Max = 85, Total = 91<br />
   Average of Min and Max = 45.50</p>
<p>$ ./numbers</p>
<p>75 21 85 19 41 73 66 88 53 81</p>
<p>   Average = 60.20<br />
   Min = 19, Max = 88, Total = 107<br />
   Average of Min and Max = 53.50</p>
<p>$ ./numbers</p>
<p>26 1 26 12 89 35 70 79 81 54</p>
<p>   Average = 47.30<br />
   Min = 1, Max = 89, Total = 90<br />
   Average of Min and Max = 45.00</p>
<p>&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211;</p>
<p>As you can see, they are close, but they ain&#8217;t the same thing.<br />
In these examples it looks like they can be up to 12% off. There<br />
can be a significant difference.</p>
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	<item>
		<title>By: Carnival of Mathematics 1000 &#171; JD2718</title>
		<link>http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-1351</link>
		<dc:creator>Carnival of Mathematics 1000 &#171; JD2718</dc:creator>
		<pubDate>Fri, 22 Feb 2008 20:27:28 +0000</pubDate>
		<guid isPermaLink="false">http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-1351</guid>
		<description>[...] - A bit old, but there is an annoying temperature discussion at Think Again. Just what does &#8220;average temperature&#8221; [...]</description>
		<content:encoded><![CDATA[<p>[...] - A bit old, but there is an annoying temperature discussion at Think Again. Just what does &#8220;average temperature&#8221; [...]</p>
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		<title>By: Richard Sabey</title>
		<link>http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-217</link>
		<dc:creator>Richard Sabey</dc:creator>
		<pubDate>Wed, 19 Dec 2007 14:23:16 +0000</pubDate>
		<guid isPermaLink="false">http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-217</guid>
		<description>By 2-point Gaussian quadrature, if c(x) is a cubic, the integral of c(x) over [-1...1] is

w_1 c(x_1) + w_2 c(x_2)

where the w_i are 1 and the x_i are ±1/sqrt(3). To convert the integral into the mean value, divide by the size of the range (2, in the above example), thus making the weights 1/2. In Jan's example, the range is [0, 24], so scale by adding 1 and multiplying by 12. Now the x_i are 12(1±1/sqrt(3)) ~~ 5.0718 and 18.9282. Jan "added f(5.071667) and f(18.928333) and divided by two" because, among the x-values for which f had been recorded, those were nearest to the values demanded by 2-point Gaussian quadrature.

Gaussian quadrature played no part in the choice of  "4, 8, 16, and 22 hours" as the times for which the values of f were requested for the purpose of approximating f by a cubic. Perhaps these times were chosen because they are a useful set of 4 for approximating the change of temperature throughout  a winter day: for some latitudes, 8 is just after sunrise, so f(4) and f(8) show how the rising sun makes the temperature rise; 4pm is about sunset, so f(16) shows how the temperature has risen as a result of the sunshine in the daylight hours.</description>
		<content:encoded><![CDATA[<p>By 2-point Gaussian quadrature, if c(x) is a cubic, the integral of c(x) over [-1...1] is</p>
<p>w_1 c(x_1) + w_2 c(x_2)</p>
<p>where the w_i are 1 and the x_i are ±1/sqrt(3). To convert the integral into the mean value, divide by the size of the range (2, in the above example), thus making the weights 1/2. In Jan&#8217;s example, the range is [0, 24], so scale by adding 1 and multiplying by 12. Now the x_i are 12(1±1/sqrt(3)) ~~ 5.0718 and 18.9282. Jan &#8220;added f(5.071667) and f(18.928333) and divided by two&#8221; because, among the x-values for which f had been recorded, those were nearest to the values demanded by 2-point Gaussian quadrature.</p>
<p>Gaussian quadrature played no part in the choice of  &#8220;4, 8, 16, and 22 hours&#8221; as the times for which the values of f were requested for the purpose of approximating f by a cubic. Perhaps these times were chosen because they are a useful set of 4 for approximating the change of temperature throughout  a winter day: for some latitudes, 8 is just after sunrise, so f(4) and f(8) show how the rising sun makes the temperature rise; 4pm is about sunset, so f(16) shows how the temperature has risen as a result of the sunshine in the daylight hours.</p>
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	<item>
		<title>By: Anceps</title>
		<link>http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-216</link>
		<dc:creator>Anceps</dc:creator>
		<pubDate>Wed, 19 Dec 2007 13:51:50 +0000</pubDate>
		<guid isPermaLink="false">http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-216</guid>
		<description>The link has a final dot.</description>
		<content:encoded><![CDATA[<p>The link has a final dot.</p>
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	<item>
		<title>By: John S.</title>
		<link>http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-213</link>
		<dc:creator>John S.</dc:creator>
		<pubDate>Wed, 19 Dec 2007 12:18:28 +0000</pubDate>
		<guid isPermaLink="false">http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-213</guid>
		<description>Strange that when I click on Jan's link above, wikipedia tells me there is no such page, but then offers to do a search, which then turns up a page with the same name.

What is interesting about the (high + low)/2 method (the "traditional" way of finding daily average temperature) is that in effect, it is saying that the integral of a function over a given range can be approximated by taking the average of the maximum and minimum values of the function on that range. I learned the trapezoid rule, and Simpson's rule, but this is an approximation technique that no one ever mentioned in Intro to Calculus. Why does it work? Is it related to Gaussian quadrature? That is, does it work because the minimum tends to occur at or near 5.071667 hours from midnight, and the maximum at 18.928333 hours from midnight? I would think the maximum occurs earlier in the day.</description>
		<content:encoded><![CDATA[<p>Strange that when I click on Jan&#8217;s link above, wikipedia tells me there is no such page, but then offers to do a search, which then turns up a page with the same name.</p>
<p>What is interesting about the (high + low)/2 method (the &#8220;traditional&#8221; way of finding daily average temperature) is that in effect, it is saying that the integral of a function over a given range can be approximated by taking the average of the maximum and minimum values of the function on that range. I learned the trapezoid rule, and Simpson&#8217;s rule, but this is an approximation technique that no one ever mentioned in Intro to Calculus. Why does it work? Is it related to Gaussian quadrature? That is, does it work because the minimum tends to occur at or near 5.071667 hours from midnight, and the maximum at 18.928333 hours from midnight? I would think the maximum occurs earlier in the day.</p>
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		<title>By: Jan Nordgreen</title>
		<link>http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-189</link>
		<dc:creator>Jan Nordgreen</dc:creator>
		<pubDate>Tue, 18 Dec 2007 12:44:35 +0000</pubDate>
		<guid isPermaLink="false">http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-189</guid>
		<description>Have a look at this Wikipedia article:
&lt;a href="http://en.wikipedia.org/wiki/Gaussian_quadrature" rel="nofollow"&gt;http://en.wikipedia.org/wiki/Gaussian_quadrature&lt;/a&gt;.</description>
		<content:encoded><![CDATA[<p>Have a look at this Wikipedia article:<br />
<a href="http://en.wikipedia.org/wiki/Gaussian_quadrature" rel="nofollow">http://en.wikipedia.org/wiki/Gaussian_quadrature</a>.</p>
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	<item>
		<title>By: Richard Sabey</title>
		<link>http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-188</link>
		<dc:creator>Richard Sabey</dc:creator>
		<pubDate>Tue, 18 Dec 2007 12:01:20 +0000</pubDate>
		<guid isPermaLink="false">http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-188</guid>
		<description>I'm still interested to learn the method to determine the 2 values at which f must be evaluated. What constraints are there in the choice of the 4 values at which the original function is evaluated?</description>
		<content:encoded><![CDATA[<p>I&#8217;m still interested to learn the method to determine the 2 values at which f must be evaluated. What constraints are there in the choice of the 4 values at which the original function is evaluated?</p>
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	<item>
		<title>By: Michael</title>
		<link>http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-175</link>
		<dc:creator>Michael</dc:creator>
		<pubDate>Mon, 17 Dec 2007 16:35:16 +0000</pubDate>
		<guid isPermaLink="false">http://easyquestion.net/thinkagain/2007/12/17/get-out-of-here/#comment-175</guid>
		<description>Neat!  Apple's Grapher application (free / included with OS X) did a fine job of this interpolation as well.  I'd never really given any thought to how average temperatures are calculated but yeah I suppose this is the only way that makes sense.</description>
		<content:encoded><![CDATA[<p>Neat!  Apple&#8217;s Grapher application (free / included with OS X) did a fine job of this interpolation as well.  I&#8217;d never really given any thought to how average temperatures are calculated but yeah I suppose this is the only way that makes sense.</p>
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