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	<title>Comments on: arXival mysteries</title>
	<atom:link href="http://bit-player.org/2008/arxival-mysteries/feed" rel="self" type="application/rss+xml" />
	<link>http://bit-player.org/2008/arxival-mysteries</link>
	<description>An amateur's outlook on computation and mathematics.</description>
	<pubDate>Thu, 17 May 2012 09:14:55 +0000</pubDate>
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		<title>By: Jim Ward</title>
		<link>http://bit-player.org/2008/arxival-mysteries#comment-1727</link>
		<dc:creator>Jim Ward</dc:creator>
		<pubDate>Mon, 14 Jul 2008 16:20:22 +0000</pubDate>
		<guid isPermaLink="false">http://bit-player.org/?p=153#comment-1727</guid>
		<description>The Baillie and Thomas Schmelzer article point to Nick Trefethen's TEN DIGIT ALGORITHMS -  "Ten digits, five seconds, and just one page"

http://www.comlab.ox.ac.uk/nick.trefethen/tda.html

The algorithms might make a fun book.</description>
		<content:encoded><![CDATA[<p>The Baillie and Thomas Schmelzer article point to Nick Trefethen&#8217;s TEN DIGIT ALGORITHMS -  &#8220;Ten digits, five seconds, and just one page&#8221;</p>
<p><a href="http://www.comlab.ox.ac.uk/nick.trefethen/tda.html" rel="nofollow">http://www.comlab.ox.ac.uk/nick.trefethen/tda.html</a></p>
<p>The algorithms might make a fun book.</p>
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		<title>By: Robert Baillie</title>
		<link>http://bit-player.org/2008/arxival-mysteries#comment-1726</link>
		<dc:creator>Robert Baillie</dc:creator>
		<pubDate>Fri, 11 Jul 2008 10:46:59 +0000</pubDate>
		<guid isPermaLink="false">http://bit-player.org/?p=153#comment-1726</guid>
		<description>to respond to craid kaplan... for kempner series: given a sum T, one can use trial and error to find a set (say, S) of digit strings such that the kempner sum of 1/n where n has no digits in S, is near T.  i don't know if that's possible for irwin series, since it is not yet known how to compute the sum of 1/n where n has, say, one occurrence of a multi-digit string like "314".

here's another interesting point.  let K = sum of kempner's "no 9" series:
K = 1/1 + 1/2 + ... + 1/8 + 1/10 + ... ~ 22.92067...
let I1 = sum of 1/n where n has one 9:
I1 = 1/9 + 1/19 + 1/29 + ... .
the abstract says that I1 &#62; K, even though the I1 series starts out being much smaller than the K series.  but that's also true for I2 = sum of 1/n where n has 2 nines:
I2 = 1/99 + 1/199 + 1/299 + ... + 1/899 + 1/909 + ... ~ 23.02604... &#62; K.
the same is true for, say, 10 nines:
I10 = 1/9999999999 + 1/19999999999 + ... ~ 23.02585092994045684022 &#62; K.

the paper doesn't prove that the sum is &#62; K for an arbitrary number of nines, but the computational results suggest that for a large number of nines, the sum is close to 10*ln(10) ~ 23.02585092994045684018 .</description>
		<content:encoded><![CDATA[<p>to respond to craid kaplan&#8230; for kempner series: given a sum T, one can use trial and error to find a set (say, S) of digit strings such that the kempner sum of 1/n where n has no digits in S, is near T.  i don&#8217;t know if that&#8217;s possible for irwin series, since it is not yet known how to compute the sum of 1/n where n has, say, one occurrence of a multi-digit string like &#8220;314&#8243;.</p>
<p>here&#8217;s another interesting point.  let K = sum of kempner&#8217;s &#8220;no 9&#8243; series:<br />
K = 1/1 + 1/2 + &#8230; + 1/8 + 1/10 + &#8230; ~ 22.92067&#8230;<br />
let I1 = sum of 1/n where n has one 9:<br />
I1 = 1/9 + 1/19 + 1/29 + &#8230; .<br />
the abstract says that I1 &gt; K, even though the I1 series starts out being much smaller than the K series.  but that&#8217;s also true for I2 = sum of 1/n where n has 2 nines:<br />
I2 = 1/99 + 1/199 + 1/299 + &#8230; + 1/899 + 1/909 + &#8230; ~ 23.02604&#8230; &gt; K.<br />
the same is true for, say, 10 nines:<br />
I10 = 1/9999999999 + 1/19999999999 + &#8230; ~ 23.02585092994045684022 &gt; K.</p>
<p>the paper doesn&#8217;t prove that the sum is &gt; K for an arbitrary number of nines, but the computational results suggest that for a large number of nines, the sum is close to 10*ln(10) ~ 23.02585092994045684018 .</p>
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		<title>By: Flocon</title>
		<link>http://bit-player.org/2008/arxival-mysteries#comment-1724</link>
		<dc:creator>Flocon</dc:creator>
		<pubDate>Thu, 03 Jul 2008 06:12:41 +0000</pubDate>
		<guid isPermaLink="false">http://bit-player.org/?p=153#comment-1724</guid>
		<description>In his book "Gamma: Exploring Euler's Constant" Julian Havil devotes a few pages to the Kempner series.  The book is worth checking out if you're interested in the harmonic -- and related -- series.</description>
		<content:encoded><![CDATA[<p>In his book &#8220;Gamma: Exploring Euler&#8217;s Constant&#8221; Julian Havil devotes a few pages to the Kempner series.  The book is worth checking out if you&#8217;re interested in the harmonic &#8212; and related &#8212; series.</p>
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		<title>By: Jack</title>
		<link>http://bit-player.org/2008/arxival-mysteries#comment-1720</link>
		<dc:creator>Jack</dc:creator>
		<pubDate>Wed, 02 Jul 2008 21:20:40 +0000</pubDate>
		<guid isPermaLink="false">http://bit-player.org/?p=153#comment-1720</guid>
		<description>My college analysis prof gave this series as an example. When you look at the first few dozen terms after removing from the harmonic series the terms having a 9 in them, it doesn't seem like you're omitting enough terms to bring the series into convergence.

When he began to explain why the series unintuitively converges, he opened with "Most integers have more than a billion digits" which was at the same time entertaining and illuminating.</description>
		<content:encoded><![CDATA[<p>My college analysis prof gave this series as an example. When you look at the first few dozen terms after removing from the harmonic series the terms having a 9 in them, it doesn&#8217;t seem like you&#8217;re omitting enough terms to bring the series into convergence.</p>
<p>When he began to explain why the series unintuitively converges, he opened with &#8220;Most integers have more than a billion digits&#8221; which was at the same time entertaining and illuminating.</p>
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		<title>By: Craig Kaplan</title>
		<link>http://bit-player.org/2008/arxival-mysteries#comment-1716</link>
		<dc:creator>Craig Kaplan</dc:creator>
		<pubDate>Wed, 02 Jul 2008 16:51:58 +0000</pubDate>
		<guid isPermaLink="false">http://bit-player.org/?p=153#comment-1716</guid>
		<description>Could you submit a dummy paper, obtain its reference number, and then update the paper in such a way that it incorportates the number?  If you had a magic formula that, given a string of digits, produced a description of the form "the sum of 1/n where n has at most k occurrences of the digit d" that contained those digits, you'd be all set...</description>
		<content:encoded><![CDATA[<p>Could you submit a dummy paper, obtain its reference number, and then update the paper in such a way that it incorportates the number?  If you had a magic formula that, given a string of digits, produced a description of the form &#8220;the sum of 1/n where n has at most k occurrences of the digit d&#8221; that contained those digits, you&#8217;d be all set&#8230;</p>
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