Fokker block: Difference between revisions

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=Example=

Consider the periodic scale S[i] with quasiperiod P = 22 whose values for i from 0 to 22 are 1, 33/32, 16/15, 11/10, 9/8, 75/64, 6/5, 5/4, 165/128, 33/25, 11/8, 45/32, 35/24, 3/2, 99/64, 8/5, 33/20, 12/7, 7/4, 231/128, 15/8, 77/40, 2. By solving for the val, or simply testing to see if the patent val works, we quickly find that v = <22 35 51 62 76| sorts the scale in ascending order. A basis for the commas of this val is {50/49, 55/54, 64/63, 99/98}, and by taking three element subsets we find a basis for the wedgies to be {<<1 9 -2 -6 12 -6 -13 -30 -45 -10||, <<2 -4 -4 -12 -11 -12 -26 2 -14 -20||, <<6 10 10 8 2 -1 -8 -5 -16 -12||, <<2 -4 -4 10 -11 -12 9 2 37 42||}, which is to say, {suprapyth, pajara, hedgehog, pajarous}. Taking Z-linear (integer coefficient) combinations, we quickly find that there are four and only four wedgies which give a Graham complexity for the scale less than 22, which are pajara, magic = pajara+hedgehog-suprapyth-pajarous, orwell = pajara+hedgehog-suprapyth, porcupine = suprapyth+pajarous; hence, S is a Fokker block.</pre></div>

Consider the periodic scale S[i] with quasiperiod P = 22 whose values for i from 0 to 22 are 1, 33/32, 16/15, 11/10, 9/8, 75/64, 6/5, 5/4, 165/128, 33/25, 11/8, 45/32, 35/24, 3/2, 99/64, 8/5, 33/20, 12/7, 7/4, 231/128, 15/8, 77/40, 2. By solving for the val, or simply testing to see if the patent val works, we quickly find that v = <22 35 51 62 76| sorts the scale in ascending order. A basis for the commas of this val is {50/49, 55/54, 64/63, 99/98}, and by taking three element subsets we find a basis for the wedgies to be {<<1 9 -2 -6 12 -6 -13 -30 -45 -10||, <<2 -4 -4 -12 -11 -12 -26 2 -14 -20||, <<6 10 10 8 2 -1 -8 -5 -16 -12||, <<2 -4 -4 10 -11 -12 9 2 37 42||}, which is to say, {suprapyth, pajara, hedgehog, pajarous}. Taking Z-linear (integer coefficient) combinations, we quickly find that there are four and only four wedgies which give a Graham complexity for the scale less than 22, which are pajara, magic = pajara+hedgehog-suprapyth-pajarous, orwell = pajara+hedgehog-suprapyth, porcupine = suprapyth+pajarous; hence, S is a Fokker block.

If Q(a,b,c,d) is the ∑(T[i] - μ)^2 quadratic form on a*suprapyth+b*pajara+c*hedgehog+d*pajarous, then explicitly we have Q = 2205.5*a^2 + 880*b^2 + 2904*c^2 + 1254*d^2 + 264*a*b + 2992*a*c - 2574*a*d - 1848*b*c - 440*b*d - 880*c*d. From this we can find Q(pajara) = 880, Q(magic) = 885.5, Q(orwell) = 885.5, and Q(porcupine) = 885.5, with the Graham complexity of S being 21 in magic, orwell and porcupine, and 20 in pajara. If we look at the extrema of a, b, c, and d separately after setting Q = 900, we find they are all less than 2 in absolute value, so we need look no farther than the 27 Z-linear combinations of suprapyth, pajara, hedgehog and pajarous with coefficients less than 2 in absolute value. Had the block not been Fokker, we could have used the analysis of extrema to show it was not.</pre></div>

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<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html"><html><head><title>Fokker blocks</title></head><body><a href="#Preliminaries">Preliminaries</a> | <a href="#First definition of a Fokker block">First definition of a Fokker block</a> | <a href="#Second definition of a Fokker block">Second definition of a Fokker block</a> | <a href="#Third definition of a Fokker block">Third definition of a Fokker block</a> | <a href="#Fourth definition of a Fokker block">Fourth definition of a Fokker block</a> | <a href="#Determining if a scale is a Fokker block">Determining if a scale is a Fokker block</a> | <a href="#Example">Example</a>

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<br />

<h1 id="toc6"><a name="Example"></a>Example</h1>

Consider the periodic scale S[i] with quasiperiod P = 22 whose values for i from 0 to 22 are 1, 33/32, 16/15, 11/10, 9/8, 75/64, 6/5, 5/4, 165/128, 33/25, 11/8, 45/32, 35/24, 3/2, 99/64, 8/5, 33/20, 12/7, 7/4, 231/128, 15/8, 77/40, 2. By solving for the val, or simply testing to see if the patent val works, we quickly find that v = &lt;22 35 51 62 76| sorts the scale in ascending order. A basis for the commas of this val is {50/49, 55/54, 64/63, 99/98}, and by taking three element subsets we find a basis for the wedgies to be {&lt;&lt;1 9 -2 -6 12 -6 -13 -30 -45 -10||, &lt;&lt;2 -4 -4 -12 -11 -12 -26 2 -14 -20||, &lt;&lt;6 10 10 8 2 -1 -8 -5 -16 -12||, &lt;&lt;2 -4 -4 10 -11 -12 9 2 37 42||}, which is to say, {suprapyth, pajara, hedgehog, pajarous}. Taking Z-linear (integer coefficient) combinations, we quickly find that there are four and only four wedgies which give a Graham complexity for the scale less than 22, which are pajara, magic = pajara+hedgehog-suprapyth-pajarous, orwell = pajara+hedgehog-suprapyth, porcupine = suprapyth+pajarous; hence, S is a Fokker block.</body></html></pre></div>

Consider the periodic scale S[i] with quasiperiod P = 22 whose values for i from 0 to 22 are 1, 33/32, 16/15, 11/10, 9/8, 75/64, 6/5, 5/4, 165/128, 33/25, 11/8, 45/32, 35/24, 3/2, 99/64, 8/5, 33/20, 12/7, 7/4, 231/128, 15/8, 77/40, 2. By solving for the val, or simply testing to see if the patent val works, we quickly find that v = &lt;22 35 51 62 76| sorts the scale in ascending order. A basis for the commas of this val is {50/49, 55/54, 64/63, 99/98}, and by taking three element subsets we find a basis for the wedgies to be {&lt;&lt;1 9 -2 -6 12 -6 -13 -30 -45 -10||, &lt;&lt;2 -4 -4 -12 -11 -12 -26 2 -14 -20||, &lt;&lt;6 10 10 8 2 -1 -8 -5 -16 -12||, &lt;&lt;2 -4 -4 10 -11 -12 9 2 37 42||}, which is to say, {suprapyth, pajara, hedgehog, pajarous}. Taking Z-linear (integer coefficient) combinations, we quickly find that there are four and only four wedgies which give a Graham complexity for the scale less than 22, which are pajara, magic = pajara+hedgehog-suprapyth-pajarous, orwell = pajara+hedgehog-suprapyth, porcupine = suprapyth+pajarous; hence, S is a Fokker block.<br />

<br />

If Q(a,b,c,d) is the ∑(T[i] - μ)^2 quadratic form on a*suprapyth+b*pajara+c*hedgehog+d*pajarous, then explicitly we have Q = 2205.5*a^2 + 880*b^2 + 2904*c^2 + 1254*d^2 + 264*a*b + 2992*a*c - 2574*a*d - 1848*b*c - 440*b*d - 880*c*d. From this we can find Q(pajara) = 880, Q(magic) = 885.5, Q(orwell) = 885.5, and Q(porcupine) = 885.5, with the Graham complexity of S being 21 in magic, orwell and porcupine, and 20 in pajara. If we look at the extrema of a, b, c, and d separately after setting Q = 900, we find they are all less than 2 in absolute value, so we need look no farther than the 27 Z-linear combinations of suprapyth, pajara, hedgehog and pajarous with coefficients less than 2 in absolute value. Had the block not been Fokker, we could have used the analysis of extrema to show it was not.</body></html></pre></div>

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 <h2>IMPORTED REVISION FROM WIKISPACES</h2>
 This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
-: This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2012-03-02 13:04:41 UTC</tt>.<br>
+: This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2012-03-02 13:24:18 UTC</tt>.<br>
-: The original revision id was <tt>307196784</tt>.<br>
+: The original revision id was <tt>307204310</tt>.<br>
-: The revision comment was: <tt>Reverted to Mar 1, 2012 11:49 pm: The Wikipspaces editor deleted everything</tt><br>
+: The revision comment was: <tt></tt><br>
 The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>
 <h4>Original Wikitext content:</h4>
@@ Line 47: / Line 47: @@
 =Example=
-Consider the periodic scale S[i] with quasiperiod P = 22 whose values for i from 0 to 22 are 1, 33/32, 16/15, 11/10, 9/8, 75/64, 6/5, 5/4, 165/128, 33/25, 11/8, 45/32, 35/24, 3/2, 99/64, 8/5, 33/20, 12/7, 7/4, 231/128, 15/8, 77/40, 2. By solving for the val, or simply testing to see if the patent val works, we quickly find that v = &lt;22 35 51 62 76| sorts the scale in ascending order. A basis for the commas of this val is {50/49, 55/54, 64/63, 99/98}, and by taking three element subsets we find a basis for the wedgies to be {&lt;&lt;1 9 -2 -6 12 -6 -13 -30 -45 -10||,  &lt;&lt;2 -4 -4 -12 -11 -12 -26 2 -14 -20||, &lt;&lt;6 10 10 8 2 -1 -8 -5 -16 -12||, &lt;&lt;2 -4 -4 10 -11 -12 9 2 37 42||}, which is to say, {suprapyth, pajara, hedgehog, pajarous}. Taking Z-linear (integer coefficient) combinations, we quickly find that there are four and only four wedgies which give a Graham complexity for the scale less than 22, which are pajara, magic = pajara+hedgehog-suprapyth-pajarous, orwell = pajara+hedgehog-suprapyth, porcupine = suprapyth+pajarous; hence, S is a Fokker block.</pre></div>
+Consider the periodic scale S[i] with quasiperiod P = 22 whose values for i from 0 to 22 are 1, 33/32, 16/15, 11/10, 9/8, 75/64, 6/5, 5/4, 165/128, 33/25, 11/8, 45/32, 35/24, 3/2, 99/64, 8/5, 33/20, 12/7, 7/4, 231/128, 15/8, 77/40, 2. By solving for the val, or simply testing to see if the patent val works, we quickly find that v = &lt;22 35 51 62 76| sorts the scale in ascending order. A basis for the commas of this val is {50/49, 55/54, 64/63, 99/98}, and by taking three element subsets we find a basis for the wedgies to be {&lt;&lt;1 9 -2 -6 12 -6 -13 -30 -45 -10||,  &lt;&lt;2 -4 -4 -12 -11 -12 -26 2 -14 -20||, &lt;&lt;6 10 10 8 2 -1 -8 -5 -16 -12||, &lt;&lt;2 -4 -4 10 -11 -12 9 2 37 42||}, which is to say, {suprapyth, pajara, hedgehog, pajarous}. Taking Z-linear (integer coefficient) combinations, we quickly find that there are four and only four wedgies which give a Graham complexity for the scale less than 22, which are pajara, magic = pajara+hedgehog-suprapyth-pajarous, orwell = pajara+hedgehog-suprapyth, porcupine = suprapyth+pajarous; hence, S is a Fokker block.
+If Q(a,b,c,d) is the ∑(T[i] - μ)^2 quadratic form on a*suprapyth+b*pajara+c*hedgehog+d*pajarous, then explicitly we have Q = 2205.5*a^2 + 880*b^2 + 2904*c^2 + 1254*d^2 + 264*a*b + 2992*a*c - 2574*a*d - 1848*b*c - 440*b*d - 880*c*d. From this we can find Q(pajara) = 880, Q(magic) = 885.5, Q(orwell) = 885.5, and Q(porcupine) = 885.5, with the Graham complexity of S being 21 in magic, orwell and porcupine, and 20 in pajara. If we look at the extrema of a, b, c, and d separately after setting Q = 900, we find they are all less than 2 in absolute value, so we need look no farther than the 27 Z-linear combinations of suprapyth, pajara, hedgehog and pajarous with coefficients less than 2 in absolute value. Had the block not been Fokker, we could have used the analysis of extrema to show it was not.</pre></div>
 <h4>Original HTML content:</h4>
 <div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html">&lt;html&gt;&lt;head&gt;&lt;title&gt;Fokker blocks&lt;/title&gt;&lt;/head&gt;&lt;body&gt;&lt;!-- ws:start:WikiTextTocRule:14:&amp;lt;img id=&amp;quot;wikitext@@toc@@flat&amp;quot; class=&amp;quot;WikiMedia WikiMediaTocFlat&amp;quot; title=&amp;quot;Table of Contents&amp;quot; src=&amp;quot;/site/embedthumbnail/toc/flat?w=100&amp;amp;h=16&amp;quot;/&amp;gt; --&gt;&lt;!-- ws:end:WikiTextTocRule:14 --&gt;&lt;!-- ws:start:WikiTextTocRule:15: --&gt;&lt;a href="#Preliminaries"&gt;Preliminaries&lt;/a&gt;&lt;!-- ws:end:WikiTextTocRule:15 --&gt;&lt;!-- ws:start:WikiTextTocRule:16: --&gt; | &lt;a href="#First definition of a Fokker block"&gt;First definition of a Fokker block&lt;/a&gt;&lt;!-- ws:end:WikiTextTocRule:16 --&gt;&lt;!-- ws:start:WikiTextTocRule:17: --&gt; | &lt;a href="#Second definition of a Fokker block"&gt;Second definition of a Fokker block&lt;/a&gt;&lt;!-- ws:end:WikiTextTocRule:17 --&gt;&lt;!-- ws:start:WikiTextTocRule:18: --&gt; | &lt;a href="#Third definition of a Fokker block"&gt;Third definition of a Fokker block&lt;/a&gt;&lt;!-- ws:end:WikiTextTocRule:18 --&gt;&lt;!-- ws:start:WikiTextTocRule:19: --&gt; | &lt;a href="#Fourth definition of a Fokker block"&gt;Fourth definition of a Fokker block&lt;/a&gt;&lt;!-- ws:end:WikiTextTocRule:19 --&gt;&lt;!-- ws:start:WikiTextTocRule:20: --&gt; | &lt;a href="#Determining if a scale is a Fokker block"&gt;Determining if a scale is a Fokker block&lt;/a&gt;&lt;!-- ws:end:WikiTextTocRule:20 --&gt;&lt;!-- ws:start:WikiTextTocRule:21: --&gt; | &lt;a href="#Example"&gt;Example&lt;/a&gt;&lt;!-- ws:end:WikiTextTocRule:21 --&gt;&lt;!-- ws:start:WikiTextTocRule:22: --&gt;
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 &lt;!-- ws:start:WikiTextHeadingRule:12:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc6"&gt;&lt;a name="Example"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:12 --&gt;Example&lt;/h1&gt;
-Consider the periodic scale S[i] with quasiperiod P = 22 whose values for i from 0 to 22 are 1, 33/32, 16/15, 11/10, 9/8, 75/64, 6/5, 5/4, 165/128, 33/25, 11/8, 45/32, 35/24, 3/2, 99/64, 8/5, 33/20, 12/7, 7/4, 231/128, 15/8, 77/40, 2. By solving for the val, or simply testing to see if the patent val works, we quickly find that v = &amp;lt;22 35 51 62 76| sorts the scale in ascending order. A basis for the commas of this val is {50/49, 55/54, 64/63, 99/98}, and by taking three element subsets we find a basis for the wedgies to be {&amp;lt;&amp;lt;1 9 -2 -6 12 -6 -13 -30 -45 -10||,  &amp;lt;&amp;lt;2 -4 -4 -12 -11 -12 -26 2 -14 -20||, &amp;lt;&amp;lt;6 10 10 8 2 -1 -8 -5 -16 -12||, &amp;lt;&amp;lt;2 -4 -4 10 -11 -12 9 2 37 42||}, which is to say, {suprapyth, pajara, hedgehog, pajarous}. Taking Z-linear (integer coefficient) combinations, we quickly find that there are four and only four wedgies which give a Graham complexity for the scale less than 22, which are pajara, magic = pajara+hedgehog-suprapyth-pajarous, orwell = pajara+hedgehog-suprapyth, porcupine = suprapyth+pajarous; hence, S is a Fokker block.&lt;/body&gt;&lt;/html&gt;</pre></div>
+Consider the periodic scale S[i] with quasiperiod P = 22 whose values for i from 0 to 22 are 1, 33/32, 16/15, 11/10, 9/8, 75/64, 6/5, 5/4, 165/128, 33/25, 11/8, 45/32, 35/24, 3/2, 99/64, 8/5, 33/20, 12/7, 7/4, 231/128, 15/8, 77/40, 2. By solving for the val, or simply testing to see if the patent val works, we quickly find that v = &amp;lt;22 35 51 62 76| sorts the scale in ascending order. A basis for the commas of this val is {50/49, 55/54, 64/63, 99/98}, and by taking three element subsets we find a basis for the wedgies to be {&amp;lt;&amp;lt;1 9 -2 -6 12 -6 -13 -30 -45 -10||,  &amp;lt;&amp;lt;2 -4 -4 -12 -11 -12 -26 2 -14 -20||, &amp;lt;&amp;lt;6 10 10 8 2 -1 -8 -5 -16 -12||, &amp;lt;&amp;lt;2 -4 -4 10 -11 -12 9 2 37 42||}, which is to say, {suprapyth, pajara, hedgehog, pajarous}. Taking Z-linear (integer coefficient) combinations, we quickly find that there are four and only four wedgies which give a Graham complexity for the scale less than 22, which are pajara, magic = pajara+hedgehog-suprapyth-pajarous, orwell = pajara+hedgehog-suprapyth, porcupine = suprapyth+pajarous; hence, S is a Fokker block.&lt;br /&gt;
+&lt;br /&gt;
+If Q(a,b,c,d) is the ∑(T[i] - μ)^2 quadratic form on a*suprapyth+b*pajara+c*hedgehog+d*pajarous, then explicitly we have Q = 2205.5*a^2 + 880*b^2 + 2904*c^2 + 1254*d^2 + 264*a*b + 2992*a*c - 2574*a*d - 1848*b*c - 440*b*d - 880*c*d. From this we can find Q(pajara) = 880, Q(magic) = 885.5, Q(orwell) = 885.5, and Q(porcupine) = 885.5, with the Graham complexity of S being 21 in magic, orwell and porcupine, and 20 in pajara. If we look at the extrema of a, b, c, and d separately after setting Q = 900, we find they are all less than 2 in absolute value, so we need look no farther than the 27 Z-linear combinations of suprapyth, pajara, hedgehog and pajarous with coefficients less than 2 in absolute value. Had the block not been Fokker, we could have used the analysis of extrema to show it was not.&lt;/body&gt;&lt;/html&gt;</pre></div>