The Riemann zeta function and tuning: Difference between revisions

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<h2>IMPORTED REVISION FROM WIKISPACES</h2>

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: This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2012-07-16 01:32:48 UTC</tt>.<br>

: This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2012-07-16 01:59:10 UTC</tt>.<br>

: The original revision id was <tt>~~353296842~~</tt>.<br>

: The original revision id was <tt>353298870</tt>.<br>

: The revision comment was: <tt></tt><br>

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\theta(t) \sim \frac{t}{2}\log \frac{t}{2\pi} - \frac{t}{2} - \frac{\pi}{8}+\frac{1}{48t}+ \frac{7}{5760t^3}+\cdots

[[math]]

From this we may deduce that θ(t)/π ≈ r ln(r) - r - 1/8, where r = t/2π = x/ln(2); hence while x is the number of equal steps to an octave, r is the number of equal steps to an "e-tave", meaning the interval of e, 1200/ln(2) = 1731.234 cents.

Recall that Gram points near to pure-octave edos, where x is an integer, can be expected to correspond to peak values of |ζ} = |Z|. We can find these Gram points by Newton's method applied to the above formula. If r = x/ln(2), and if n = floor(rln(r) - r + 3/8) is the nearest integer to θ(2πr)/π, then we may set r⁺ = (r + n + 1/8)/ln(r). This is the first iteration of Newton's method, which we may repeat if we like, but in fact no more than one iteration is really required. This is the first black magic formula, giving an adjusted "Gram" tuning from the orginal one.

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\theta(t) \sim \frac{t}{2}\log \frac{t}{2\pi} - \frac{t}{2} - \frac{\pi}{8}+\frac{1}{48t}+ \frac{7}{5760t^3}+\cdots&lt;br/&gt;[[math]]

--><script type="math/tex">\theta(t) \sim \frac{t}{2}\log \frac{t}{2\pi} - \frac{t}{2} - \frac{\pi}{8}+\frac{1}{48t}+ \frac{7}{5760t^3}+\cdots</script><br />

From this we may deduce that θ(t)/π ≈ r ln(r) - r - 1/8, where r = t/2π = x/ln(2); hence while x is the number of equal steps to an octave, r is the number of equal steps to an &quot;e-tave&quot;, meaning the interval of e, 1200/ln(2) = 1731.234 cents.<br />

<br />

Recall that Gram points near to pure-octave edos, where x is an integer, can be expected to correspond to peak values of |ζ} = |Z|. We can find these Gram points by Newton's method applied to the above formula. If r = x/ln(2), and if n = floor(rln(r) - r + 3/8) is the nearest integer to θ(2πr)/π, then we may set r⁺ = (r + n + 1/8)/ln(r). This is the first iteration of Newton's method, which we may repeat if we like, but in fact no more than one iteration is really required. This is the first black magic formula, giving an adjusted &quot;Gram&quot; tuning from the orginal one.<br />

<br />

@@ Line 1: / Line 1: @@
 <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>2012-07-16 01:32:48 UTC</tt>.<br>
+: This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2012-07-16 01:59:10 UTC</tt>.<br>
-: The original revision id was <tt>353296842</tt>.<br>
+: The original revision id was <tt>353298870</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>
@@ Line 143: / Line 143: @@
 \theta(t) \sim \frac{t}{2}\log \frac{t}{2\pi} - \frac{t}{2} - \frac{\pi}{8}+\frac{1}{48t}+ \frac{7}{5760t^3}+\cdots
 [[math]]
+From this we may deduce that θ(t)/π ≈ r ln(r) - r - 1/8, where r = t/2π = x/ln(2); hence while x is the number of equal steps to an octave, r is the number of equal steps to an "e-tave", meaning the interval of e, 1200/ln(2) = 1731.234 cents.
+Recall that Gram points near to pure-octave edos, where x is an integer, can be expected to correspond to peak values of |ζ} = |Z|. We can find these Gram points by Newton's method applied to the above formula. If r = x/ln(2), and if n = floor(rln(r) - r + 3/8) is the nearest integer to θ(2πr)/π, then we may set r⁺ = (r + n + 1/8)/ln(r). This is the first iteration of Newton's method, which we may repeat if we like, but in fact no more than one iteration is really required. This is the first black magic formula, giving an adjusted "Gram" tuning from the orginal one.
@@ Line 314: / Line 317: @@
 \theta(t) \sim \frac{t}{2}\log \frac{t}{2\pi} - \frac{t}{2} - \frac{\pi}{8}+\frac{1}{48t}+ \frac{7}{5760t^3}+\cdots&amp;lt;br/&amp;gt;[[math]]
   --&gt;&lt;script type="math/tex"&gt;\theta(t) \sim \frac{t}{2}\log \frac{t}{2\pi} - \frac{t}{2} - \frac{\pi}{8}+\frac{1}{48t}+ \frac{7}{5760t^3}+\cdots&lt;/script&gt;&lt;!-- ws:end:WikiTextMathRule:13 --&gt;&lt;br /&gt;
+From this we may deduce that θ(t)/π ≈ r ln(r) - r - 1/8, where r = t/2π = x/ln(2); hence while x is the number of equal steps to an octave, r is the number of equal steps to an &amp;quot;e-tave&amp;quot;, meaning the interval of e, 1200/ln(2) = 1731.234 cents.&lt;br /&gt;
+&lt;br /&gt;
+Recall that Gram points near to pure-octave edos, where x is an integer, can be expected to correspond to peak values of |ζ} = |Z|. We can find these Gram points by Newton's method applied to the above formula. If r = x/ln(2), and if n = floor(rln(r) - r + 3/8) is the nearest integer to θ(2πr)/π, then we may set r⁺ = (r + n + 1/8)/ln(r). This is the first iteration of Newton's method, which we may repeat if we like, but in fact no more than one iteration is really required. This is the first black magic formula, giving an adjusted &amp;quot;Gram&amp;quot; tuning from the orginal one.&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;