EFD: Difference between revisions

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An EFD will be equivalent to some [[AFS|AFS, or arithmetic frequency sequence]], which has had its count of pitches specified by prefixing "n-"; specifically, n-EFDp = n-AFS((p-1)/n).

To find the steps for an n-EFDp, begin by recognizing that while the multiplicative interval relating your root position to the end position is <math>p</math> (or <math>\frac p1</math>), if you are going to move arithmetically (by repeated addition) from <math>1</math> to <math>p</math>, then the difference in frequency space that you are dividing up is not actually <math>p</math>, but <math>p - 1</math>. And because you are dividing it into <math>n</math> parts, each step will have a size of <math>\frac{p-1}{n}</math>. So, the formula for the frequency of step <math>k</math> of an n-EFDp is:

<math>

f(k) = 1 + (\frac kn)(p-1)

</math>

This way, when <math>k</math> is <math>0</math>, <math>f(k)</math> is simply <math>1</math>. And when <math>k</math> is <math>n</math>, <math>f(k)</math> is simply <math>1 + (p-1) = p</math>.

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 An EFD will be equivalent to some [[AFS|AFS, or arithmetic frequency sequence]], which has had its count of pitches specified by prefixing "n-"; specifically, n-EFDp = n-AFS((p-1)/n).
+To find the steps for an n-EFDp, begin by recognizing that while the multiplicative interval relating your root position to the end position is <span><math>p</math></span> (or <span><math>\frac p1</math></span>), if you are going to move arithmetically (by repeated addition) from <span><math>1</math></span> to <span><math>p</math></span>, then the difference in frequency space that you are dividing up is not actually <span><math>p</math></span>, but <span><math>p - 1</math></span>. And because you are dividing it into <span><math>n</math></span> parts, each step will have a size of <span><math>\frac{p-1}{n}</math></span>. So, the formula for the frequency of step <span><math>k</math></span> of an n-EFDp is:
+<math>
+f(k) = 1 + (\frac kn)(p-1)
+</math>
+This way, when <span><math>k</math></span> is <span><math>0</math></span>, <span><math>f(k)</math></span> is simply <span><math>1</math></span>. And when <span><math>k</math></span> is <span><math>n</math></span>, <span><math>f(k)</math></span> is simply <span><math>1 + (p-1) = p</math></span>.
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