Logharmonic series: Difference between revisions

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with Mike's blessing, modifying the title of his section to clarify its purpose and relation to the other info, and reordering the section to improve the flow of the material on the page
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If a natural number is chosen as <span><math>b</math></span>, the resulting series will be a superset of the harmonic series, inserting extra pitches. For example, the 2-logharmonic series inserts an extra step in between the fundamental and the 2nd harmonic, so that it takes <span><math>2^1 = 2</math></span> steps to reach the 2nd harmonic instead of one. Then it inserts 3 extra steps in between the 2nd harmonic and 3rd harmonic so that it takes <span><math>2^2 = 4</math></span> steps instead of one. Then 7 extra steps before the 4th harmonic so it takes <span><math>2^3 = 8</math></span> steps instead of one.
If a natural number is chosen as <span><math>b</math></span>, the resulting series will be a superset of the harmonic series, inserting extra pitches. For example, the 2-logharmonic series inserts an extra step in between the fundamental and the 2nd harmonic, so that it takes <span><math>2^1 = 2</math></span> steps to reach the 2nd harmonic instead of one. Then it inserts 3 extra steps in between the 2nd harmonic and 3rd harmonic so that it takes <span><math>2^2 = 4</math></span> steps instead of one. Then 7 extra steps before the 4th harmonic so it takes <span><math>2^3 = 8</math></span> steps instead of one.
== Sound ==
The usual harmonic series can be thought of as being derived from a periodic timbre, such as a sawtooth wave. In general, we can derive the sawtooth wave as
<math>\sum_{n=1}^\infty \frac{\sin(nt)}{n}</math>
We can change the "rolloff" on the harmonics to get different derived waveforms which instead have the partials rolling off at a rate of <math>n^s</math>:
<math>\sum_{n=1}^\infty \frac{\sin(nt)}{n^s}</math>
As <math>s</math> increases, so does the rolloff (and hence the brightness of the timbre), so that as we get <math>s \to \infty</math>, our sawtooth wave goes to a sine wave.
If we instead replace our <math>\sin(nt)</math> term with <math>\sin(\log(n) t)</math>, we instead get the Riemann Zeta function, treated as an audio sound wave:
<math>\sum_{n=1}^\infty \frac{\sin(\log(n) t)}{n^s} = \mathbf{Im}[-\zeta(s+it)]</math>
This is easy to see using the Dirichlet series representation:
<math>\zeta(s+it) = \sum_n \frac{1}{n^{s+it}} = \sum_n \frac{n^{-it}}{n^s} = \sum_n \frac{\exp(-it \log(n))}{n^s}</math>
And since we have <math>-\sin(t \log(n)) = \exp(-it \log(n))</math>, this completes our proof.
Note that the above only converges if we have <math>s > 1</math>, e.g. if the rolloff is greater than that of a sawtooth. However, we can use the analytic continuation of the Riemann Zeta function to procure a function even for <math>s < 1</math>. Here is an example of the critical line of the Riemann zeta function (e.g. <math>s = 0.5</math> played as a sound wave):
<youtube>https://www.youtube.com/watch?v=-iZhaAlmoFE</youtube>
(Mike Battaglia's note: I believe this is technically the real part of the critical line, which sounds basically the same as the imaginary part, only with the phases shifted from sine waves to cosine waves.)


== Frequencies ==
== Frequencies ==
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An analogous [https://en.xen.wiki/w/Powharmonic_series#Emulatory_edharmonic_series emulatory edharmonic series] exists.
An analogous [https://en.xen.wiki/w/Powharmonic_series#Emulatory_edharmonic_series emulatory edharmonic series] exists.
== Sound of the logharmonic series as a timbre ==
The usual harmonic series can be thought of as being derived from a periodic timbre, such as a sawtooth wave. In general, we can derive the sawtooth wave as
<math>\sum_{n=1}^\infty \frac{\sin(nt)}{n}</math>
We can change the "rolloff" on the harmonics to get different derived waveforms which instead have the partials rolling off at a rate of <math>n^s</math>:
<math>\sum_{n=1}^\infty \frac{\sin(nt)}{n^s}</math>
As <math>s</math> increases, so does the rolloff (and hence the brightness of the timbre), so that as we get <math>s \to \infty</math>, our sawtooth wave goes to a sine wave.
If we instead replace our <math>\sin(nt)</math> term with <math>\sin(\log(n) t)</math>, we instead get the Riemann Zeta function, treated as an audio sound wave:
<math>\sum_{n=1}^\infty \frac{\sin(\log(n) t)}{n^s} = \mathbf{Im}[-\zeta(s+it)]</math>
This is easy to see using the Dirichlet series representation:
<math>\zeta(s+it) = \sum_n \frac{1}{n^{s+it}} = \sum_n \frac{n^{-it}}{n^s} = \sum_n \frac{\exp(-it \log(n))}{n^s}</math>
And since we have <math>-\sin(t \log(n)) = \exp(-it \log(n))</math>, this completes our proof.
Note that the above only converges if we have <math>s > 1</math>, e.g. if the rolloff is greater than that of a sawtooth. However, we can use the analytic continuation of the Riemann Zeta function to procure a function even for <math>s < 1</math>. Here is an example of the critical line of the Riemann zeta function (e.g. <math>s = 0.5</math> played as a sound wave):
<youtube>https://www.youtube.com/watch?v=-iZhaAlmoFE</youtube>
(Mike Battaglia's note: I believe this is technically the real part of the critical line, which sounds basically the same as the imaginary part, only with the phases shifted from sine waves to cosine waves.)


== Listening ==  
== Listening ==