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The fact that noise doesn't repeat makes it useful the way a paint brush is useful when painting. You use a particular paint brush because the bristles have a particular statistical quality - because of the size and spacing and stiffness of the bristles. You don't know, or want to know, about the arrangement of each particular bristle. In effect, oil painters use a controlled random process (centuries before John Cage made such a big deal about it).

Ken Perlin

Perlin Noise Visualizer (made by me)


Desmos

Resources

Let \(s\) be an irrational real number. Take the sequence of values \(\displaystyle m\) such that, given \(\displaystyle a_{k} =\sum _{n=0}^{k}( -1)^{\lfloor ns\rfloor }\),

\begin{equation} a_{k} =a_{2m-k} \end{equation}

For \(\displaystyle k=0,1,\dotsc , m\).

Alternatively, write \(\displaystyle k=m-j\) for \(\displaystyle j=0,1,\dotsc m\). Then (1) can also be written as: \begin{equation*} a_{m-j} =a_{m+j} \end{equation*} We call these \(\displaystyle m_{k}( s)\) the symmetries of \(\displaystyle s\).

Some symmetry formulas

\(\displaystyle m_{k}\left(\sqrt{5}\right) =\begin{cases} -\frac{1}{2} +\frac{\left( 9-4\sqrt{5}\right)^{\frac{k}{2} +1} +\left( 9+4\sqrt{5}\right)^{\frac{k}{2} +1}}{4} & k=0\bmod 2\\ -\frac{1}{2} +\frac{\left( 9-4\sqrt{5}\right)^{\frac{k-1}{2}}\left( 85-38\sqrt{5}\right) +\left( 9+4\sqrt{5}\right)^{\frac{k-1}{2}}\left( 85+38\sqrt{5}\right)}{20} & k=1\bmod 2 \end{cases}\)


\(\displaystyle m_{r}( s) =-\frac{1}{2} +\frac{\left( 1-ns^{*}\right)^{\left\lfloor \frac{r}{nk}\right\rfloor } -( 1-ns)^{\left\lfloor \frac{r}{nk}\right\rfloor }}{2( s+k)}\lceil m_{0}( r) s\rceil +\left[( 1-ns)^{\left\lfloor \frac{r}{nk}\right\rfloor }( 2-ns) +\left( 1-ns^{*}\right)^{\left\lfloor \frac{r}{nk}\right\rfloor }\left( 2-ns^{*}\right)\right] \cdotp \frac{m_{0}( r) +1/2}{2nk+4}\) where \(\displaystyle s=-k+\sqrt{k^{2} +\frac{2k}{n}} ;\ s^{*} =-k-\sqrt{k^{2} +\frac{2k}{n}} ;\ m_{0}( r) =n(\bmod( r,\ k) +1) \ =n,2n,...kn\)


\(\displaystyle \begin{array}{{>{\displaystyle}l}} m_{n}\left(\sqrt{19}\right) =-\frac{1}{2}\\ +\left( 57799-13260\sqrt{19}\right)^{\left\lfloor \frac{n}{10}\right\rfloor }\left[ -\frac{\left\lceil m_{0}( n)\left( -4+\sqrt{19}\right)\right\rceil }{2\sqrt{19}} +\left(\frac{1}{2} -\frac{2}{\sqrt{19}}\right)\Bigl[ m_{0}( n) +1/2\Bigr]\right]\\ +\left( 57799+13260\sqrt{19}\right)^{\left\lfloor \frac{n}{10}\right\rfloor }\left[\frac{\left\lceil m_{0}( n)\left( -4+\sqrt{19}\right)\right\rceil }{2\sqrt{19}} +\left(\frac{1}{2} +\frac{2}{\sqrt{19}}\right)\Bigl[ m_{0}( n) +1/2\Bigr]\right]\\ \\ m_{0}( n) =\begin{cases} 2 & n=0\bmod 10\\ 5 & n=1\bmod 10\\ 19 & n=2\bmod 10\\ 345 & n=3\bmod 10\\ 1362 & n=4\bmod 10\\ 2379 & n=5\bmod 10\\ 15639 & n=6\bmod 10\\ 28899 & n=7\bmod 10\\ 42159 & n=8\bmod 10\\ 55149 & n=9\bmod 10 \end{cases} \end{array}\)


Unorganized ideas (to be cleaned up!)

Check out Joseph Monzo's Tonalsoft Encyclopedia here!

Or check out the xenharmonics wiki!


Let \( f( x) =\left(\frac{3}{2}\right)^{x} 2^{-\lfloor x\log_{2}( 3/2)\rfloor }\).

Then:

  • \( a_0 = 1 \)
  • \( a_n \) is the smallest natural number such that \( f(a_n) < f(a_{n-1}) \)

Some values:

n 0 1 2 3 4 5
\(a_n\) 1 2 7 12 53 359

These correspond to \(a_n\)-ET (equal temperament) tunings that are each successively closer to a Pythagorean tuning.

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