S-expression: Difference between revisions

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== Equivalent S-expressions ==

~~Curiously, some~~ S-expressions have other equivalent S-expressions. This is easiest to observe in semiparticulars. Here is an incomplete list of examples (feel free to expand):

Some S-expressions have other equivalent S-expressions. This is easiest to observe in semiparticulars. Here is an incomplete list of examples (feel free to expand):

{| class="wikitable center-all

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Note: Where a comma written in the form a/b is used in an S-expression, this means to replace that comma with any equivalent S-expression. This is done in the case of [[3025/3024]] as there are many S-expressions for it so restating them each time it appears seems inconvenient.

A proof that every positive rational number (and thus every JI interval) can be written as an S-expression follows.

It suffices to show every superparticular number including 2/1 has an expression using square-particulars:

<math>

\small 2/1 = S_2 \cdot S_2 \cdot S_3\ ,\\

\small 3/2 = S_2 \cdot S_3\ ,\\

\small 4/3 = S_2\ ,\\

\normalsize \frac{a/(a-1)}{(b+1)/b} = \prod_{k=a}^b \left( S_k = \frac{k/(k-1)}{(k+1)/k} \right) \\

\ \ \ = \frac{a/(a-1)}{(a+1)/a} \cdot \frac{(a+1)/a}{(a+2)/(a+1)} \cdot \frac{(a+2)/(a+1)}{(a+3)/(a+2)} \cdot\ ...\ \cdot \frac{b/(b-1)}{(b+1)/b} = \frac{a/(a-1)}{(b+1)/b} \\

\implies \frac{a/(a-1)}{(b+1)/b} = S_a \cdot S_{a+1} \cdot S_{a+2} \cdot\ ...\ \cdot S_b \\

\implies \frac{2/1 = S_2 \cdot S_2 \cdot S_3}{\prod_{a=2}^k S_a} = (k+1)/k

</math>

From here it should not be hard to see how to make any positive rational number. For 11/6, for example, we can do (11/10)(10/9)(9/8)...(2/1) = 11 and then divide that by (6/5)(5/4)(4/3)(3/2)(2/1), meaning 11/6 = (11/10)(10/9)(9/8)(8/7)(7/6) because of the cancellations, then each of those superparticulars we replace with the corresponding S-expression to get the final S-expression.

== Glossary ==

@@ Line 207: / Line 207: @@
 == Equivalent S-expressions ==
-Curiously, some S-expressions have other equivalent S-expressions. This is easiest to observe in semiparticulars. Here is an incomplete list of examples (feel free to expand):
+Some S-expressions have other equivalent S-expressions. This is easiest to observe in semiparticulars. Here is an incomplete list of examples (feel free to expand):
 {| class="wikitable center-all
@@ Line 231: / Line 231: @@
 Note: Where a comma written in the form a/b is used in an S-expression, this means to replace that comma with any equivalent S-expression. This is done in the case of [[3025/3024]] as there are many S-expressions for it so restating them each time it appears seems inconvenient.
+A proof that every positive rational number (and thus every JI interval) can be written as an S-expression follows.
+It suffices to show every superparticular number including 2/1 has an expression using square-particulars:
+<math>
+\small 2/1 = S_2 \cdot S_2 \cdot S_3\ ,\\
+\small 3/2 = S_2 \cdot S_3\ ,\\
+\small 4/3 = S_2\ ,\\
+\normalsize \frac{a/(a-1)}{(b+1)/b} = \prod_{k=a}^b \left( S_k = \frac{k/(k-1)}{(k+1)/k} \right) \\
+\ \ \ = \frac{a/(a-1)}{(a+1)/a} \cdot \frac{(a+1)/a}{(a+2)/(a+1)} \cdot \frac{(a+2)/(a+1)}{(a+3)/(a+2)} \cdot\ ...\ \cdot \frac{b/(b-1)}{(b+1)/b} = \frac{a/(a-1)}{(b+1)/b} \\
+\implies \frac{a/(a-1)}{(b+1)/b} = S_a \cdot S_{a+1} \cdot S_{a+2} \cdot\ ...\ \cdot S_b \\
+\implies \frac{2/1 = S_2 \cdot S_2 \cdot S_3}{\prod_{a=2}^k S_a} = (k+1)/k
+</math>
+From here it should not be hard to see how to make any positive rational number. For 11/6, for example, we can do (11/10)(10/9)(9/8)...(2/1) = 11 and then divide that by (6/5)(5/4)(4/3)(3/2)(2/1), meaning 11/6 = (11/10)(10/9)(9/8)(8/7)(7/6) because of the cancellations, then each of those superparticulars we replace with the corresponding S-expression to get the final S-expression.
 == Glossary ==