User:TallKite/The delta method: Difference between revisions

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Adding 1 to both numerator and denominator is called '''bumping up'''. Subtracting 1 from both is called '''bumping down'''. Note that bumping up increases the integer limit, but ''decreases'' the size in cents. The basic process is:

* possibly octave-reduce, see below

* possibly unsimplify, see below

*bump the ratio up or down to get a new ratio in which both the numerator and the denominator are multiples of the delta

*simplify by dividing both numerator and denominator by the delta to get the simpler ancestor

*subtract the simpler ancestor from the original ratio to get the more complex ancestor

*~~optional:~~ confirm the answer by multiplying the numerator of one ancestor by the denominator of the other. The two products should differ by 1.

*confirm the answer by multiplying the numerator of one ancestor by the denominator of the other. The two products should differ by 1.

For example, 7/4 is delta-3. Bumping up, we get 8/5. But neither 8 nor 5 is divisible by 3. So instead we bump 7/4 down to 6/3. This simplifies to 2/1, which is the simpler ancestor. Subtract 2/1 from 7/4 to get 5/3 (because 7-2=5 and 4-1=3), which is the more complex ancestor. Optional confirmation: the two products are 2*3=6 and 7*1=7, which do indeed differ by 1.

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For very large deltas, one might want to use the octave complement. For example, 27/16 is delta-11, but 32/27 is only delta-5. The ancestors of 32/27 are 13/11 and 19/16. Thus 27/16 is between 22/13 and 32/19. ~~Its~~ actual ancestors ~~are~~ 5/3 ~~and~~ 22/13.

For very large deltas, one might want to use the octave complement. For example, 27/16 is delta-11, but 32/27 is only delta-5. The ancestors of 32/27 are 13/11 and 19/16. Thus 27/16 is between 22/13 and 32/19. But 32/19 is more complex than 27/16, not simpler. To find the actual ancestors, "subtract" one from another: 32/19 "minus" 22/13 = 10/6 = 5/3. indeed, the actual ancestors of 27/16 are 22/13 and 5/3.

...

The two numbers in the ratio can be thought of as two edos (see "comparing edos" below). This suggests another way of dealing with very large deltas. To solve 27/16, we think of it as a 16edo interval 27\16 and '''octave-reduce''', getting 11\16. Since that's larger than a half-octave, we can further reduce by finding the octave complement 5\16. The question becomes, how many 5\16 steps equals ±1 edostep of 16edo when octave-reduced? Three steps is 15\16, one short of an octave, so the answer is 3. We multiply the larger edo (27) by 3 and get 81. We divide 81 by the smaller edo (16) and get about 5. Thus the simpler stern-brocot ancestor of 27/16 is 5/3.

Another example: 72/41 --> 72\41 --> 31\41 --> 10\41 --> 4*(10\41)=40\41 --> 4*72=288 --> round(288/41)=7 --> 7/4

== Applications ==

=== Approximating ratios ===

Two ratios can be combined to make a 3rd ratio via the [[mediant]] ~~or "~~freshman sum". The 3rd ratio is always intermediate in cents between the other two. For example 8/5 "plus" 15/8 equals 23/13. One can work backwards and decompose any ratio into two simpler ratios, one larger and one smaller. In this example, knowing that 23/13 lies between 8/5 and 15/8 isn't very useful. Far better to find the two stern-brocot ancestors. The delta method gives 7/4 and 16/9, telling us that 23/13 sounds like a slightly flat minor 7th. Furthermore, because 16/9's integer limit is about double that of 7/4, 23/13 is about twice as close to 16/9 than 7/4. If one knows that 7/4 = 969¢ and 16/9 = 996¢, one can estimate 23/13 to be about 985¢ (actual size is 988¢).

Two ratios can be combined to make a 3rd ratio via the [[mediant]] aka freshman sum. The 3rd ratio is always intermediate in cents between the other two. For example 8/5 "plus" 15/8 equals 23/13. One can work backwards and decompose any ratio into two simpler ratios, one larger and one smaller. In this example, knowing that 23/13 lies between 8/5 and 15/8 isn't very useful. Far better to find the two stern-brocot ancestors. The delta method gives 7/4 and 16/9, telling us that 23/13 sounds like a slightly flat minor 7th. Furthermore, because 16/9's integer limit is about double that of 7/4, 23/13 is about twice as close to 16/9 than 7/4. If one knows that 7/4 = 969¢ and 16/9 = 996¢, one can estimate 23/13 to be about 985¢ (actual size is 988¢).

=== Comparing edos ===

@@ Line 28: / Line 28: @@
 Adding 1 to both numerator and denominator is called '''bumping up'''. Subtracting 1 from both is called '''bumping down'''. Note that bumping up increases the integer limit, but ''decreases'' the size in cents. The basic process is:
+* possibly <u>octave-reduce</u>, see below
 * possibly <u>unsimplify</u>, see below
 *<u>bump</u> the ratio up or down to get a new ratio in which both the numerator and the denominator are multiples of the delta
 *<u>simplify</u> by dividing both numerator and denominator by the delta to get the simpler ancestor
 *<u>subtract</u> the simpler ancestor from the original ratio to get the more complex ancestor
-*optional: <u>confirm</u> the answer by multiplying the numerator of one ancestor by the denominator of the other. The two products should differ by 1.
+*<u>confirm</u> the answer by multiplying the numerator of one ancestor by the denominator of the other. The two products should differ by 1.
 For example, 7/4 is delta-3. Bumping up, we get 8/5. But neither 8 nor 5 is divisible by 3. So instead we bump 7/4 down to 6/3. This simplifies to 2/1, which is the simpler ancestor. Subtract 2/1 from 7/4 to get 5/3 (because 7-2=5 and 4-1=3), which is the more complex ancestor. Optional confirmation: the two products are 2*3=6 and 7*1=7, which do indeed differ by 1.
@@ Line 71: / Line 72: @@
 |quintuple
 |}
-For very large deltas, one might want to use the octave complement. For example, 27/16 is delta-11, but 32/27 is only delta-5. The ancestors of 32/27 are 13/11 and 19/16. Thus 27/16 is between 22/13 and 32/19. Its actual ancestors are 5/3 and 22/13.
+For very large deltas, one might want to use the octave complement. For example, 27/16 is delta-11, but 32/27 is only delta-5. The ancestors of 32/27 are 13/11 and 19/16. Thus 27/16 is between 22/13 and 32/19. But 32/19 is more complex than 27/16, not simpler. To find the actual ancestors, "subtract" one from another: 32/19 "minus" 22/13 = 10/6 = 5/3. indeed, the actual ancestors of 27/16 are 22/13 and 5/3.
-...
+The two numbers in the ratio can be thought of as two edos (see "comparing edos" below). This suggests another way of dealing with very large deltas. To solve 27/16, we think of it as a 16edo interval 27\16 and '''octave-reduce''', getting 11\16. Since that's larger than a half-octave, we can further reduce by finding the octave complement 5\16. The question becomes, how many 5\16 steps equals ±1 edostep of 16edo when octave-reduced? Three steps is 15\16, one short of an octave, so the answer is 3. We multiply the larger edo (27) by 3 and get 81. We divide 81 by the smaller edo (16) and get about 5. Thus the simpler stern-brocot ancestor of 27/16 is 5/3.
+Another example: 72/41 --> 72\41 --> 31\41 --> 10\41 --> 4*(10\41)=40\41 --> 4*72=288 --> round(288/41)=7 --> 7/4
 == Applications ==
 === Approximating ratios ===
-Two ratios can be combined to make a 3rd ratio via the [[mediant]] or "freshman sum". The 3rd ratio is always intermediate in cents between the other two. For example 8/5 "plus" 15/8 equals 23/13. One can work backwards and decompose any ratio into two simpler ratios, one larger and one smaller. In this example, knowing that 23/13 lies between 8/5 and 15/8 isn't very useful. Far better to find the two stern-brocot ancestors. The delta method gives 7/4 and 16/9, telling us that 23/13 sounds like a slightly flat minor 7th. Furthermore, because 16/9's integer limit is about double that of 7/4, 23/13 is about twice as close to 16/9 than 7/4. If one knows that 7/4 = 969¢ and 16/9 = 996¢, one can estimate 23/13 to be about 985¢ (actual size is 988¢).
+Two ratios can be combined to make a 3rd ratio via the [[mediant]] aka freshman sum. The 3rd ratio is always intermediate in cents between the other two. For example 8/5 "plus" 15/8 equals 23/13. One can work backwards and decompose any ratio into two simpler ratios, one larger and one smaller. In this example, knowing that 23/13 lies between 8/5 and 15/8 isn't very useful. Far better to find the two stern-brocot ancestors. The delta method gives 7/4 and 16/9, telling us that 23/13 sounds like a slightly flat minor 7th. Furthermore, because 16/9's integer limit is about double that of 7/4, 23/13 is about twice as close to 16/9 than 7/4. If one knows that 7/4 = 969¢ and 16/9 = 996¢, one can estimate 23/13 to be about 985¢ (actual size is 988¢).
 === Comparing edos ===