Taylor series - Revision history

Lériendil: /* Bimodular approximants */

2025-12-19T08:37:16Z

Bimodular approximants

← Older revision		Revision as of 08:37, 19 December 2025
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	Taking the Taylor series of its logarithm, we obtain (''h''/''s'') (2/''t'') (1 + 1/(3''t''<sup>2</sup>)) - (''h''/''t'') (2/''s'') (1 + 1/(3''s''<sup>2</sup>)) = 2''h''/(''st'') (1 + 1/(3''t''<sup>2</sup>) - 1 - 1/(3''s''<sup>2</sup>)) = '''(2/3) ''h''/(''st'') [1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>)]''', to third order in 1/''s'' and 1/''t'', which is why these are called "third-order" commas - the first-order terms cancel out.		Taking the Taylor series of its logarithm, we obtain (''h''/''s'') (2/''t'') (1 + 1/(3''t''<sup>2</sup>)) - (''h''/''t'') (2/''s'') (1 + 1/(3''s''<sup>2</sup>)) = 2''h''/(''st'') (1 + 1/(3''t''<sup>2</sup>) - 1 - 1/(3''s''<sup>2</sup>)) = '''(2/3) ''h''/(''st'') [1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>)]''', to third order in 1/''s'' and 1/''t'', which is why these are called "third-order" commas - the first-order terms cancel out.

	Let ''h''/''s'' = ''a'' and ''h''/''s'' = ''b''. In that case, ''h''/(''st'') = sqrt(''ab''/''st''). We also note 1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>) = (''s''-''t'')(''s''+''t'')/(''s''<sup>2</sup>''t''<sup>2</sup>), so the expression for the third-order comma becomes '''(2/3) sqrt(''ab'')(''s''-''t'')(''s''+''t'')/(''st'')<sup>5/2</sup>'''. Note that generically, sqrt(''ab''), ''s''-''t'', and ''s''+''t'' are all first-order in the sizes of ''s'' and ''t'', and in that case the expression is in fact only ''second-order'' in ''s'' and ''t''. The families of commas that are truly third-order have either fixed ''ab'', or fixed ''s''-''t''.		But the expression is not necessarily truly third-order. Let ''h''/''s'' = ''a'' and ''h''/''s'' = ''b''. In that case, ''h''/(''st'') = sqrt(''ab''/''st''). We also note 1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>) = (''s''-''t'')(''s''+''t'')/(''s''<sup>2</sup>''t''<sup>2</sup>), so the expression for the third-order comma becomes '''(2/3) sqrt(''ab'')(''s''-''t'')(''s''+''t'')/(''st'')<sup>5/2</sup>'''. Note that generically, sqrt(''ab''), ''s''-''t'', and ''s''+''t'' are all first-order in the sizes of ''s'' and ''t'', and in that case the expression is in fact only ''second-order'' in ''s'' and ''t''. The families of commas that are truly third-order have either fixed ''ab'', or fixed ''s''-''t''.

Lériendil: /* Bimodular approximants */

2025-12-19T08:36:42Z

Bimodular approximants

← Older revision		Revision as of 08:36, 19 December 2025
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	Taking the Taylor series of its logarithm, we obtain (''h''/''s'') (2/''t'') (1 + 1/(3''t''<sup>2</sup>)) - (''h''/''t'') (2/''s'') (1 + 1/(3''s''<sup>2</sup>)) = 2''h''/(''st'') (1 + 1/(3''t''<sup>2</sup>) - 1 - 1/(3''s''<sup>2</sup>)) = '''(2/3) ''h''/(''st'') [1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>)]''', to third order in 1/''s'' and 1/''t'', which is why these are called "third-order" commas - the first-order terms cancel out.		Taking the Taylor series of its logarithm, we obtain (''h''/''s'') (2/''t'') (1 + 1/(3''t''<sup>2</sup>)) - (''h''/''t'') (2/''s'') (1 + 1/(3''s''<sup>2</sup>)) = 2''h''/(''st'') (1 + 1/(3''t''<sup>2</sup>) - 1 - 1/(3''s''<sup>2</sup>)) = '''(2/3) ''h''/(''st'') [1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>)]''', to third order in 1/''s'' and 1/''t'', which is why these are called "third-order" commas - the first-order terms cancel out.

	Let ''h''/''s'' = ''a'' and ''h''/''s'' = b. In that case, ''h''/(''st'') = sqrt(''ab''/''st''). We also note 1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>) = (''s''-''t'')(''s''+''t'')/(''s''<sup>2</sup>''t''<sup>2</sup>), so the expression for the third-order comma becomes '''(2/3) sqrt(''ab'')(''s''-''t'')(''s''+''t'')/(''st'')<sup>5/2</sup>'''. Note that generically, sqrt(''ab''), ''s''-''t'', and ''s''+''t'' are all first-order in the sizes of ''s'' and ''t'', and in that case the expression is in fact only ''second-order'' in ''s'' and ''t''. The families of commas that are truly third-order have either fixed ''ab'', or fixed ''s''-''t''.		Let ''h''/''s'' = ''a'' and ''h''/''s'' = ''b''. In that case, ''h''/(''st'') = sqrt(''ab''/''st''). We also note 1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>) = (''s''-''t'')(''s''+''t'')/(''s''<sup>2</sup>''t''<sup>2</sup>), so the expression for the third-order comma becomes '''(2/3) sqrt(''ab'')(''s''-''t'')(''s''+''t'')/(''st'')<sup>5/2</sup>'''. Note that generically, sqrt(''ab''), ''s''-''t'', and ''s''+''t'' are all first-order in the sizes of ''s'' and ''t'', and in that case the expression is in fact only ''second-order'' in ''s'' and ''t''. The families of commas that are truly third-order have either fixed ''ab'', or fixed ''s''-''t''.

Lériendil: /* Bimodular approximants */

2025-12-19T08:35:28Z

Bimodular approximants

← Older revision		Revision as of 08:35, 19 December 2025
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	Taking the Taylor series of its logarithm, we obtain (''h''/''s'') (2/''t'') (1 + 1/(3''t''<sup>2</sup>)) - (''h''/''t'') (2/''s'') (1 + 1/(3''s''<sup>2</sup>)) = 2''h''/(''st'') (1 + 1/(3''t''<sup>2</sup>) - 1 - 1/(3''s''<sup>2</sup>)) = '''(2/3) ''h''/(''st'') [1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>)]''', to third order in 1/''s'' and 1/''t'', which is why these are called "third-order" commas - the first-order terms cancel out.		Taking the Taylor series of its logarithm, we obtain (''h''/''s'') (2/''t'') (1 + 1/(3''t''<sup>2</sup>)) - (''h''/''t'') (2/''s'') (1 + 1/(3''s''<sup>2</sup>)) = 2''h''/(''st'') (1 + 1/(3''t''<sup>2</sup>) - 1 - 1/(3''s''<sup>2</sup>)) = '''(2/3) ''h''/(''st'') [1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>)]''', to third order in 1/''s'' and 1/''t'', which is why these are called "third-order" commas - the first-order terms cancel out.

			Let ''h''/''s'' = ''a'' and ''h''/''s'' = b. In that case, ''h''/(''st'') = sqrt(''ab''/''st''). We also note 1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>) = (''s''-''t'')(''s''+''t'')/(''s''<sup>2</sup>''t''<sup>2</sup>), so the expression for the third-order comma becomes '''(2/3) sqrt(''ab'')(''s''-''t'')(''s''+''t'')/(''st'')<sup>5/2</sup>'''. Note that generically, sqrt(''ab''), ''s''-''t'', and ''s''+''t'' are all first-order in the sizes of ''s'' and ''t'', and in that case the expression is in fact only ''second-order'' in ''s'' and ''t''. The families of commas that are truly third-order have either fixed ''ab'', or fixed ''s''-''t''.

Lériendil: /* Bimodular approximants */

2025-12-18T20:34:06Z

Bimodular approximants

← Older revision		Revision as of 20:34, 18 December 2025
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	Therefore, ''q''<sup>''h''/''s''</sup> ''p''<sup>-''h''/''t''</sup> is a rational number that represents the difference between a stack of interval ''q'' and another stack of interval ''p''. This defines the third-order comma between the intervals ''p'' and ''q''. In this example, our comma is (7/5)<sup>2</sup> (5/4)<sup>-3</sup> = [[3136/3125]], the [[Didacus]] comma.		Therefore, ''q''<sup>''h''/''s''</sup> ''p''<sup>-''h''/''t''</sup> is a rational number that represents the difference between a stack of interval ''q'' and another stack of interval ''p''. This defines the third-order comma between the intervals ''p'' and ''q''. In this example, our comma is (7/5)<sup>2</sup> (5/4)<sup>-3</sup> = [[3136/3125]], the [[Didacus]] comma.

	Taking the Taylor series of its logarithm, we obtain (''h''/''s'') (2/''t'') (1 + 1/(3''t''<sup>2</sup>)) - (''h''/''t'') (2/''s'') (1 + 1/(3''s''<sup>2</sup>)) = 2''h''/(''st'') (1 + 1/(3''t''<sup>2</sup>) - 1 - 1/(3''s''<sup>2</sup>)) = (2/3) ~~'''~~''h''/(''st'') [1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>)]''', to third order in 1/''s'' and 1/''t'', which is why these are called "third-order" commas - the first-order terms cancel out.		Taking the Taylor series of its logarithm, we obtain (''h''/''s'') (2/''t'') (1 + 1/(3''t''<sup>2</sup>)) - (''h''/''t'') (2/''s'') (1 + 1/(3''s''<sup>2</sup>)) = 2''h''/(''st'') (1 + 1/(3''t''<sup>2</sup>) - 1 - 1/(3''s''<sup>2</sup>)) = '''(2/3) ''h''/(''st'') [1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>)]''', to third order in 1/''s'' and 1/''t'', which is why these are called "third-order" commas - the first-order terms cancel out.

Lériendil: /* Bimodular approximants */

2025-12-18T20:33:53Z

Bimodular approximants

← Older revision		Revision as of 20:33, 18 December 2025
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	Therefore, ''q''<sup>''h''/''s''</sup> ''p''<sup>-''h''/''t''</sup> is a rational number that represents the difference between a stack of interval ''q'' and another stack of interval ''p''. This defines the third-order comma between the intervals ''p'' and ''q''. In this example, our comma is (7/5)<sup>2</sup> (5/4)<sup>-3</sup> = [[3136/3125]], the [[Didacus]] comma.		Therefore, ''q''<sup>''h''/''s''</sup> ''p''<sup>-''h''/''t''</sup> is a rational number that represents the difference between a stack of interval ''q'' and another stack of interval ''p''. This defines the third-order comma between the intervals ''p'' and ''q''. In this example, our comma is (7/5)<sup>2</sup> (5/4)<sup>-3</sup> = [[3136/3125]], the [[Didacus]] comma.

	Taking the Taylor series of its logarithm, we obtain (''h''/''s'') (2/''t'') (1 + 1/(3''t''<sup>2</sup>)) - (''h''/''t'') (2/''s'') (1 + 1/(3''s''<sup>2</sup>)) = 2''h''/(''st'') (1 + 1/(3''t''<sup>2</sup>) - 1 - 1/(3''s''<sup>2</sup>)) = (2/3) ''h''/(''st'') [1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>)], to third order in 1/''s'' and 1/''t'', which is why these are called "third-order" commas - the first-order terms cancel out.		Taking the Taylor series of its logarithm, we obtain (''h''/''s'') (2/''t'') (1 + 1/(3''t''<sup>2</sup>)) - (''h''/''t'') (2/''s'') (1 + 1/(3''s''<sup>2</sup>)) = 2''h''/(''st'') (1 + 1/(3''t''<sup>2</sup>) - 1 - 1/(3''s''<sup>2</sup>)) = (2/3) '''''h''/(''st'') [1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>)]''', to third order in 1/''s'' and 1/''t'', which is why these are called "third-order" commas - the first-order terms cancel out.

Lériendil at 20:33, 18 December 2025

2025-12-18T20:33:35Z

← Older revision		Revision as of 20:33, 18 December 2025
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	== Bimodular approximants ==		== Bimodular approximants ==
			A [[comma]] is a ratio (or logarithmic difference) between stacks of intervals. Consider two intervals, ''p'' < ''q'', with bimodular approximants ''s'' and ''t'' respectively. Let ''h'' be the least common multiple of ''s'' and ''t''. For example, if our two intervals are ''p'' = [[5/4]] and ''q'' = [[7/5]], ''s'' = 9 and ''t'' = 6, so that ''h'' = 18. ''h''/''s'' and ''h''/''t'' are by definition always integers; in this case ''h''/''s'' = 2 and ''h''/''t'' = 3.

			Therefore, ''q''<sup>''h''/''s''</sup> ''p''<sup>-''h''/''t''</sup> is a rational number that represents the difference between a stack of interval ''q'' and another stack of interval ''p''. This defines the third-order comma between the intervals ''p'' and ''q''. In this example, our comma is (7/5)<sup>2</sup> (5/4)<sup>-3</sup> = [[3136/3125]], the [[Didacus]] comma.

			Taking the Taylor series of its logarithm, we obtain (''h''/''s'') (2/''t'') (1 + 1/(3''t''<sup>2</sup>)) - (''h''/''t'') (2/''s'') (1 + 1/(3''s''<sup>2</sup>)) = 2''h''/(''st'') (1 + 1/(3''t''<sup>2</sup>) - 1 - 1/(3''s''<sup>2</sup>)) = (2/3) ''h''/(''st'') [1/(''t''<sup>2</sup>) - 1/(''s''<sup>2</sup>)], to third order in 1/''s'' and 1/''t'', which is why these are called "third-order" commas - the first-order terms cancel out.

Inthar at 20:55, 16 December 2025

2025-12-16T20:55:33Z

← Older revision		Revision as of 20:55, 16 December 2025
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	~~''This is an advanced page dealing with detailed mathematical topics, and should not be referred to for guidance on aspects of xenharmony mentioned here that can be described more simply.''~~		{{technical}}

	A '''Taylor series''' is a method of approximating a function within a certain range by means of adding successive powers of a small parameter, such that each approximation is increasingly precise by a factor of that parameter. While these can be defined for essentially any function, the most useful series in xenharmony are the Taylor series for the logarithm, as the size of an [[interval]] is the logarithm of its [[frequency]] ratio.		A '''Taylor series''' is a method of approximating a function within a certain range by means of adding successive powers of a small parameter, such that each approximation is increasingly precise by a factor of that parameter. While these can be defined for essentially any function, the most useful series in xenharmony are the Taylor series for the logarithm, as the size of an [[interval]] is the logarithm of its [[frequency]] ratio.

Lériendil: /* Logarithmic Taylor series */

2025-12-16T19:22:03Z

Logarithmic Taylor series

← Older revision		Revision as of 19:22, 16 December 2025
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	'''ln((''k''+1)/(''k''-1)) = 2[1/''k'' + 1/(3''k''<sup>3</sup>) + 1/(5''k''<sup>5</sup>) + ...].''' This series is particularly important as this series can compute the logarithm of all intervals from 1 to infinity while ''x'' = 1/''k'' only varies from 0 to 1, as well as due to its rapid convergence.		'''ln((''k''+1)/(''k''-1)) = 2[1/''k'' + 1/(3''k''<sup>3</sup>) + 1/(5''k''<sup>5</sup>) + ...].''' This series is particularly important as this series can compute the logarithm of all intervals from 1 to infinity while ''x'' = 1/''k'' only varies from 0 to 1, as well as due to its rapid convergence.

	To take the logarithm of any arbitrary ratio ''a''/''b'', let ''k'' = (''a''+''b'')/(''a''-''b''); then (''k''+1)/(''k''-1) = ''a''/''b''. We call ''k'' the '''bimodular approximant''' of the fraction ''a''/''b''.		To take the logarithm of any arbitrary ratio ''a''/''b'', let ''k'' = (''a''+''b'')/(''a''-''b''); then (''k''+1)/(''k''-1) = ''a''/''b''. We call ''k'' the '''bimodular approximant''' of the fraction ''a''/''b'', and we will concern ourselves with these as we study '''third-order commas'''.

	== Bimodular approximants ==		== Bimodular approximants ==

Lériendil: Created page with "''This is an advanced page dealing with detailed mathematical topics, and should not be referred to for guidance on aspects of xenharmony mentioned here that can be described more simply.'' A '''Taylor series''' is a method of approximating a function within a certain range by means of adding successive powers of a small parameter, such that each approximation is increasingly precise by a factor of that parameter. While these can be defined for essentially any function,..."

2025-12-16T19:13:50Z

Created page with "''This is an advanced page dealing with detailed mathematical topics, and should not be referred to for guidance on aspects of xenharmony mentioned here that can be described more simply.'' A '''Taylor series''' is a method of approximating a function within a certain range by means of adding successive powers of a small parameter, such that each approximation is increasingly precise by a factor of that parameter. While these can be defined for essentially any function,..."

New page

''This is an advanced page dealing with detailed mathematical topics, and should not be referred to for guidance on aspects of xenharmony mentioned here that can be described more simply.''

A '''Taylor series''' is a method of approximating a function within a certain range by means of adding successive powers of a small parameter, such that each approximation is increasingly precise by a factor of that parameter. While these can be defined for essentially any function, the most useful series in xenharmony are the Taylor series for the logarithm, as the size of an [[interval]] is the logarithm of its [[frequency]] ratio.

== Logarithmic Taylor series ==
We will use certain properties of logarithms in what follows: as a reminder, the sum of two logarithms is the logarithm of the product and the difference of two logarithms is the logarithm of the fraction, and the logarithm of 1 is 0.

For a value of ''x'' between -1 and 1, exclusive, the following series can be found for the natural logarithm (ln) function:

ln(1+''x'') = ''x'' - ''x''2/2 + ''x''3/3 - ''x''4/4 + .... If ''x'' here is a fraction, 1/''k'', this series gives the logarithm of the [[superparticular]] interval (''k''+1)/''k''. In fact, we will generally write these series, and those that follow, in terms of ''k''.

ln(1 + 1/''k'') = ln((''k''+1)/''k'') = 1/''k'' - 1/(2''k''2) + 1/(3''k''3) - 1/(4''k''4) + .... Reversing the sign of ''k'' here, and flipping the sign of the entire logarithm, gives us:

ln(1/(1 - 1/''k'')) = ln(''k''/(''k''-1))= 1/''k'' + 1/(2''k''2) + 1/(3''k''3) + 1/(4''k''4) + ....

Note that we can add these series or subtract these series. Subtracting, and flipping the overall sign, provides us with the expression for -ln((''k''+1)/''k'') + ln(''k''/(''k''-1)) = ln(''k''2/(''k''2 - 1)), the logarithmic size of a [[Mathematics of commas#Square superparticulars|square superparticular]] comma:

ln(''k''2/(''k''2 - 1)) = 1/''k''2 + 1/(2''k''4) + 1/(3''k''6) + .... As the series' first term has 1/''k'' to the second power, square superparticulars (and triangle-particulars, etc.) can be called '''second-order commas'''.

If we add the two expressions, we get the so-called Euler series:

'''ln((''k''+1)/(''k''-1)) = 2[1/''k'' + 1/(3''k''3) + 1/(5''k''5) + ...].''' This series is particularly important as this series can compute the logarithm of all intervals from 1 to infinity while ''x'' = 1/''k'' only varies from 0 to 1, as well as due to its rapid convergence.

To take the logarithm of any arbitrary ratio ''a''/''b'', let ''k'' = (''a''+''b'')/(''a''-''b''); then (''k''+1)/(''k''-1) = ''a''/''b''. We call ''k'' the '''bimodular approximant''' of the fraction ''a''/''b''.

== Bimodular approximants ==