User talk:Contribution: Difference between revisions

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May we know how you arrive at these values? I feel that without any context or explanation people are going to have a hard time understanding why you are editing these pages. Thank you! [[User:2^67-1|2^67-1]] 05:50, 28 August 2024 (UTC)

: Hi. For each pair of superparticular ratios <math>{s1}/{s2}</math> and <math>{s2}/{s3}</math>, there exists a ratio <math>{a}/{b}</math> such that <math>{s1}/{s2}</math> and <math>{s2}/{s3}</math> are <math>{a}/{b}</math> complementary; it is observed that <math>a−b=1</math> or <math>a−b=2</math>.

: In other words, for each ratio <math>a/b</math> where <math>a−b=1</math> or <math>a−b=2</math>, there exists a pair of superparticular ratios <math>{s1}/{s2}</math> and <math>{s2}/{s3}</math> that are <math>{a}/{b}</math> complementary.

: Bellow is a Python code that show for equal divisions of <math>a/b</math> the cent error in the mapping of superparticular ratios <math>{s1}/{s2}</math> and <math>{s2}/{s3}</math> that are <math>a/b</math> complementary.

: When running the tests sequentially for equal divisions of 2/1, 5/3, 3/2, 7/5, 4/3, 9/7, 5/4, 11/9, 6/5, and so on, we observe a converging sequence and pattern for low errors: 5, 7, 12; then 7, 9, 16; then 9, 11, 20; then 11, 13, 24; then 13, 15, 28; then 15, 17, 32; then 17, 19, 36; then 19, 21, 40; then 21, 23, 44; etc. --[[User:Contribution|Contribution]] ([[User talk:Contribution|talk]]) 11:08, 28 August 2024 (UTC)

<pre>

# Enter the values for the X ratio of EDX. Ensure the numerator and denominator differ by 1 or 2.

numerator, denominator = 3, 2

# Program

from math import log

def calculate_complementary_pair(numerator, denominator):

"""

Calculates the superparticular complementary pair based on the difference

between the numerator and denominator.

"""

diff = numerator - denominator

if diff == 1:

# Case where numerator and denominator differ by 1

numerator_2 = numerator * 2

denominator_2 = numerator + denominator

numerator_3 = numerator + denominator

denominator_3 = denominator * 2

elif diff == 2:

# Case where numerator and denominator differ by 2

numerator_2 = numerator

denominator_2 = int((numerator + denominator) / 2)

numerator_3 = int((numerator + denominator) / 2)

denominator_3 = denominator

else:

raise ValueError("Invalid input: numerator and denominator must differ by 1 or 2.")

return (numerator_2, denominator_2), (numerator_3, denominator_3)

def calculate_log_ratio(ratio):

"""

Calculates the logarithmic value of a ratio in cents.

"""

return 1200 * log(ratio) / log(2)

def print_mappings(ratio_log, superparticular_log, numerator, denominator):

"""

Prints the mapping of ratios and their differences in logarithmic cents.

"""

for ed in range(1, 101):

mapping = int(superparticular_log / (ratio_log / ed) + 0.5) # Rounds to the nearest integer

error = mapping * (ratio_log / ed) - superparticular_log

print(f"scale: {ed}ed{numerator}/{denominator}, error: {error:.5f}")

# Calculate the superparticular complementary pairs

(pair_1, pair_2) = calculate_complementary_pair(numerator, denominator)

print(f"Successive superparticular complementary pair of {numerator}/{denominator}: {pair_1[0]}/{pair_1[1]} and {pair_2[0]}/{pair_2[1]}")

# Calculate ratios

ratio = numerator / denominator

superparticular_complementary_ratio = pair_1[0] / pair_1[1]

# Calculate the logarithmic values for the ratios in cents

ratio_log_cents = calculate_log_ratio(ratio)

superparticular_complementary_log_cents = calculate_log_ratio(superparticular_complementary_ratio)

# Output the mappings

print_mappings(ratio_log_cents, superparticular_complementary_log_cents, numerator, denominator)

</pre>

@@ Line 180: / Line 180: @@
 May we know how you arrive at these values? I feel that without any context or explanation people are going to have a hard time understanding why you are editing these pages. Thank you!     [[User:2^67-1|2^67-1]] 05:50, 28 August 2024 (UTC)
+: Hi. For each pair of superparticular ratios <math>{s1}/{s2}</math> and <math>{s2}/{s3}</math>, there exists a ratio <math>{a}/{b}</math> such that <math>{s1}/{s2}</math> and <math>{s2}/{s3}</math> are <math>{a}/{b}</math> complementary; it is observed that <math>a−b=1</math> or <math>a−b=2</math>.
+: In other words, for each ratio <math>a/b</math> where <math>a−b=1</math> or <math>a−b=2</math>, there exists a pair of superparticular ratios <math>{s1}/{s2}</math> and <math>{s2}/{s3}</math> that are <math>{a}/{b}</math> complementary.
+: Bellow is a Python code that show for equal divisions of <math>a/b</math> the cent error in the mapping of superparticular ratios <math>{s1}/{s2}</math> and <math>{s2}/{s3}</math> that are <math>a/b</math> complementary.
+: When running the tests sequentially for equal divisions of 2/1, 5/3, 3/2, 7/5, 4/3, 9/7, 5/4, 11/9, 6/5, and so on, we observe a converging sequence and pattern for low errors: 5, 7, 12; then 7, 9, 16; then 9, 11, 20; then 11, 13, 24; then 13, 15, 28; then 15, 17, 32; then 17, 19, 36; then 19, 21, 40; then 21, 23, 44; etc. --[[User:Contribution|Contribution]] ([[User talk:Contribution|talk]]) 11:08, 28 August 2024 (UTC)
+<pre>
+# Enter the values for the X ratio of EDX. Ensure the numerator and denominator differ by 1 or 2.
+numerator, denominator = 3, 2
+# Program
+from math import log
+def calculate_complementary_pair(numerator, denominator):
+    """
+    Calculates the superparticular complementary pair based on the difference
+    between the numerator and denominator.
+    """
+    diff = numerator - denominator
+    if diff == 1:
+        # Case where numerator and denominator differ by 1
+        numerator_2 = numerator * 2
+        denominator_2 = numerator + denominator
+        numerator_3 = numerator + denominator
+        denominator_3 = denominator * 2
+    elif diff == 2:
+        # Case where numerator and denominator differ by 2
+        numerator_2 = numerator
+        denominator_2 = int((numerator + denominator) / 2)
+        numerator_3 = int((numerator + denominator) / 2)
+        denominator_3 = denominator
+    else:
+        raise ValueError("Invalid input: numerator and denominator must differ by 1 or 2.")
+    return (numerator_2, denominator_2), (numerator_3, denominator_3)
+def calculate_log_ratio(ratio):
+    """
+    Calculates the logarithmic value of a ratio in cents.
+    """
+    return 1200 * log(ratio) / log(2)
+def print_mappings(ratio_log, superparticular_log, numerator, denominator):
+    """
+    Prints the mapping of ratios and their differences in logarithmic cents.
+    """
+    for ed in range(1, 101):
+        mapping = int(superparticular_log / (ratio_log / ed) + 0.5)  # Rounds to the nearest integer
+        error = mapping * (ratio_log / ed) - superparticular_log
+        print(f"scale: {ed}ed{numerator}/{denominator}, error: {error:.5f}")
+# Calculate the superparticular complementary pairs
+(pair_1, pair_2) = calculate_complementary_pair(numerator, denominator)
+print(f"Successive superparticular complementary pair of {numerator}/{denominator}: {pair_1[0]}/{pair_1[1]} and {pair_2[0]}/{pair_2[1]}")
+# Calculate ratios
+ratio = numerator / denominator
+superparticular_complementary_ratio = pair_1[0] / pair_1[1]
+# Calculate the logarithmic values for the ratios in cents
+ratio_log_cents = calculate_log_ratio(ratio)
+superparticular_complementary_log_cents = calculate_log_ratio(superparticular_complementary_ratio)
+# Output the mappings
+print_mappings(ratio_log_cents, superparticular_complementary_log_cents, numerator, denominator)
+</pre>