Operations on intervals

The following are common arithmetic operations on musical intervals. If you're unfamiliar with the operations, you're encouraged to verify the examples for interval operation rules with a calculator.

Stacking and unstacking

Stacking two intervals feels perceptually like we are adding two distances, but it corresponds to multiplying two frequency ratios. The logarithm function is the bridge between frequency space and pitch space:

${\log }_{2}a+{\log }_{2}b={\log }_2(ab)$

for any two frequency ratios a and b. The above equation tells us that the sum of the size (in octaves) of a and the size (in octaves) of b is equal to the size in octaves of the product of two frequency ratios, ab. Hence, stacking corresponds to multiplying frequency ratios in linear (frequency) space, and adding cent values in logarithmic (pitch) space.

To convert octaves to cents, we simply multiply both sides by 1200:

$1200{\log }_{2}a+1200{\log }_{2}b=1200{\log }_2(ab).$

As a consequence, since the inverse operation of multiplication is division, dividing frequency ratios corresponds to subtracting their perceptual sizes:

$1200{\log }_{2}a-1200{\log }_{2}b=1200{\log }_2(a/b).$

Examples

1. $1200{\log }_2(3/2)-1200{\log }_2(5/4)=1200{\log }_2(6/5)$
2. $1200{\log }_2(3/2)+1200{\log }_2(3/2)=1200{\log }_2(9/4)$
3. $1200{\log }_2(16/9)-1200{\log }_2(3/2)=1200{\log }_2(32/27)$

Logarithmic multiplication and division

From the log rule

$n{\log }_{2}r={\log }_2(r^n)$

we see that stacking n copies of the ratio r corresponds to taking the nth power of r:

$n(1200{\log }_{2}r)=1200{\log }_2(r^n).$

The analogous holds for dividing a ratio into n equal parts:

$\frac{1}{n}(1200{\log }_{2}r)=1200{\log }_2(r^{1/n}).$

There is another operation that can be called "logarithmic division", which is finding the perceptual ratio of two intervals. This is given by the base change operation:

$\frac{1200{\log }_{2}r}{1200{\log }_{2}s}={\log }_{s}r.$

Examples

1. $2\cdot 1200{\log }_2(4/3)=1200{\log }_2((4/3{)}^2)=1200{\log }_2(16/9)$
2. $2\cdot 1200{\log }_2(9/8)=1200{\log }_2((9/8{)}^2)=1200{\log }_2(81/64)$
3. $\frac{1}{13}\cdot 1200{\log }_2(2/1)=1200{\log }_2(2^{1/13})$
4. $\frac{1200{\log }_2(81/64)}{1200{\log }_2(9/8)}={\log }_{9/8}(81/64)=2$