Mathematics and Music
Partial Solutions for Final Exam Review General Questions

In 5  4 time you need 20 sixteenth notes to fill a measure (5 times 4).
In 6  8 time you need 12 sixteenth notes to fill a measure (6 times 2).

In 4  4 time, a tripledotted eighth note gets 15/16 beats (1/2 + 1/4 + 1/8 + 1/16).
In 3  8 time, a tripledotted eighth note gets 15/8 beats (1+ 1/2 + 1/4 + 1/8).

5  5/4 + 5/16  5/64. The sum of the infinite series is 5/(1  1/4) = 4.

The lcm(a,b) = ab if and only if a and b are relatively prime (ie. the greatest common divisor of
a and b is 1.)

D flat major has five flats using the circle of fifths.

The Pythagorean Comma is 3^(12)/2^(19). It comes from the fact that H^2 does not equal W or the fact that
the circle of fifths does not close up (as explained in the review session.)

Since (2^(7/12))^12 = 2^7, we have that 12 perfect fifths (going around the circle of fifths once) equals 7 octaves
(this is the number of octaves in the circle of fifths going around once.)

52 white keys (88 total  36 black = 52 white  see Exam 1.)

The two notes have frequency 330 Hz and 326 Hz. We "hear" their average, 328 Hz, with 4 beats per second (their difference.)

The brain can distinguish between different instruments. It can also distinguish different frequencies of the
overtone series. It can memorize sounds and filter out unwanted noises.

7Pi is 3.5 times around since 2Pi is equivalent to 1 time around. The cos(7Pi) = 1.

A minor 3rd. The ratio 30/25 = 6/5. In just intonation this is a minor third since a major triad starting
on the tonic will be in the ratio 4:5:6 and the top half of this chord, the interval between 3 and 5 is a
minor third. Alternatively, one could write out the overtone series and realize that the interval between
5f and 6f (ratio 6/5) is a minor third.

Note that the A below middle C will have frequency 220 Hz because this is half 440. The interval from A below middle
C to F sharp above middle C is a Major 6th. Thus, in Pythagorean Tuning we have 220 times 27/16 = 371.25 Hz.
For Just Intonation, a Major 6th is obtained by multiplying by 5/3 so we have 220 times 5/3 = 366.6666 Hz.
Finally, for Equal Temperament we either have 440 divided by 2^(3/12) = 369.994 (down three half steps from A 440) or
220 times 2^(9/12) = 369.994 (up nine half steps from A 220.)

See notes or the text.

See notes. The only note which does not fit well is the 7th overtone which is not in the scale of the
tonic (in this case 7f gives C which is not in the D major scale.) This note falls between two notes on the piano.

2^(4/12) = 2^(1/3) (M3 is four halfsteps.) To show this number is irrational, set 2^(1/3) = p/q, with
p and q integers. Raise both sides to the 3rd power to obtain 2 = p^3/q^3. Cross multiplying
gives 2 q^3 = p^3. The lefthand side of this equation is an integer which when factored into
its prime factorization will have a number of 2's one greater than a multiple of 3 (ie. mod 3 the number
of 2's is 1.) But the righthand side has a prime
factorization with a number of 2's that is exactly a multiple of 3 (ie. mod 3 the number of 2's is 0.)
This violates the Fundamental Theorem of Arithmetic
which states that a positive integer has a unique factorization into its prime factors.

See the handout from class on change ringing. You could write out Plain Bob Minimus or perhaps an extent you composed for homework.

There are 88 possible moves for n = 10 bells. Build the table of allowable moves using the rule that the next entry is the
sum of the previous two entries plus one.

a*b = (4 2 1 5 3), b*a = (2 5 1 4 3), a^3 = (5 3 4 1 2), b^(1) = b.

k = 120 will definitely work. Here's why: Start looking at higher and higher powers
of a. Call this set A = {a, a^2, a^3, a^4, ...}. This set is finite because A is
a subset of S_5 which has 5! = 120 elements. Either we keep multiplying until we find
a k value where a^k = e, or we will exhaust all 120 elements and a^120 = e. Now,
if we do find a k before 120 such that a^k = e, it will have to be a factor of 120.
That's because the set
A = {a, a^2, a^3, a^4, ..., a^k = e}
will be a subgroup! It has closure since any power of a times another power of a can be
reduced using a^k = e to be one of the elements in the group. It is associative. It has
the identity element and every element has an inverse in A since a^j * a^(kj) = a^k = e.
But subgroups have a number of elements which is a factor of the order
of the full group. Thus k is a factor of 120 and if we write k times j = 120 for some
positive integer j, we then have that
a^120 = a^(kj) = (a^k)^j = e^j = e
Thanks to Kate Schelzi who gets bonus points on the final for figuring out an easier
answer to this problem in the review session than I did!

As stated in the last question, subgroups have a number of elements which is a factor of the order
of the full group. Thus A can have either 1, 2, 5, 10, 25 or 50 members.

a^(1) = a^4, b^(1) = b, bab = a implies ab = ba by multiplying by b on the left of both sides, a^(2000) = e,
(ab)^5 = b.

See class notes or the last exam solutions for an example.

The key will be E flat because you will end up having the notes C, F, E flat, G, A flat and B flat (3 flats.)
It is possible this is the key of A flat as well.

Bach  nearly any fugue he wrote did this, "I Got Rhythm" by George Gershwin, "Mikrokosmos No. 141" by Bartok.
A horizontal reflection is known as an inversion in music.

48. There are 12 transpositions, 12 inversions, 12 retrogrades and 12 retrogradeinversions.

Yes. Both 3 and 11 when divided by 4 have a remainder of 3 so they are equivalent. You just add 4 repeatedly
to a number to get other numbers equivalent to it mod 4. So, for example, 3, 7, 11, 15, 19, 23, 27, etc.
are all equivalent mod 4. This is the same as with twelvetone music or thinking of notes on the piano mod 12, ie.
1 == 13 mod 12, 2 == 14 mod 12, 3 == 15 mod 12, etc.