Wikipedia:Reference desk/Archives/Mathematics/2008 September 5

= September 5 =

How many different song variations are there?
It's probably an astronomical number, but the question came to me as I was reading about how one star supposedly was "inspired by" another star to write a similar song and successfully was sued. (It was the one with My Sweet Lord by George Harrison, and He's So Fine)

My question - which is why it's in the math section - is how many possible variations are there on a song? If 10,000 performers/bands each wrote, say, 100 songs a year, thus making a million songs a year, how long before they ran out of variations that were suitably different from each other, so no one performer could say, "Well, that sounds almost exactly like my song." Note that I'm not saying "the exacty same" - obviously, there are subtle differences in songs, but yet "My Sweet Lord" and "He's So Fine" were seen by a court to be similar enough.

I'm just amazed at how people on Name That Tune can guess a song after one note, or even seven; I'm part deaf, but even so, I'm amazed anormal hearing person can tell the difference between songs that seem to start so similarly.Somebody or his brother (talk) 00:22, 5 September 2008 (UTC)


 * There are enormous numbers of possibilities, the vast majority of which, however, will sound absolutely terrible. Calculating the total number of songs that anyone is likely to enjoy listening too is probably extremely difficult due to problems defining what people like. --Tango (talk) 00:59, 5 September 2008 (UTC)


 * Quite true. If the OP wants an idea of the magnitude, let's assume that there are 88 notes (the number of keys on a piano), that only a single note is played at any given time, that every note is held for the same amount of time, that there are no rests, and that the song is exactly 100 notes long. Given these restrictions, more "songs" can be composed than there are atoms in the known universe. Wikiant (talk) 01:36, 5 September 2008 (UTC)
 * There are more "songs" than the number of atoms in the observable universe squared. --Tango (talk) 01:47, 5 September 2008 (UTC)


 * Awww, but what fraction of those songs seems similar? Let's be more concrete.  Lets songs be defined by sequences of notes (ai) and (bi), and define two songs as "similar" if there exists p,q,r such that (ap,...,ap+9) = (bq + r,...,bq+9 + r).  In other words, they are similar if the same pattern of 10 notes occur in both songs modulo a change in pitch.  Now for each real song there would be quite large space of disallowed similar songs.  I'll leave estimating that number to someone with more free time ;-).  Dragons flight (talk) 01:59, 5 September 2008 (UTC)
 * My extremely rough estimation (so rough I daren't share the details!) puts it somewhere in the trillions, at least. --Tango (talk) 02:17, 5 September 2008 (UTC)
 * Assume that a song can be recorded on a CD containing N bits. The total number of possible songs cannot exceed 2N. A lower bound is obtained by using the observation that skilled people can identify a song after hearing M≈7 notes. Let A≈24 be the number of different pitches and B≈5 the number of different durations of a single note. Then the number of different songs are ≤(AB)M≈1207=358318080000000. Bo Jacoby (talk) 07:23, 5 September 2008 (UTC).
 * I believe that part of what makes a song easily recognizable after just a few notes is its general "sound" (instrumentation, accompanying chords, tempo, echoes or lack of them and so on). So even songs that are "formally" very similar (say, sharing a subsequence of notes in their main melody) can be easily told apart. (On the other hand, they are not different in the sense of the OP.) If we could strip a song of all these factors, probably more notes would be necessary to recognize them. Goochelaar (talk) 11:28, 5 September 2008 (UTC)
 * You might want to reduce the number of songs by taking conventions of western music into account. Much of western music is in four-four time and the range of notes maybe span two/three octaves, further if we restrict to songs in a major key, the notes will tend to comprise the root note, the 3rd and the 5th, which sound more harmonious. (This is due to fact that 24/12≈5/4, 27/12≈3/2 see Just intonation). So if we ignore all the fancy jazz stuff and stick to major keys in four four time we will get a much reduced but still large number. I've read somewhere that there is actually quite a small number of common baselines or grooves which tend to be more repetative.
 * Maybe the correct mathematical way to workout the chance of two songs sounding the same would be more a bassian approach. Sample songs in a given genera, rescale so they are all in the same key, find frequency of each note and do the sums. --Salix alba (talk) 11:30, 5 September 2008 (UTC)

What does "bassian" mean here? Cuddlyable3 (talk) 10:34, 10 September 2008 (UTC)


 * All you need now is a way to extrapolate the tune from the first seven notes taking into account these various rules and put the lot on the web so when a person asks for tune number 52933031492 your web server returns the tune with an appropriate copyright. Having filtered out all the current tunes of course. You're gong to make a fortune using the RIAA. Dmcq (talk) 20:03, 5 September 2008 (UTC)
 * For an estimate of how many different themes there are in real music, see Dictionary of musical themes. The book has 655 pages, and lists themes from classical music, transposed to the key of C. I don't know how many themes there are per page, but can hardly imagine there being room for more than 15 (it's written in music notation). This would imply that the book contains about 10,000 entries. The book has a companion volume with themes from opera and songs (lieder?), which has about 8000 entries. So I would guess that the number of reasonably well known themes in classical music is probably smaller than 50,000. --NorwegianBluetalk 20:08, 5 September 2008 (UTC)
 * Consider coding melodies this way: 1) Decide how many notes distinguish melodies, say $$N$$. 2)Give each note one of these 3 values: +, -, or =. The value says whether the note is above, below or equal to the preceding note. There will be 3$N$-1 possible codes since the first note is not coded. Example: Happy Birthday song is =+-+-. This code, which could also be expressed as a number by giving the notes arbitrary numerical weights, is not unique to this song but can serve as a hash for compactly cataloging the vast number of possible melodies.


 * Further, no human can sing all the notes of a piano and we can restrict our estimae to melodies that stay on key i.e. can be played on just the white piano keys, key of C, and have maximum one octave difference between successive notes. Our 3$N$-1 hash codes then suggest (3 x 8)$N$-1 different songs. Error sources in that estimate can be that note changes coded "=" are just that (rather than 8 possible sizes), and that we ignore that melodies can have off-key notes, exceed the octave difference or vary note lengths and intermissions. But since the first error is in opposite sense to the others we can expect some cancellation of errors.


 * If I suppose $$N$$ = 7 then my estimate is (3 x 8)$7$-1 = 2E8 different songs. The 10,000 bands would have work for 20 years.


 * However the contestants identifying songs work with a much smaller number of songs encompassed by culture, known recordings and suitability for the TV show, its singers, orchestra, etc. Cuddlyable3 (talk) 11:27, 10 September 2008 (UTC)

Differential equation
Hello. I've done another exam question and, as usual, I would like you kind people to check it for me. Same old story, please tell me if I'm right but if I'm wrong, let me know but don't tell me how, when, where or why.

Find functions f, g and h such that the equation

$$\frac{d^2y}{dx^2}+f(x)\frac{dy}{dx} +g(x)y=h(x)$$

is satisfied by $$y=x$$, $$y=1$$, $$y=x^{-1}$$ for 0<x<1.

If f, g and h are the functions from above, what condition must the real numbers a, b and c satisfy in order that

$$y=ax+b+\frac{c}{x}$$

should be a solution of the above differential equation?

I have

$$f(x)= -\frac{2}{x(x-1)}$$

$$g(x)=h(x)=\frac{2}{x(x-1)^2}$$

and the final condition is a+b+c=1. Thanks 92.3.59.243 (talk) 19:09, 5 September 2008 (UTC)
 * Correct. Algebraist 19:19, 5 September 2008 (UTC)
 * Cheers. Really appreciate it. 92.3.59.243 (talk) 19:21, 5 September 2008 (UTC)