3/9/98
- Understanding the mean and standard deviation of a data set (such as examination scores): the mean is the average, that is, the sum of the scores divided by the number of scores. The standard deviation measures how the data set is spread from the mean. In essence, it measures the average distance from the mean. It can be computed by subtracting each datum from the mean, squaring the result, averaging those squares, and then computing the square root of the answer.
- Now that we have passed the mid-point of the semester, we will be using our previously established skills to gain a much deeper understanding of linguistics and elementary mathematics.
- What is a process? Recall that our model had a speaker with a statement to communicate using a language (grammar) to create sounds, which are heard by the listener's grammar to create an understanding of the speaker's statement. We had considered the language L as a black box, in which items (the input) were operated on by processes to create new items (the output).
A similar process takes place in mathematics, leading us to ponder over the remainder of the semester the following questions:
- How are the items of language like the items of mathematics? How do they differ?
- How do the processes of language resemble those of mathematics? How do they differ?
- Today, we will be speaking about linguistic correspondences. Remember Grimm's Law, or the First Germanic Sound Shift. Consider one example of a correspondence:
Remember that "five" does not derive from "pente" or "quinque," but that instead both derive from a common root in proto-Indo-European.
Greek Latin Sanskrit Old Slavic Germanic p- p- p- p- f-
This correspondence indicates that a process has taken place. We formulate this process in terms of laws (for major processes) and rules (for minor ones).
- Our example for today deals with words for five in various IE languages. The set of information we will use is called a "corpus."
The reconstructed form of the word for 5 in proto-Indo-European is *penkwe. The initial asterisk means that this is a reconstructed form, and the letter "w" is superscripted to show a particular [kw] sound.
- We need to talk a bit about this [kw] sound. Recall that we classified consonant sounds in part on their place of articulation. Consider the voiceless plosives:
There are actually two different "k" sounds, corresponding to the English words "kill" and "cool." The "k" of "kill" has a higher apparent pitch than that of "cool." This corresponds to a slightly different place of articulation. The "k" of "kill" is written [k'] (the accent mark should be over the letter k), and is called palatal; the "k" of "cool" is velar. The chart should therefore look like this:
Labial Dental Velar p t k
If we add voiced plosives, we actually get this:
Labial Dental Palatal Velar p t k' k
The sound [g'] means the "g" of "gill," and the sound [g] means the "g" of "ghoul." In English, these sounds function similarly, and which to use is determined by the following vowel sound. If the succeeding vowel sound is front, we use the palatal consonant; otherwise, we use the velar one.
Labial Dental Palatal Velar Voiceless plosives p t k' k Voiced plosives b d g' g
But that is not the whole story. There are actually three types of "k" and "g" sounds. The other one is called "labiovelar", and is written [kw] and [gw] (where the letter "w" should be superscripted). This is a velar sound made with rounded lips, which explains the name. So the chart actually is:
Labial Dental Palatal Labiovelar Velar Voiceless plosives p t k' kw k Voiced plosives b d g' gw g
- Now return to our discussion of the words for 5 in various Indo-European languages. We have
Some examples of some relatives of some of these words: "punch" has five ingredients. "Punjab" is a region with five rivers. "Parcheesi(tm)" is a game with a top score of 25. In the game of Keno, a player picks 5 numbers (derived from the French quine, from Latin quinies `5 apiece').
Sanskrit pañca Greek pénte/pémpe Latin quinque Gothic fimf Old English fif (long i) Old High German finf/funf Old Church Slavonic peti Lithuanian penkí Armenian hing Old Irish cóig
- What processes lead from *penkwe to all of these? First look at vowels. Proto-IE has five vowel sounds, [a], [e], [i], [o], and [u]. Sanskrit has only three: [i], [u], and [a] (all long or short). What has happened? Vowel merger. The [e], [o], and [a] sounds all merge into the Sanskrit [a].
So as *penkwe becomes pañca, the [e] sounds become [a]. (NB: This is the sound of u in the English word "but.")
The schwa and [i] merged into [i] in Sanskrit as well. For example, the word for "father" in many IE languages has an [a] sound, but it has an [i] sound in Sanskrit; this shows that the sound in Proto-IE must have been a schwa.
- But what about the Sanskrit word "ekadhena?" This [e] is of secondary origin. It is derived from a diphthong. Consider the following table:
There are other examples of this vowel change before consonants. *nai means "lead," but *nayati is "he leads", as the "ai" becomes "ay" before the following vowel sound [a]. However, the infinitive form is *netum, because the sound following [e] is [t], a consonant.
ei
oi
aiai e (before consonant)
ay (before vowel)eu
ou
auau o (before consonant)
av (before vowel)
So the "a" sounds in pañca are the result of vowel merger.
- Another process that takes place is palatalization. In this process, a plosive sound becomes an affricate.
Recall that [t] is a plosive, and [s] is a fricative consonant. A plosive followed by a fricative is an affricate, such as "ts" in "tsetse." In English, there are two of these. The "ch" in "chill" is actually [t] followed by [sh], and the "J" in "Jack and Jill" is actually [d] followed by [sh] (the voiced version of the first one).
So the labiovelar plosive [kw] of Sanskrit became the palatal affricate [ch] of pañca.
- What triggers this process? More broadly, what is the context of this change? A front vowel:
A labiovelar plosive or velar plosive becomes a palatal or prepalatal affricate before a front vowel. In the word *penkwe, the [kw] is followed by e, a front vowel, the [kw] becomes the ch of pañca..
- But there is something more going on here. These two processes must have taken place in a particular order. If the vowel merger had occurred first, then the palatalization would have have happened, because the vowel following [kw] wouldn't have been a front vowel. So we see that palatalization had to occur before the vowel merger.
3/11/98
- Recall that we are learning how "penkwe" became "pañca." The first process was vowel merger, in which we had the following vowels merge:
Recall how we can classify these by distinctive features:
i
schwai u u e
a
oa
Now that we have only three vowel sounds, we need only two distinctive features:
i e a o u HI + - - - + LOW - - + - - BACK - - + + +
i a u HI + - + BACK - + +
- The second process we spoke about was palatalization, in which the plosives [k] and [kw] became the affricate spelled c (pronounced something like [ch]; the exact symbol is not possible in HTML) before [i] and [e].
- Finally, we need to talk about the tilde over the n. A consonant feature that we have not talked about yet is "nasals." They can be labial [m], dental [n], palatal (the symbol is not part of HTML, but it is pronounced something like "nya") and velar (again, the symbol is not part of HTML, but it is pronounced something like the "ng" of "rang"). The palatal nasal is written ñ.
- How did the dental nasal in "penkwe" become the palatal nasal of "pañca?" The answer is assimilation, the third process we will learn about today.
The word "assimilation" is itself an example of the process of assimilation. The Latin roots are "ad"+"similis" (meaning to+like), making something like something else. In this case, the "d" in "ad" became another "s" from "similis."
In words like "impossible" and "incredible," the "n" sound of "in" has changed. In "impossible," it became a labial nasal; in "incredible," it became a velar nasal. These are examples of partial assimilation, in which the nasal adjusts to the following consonant sound. In contrast, in the Hindi word "pacisi" `five', the "n" has been totally assimilated into the "c."
What happened here? The palatalization of "penkwe" into "panca" then triggered assimilation, and the "nc" became "ñc."
Assimilation can be total or partial. It can also be progressive or regressive. In the word "assimilation," the assimilation was regressive, in that the "d" sound was absorbed by the "s." Similarly, the assimilation in "pañca" was regressive. Assimilation can also be contiguous (neighboring letters) or discontiguous; the examples above are contiguous.
- Recall our chart of words meaning five in various languages. Look in particular at Latin "quinque" and Gothic "fimf." Grimm's Law tells us that "p" becomes "f". So pre-Grimm, the Gothic word must have been "pimp." The transformation of "penkwe" into "penpe" is an example of a progressive discontiguous assimilation, as the initial [p] assimilated the later [kw] into another [p]. (This is discontiguous because there are intervening letters.) Then "penpe" must have become "pempe," an example of regressive assimilation.
On the other hand, in Latin, "penkwe" must have become "kwenkwe," an example of regressive discontiguous assimilation.
- What determines whether the assimilation is regressive or progressive? There is no over-riding pattern.
- Now look at the Greek "pente" or "pempe" (depending on dialect). Greek treats labiovelars strangely. [kw] can become [p], [t], or [k], depending on the preceding vowel sound. (This is an example of a split, which is the opposite of a merger.)
Another example is the Indo-European "kwel" (meaning "to dwell," as in the Latin "colere" and the Sanskrit "colere"). In Greek, forms are both "aipolos" and "boukoulos," so [kw] can become either [p] or [k] depending on the preceding vowel. The Indo-European word "kwi" became Greek "ti."
The IE "penkwe" has become first "penpe" and then by assimilation "pempe."
- Comments on the examination:
- When calculating the complement of a number using nikhilam, you must use the sutra and take each from 9, the last from 10. It isn't enough to simply write that you subtracted the number from 1000 (or whatever the base is).
- "Spare the rod and spoil the child" can be interpreted as a command (i.e., "you should spare the rod and spoil the child"). Similarly, consider the phrase "move and I'll shoot."
However, we know that the correct interpretation is "If you spare the rod, then you will spoil the child." Other phrasings are "He who spares the rod, spoils the child" or "Whomever spares the rod will spoil the child."
A related example is "feed a cold and starve a fever." This is not a two-part command, but instead a conditional statement: "If you feed a cold, then you will starve of [e.g., die of] fever."
- Comments about primality testing: To see if the number 209 is prime, one could do several things:
When you do this, you find that 209=11*19, and so it isn't prime. But the point here is that the number of possible factors that need to be tried is much smaller than you might expect at first.
- Try to divide 209 by 2, 3, 4, ..., 208.
This is terrible, since clearly, one can see that no number bigger than 104 (i.e, half of 209) can divide 209.
- Try to divide 209 by 2, 3, 4, ..., 104.
This is better, but not much. Notice that if two numbers a and b multiply to 209, then one of them must be less than 15. (Why? Because 15*15 is 225, which is bigger than 209.) So we need only try dividing by numbers less than 15.
- Try to divide 209 by 2, 3, 4, 5, ..., 14.
This is almost as good as we can do, but not quite: If 209 is not divisible by it also can't be divisible by 6, 9, or 12. We need only try dividing 209 by primes. So our final answer is:
- Try to divide 209 by 2, 3, 5, 7, 11, and 13 (the only primes under 15).
3/13/98
For a detailed presentation of this lecture material see
Musical intervals, harmonics and the pure-tone scale
- Recall that dividing strings in a ratio of 2:1 produces an octave. Repeating this idea, we get the following set of ratios, assuming that the full string makes a C:
Roughly, G to c'' is the range of a human voice. Side note: The note G used to be denoted "Gamma," and the high note used to be called (and still is in some countries) "ut." We get the word "gamut" from a contraction of "gamma-ut."
C c c' c'' c''' 1 2 4 8 16
- Other ratios give other tones:
The ratio of 7:6 is a weird sound, somewhere between A and B-flat. Similarly for 11:10, and 13:12.
C c g c' e' g' [i'] c'' d'' e'' [k''] g'' [l''] [i''] b'' c''' 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Side note: in German, the letter B refers to the note B-flat, and the letter H is used for B-natural (the note that we call B in English). Hence, BACH translated his name into the notes we call B-flat, A, C, B-natural.- The ratio of 4:5 is a major third, and we divide 8:10 into 8:9 and 9:10, giving two types of seconds.
- To subtract intervals, treat them as fractions and divide. To add intervals, treat them as intervals and multiply.
3/16/98
- Let's look at the scale again in terms of intervals:
In this chart, II refers to a major second of 9:8, and 2 refers to a minor second of 10:9. s is a semi-tone.
c d e f g a b c' II 2 s II 2 II s
Notice that you can think of this as two fourths from C-F and G-C. In the first fourth, the intervals are II-2-s, while in the second fourth, the intervals are 2-II-s. You can also think of this as two overlapping fifths of C-G and F-C.
- Another important concept related to this is the circle of fifths. In the following table, think of C as being at the top of a clock, and each succeeding note proceeds around the clock in a clock-wise direction, ending with B# at the top of the clock again:
O'Clock Key Key signature 12 or 0 C 0 sharps, 0 flats 1 G 1 sharp 2 D 2 sharps 3 A 3 sharps 4 E 4 sharps 5 B 5 sharps 6 F# or G-flat 6 sharps or 6 flats 7 C# or D-flat 5 flats 8 G# or A-flat 4 flats 9 D# or E-flat 3 flats 10 A# or B-flat 2 flats 11 E# or F 1 flat 12 B# or C 0 sharps, 0 flats
We call B# enharmonic to C, and E# enharmonic to F. Similarly, C-flat is enharmonic to B, and F-flat is enharmonic to E.- More about *penkwe and its descendants. Recall that last time we saw that there were at least three processes acting:
- Merger
- Palatalization
- Assimilation
Today we will discuss three more processes:- Raising
- Nasalization
- Compensatory lengthening
- Raising is a process in which mid-vowel sounds become high-vowel sounds. [e] can become [i], and [o] can become [u].
Consider the Gothic word "fimf." How did this derive from *penkwe? First, "apocope" (the loss of the final vowel sound) occurred. Then "penkw" became "penp" (assimilation), which became "pemp" (more assimilation), which became "femf" (Grimm's Law). But what about the change in the vowel? The transition from [e] to [i] is an example of vowel raising.
What is the motivation? The vowel is raised when it is before a nasal+consonant combination.
- Look at the Latin "quinque." Here, we have *penkwe turning into kwenkwe (regressive discontiguous assimilation), and then vowel raising turned the "enk" into "ink".
One more bit of an example. The Armenian word "hing" has obviously had man changes, but in particular, the vowel [e] has become [i].
This didn't occur in all languages: it didn't happen in the words for 5 in Greek, Lithuanian, and Sanskrit.
- Slavic and French show the process of nasalization. This means that a vowel is nasalized when followed by a nasal+consonant combination. The nasal then disappears. The Slavic word "peti" (should be written differently, but HTML limits our options) has a nasal [e], and the [n] has disappeared.
- Sometimes the disappearance of a sound is followed by compensatory lengthening, in which a nasal disappears, but the vowel is lengthened, rather than becoming nasalized.
- More about gaps between primes. In the following chart, the left-most column indicates the gap between prime pairs, and the column-header indicates the largest prime considered. For example, the row labeled "2" and the column labeled "300" indicate that primes differing by 2 were searched for, up to a maximum of 300. My convention was to include a prime pair even if only the smaller of the two primes was less than the column-header.
Largest number considered Gap between primes 100 300 1000 10,000 100,000 1,000,000 2 8 19 35 205 1224 8169 4 9 17 41 203 1216 8144 6 16 34 74 411 2447 16386 8 9 17 38 208 1260 8242 10 11 22 51 270 1624 10934 12 15 33 70 404 2421 16378
One can conjecture without too much fear that the prime pairs with gaps of 2, 4, and 8 occur with about the same frequency, while prime pairs with gaps of 6 and 12 occur about twice as often. The prime pairs with a gap of 10 occur around 4/3 as often as prime pairs with a gap of 2, though this is not as clear from the data.
The interesting thing about these patterns is that they were first noticed around the turn of the century (when good prime tables became available), and yet no one has the faintest idea of how to prove any of them.- Another interesting thing about 5 is that it is 1+4. That raises the question of how to find a formula for 12 + 22 + ... + n2. Trying the trick that we did for adding up the first few integers doesn't work. If we try
and nothing interesting seems to be going on.
S = 1 + 4 + 9 + 16 + 25 S = 25 + 16 + 9 + 4 + 1 2S = 26 + 20 + 18 + 20 + 26
- Let's try another approach, based on the idea that (a+1)2=a2+2a+1. Take 12 + 22 + 32 + ... + n2, and rewrite it as (0+1)2 + (1+1)2 + (2+1)2 + ... + ((n-1)+1)2 = 02 + 2*0 + 1 + 12 + 2*1 + 1 + 22 + 2*2 + 1 + ... + (n-1)2 + 2*(n-1) + 1. Cancel out the terms 12+22+...+(n-1)2 from both sides of the equation, and we get
n2 We are adding up 1 n times so if we do that addition, we get
n2 = n + 2*(1+2+...+(n-1)) Bring the n to the other side of the equation, and divide by 2, and we have
(n2 - n)/2 = 1+2+...+(n-1) This is a true formula, but of limited use. We can, however, draw one conclusion for next time: We are going to need better ways to handle summations. Next time, we will talk about "sigma notation," which will simplify a lot of our algebraic problems.
3/18/98
- A bit more about the circle of fifths, this time from a harmonic point of view. Consider the piece that looks like this:
This corresponds to a bit of the left-hand side of an accordion.
F C G D A E B
The harmonic underpinnings of a piece of Western music---that is, the note or notes played by the tuba or the bass player in the band or the left-hand of the pianist---tend to move along this circle of fifths in a particular way. If we think of C as home, then the most traditional songs won't move very far away from C. For example, "The Star-Spangled Banner" mostly has harmonies of C and G, with one or two moves to F late in the piece. A more complicated piece such as "The Stars and Stripes Forever" is also mostly contained within this restricted range.
A relatively simple Beatles song such as "Let It Be" mostly is within the C-G-F range, but there is one leap, at the start of the chorus, out to A. The harmonies then move back A-G-F before landing on C.
A more harmonically adventuresome piece such as "Yesterday" is more interesting to study. Right at the start, on the words "all my troubles seemed so far away," the harmony jumps from C all the way to B. It then moves back mostly step-wise through the circle of fifths before landing on C at the end of the verse. The section "Why she had to go" remains for quite some time in the B-E-A region before returning home to C at the end of the bridge.
- A fundamental principle of Western music seems to be that pieces jump away from C in a sudden shift, and then return to C stepwise moving to the left along the chart. Sometime, the steps over-shoot C to F, and then return to C (so that the harmony might proceed in a pattern such as A-D-G-F-C). Almost no piece of music moves away from C gradually, and then suddenly jumps back to home; there is something satisfying about the gradual return through the circle of fifths to home.
- More about interval arithmetic: we can think of larger intervals as being composed of combinations of II (9:8) and 2 (10:9), along with semi-tones. Consider the following chart:
In our conventions, the interval from d-a is not a fifth, because it consists of 1 step of II, 2 steps of size 2, and one step of size s.
II 2 s 6:5 = 1.20 3 1 0 1 5:4 = 1.25 III 1 1 0 4:3 = 1.33 IV 1 1 1 3:2 = 1.50 V 2 1 1
- Greek mathematicians studied regular arrangements of points in shapes other than squares. One favorite was triangles, giving arrangements such as
We can make a chart of triangular numbers along side those of squares. Call the square numbers sq(n) (so that, for example, sq(2)=2*2=4), and call the triangular numbers tr(n) (so that we can see from the above diagrams that tr(3)=1+2+3=6).
* *
* **
* *
* * **
* *
* * *
* * * *
We can rapidly guess that sq(n) = tr(n) + tr(n-1), and some work with a square arrangement of dots, divided along the diagonal into two nearly-equal triangles, confirms this formula. (Sorry; the picture is too complicated for HTML.)
n tr(n) sq(n) 1 1 1 2 3 4 3 6 9 4 10 16 5 15 25 6 21 36
- We need to talk about sigma notation to be better equipped to handle summations. In this section, we will write sigma_(n=1)^5 to mean the letter capital sigma, with n=1 underneath the sigma, and 5 above.
The notation sigma_(k=2)^7 (k2+k) means that we set the variable k in turn to the values 2, 3, 4, 5, 6, and 7, and substitute that value of k into the formula following the sigma. We then add up the results. In this case, we get the summation (22+2) + (32+3) + (42+4) + (52+5) + (62+6) + (72+7).
The first interesting complication is that the variable which is used is irrelevant. In other words,
sigma_(k=2)^7 (k2+k) = sigma_(t=2)^7 (t2+t). The second tricky point idea is that we have to think of sigma_(r=2)^4 1 as the sum 1+1+1, where the first 1 is for r=2, the second one is for r=3, and the third 1 is for r=4. Regardless, we still have that this sum is just another way to write the number 3.
Another tricky point is that sometimes, sums which aren't obviously equal will in fact be equal. For example,
sigma_(r=2)^4 sqrt(r+1) = sigma_(m=3)^5 sqrt(m).
- Remember that we started the semester by studying sums of powers of 2, and we saw that
1 + 2 + 4 + 8 + ... + 2k = 2k+1 - 1. We can express this in sigma notation as
sigma_(p=0)^k 2p = 2k+1 - 1. This is the same formula as before, but it looks a whole lot trickier.
- We also have the formula
1 + 2 + ... + n = n(n+1)/2. We can also express that in sigma notation as
sigma_(i=1)^n i = n(n+1)/2.
3/20/98
- Class started with an analysis of symmetries and patterns in the melody of "Yesterday."
- A bit more about the homework assignment: we were talking about the tuning of a violin, which is in perfect fifths:
Transpose the e'' down an octave, and we get e' with a ratio of 27:16. But the ratio of e' to g should be a sixth, which is a ratio of 5:3. The fact that these two are unequal is an unavoidable problem.
g d' a' e'' 1 3:2 9:4 27:8
- The Pythagoreans tried to avoid this problem by tuning in a circle of fifths, and then transposing downwards. That gives
If we re-arrange these into a scale, we get the following intervals:
F C G D A E B c 4:3 1 3:2 9:8 27:16 81:64 243:128 2:1
What happens at the intervals marked with question marks? Because we have expanded some intervals from 2 to II, the semi-tones must shrink by a corresponding amount. The Pythagorean semi-tone is in fact 256:243 (B:c in the above chart). This is often rounded to 25:24.
C D E F G A B c II II ?? II II II ??
The effect is that the interval C:E has now become larger, and the intervals A:c and E:G have become smaller.
- More about summation notation. We started with the observation that (sigma) (k2 + 3k) = (sigma) k2 + 3 (sigma) k. Then we showed how to sum up the integers from 1 to n using sigma notation.
3/23/98
- This week is sponsored by the number 8 (though we will continue with applications involving the number 5 as well). Note that 8 and 5 are both examples of Fibonacci numbers.
- Tom Lehrer's "New Math" was played in class.
- More about adding up squares. We showed in class last time that to compute (sigma)_(i=1)^n i, we could use a trick:
S = 1 + 2 + ... + n = 0 + 1 + ... + n = (sigma)_(i=0)^n i But what happens when we try this same trick to add up squares?
S = (sigma)_(i=0)^n (n-i)
2S = (sigma)_(i=0)^n i + (n-i) = (sigma)_(i=0)^n n = n(n+1)
S = n(n+1)/2
S = 12 + 22 + ... + n2 = 02 + 12 + ... + n2 = (sigma)_(i=0)^n i2 And disaster strikes, since i2 occurs twice with a plus sign.
S = (sigma)_(i=0)^n (n-i)2
2S = (sigma)_(i=0)^n i2 + (n-i)2 = (sigma)_(i=0)^n i2 + n2 - 2ni + i2
- Here's another method which does work for adding up squares. Unfortunately, it's very graphical, so this explanation only supplements what happened in class.
Draw rectangles with base 1 and height 1, 2, and so on up to n. Put these rectangles together, and complete the picture into a large rectangle (actually a square, in this case). Then we see that
1 + 2 + 3 + ... + n + (missing area) = area of rectangle = n*n = n2 because the large rectangle happens to have height n and base n.
What about the missing area? We can think of it as being made up of thin strips, each of height 1. The first one has base 1, the second one has base 2, and so on until the last one has base (n-1). So the formula becomes
1 + 2 + ... + n + (1+2+...+n-1) = n2 Now add n to both sides, and we get
1 + 2 + ... + n + (1+2+...+n-1+n) = n2+n.
2(1+2+...+n) = n2+n.
1+2+...+n = (n2+n)/2.
- Now this approach can generalize to adding up squares. Take a square of side 1, a square of side 2, and so on up to a square of side n, and place them side by side. Put them all into a large rectangle, and we get the formula
12 + 22 + ... + n2 + (missing area) = area of rectangle How do we get the area of the rectangle? This time, the base is 1+2+...+n, and the height is n, so we get
12 + 22 + ... + n2 + (missing area) = n*(n2+n)/2. What about the missing area? Think of it as thin rectangular strips, of height 1. The first strip has base 1, the second one has base 1+2, the third one has base 1+2+3, and so on. We get
12 + 22 + ... + n2 + (1 + [1+2] + [1+2+3] + ... + [1+2+3+...+n-1]) = n*(n2+n)/2. What about the quantity in parentheses? We can rewrite it as (sigma)_(i=1)^(n-1) (i2+i)/2. So we get
12 + 22 + ... + n2 + (sigma)_(i=1)^(n-1) (i2+i)/2 = n*(n2+n)/2. We can also write the first sum using sigma notation, and we get
(sigma)_(i=1)^n i2 + (sigma)_(i=1)^(n-1) (i2+i)/2 = n*(n2+n)/2. Now we play with the sigma notation:
(sigma)_(i=1)^n i2 + (sigma)_(i=1)^(n-1) i2/2 + (sigma)_(i=1)^(n-1) i/2 = n*(n2+n)/2. The first two sums are very close, but they differ by n2/2. So add n2/2 to both sides of the equation:
(sigma)_(i=1)^n i2 + (sigma)_(i=1)^(n) i2/2 + (sigma)_(i=1)^(n-1) i/2 = n*(n2+n)/2 + n2/2. Now we can combine the first two sums, and get
3/2(sigma)_(i=1)^n i2 + (sigma)_(i=1)^(n-1) i/2 = n*(n2+n)/2 + n2/2. The second sum can be added up, and we get (n-1)n/2. All told, then, we have:
3/2(sigma)_(i=1)^n i2 + (n-1)n/2 = n*(n2+n)/2 + n2/2. Now this can be solved for (sigma)_(i=1)^n i2.
3/25/98
- Finishing the sums of squares: We get the formula 12+22+...+n2 = n(n+1)(2n+1)/6. This can be checked in small cases (n=1, 2, and 3). One miracle is that the expression on the right-hand side of the equals sign is always an integer.
- More discussion about musical intervals, and the difference between the natural (harmonic) and Pythagorean tunings.
3/27/98
- Discussion of the distinction between Pythagorean and natural tuning. Introduction of diatonic scale.
- Comparisons between mathematics and linguistics.
3/30/98
- Discussion of the music homework due today. One motivating question was to define F#. One possibility is to set F# to be a fourth below B. The ratio of B to C is 243:128. Since F# is a fourth below B, we can subtract the interval of 4:3, giving a result of 243:128 * 3:4 = 728:512.
We can now compute the distance from F to F#, since we know that C:F is a ratio of 4:3. The distance between F and F# is therefore 728:512 * 3:4 = 2187:2048. We can call this a "sharp" Pythagorean semi-tone sp#. The other semi-tone to compute is the distance from F# to G. Since C:G is 3:2, that distance is 3:2 * 512:729 = 256:243, the Pythagorean semi-tone sp. Notice that these two semi-tones are not equal.- We can also think of the scale as made up of six whole-tones, C-D-E-F#-G#-A#-C. That ratio is (9:8)6 = 531441:262144. If we wish to see how much this differs from a true octave of 2:1, we can compute the difference, and get the ratio 531441:524288.
- This ratio can also be computed as the difference of sp# and sp: 2187:2048 * 243:256 = 531441:524288. We call this the Pythagorean comma.
- The summation homework was reviewed. The last question on the homework suggested the formula
sigma r3 = n2 (n+1)2/ 4.
We can derive that by starting with the identity
sigma r3 = sigma (n-r)3
and then using the formula (n-r)3 = n3 - 3n2r + 3nr2 - r3, and then bringing the last term over to the left hand side of the equation to get
2 sigma r3 = sigma n3 - 3 sigma n2r + 3 sigma nr2
We can now add up each of the three terms on the right-hand side of the equation, and get the formula as promised.- Among the features that language and numbers share are creativity. Both language and numbers give us a finite means of rendering the infinite. Language can produce an infinite number of sentences using a limited number of objects, and mathematics can similarly produce an infinite sequence of numbers. Language can also decode an infinite number of sentences.
- We can model this process (simplistically, to be sure) as a finite state process. We can start with a decision node, and then have ten paths (the digits from 0 to 9) as routes to the next node. At the second node, we can also have the option of stopping, or looping back to the first node to continue.
Language similarly can have this property. We can generate sentences in this manner. Consider the following sort of table:
Article Adjective Noun Verb Article Object The
Atimid
good-looking
brave
disgustingmailman
professorbit
ate
likesthe
adog
bear
bumblebee
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February notes
April notes
Last update: April 1, 1998