Today on her Mathematics Calendar 2018, Theoni Pappas writes:
What is the surface area of a sphere with radius equal to 2sqrt(pi)/pi?
Well, the surface area of a sphere is given in Lesson 10-9 of the U of Chicago text:
A = 4pi r^2
A = 4pi (2sqrt(pi)/pi)^2
A = 4pi (4pi/pi^2)
A = 4(4)
A = 16
So the surface area of the sphere is 16 square units -- and of course, today's date is the sixteenth. It's another one of those Pappas questions where the radius must be some strange-looking irrational number in order to make the surface area come out to a whole number.
By the way, notice that Pappas could have given the radius as 2/sqrt(pi). Then we can almost find the surface area in our heads -- let's see, r^2 is 4/pi, so pi r^2 is 4, so 4pi r^2 is 16.
But as we've seen she prefers to give the radius as 2sqrt(pi)/pi. This is the idea behind "rationalizing the denominator" -- no radical must be left in the denominator. Of course, calling it "rationalizing the denominator" in this case is awkward, since the denominator pi still isn't rational.
Let's return to music. Today is the second day of our look at the 14EDL scale. We wish to explore the properties of the scale to prepare for composing music in this scale.
The 14EDL scale:
Degree Ratio Note
14 1/1 red F#
13 14/13 ocher G
12 7/6 white A
11 14/11 amber B
10 7/5 green C
9 14/9 white D
8 7/4 white E
7 2/1 red F#
Yesterday, I mentioned how this scale -- just like our common major and minor scales -- contains exactly seven notes (that is, it's heptatonic). But the scale doesn't sound like major or minor -- instead, the closest classical mode is the Locrian mode. And so it might help to do some research on the Locrian mode, since 14EDL is very similar to this mode.
The Locrian mode is a rare mode indeed. The most common modes of course are major (Ionian) and minor (Aeolian). Occasionally we might hear songs played in the Mixolydian or Dorian modes.
But the Locrian mode is probably the rarest mode of all. The Locrian mode starts with a semitone and ends with a whole tone, while the major scale starts with a whole tone and ends with a semitone. In fact, other scales are modified in order to end with a semitone (leading tone) -- for example, the natural minor scale is replaced with the harmonic minor scale, which ends with a leading tone (for example, G in the key of A natural minor, G# in the key of A harmonic minor). So this Locrian mode is awkward because it starts and ends the opposite of what we expect scales to sound like.
By the way, the Dorian mode with a leading tone becomes (ascending) melodic minor. The term "ascending" is needed because the descending melodic minor scale becomes natural minor. Once again, this shows how important leading tones are to ascending scales, and thus EDL scales sound better descending than ascending.
So here are some websites revealed in a Google search for Locrian mode:
When we click on this link, we're immediately faced with a warning: "Hold on, this gets weird." This is because compared to the other modes, the Locrian mode sounds weird:
Notice that our fundamental Locrian scale is more like a G major scale starting on F#, but besides that it's the same scale. At the link above, Seymour Duncan continues:
Sounds weird, huh? This is why the Locrian mode is sometimes called a ‘theoretical’ mode: it isn’t used much in our Western harmony system. Yes, there are a few composers who have used it, but both rock and jazz prefers chords to resolve, more than anything. One use is the bass line in the Bjork song
At the following link, blogger Keith Freund goes on to claim that the Locrian mode "doesn't exist":
Most common examples of Locrian are riffs (short melodies which are repeated), not songs. The reasons why our ears tend to drift astray when hearing Locrian only apply to chords and harmony. Riffs are not like chords. They are more flexible. Because the notes are not occurring simultaneously (in the case of many rock riffs), our ear does not hear all of the same tendencies that intervals might suggest. All this being said, it’s hard not to hear YYZ as Locrian with the lead riff constantly reinforcing the root.
For example, when I was a young child, I used to hear the following Elephant song:
Sometimes this was sung with a tune. Unfortunately, this link doesn't provide the tune, and I can't find a single website providing me with a tune. I just remember that the notes F#, A, B, C repeating throughout the song. So this could be considered a song in Locrian mode -- but since the same notes repeat, it's more like a riff than a song.
20 FOR X=1 TO 15
30 READ A
40 SOUND 261-N*A,4
50 NEXT X
60 FOR I=1 TO 400: NEXT I
80 GOTO 20
90 DATA 14,14,12,12,14,14,12,12
100 DATA 14,14,12,11,10,12,14
This program repeats the same riff indefinitely until the Break (Esc) key is pressed. Again, I wish that I had the actual score instead of being forced to rely on my memory for the tune.
As I wrote in a previous post, the so-called fourth might be spelled like a perfect fourth (F#-B), but it sounds more like a diminished fourth or major third. This is because it's actually red F#-amber B, with red F# sounding higher than white F# and amber B sounding lower than white B. We know that ocher G is closer to white G# than to white G, but red F#-ocher G is nonetheless closer to 100 cents than to 200 cents. Thus our spelling problem occurs only with the fourth, not the second.
If we change B to Bb, the resulting scale is called "Super Locrian":
By the way, recall that Kite sometimes gives the ocher and amber notes as G# and B respectively -- alternatives to ocher G and amber Bb. The resulting scale using G# and B is sometimes referred to as "Locrian Sharp 2":
Note: this links to the basic Locrian Mode. Follow the links to both Locrian Sharp 2 and the Altered Scale (another name for Super Locrian).
But today I'm going to stick to the basic Locrian mode. The reason is that I plan on playing songs on the guitar, and both G and B are open strings, while G# and Bb are not open. So our basic assumption is that G and B will be easier to play. In general, when converting from EDL to guitar, I'll just ignore all of the Kite colors (including ocher and amber).
Here's one more Locrian link:
Notice that Bridget Mermikides, the musician behind this website, provides five different phrases in the Locrian mode. So let's try to convert one of her phrases to 14EDL. Recall that our goal is not merely to convert tunes to the new scale, but to compose tunes in the new scale. Nonetheless, tune conversion demonstrates the power of the new scale.
The phrases are all written in A Locrian (the relative Locrian of Bb major). Let's look at the 18 Mocha-playable 14EDL scales:
Possible 14EDL root notes in Mocha:
14 red F#
28 red F#
42 red B
56 red F#
70 greenish D
84 red B
98 deep red G#
112 red F#
126 red E
140 greenish D
154 amber-red C#
168 red B
182 ocher-red A
198 deep red G#
210 greenish G
224 red F#
238 17esque-red E#
252 red E
Notice that the root of the basic 14EDL scale is red, the root of any playable 14EDL scale must contain the color red (including "greenish," which is green-red). The note "ocher-red A" is a very sharp A, since both "ocher" and "red" raise the tone. In fact, we see that white A (Degree 192) is actually closer to deep red G# than to amber-red A -- which is near green Bb at Degree 180. (By the way, I'll explain the name "17esque-red E#" -- why E# and not F? -- in a subsequent post.)
We might use N=13 for ocher-red A, even though we might wish to change it to N=14 for deep red G#, a note that sounds closer to concert A.
Question 5 of the SBAC Practice Exam is on solution steps:
A student solved 3/(x - 4) = x/7 in six steps, as shown.
Step 1: 3 = x(x - 4)/7
Step 2: 21 = x(x - 4)
Step 3: 21 = x^2 - 4x
Step 4: 0 = x^2 - 4x - 21
Step 5: 0 = (x - 7)(x + 3)
Step 6: x = -3, x = 7
Which statement is an accurate interpretation of the student's work?
A) The student solved the equation correctly.
B) The student made an error in step 2.
C) The student made an error in step 5.
D) Only x = 7 is a solution to the original equation.
We notice that this is definitely a second semester Algebra I question, since we obtain a quadratic equation in Step 4. The problem is asking us to check the solution.
As far as I can tell, there are no errors in the solution. We can double-check the specific steps mentioned in the answer choices. Step 2 is correct -- instead of dividing by 7, we multiply both sides by 7 -- and Step 5 correctly factors the quadratic polynomial. We can also check whether -3 is an extraneous root or not:
3/(-3 - 4) = -3/7
3/-7 = -3/7
-3/7 = -3/7
Therefore both values of x are correct solutions, and so the answer is A).
Question 6 of the SBAC Practice Exam is on parallel lines in Geometry:
When a transversal intersects a pair of parallel lines it will create two pairs of alternate exterior angles.
Ricky claims the angles within each pair are congruent to each other, but not congruent to either angle in the other pair.
Draw a transversal through the point that supports Ricky's claim, or select NONE if there is not a situation to support the claim.
Draw a transversal through the point that refutes Ricky's claim, or select NONE if there is not a situation to refute the claim.
Well, we finally reach a Geometry question. So now I switch from "preparing for summer school" to "teaching my favorite subject."
But this question is tricky for Ricky. We notice that this question mentions alternate exterior angles, which don't appear in the Second Edition of the U of Chicago text. They do, however, appear in Lesson 5-4 of the modern Third Edition of the text.
Let's assume that Ricky knows that alternate exterior angles are congruent. But notice that Ricky is making two claims -- one is about which angles are congruent and the other about which angles are not congruent. In Part A, the transversal must satisfy both claims, while in Part B, the transversal must fail at least one of the claims. Since Ricky's first claim is always true, any counterexample for Part B must fail his second claim.
A good question to ask is, how are the angles in each pair related to each other? Well, it's easy to see that one angle in each pair forms a linear pair with an angle in the other pair. Thus the angles in the other pair are supplementary. If a counterexample to Ricky's second claim exists, the angles must be both congruent and supplementary. Such angles are called "right angles."
Therefore the transversal for Part B must be perpendicular to the parallel lines. It follows that the transversal for Part A must be oblique to the parallel lines.
SBAC Practice Exam Question 5
Common Core Standard:
Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
SBAC Practice Exam Question 6
Common Core Standard:
Prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment's endpoints.
Commentary: Solving quadratic equations by factoring appears in Lesson 12-8 of the U of Chicago Algebra I text, while proportions appear in Lesson 5-7 -- but notice that the steps above don't simply cross-multiply a proportion. Alternate interior angles appear in Lesson 5-6 of the U of Chicago Geometry text, but alternate exterior angles appear only in the modern Third Edition of the text. But students unfamiliar with alternate exterior angles might misread "exterior" as "interior" -- which would nonetheless lead them to the correct transversals for Parts A and B anyway.