Understanding Section 18 of the 1799 FTA

Take a shower!

For us to understand section 18, take a shower!  In fact, make a schedule of how you can take three showers a week.  You could choose to combine Monday, Wednesday, and Friday, or perhaps Tuesday, Wednesday, and Sunday.  How many different choices of schedules do you have?

As you look at your empty weekly calendar, you have seven choices for the first day you will check off to shower on.  M, T, W, Th, F, Sa, Su are all possibilities.  For each initial day that you choose, you have six possibilities for the second day you choose to take a shower: if you chose Monday, you now can pick between T, W, Th, F, Sa, and Su; and if you chose Thursday, you can pick between M, T, W, F, Sa, and Su, and so forth.  This can be pictured as a tree:

((Tree graphic.))

So we have 7•6 = 42 "leaves," or ways of having chosen one day, and then another.

Take a break now, and think about how many ways we can choose two days to shower.  Is it 42 ways?  Think...  The tree includes both MT and TM; ThSu and Su Th; WF and FW; and two of every possibility.  Wouldn't it be necessay to find each twice?  So instead of 42, we must have 42 / 2 = 21 ways of picking two days to shower.  Check it out if you like by figuring out all the combinations.

Now let's try looking at showering three days of the week.  Add another layer of "stems" to our earlier tree.  Each way of having chosen two days can have five possibilities for the third day: MT can choose W, Th, F, Sa, or Su; and WF can choose M, T, Th, Sa, or Su.

((Tree graphic.))

Now we have 7•6•5 = 210 leaves on our tree.  But are there really that many ways of choosing three days of the week?  How many leaves are duplicated?  A Monday, Wednesday, Friday showerer would be leaves
  MWF, MFW, WMF, WFM, FMW, and FWM.
  Six ways, but why six?  Think of another, smaller tree.  How many three-day leaves will have MWF in them?  The leaf could start with a M, W, or F (three choices), followed by W or F (if M), M or F (if W), M or W (if F) -- two more choices.  So we get three•two = six duplicate leaves.

  ((Tree graphic.))

This leaves us with 7•6•5 / 3•2 = 35 ways of choosing three shower days.  Write out the possibilities to see if we are right.

Think of holding up fingers.  How many ways can you hold up two fingers?  How many ways can you hold down two fingers?  How many ways can you hold up three fingers, and how many ways can you hold down three fingers?  Do you know how many ways there are to arrange a four-day-a-week shower schedule?

Pascal's Triangle

• Go through the triangle
• Why the addition works in terms of combinations.
• Why the combination formula applies everywhere.
• Why this is a binomial expansion. (Choosing fingers.)

The Complex Domain

Look at sines and cosines from the standpoint of the complex domain. A position on the circumference of the circle could be given by the angle of rotation required to reach it from a given horizon. It could also be described in terms of the cosine and sine components of the triangle it forms. This is where sines and cosines come from: the two ways of designating a location: by angle, or by two independent directions. Doubling the distance along the circumference of the circle is easy by doubling the angle, but is difficult to represent in terms of the two-direction method of looking at location. (See Riemann for Anti-Dummies number ____?____).
Gauss's complex domain gives a fuller way of looking at the cosine-sine representation. Think of a position as cos φ + i sin φ, using i = to signify the vertical direction.

[Graphics:Images/index_gr_3.gif]

Now, looking at any complex number (a+bi) and multiplying it by itself,

will double the angle and square the length. See Bruce Director's "Bringing the Invisible to the Surface" in the Summer/Fall 2002 Fidelio for the reason for this from the equi-angular spiral. There is no reason that we cannot use (cos φ + i sin φ) similarly. This means that

will have the effect of doubling the angle φ and squaring the length of the distance from the origin.  But since we are on a circle, squaring the length is 1•1 = 1, leaving us with only the doubling of angle.
  Let's do it!
  (cos φ + i sin φ) (cos φ + i sin φ)

+ i cosφ sinφ + i cosφ sinφ +

+ 2i cosφ sin φ -

Grouping this into separate pieces by i gives:

+ i (2 cosφ sinφ)
So the vertical (sine) part of the position at angle 2φ must be 2 cosφ sinφ, which is exactly what we get from looking at this question geometrically! (See Section 13 FTA pedagogy). The horizontal part is

which you should compare to the geometrical result of looking for cos 2φ. See Jonathan Tennenbaum's "From Cardan's Paradox to the Complex Domain" for this.

Expanding the Complex Domain

• on Pascal's triangle.

Applying to Section 18

• Bring in Gauss's terminology in Section 18.

Jason Ross jross@wlym.com

Converted by Mathematica July 27, 2004