Exercise 5.5

     Penrose is giving us the following definition for e^z, where z is a complex number:

And says that we must use (1) to show the following:

e^a+b=e^ae^b

(2)

      In order to do this, we have to apply (1) to both sides of (2) and show that they give equivalent quantities.  Let's first consider e^a+b, which according to (1) expands as follows:

    This can be expanded as follows for the first few terms:

Which suggests a general pattern for the expansion up to the nth term, and continuing onwards (using Penrose's hint for the coefficient of the nth term of the expansion of (a+b)ⁿ:

Which in turn can be written more concisely as:

Now we look at the right hand side of (2).  Using (1):

Which expands as:

Comparing (6) and (7) we see that the overall equality boils down to whether:

The easiest way I've found to show (8) is to re-examine the products on the left hand side in a different way:

This appears to be exactly the operation described by the right hand side of (8), which we derived from the expression for e^(a+b).  In this way we have demonstrated (2).  

Here Goes...

This is as far as I got in Road when I was initially working through it in March and April of last year.  I was stumped and remain stumped by the second part of the challenge Penrose sets in Exercise 5.4, which is to show without detailed calculation or trigonometry that multiplication of complex number z by another complex number w corresponds to two geometric transformations of the representation of z on the complex plane: (a) a uniform expansion/contraction and (b) a non-reflective rotation.  Penrose hints that this is due to two properties of complex numbers z and w:

w(z₁+z₂) = wz₁+wz₂

(1)

w(iz) = i(wz)

(2)

    Penrose also tells us that (1) implies the preservation of linearity in the multiplication of complex numbers; and that (2) implies the preservation of right angles in the multiplication of complex numbers.  Which leaves me confused as to whether we are allowed to use the fact that multiplication of a complex number by i corresponds to a rotation of z through a right angle in the complex plane, or whether we still have to prove that.  My guess is that this is simply "guilty knowledge" that will have to be shown logically, but that ends up sucking for me because while (a) is fairly easy to show in a way that I think is consistent, (b) has proven very difficult to show without making use of trigonometry and/or "detailed calculation," whatever the hell "detailed calculation" actually means. Anyhoo, I'll talk more about this when I get around to talking about possible approaches to showing (b). 

     In general, Penrose seems to want for us to be able to think about what's happening with these complex number in terms of geometry, not algebra, and to use "mostly" geometric arguments to demonstrate these properties of complex number multiplication.  Since I've always been a more rote-calculation person it's a good exercise for my brain to try the harder, or at least different, geometric approach.

Setup

      Just to make sure I know what I'm talking about throughout, here are some basic definitions.  For the rest of the time we'll be talking about two arbitrary complex numbers z and w defined as follows:

z = a+bi

w = c+di

  

      Where a, b, c, and d are all real numbers.  Also we will be talking about representing z and w as vectors on the "complex plane" which is defined by a horizontal "real" axis and a vertical "imaginary" axis, as shown in the following diagram:

    We are also going to need to assume some givens about how these vectors can be "added" geometrically.  For example, z can be represented in the complex plane as the "resultant" vector z of vectors a and b, whose magnitudes correspond to the real numbers a and b respectively.  Finding the resultant vector from component vectors on the complex plane works according to the same parallelogram rule as works for vectors on the real plane:

Uniform Expansion/Contraction

     Show that multiplying z by w involves a uniform expansion/contraction.  Not sure what Penrose means by "uniform, " but I assume that he means an expansion/contraction by a constant factor.  In any case he seems to be asking about how multiplication affects the geometric length of the vector z in the complex plane  Let's represent this length as |z|.  Penrose is asking us to look at |wz|, i.e. the magnitude of the length of the geometric representation of the complex product wz, and see whether it differs from |z| by a constant factor.

      We begin by looking at a bit of algebra (which hopefully doesn't count as the sort of "detailed calculation" which Penrose is disallowing):

|wz| = |z(c+di)| = |cz| + z(di)| = c|z| + id|z|

(3)

     This assumes some things about two complex numbers z and w that seem to me to be true:

|zw| = |z||w|

|z+w| = |z| + |w|

|i| = i

     The first of these facts is the most important for us.  In particular, we need to show in a simpler case that |bz| = b|z| where b is a real number.  

     We can show this quite easily by interpreting bz as z added to itself b times, which can be represented in the complex plane as a simple geometric addition of the vector z to itself b times (b is a scalar value):

     Inspecting this we can see that |bz| = b|z|, which makes me feel better about saying |bz|=b|z|.

   

     Now according to (3) |zw| can be rewritten as c|z| + id|z|.  Since |z| is a scalar value, we can interpret this sum as a third complex number whose geometric interpretation is::

     Which means that the magnitude |z(c+di)| depends on the magnitude |z| multiplied by constant factors c and d.  Given (3), this means that |wz| also depends on these constant factors.  Which means that the multiplication of z by w results in a third complex number, whose magnitude is related to the magnitude of z by a constant factor.  Which is (I think) what Penrose wanted us to prove.

Rotation

      Now we have to show that multiplying z by w also involves a "non-reflective" rotation.  Since (as will be clear shortly) multiplying z by the factor i² = -1 is analogous to a geometric reflection of z through the origin, I guess that Penrose did not mean for us to consider "degenerate" cases such as when w equals i, -1, 1, or 0.  Even so, we will need to start out by considering what happens, geometrically in the complex plane, when you multiply z by the "special" value of i.  Algebraic multiplication by i of z = a+bi results in a new number z' = -b+ai.  A relationship between z and z' becomes evident upon examinations of their analagous vectors z and z' in the complex plane:

      We are most interested in the angle between z and z'.  If you imagine the two vectors defining triangles with one side along the real and imaginary axes respectively you will see that the triangles are similar.  Therefore the angle γ between z and the real axis is equal to the angle between z' and the imaginary axis.  Define β as the angle between z and the imaginary axis.  We know that γ + β equals 90 degrees.  Then note that angle between z and z' is equal to β plus the angle between z' and the imaginary axis.  But this is equal to β+γ as well, therefore the angle between z and z' is equal to 90 degrees.  And so multiplication of the complex number z by i is analogous to a counterclockwise 90-degree rotation of  z in the complex plane.  

      Now consider multiplication of z  by another arbitrary complex number w.  We can represent w by the corresponding vector w on the complex plane.  Suppose that w has components w_R and w_I:

     Examining the above, and using (1), we see that:

zw = z(w_R+w_I) = zw_R+zw_I

     Or algebraically:

zw = zw_R+zw_I

(4)

     However again by examination we know that w_I is nothing more than i multiplied by some real constant C, while w_R is nothing more than a real constant D.  Therefore (4) can be rewritten as:

zw = z(iC)+zD

     Using (2) we slightly rewrite this as:

zw = iCz+Dz

zw = iCz+Dz

(5)

     We have shown that multiplication of a vector such as z by a constant such as C or D results in  a dilation or contraction.  We have also shown that a multiplication of a vector in the complex plane by i results in a 90-degree counterclockwise rotation.  Therefore, the vector product zw is equivalent to combining a counterclockwise 90-degree rotation and dilation/contraction of z with another, not necessarily identical contraction/dilation of z:

     One component of the resultant vector equal to zw will be a stretched out/compressed copy of z rotated ninety degrees, and the other will be a non-rotated copy of z stretched out/compressed by a different amount.  Assuming that w is not one of various "degenerate" cases, the product of the two will therefore be a contraction/expansion combined with a rotation which is what we were supposed to show.

Exercise 4.6.  Given a general complex polynomial equation P = 0, where polynomial P is equal to

and z and a₀...a_n are all complex expressions, show that (1) must be factorable into the following form:

Where b₁...b_n are also complex numbers.

     I guess one way to start is by showing that if b_x is a solution to (1), then any arbitrary (z-b_x) divides evenly into (1).  First substitute b_x into (1) and solve for a₀:

     Now divide P by z-b_x to arrive at the following:

     Substitute a₀ by the result in (3) to obtain:

     Which can be rearranged into:

     Or:

      Which shows that any z-b_x will divide evenly into all terms of the quotient in (6).  Now, using (3), 

      So z-b_x also divides into P.  Since this is true for any b_x, we can say that

      Q is the remainder which, when multiplied with the divisor D in (8), will provide a fully factored expression for P, i.e. QD = P.  To find Q, note that a sequence of products such as found in D can be expanded as follows:

      Where all A_x incorporate products and sums of the solutions b_x.  It's worth noting that all A_x as well as the final standalone constant (the product of all the b_x) correspond to the various A_x found in (1).  However (1) also has a factor a_n that is a coefficient for z_n.  a_n is therefore the missing factor Q needed to ensure that P perfectly factors Q. (DUBIOUS!)  Upon multiplying (9) by a_n, each coeffficient and the final standalone constant will equal each coefficient andd standalone constant in (1).  Therefore the final factorization for (1) is:

Shit's Messed Up

      Everything about this solution seemed to go smoothly until, unfortunately, the key moment when I had to figure out wtf happened to a_n.  I am not very happy that a_n did not arise naturally but instead had to be artificially multiplied in to (9) based on examination of (1): it seems too much like I am using my guilty knowledge of what the answer is supposed to be and then jerry-rigging the solution accordingly.  On the other hand, since z_n has a coefficient of 1 in (9), I suppose that you could always factor out the term a_n with the result that that coefficient becomes 1/a_n.  It still seems shitty and I'd be grateful if anyone knows how to go through these steps in a way that a_n pops up naturally as a coefficient for z_n.  Another potentially much more serious problem with the whole thing is that it doesn't use any of the properties of complex numbers.  Which is troubling.

     I apologize for the tininess of most of the inline math, it's ridiculously hard to read. Blogspot's image upload system is pretty crappy (as is their too-small text window, grah!).  But the only alternative I have is to basically buy  another box and start up my own, fully customizable blogging engine which would just take too much work.  Oh, if there was just a way to upload these things as inline-viewable PDFs!

     Exercise 2.8.  Show that given a sphere with radius R, the surface area of a triangle T with vertices A, B, C and interior angles α, β, and γ inscribed on the surface of the sphere is R²(α+β+γ-π).

     We can start about by examining the surface area swept out by a single pair of great circles forming one of the interior angles of T.  For instance, the surface area swept out by the pair of great circles intersecting to form angle β:

     When β equals π, the area swept out covers the entire surface area of the sphere, or 4πR².  Therefore we can infer (dubious, see below discussion) the following relationship between the great circles intersecting to form angle β and the surface area A_β swept out by these great circles:

A_β = 4βR²

(1)

    

     Armed with this knowledge, let's look at all of the surface areas swept out by the great circles intersecting to form all of T's interior angles:

     Using equation (1), the total surface area swept out by the great circles intersecting to form the three interior angles for T is:

A_α,β,γ = 4R²(α+β+γ)

     Then note that the area of T, A_T, is swept out three times by each pair of great circles, and is mirrored on opposite poles of the sphere, so that triangle ABC is congruent to triangle A'B'C' on the sphere.  Therefore A_α,β,γ covers the entire surface area of the sphere plus extra area equal to a multiple of A_T:

4R²(α+β+γ)=4πR²+4A_T

       (2)

      

     Four times A_T because A_T is swept out two "extra" times on each pole of the sphere.  Using (2) it is easy to solve for the correct value of A_T.  (2) also indicates that the sum of the angles of a triangle inscribed on the surface of a sphere must always be greater than π.

Dubiousness...

      I'm not particularly happy with the way I just kinda closed one eye and went ahead with "inferring" expression (1) for the surface area swept out by the great circle intersections.  It seems right but I dunno whether there's something more I'm supposed to be doing so that I could nail that down with more rigor.  Or maybe that's par for the course given that Road is not after all a full-blown math course.  

     Anyway I got to the right answer without too much B.S. so I guess I should count myself lucky.  Working to finish up a beast right now that I may have leave unfinished and ask for help from whoever the hell might eventually one day actually read this stupid blog.  (hi, unlucky historian of the Internet reading this a thousand years in the future!)

Understanding Penrose's Discussion of the Dedekind Cut

What Does He Mean?

Penrose's discussion of what he calls the "Dedekind Cut" in Chapter 2 of Road is tantalizing but unfortunately brief to the point of being almost impossible for me to understand.   I was particularly confused by his reference to how irrational numbers such as the square root of 2 are defined by Dedekind cuts when "those to the left have no actual largest number and those to the right, no actual smallest one" (3.3).  Huh?

But with some reference to various pages on Wikipedia (which I fear I'll be making much use of as I slog through this tome) I believe I have pieced together what Penrose is talking about.  And it is quite interesting, and worth working through here.  

Basically, what Penrose seems to be talking about is defining irrational numbers as the "gaps" between rational numbers on the number line.  Let's define Q as the set of all rational (not real!) numbers.   We then "cut" Q into subsets A and B.  The trick is to define the point γ which divides A from B in such a way that γ belongs to neither subset.  Since A and B are subsets of Q and hence also comprised of rational numbers, but γ is not a member of A or B, γ cannot be a rational number.   

Suppose for example that you wanted to use such a cut to define the square root of 2.  You would cut Q as follows:

The notation is stolen from Wikipedia: the vee-shaped thingy means logical OR, and the wedge-shaped thingy means logical AND.  So A is the set of all rationals a in Q whose square is less than 2 OR all rationals less than or equal to 0.  Similarly, B is the set of all rationals b in Q whose square is greater than or equal to 2 AND all rationals greater than 0.  

Well what does this all add up to.  It is pretty obvious that all a must be less than γ, since we are either talking about negative numbers or positive numbers (including 0) whose square is less than 2.  It is not quite so obvious that all b must also be greater than γ. This is because the square root of 2 is not a rational number and hence cannot be a member of B.  All members of B must therefore have squares greater than 2.  

Looking at this we start to get some idea of what Penrose means by his mysterious references to "no actual largest/smallest numbers" in the two sides of the cut.  However I think it really helps to get a bit more precise in this case so that we can understand what is and is not being discussed.  Apparently Penrose is referring to the concepts of a supremum and infimum of a set.  You can define the supremum S (my symbol) of a set T of real numbers to be the smallest real number greater than or equal to every member of T; similarly, you can define the infimum I (again my symbol) of T to be the largest real number less than or equal to ever member of T. Examples:

We can now state much more concisely (and precisely, and clearly) that while γ is not a member of A or B, γ = S{A} and γ = I{B}.  The supremum and infimum for A and B respectively fall outside A and B: this is what Penrose is saying, I think (read: I hope).  

Additional Thoughts

Once I managed to piece all of this together in a reasonably coherent way I skimmed a page in Wikipedia concerning the relationship of suprema and infima to the completeness or lack thereof of sets.  I didn't really read closely but it was indeed interesting that it is possible to define subsets of rational numbers in a way that appears to violate the completeness requirements (since there are suprema and infima that fall outside these subsets, as we have seen).  So the irrationals arise almost as a necessity (?) in order to complete the set of rationals and form the larger (complete?) set of all real numbers.  Assuming that it even makes sense to talk about irrationals as a "necessary" complement to the rationals, one wonders whether there are other ways to cut the reals that "necessitates" complex numbers.  I dunno.  *shrug*

Halp!

Putting together entries with equations in them such as the last post is proving to be a MAJOR pain. Been spending some time online looking for a fix that would ideally provide fairly quick translation from TeX source to text and image files friendly for this blog format...

Exercise 2.4

Exercise 2.4.  Show that the projective representation of a hyperbolic straight line may be obtained from the conformal representation of a hyperbolic straight line by expansion from the center by a factor of M, and that:

Solution:

Above is a conformal pro jection on a minor circle c, with radius r , that we project onto a projective representation on the great circle C, which has radius R and origin O. Let r_c be the Euclidean distance from O to any point on the conformal repreesentation on C , and let b be the Euclidean distance from r_c to its corresponding point on the pro jective representation, so that the additive Euclidean distance from O to any point on the projective representation is b + r_c. Let M be the expansion factor from the Euclidean conformal distance to the Euclidean projective distance, so that:

                   (1)

Now, it is evident from the similarity of the two triangles that:

The problem is how to restate d in terms of R, r_c,  and b.  Given that the segment extending from O and joining d is equal to R, by the Pythagorean Theorem we know that 

So we rewrite the previous expression as 

And solve for b:

With this we can rewrite (1) and solve for M: