A comment on my post about complex numbers has made me want to go back and discuss exponentials in more detail even though it feels off-topic to this series as a whole. There are two problems this discussion will run into. The first is that I've been trying to build up the math I've been using systematically. Even though I leave out a lot of the details, I've avoided statements along the lines of "this is true, but I can't show that right now. You'll just have to trust me." I think I will get to all of the math involved in this subject eventually, but it may not be until June or so. Even then, there's one step I'm not sure I will be able to fully justify.
The second issue is more practical. I have been writing all of my math in plain HTML. I think this has been basically working in previous posts, although I haven't always been able to write everything as clearly as I would have liked. In this post, there will be some expressions where I can either write them in a way that's comprehensible or a way that's legible, but I can't do both. I'm going to aim for legible and hope that anyone who is following along will be able to decode the meaning.
Going back to raising negative numbers to rational exponents, there's a problem I glossed over. The equation (−1)1/2=x is equivalent to −1=x2, and there are two (complex) solutions to that equation. In general, for any real number a (other than 0) and any natural number n, there are n complex solutions to the equation a1/n=x. In some cases there is an obvious preferred solution. If a is a positive real number, then there is always a positive real number x which satisfies the equation. If a is a negative real number and n=2, the two solutions are of the form bi and −bi, where b is a positive real number, so it makes sense to prefer the positive value.
However, the four solutions to (−1)1/4 do not have an obvious preferred solution. The solutions are 2−1/2+i2−1/2, −2−1/2+i2−1/2, 2−1/2−i2−1/2, and −2−1/2−i2−1/2. You could choose the solution with positive components, but that poses a problem, because (−1)3/4 also has four solutions, and if we also choose the solution to that with positive components, then [(−1)1/4]3≠(−1)3/4.
One resolution to this is to write the complex numbers in polar form. Given a complex number a+bi, the magnitude (or absolute value) r of the number is √a²+b². If we think of a+bi as a right triangle with legs a and b and hypotenuse r, then there is an angle θ such that cos θ=a/r and sin θ=b/r. Then a+bi=r(a/r)+ir(b/r)=r cos θ + ir sin θ=r(cos θ + i sin θ).
There is not a unique value of θ in this representation. If a+bi=r(cos θ + i sin θ), if θ is a solution then any value of the form θ+2πk is a solution for any integer k. We can define the preferred value of θ such that 0≤θ<2π. (If it's not obvious, we are measuring the angle θ in radians. We could measure it in degrees, but things would get really messy soon.)
Here's the step I can't fully justify. I am going to assert that cos θ + i sin θ = eiθ. I can offer a partial justification for this, but this justification has a hole, at least in the scope of my math experience. The values for ex, cos x, and sin x can be written as Taylor series, or infinite sums of polynomial terms. I expect to cover Taylor series in depth eventually, but for now I'm just going to assert that this works. However, in my experience, Taylor series are only defined for real values of x.
If we write eiθ as a Taylor series, I'm not convinced that's valid since iθ is not a real number. Regardless, the series has a bunch of real terms and a bunch of imaginary terms. We can separate the series into two series, one consisting of real terms and one consisting of imaginary terms. The real series is the Taylor series of cos θ. The imaginary series is i times the Taylor series of sin θ. So eiθ=cos θ + i sin θ, as long as we accept the Taylor series expansion of eiθ as valid.
Then we can write any complex number a+bi in the form reiθ, where r is a non-negative real number and θ is between 0 and 2π. θ can equal 0, but cannot equal 2π. As stated earlier, reiθ=rei(θ+2πk) for any integer k. With the laws of exponents for real numbers and the identity a=eln a, we can now find a unique value for any complex number raised to any complex power, except for 00.
Looking at complex numbers raised to rational powers, this gives us a preferred unique solution for (a+bi)1/n, but it also allows us to find the other solutions. (a+bi)1/n=(reiθ)1/n=r1/neiθ/n. However, reiθ is equivalent to rei(θ+2πk) for any integer k, and (rei(θ+2πk))1/n=r1/nei(θ/n+2πk/n). This has the same magnitude as the original solution, but only has the same angle if k is a multiple of n, so by taking all k from 0 to n-1, we can find all of the solutions.
In the general case of any real number x for (rei(θ+2πk))x, if x is irrational, then the set of angles produced by all integers k is infinite. However, the magnitude is always the same. We can still be assured of a unique value by requiring θ to be between 0 and 2π and k to be 0. Weirder, if the exponent is a complex number, then there is also an infinite set of magnitudes. I find this disconcerting, even though we can again specify a unique magnitude (and angle) by constraining the range of θ and requiring k to be 0.
It's time for the Friday Random Ten.
- Rush - Cinderella Man
- Emerson, Lake & Palmer - Karn Evil 9 (First Impression, Part 2)
- Guarneri Quartet - Ludwig van Beethoven: String Quartet No. 2, Third Movement
- The Be Good Tanyas - Scattered Leaves
- Chor Des Norddeutschen Rundfunks - Johannes Brahms: German Folksongs WoO 35, Mit Lust tät ich ausreiten
- NDR Chorus Hamburg - Johannes Brahms: Lieder Und Romanzen, Op. 93A, Beherzigung
- Brian May - 39/Let Your Heart Rule Your Head
- William Orbit - The Story of Light
- Mozart Akademie Amsterdam - Wolfgang Amadeus Mozart: Symphony #23, First Movement
- Gravity Kills - Goodbye (Demo)
Euclidean spaces are multidimensional spaces where each point in the space can be represented by a unique tuple of real numbers that can be called a vector. Addition, the inner product, and the norm of vectors are all defined. Now we can look at inequalities involving vectors.
When we looked at complex numbers, we defined the triangle inequality, |z+w|≤|z|+|w| for any two complex numbers z and w. This is also true for vectors. If x and y are vectors in a k-space and ||x|| is the norm of the vector, then ||x+y||≤||x||+||y||. It follows from this that ||x−z||≤||x−y||+||y−z||.
A second related inequality is the Cauchy-Schwarz inequality, which states that |x⋅y|≤||x|| ||y||. Essentially, the triangle inequality applies to the addition of vectors and the Cauchy-Schwarz inequality is the same result for the dot product.
A big part of mathematics is taking a specific result and creating a more general rule. One example of that that we've used several times is in extending number systems, starting with the natural numbers, extended up to the real numbers and then complex numbers and euclidean spaces. A second form of that occurred when we looked at the addition and multiplication properties for the rational numbers and declared that these properties define a field, and therefore the rational numbers are a field.
I find wrapping my head around defining fields from the rational numbers significantly harder than understanding defining euclidean spaces from the real numbers. This says to me that the transformation behind the definition of a field is more significant than the transformation behind defining euclidean spaces. And that's a good thing, because it's time for another generalization like the one from rational numbers to fields. Unfortunately, it will have to wait until next time.
Last time, we looked a little bit at the ordered pair representation for complex numbers. An obvious followup is to ask whether we can represent numbers as ordered triples or ordered tuples in general.
The answer is that of course we can. A euclidean space is the set of all ordered tuples of real numbers where the tuples all have the same number of members. Okay, that probably made no sense. Choose some natural number k. Choose k real numbers, x1,x2,…,xk. Then x=(x1,x2,…,xk) is a vector in euclidean k-space. We can write that x is a member of ℜk.
It should be clear that the real numbers are equivalent to ℜ1 and the complex numbers are equivalent to ℜ2. Every point in the interior of a cube can be represented as a number in ℜ3, so that's not too hard to visualize. Higher dimensions are hard to visualize, but fortunately they are not too hard to work with mathematically.
Addition and multiplication by real numbers are defined in euclidean spaces. Addition is defined to be consistent with addition of real numbers or complex numbers, just by adding the components in each direction. Multiplication by a real number multiplies each component by that number, so it is again consistent with multiplication of real numbers and complex numbers. Visually, multiplication by a real number changes a vector's length, but it does not change the direction the vector points in.
Multiplication of two vectors by each other to produce another vector of the same dimension is not defined. (Therefore, euclidean spaces are not fields.) Depending on what you are doing with them, it may be better to consider real numbers and complex numbers as numbers rather than as points in euclidean space, because the numbers can be multiplied by the points cannot be.
However, multiplying two vectors to produce a real number is defined. This is called the dot product or inner product, and is defined by x⋅y=∑xiyi. The norm of a vector is defined by ||x||=(x⋅x)1/2. The important thing here is that the norm of a vector is equivalent to the absolute value of a complex or real number.
Now that my rant is out of the way, we can go back to the question I left hanging a couple of posts ago. What is the value of a negative number raised to a rational exponent? We can reduce this question down to the simple question, what is the square root of −1?
It's obviously not a real number, because the square of any real number is a positive number or 0. But that doesn't mean it doesn't exist. We can extend the real numbers to get the complex numbers. (Note that we are throwing out ∞ as a number and going back to the reals. Being a field is more important than having a greatest number.) By definition, i is a number such that i2=−1. Then the complex numbers are defined as numbers in the form a+bi, where a and b are both real numbers.
There is an alternative definition, which is equivalent. We can define z=(a,b), where (a,b) is an ordered pair of real numbers. The fact that it is an ordered pair means that (a,b)≠(b,a) if a≠b. We can then define addition and multiplication of the ordered pairs so that we get the same results as the previous definition. One advantage of this approach is that it encourages us to see complex numbers as points in a 2-dimensional plane.
The lecture notes begin with the definition as a+bi and then show the equivalence of the ordered pair (a,b). The book does the reverse. This is one subtle example of how the book and the lecture notes choose to emphasize different things, even when they cover the same material.
With either definition, the complex numbers are a field, and so retain all of the properties of the real numbers for addition and multiplication. In addition, for any complex number z and any real number x, except for z=x=0, zx is a complex number. However, the complex numbers are not ordered. One of the properties of ordered numbers is that x2>0 for any x≠0, and the whole point of the complex numbers is to define a number for which that is not true.
However, we can define the absolute value of a complex number. If z is a complex number, it can be written either as a+bi or as (a,b) where a and b are real numbers. Either way, the absolute value of z, |z|, is equal to √a²+b². Note that if we think of z as a point in a plane, then we can also think of z as a vector, in which case the absolute value of z is the length of the vector.
We can't speak of a complex number being greater or less than another complex number, but we can compare their absolute values. Note also that the real numbers are a subset of the complex numbers (any complex number where b=0 is a real number), and so the same definition for absolute value applies. In practice, for a real number a, if a>0 then |a|=a, and if a<0, then |a|=−a.
There are two useful inequalities that we can use with the absolute values of complex numbers. These inequalities hold for real numbers, but are almost too obvious to state, even though they are often useful. The first is the triangle inequality. Given two complex numbers z and w, |z+w|≤|z|+|w|. The name comes from a visual representation of z, w, and z+w as vectors in the plane. Visually, this is saying that with a triangle with sides of length z, w, and z+w, z+w must be less than or equal to the lengths of the other two sides. (They are only equal if all three vectors point in the same direction.) Of course, we can also rigorously prove the inequality by going back to the definitions for absolute value and for complex numbers.
The second inequality is essentially a restatement of the triangle inequality, but it is still useful as an independent statement. The inequality is |x-y|≤|x-z|+|z-y| for any three complex numbers x, y, and z. Again, since all real numbers are also complex numbers, these inequalities are true for the real numbers as well, and are more useful than they might seem based on how obvious they are.
This time I'm just going to complain about the term, "the real numbers". There are two problems with this term. First, the real numbers aren't "real" in any meaningful sense. Second, the fact that some numbers are called "real" implies that all other numbers are somehow "unreal", when they are not any more unreal than the real numbers themselves.
The real numbers aren't real. We have lots of experience with the real world. That's where we live, after all. We spend hours there every day. We use numbers all the time, whether we're spending money, or measuring something, or checking the time, or whatever else we might be doing. The numbers we use are never real numbers.
We always round off numbers. Even if we're being precise about the time, we only give it to the nearest second, not the nearest hundredth. Even careful physics experiments which measure the speed of light or the mass of the Higgs boson or whatever have to always state the uncertainty in the results. The real numbers imply that numbers are never rounded off, and beyond the fact that we don't do that, we can't do that.
You could make the claim that the natural numbers exist in some way in the real world. After all, we count things all the time, and the natural numbers are the counting numbers. Even then, there are limits. We may care about an exact count for small numbers but once the numbers get big enough, we round them off. A chemist can tell you that there are 6.02×1023 carbon atoms in 12 g of graphite, but probably can't tell you all of the next 20 digits of that number, and even if they could, they wouldn't bother most of the time.
And that's before we get into the fact that there are more natural numbers than there are things in the universe to count. I'd like to post sometime about really big numbers, and the fact that it's easy to come up with numbers that are so big that they are both incomprehensible and useless. It's a fun topic, but in the real world we don't talk about incomprehensibly big numbers on that scale.
The point, simply, is that thinking of even the natural numbers as real world numbers is misleading, and our experience with real world numbers can lead us to sloppy or incorrect thinking about numbers in a purely mathematical context.
At the same time, the fact that some numbers are "real" implies that other numbers are "unreal". Since my next post will discuss numbers which are not real numbers, I feel a need to preemptively stick up for them. Not real should not imply not useful. Although I won't get into the applications of complex numbers, the applications do exist. Don't think that because the complex numbers aren't real numbers, you can just ignore them.
In short, the real numbers is a stupid term for stupid people and we should change it. So there.