Powers!

Quite often, what mathematicians do is generalize from simple ideas to more and more complicated ones, so that what starts out clear and obvious little by little enters the realm of WTF? It’s important to remember that the goal isn’t to mess with people; it’s to follow logic wherever it leads without letting “seems crazy” get in the way. If the result is to make people shake their heads and think that understanding math requires being a super-genius, that’s all well and good, but in strict point of fact it’s not true. (Though it would be cruel to tell that to our throngs of beautiful and sexually ravenous groupies; we magnanimously allow them their comforting illusions.)

Exponents are a good example. When they’re positive whole numbers, their meaning is completely straightforward: xⁿ means n x’s multiplied together, so 4³ is 4*4*4 or 64. 4³ times 4² would be (4*4*4) * (4*4) or 4⁵; generalizing this, we see that we can multiply by adding exponents. If we now consider the value of x⁰, we see that x⁰ * xⁿ equals xⁿ, since 0 + n = n. Thus, though this has confused beginning algebra student for centuries, x⁰ must be given the value 1.

Negative numbers aren’t much harder. xⁿ * x^-n = x⁰ = 1, so x^-n = 1/xⁿ (e.g. 4^-2=1/16). Fractions are also straightforward. x^1/2 * x^1/2 = x¹ = x, so
x^1/2 = √ x  e.g. 4^1/2 = 2 and 4^5/2 = 32. Irrational exponents are a bit tricker, but still clear. The sequence

x³, x^3.1, x^3.14, x^3.141, x^3.1415, x^3.14159 …

is increasing and bounded above by x^3.2, so it has a limit, and we call the value of that limit x^π. What else could you do? (Particularly if you want f(x) = a^x to be continuous.)

Now that we’ve defined xⁿ for all real n, we’re ready to think about complex n. And there are a few things we’re ready to say:

x²ⁱ = xⁱ * xⁱ
x^-i = 1/xⁱ
x¹⁺ⁱ = x * xⁱ

But none of them help us to define xⁱ in the first place. We need to do some sideways thinking to find the right approach. Let’s see …

Our favorite number to raise to powers is e, because f(x) = e^x has the nice property of being its own derivative. (In general, the derivative of a^x is a^xln(a), so it’s best to choose the a where ln(a) is 1. Likewise, we let the arguments to sin(x) and cos(x) be in radians so that sin'(x) is simply cos(x), not (π/180)cos(x), as it would be if x were in degrees.) The first derivative of g(x) = e^-x is, of course, -e^-x or -g(x), by a simple application of the chain rule. Again by the chain rule, the derivatives of h(x) = e^ix would be:

First: (i)h(x)
Second: -h(x)
Third: (-i)h(x)
Fourth: h(x)

The second derivative being the negative of the original function should ring a bell. The derivatives of sin(x) are:

First: cos(x)
Second: -sin(x)
Third: -cos(x)
Fourth: sin(x)

So perhaps sin(x) is somehow related to e^ix. Let’s look at the power series expansions.

Because e^x is its own derivative, and e⁰=1, its power series is very simple:

e^x = 1 + x + x²/2! + x³/3! + x⁴/4! + …

sin(x) and cos(x) are almost as simple. Recall that their values and derivatives at 0 have the pattern 0, 1, 0, -1, 0, 1, … (cos(x) starting at 1 and sin(x) at 0), so

cos(x) = 1 – x²/2! + x⁴/4! – x⁶/6! + …
sin(x) = x – x³/3! + x⁵/5! – x⁷/7! + …

Ordinarily we think of x in these expansions as being real. But if we use them to define these functions for complex x too, there are some nice consequences:

• They converge for all complex x, because factorials grow so much faster than exponentials do, so we’re defining them over the whole complex plane.
• It’s easy to verify that e(x) is still its own derivative, but now over the whole complex plane.
• It’s similarly easy to verify that sin'(x) = cos(x) over the whole plane.

Given that, let’s look at what happens if we apply the expansion of e^x to ix:

e^ix = 1 + ix – x²/2! – ix³/3! + x⁴/4! + …

or, rearranging a bit (which we can do because the series converges absolutely)

e^ix = (1 – x²/2! + x⁴/4! – …) + i(x – x³/3! + x⁵/5! – …)

or

e^ix = cos(x) + (i)sin(x)

And this is true for all x, by simple algebra. If we restrict x to real numbers, so that we already know the value of the right-hand side, we get a picture of e^x when x is a pure imaginary number:

• e⁰ = 1 (of course)
• e^x has a period of 2πi, so that e^2πi = e^4πi = e^6πi = e^-2πi = … = 1
• Similarly e^πi = e^3πi = e^5πi = e^-πi = … = -1
• Similarly e^πi/2 = e^5πi/2 = … = i, while e^3πi/2 = e^7πi/2 = … = -i

And since the rules for multiplying by adding exponents still apply, e^a+bi (where a and b are real) equals e^a * (cos(b) + (i)sin(b[/efn_note]. So you can picture (if you’re good at visualizing things in four dimensions) that, along the imaginary axis, e^x is a sort of three-dimensional sine curve (it curves in both of the other two dimensions) always lying on a cylinder of radius 1. Along the line x = k + yi that lie parallel to the imaginary axis, (k fixed, y varying), it has the same shape, but gets bigger for positive k and smaller for negative k (the cylinder’s radius being e^k.) Along the real axis, e^x has the familiar exponential shape. Along a parallel line x = y + ki, (k fixed, y varying), it has the same shape, but rotated according to the value of e^ki. (Of course, it rotates back to its familiar location whenever k is a multiple of 2π.)

If that’s too much to try to see in your mind’s eye (and the best I ever do is to frustratingly get almost there), at least you have the most commonly stated consequence of all this:

e^πi = cos(π) + i(sin(π[/efn_note] = -1