Weekend Roundup

A sudden and mysterious fever left me unable to post my usual weekend roundup on time today. Now that I’m feeling human again, here it is, a few hours late:

We started the week with my best attempt to explain the intuition underlying the spectacular formula e = -1, frequently described as the most beautiful and astonishing equation in all of mathematics. Gauss reportedly once said that if this formula is not immediately obvious to you, you have no hope of being a mathematician—but I’ve heard more than one Fields Medalist say he’d been dumbstruck when he first encountered it.

We reviewed Yale professor Gary Gorton‘s account of the financial crisis; he says it was a bank run, and if you’re going to have banks, bank runs go with the territory.

On the lighter side, we talked about web comics; on the less-light side we talked about the advantages of genocide over other forms of mass murder, and about moral paradoxes.

Barring a relapse, I’ll see you Monday.

5 Responses to “Weekend Roundup”

1. 1 1 Bennett Haselton

Some thoughts on e**pi(i) = -1:

When I was in high school I became minorly obsessed with the question of whether imaginary exponents were really meaningful. What bothered me is that you couldn’t extend the definition of exponentiation to imaginary numbers the way you could extend it to negative numbers and non-whole numbers, by “relating them back” to the negative numbers.

What is x**-1? Well we know that x**(a+b) = x**a * x**b for all natural numbers. To extend this consistently to negatives, you’d have to have x**-1 = 1/x.

Similarly, to find x**1/2, reason that x**1/2 * x**1/2 = x**1, therefore x**1/2 = +/- sqrt(x).

But what about x**i? I reasoned that since (x**a)**b = x**(a*b) for all natural numbers, then if x**i exists at all, it should be the case that (x**i)**i = 1/x (with x**0 being undefined). But where do you go from there? And the textbook claimed that e**ix = cos(x) + isin(x), so presumably x**i = cos(lnx(x)) + isin(ln(x)). But it’s far from obvious that if you apply that function twice, you get 1/x!

Much later, I realized that you can’t have a continuous *single-valued* function f such that f(f(x)) = 1/x for all nonzero complex numbers, for a simple reason. Consider the image of the unit circle (going clockwise). That image IMG has to lie somewhere in the plane, and the image of IMG is the unit circle going counter-clockwise. Now first, IMG has to loop around zero exactly once. If it looped more than once, it would self-intersect, and if it looped zero times, then you could shrink it down continuously to a single nonzero point, but the image of IMG (which is the unit circle) wouldn’t be able to shrink down to a single point without crossing zero. So since IMG loops around zero exactly once, then if IMG goes clockwise, you could continuously transform the unit circle *to* IMG, and in the process the image of that shifting loop would have to be continuously transformed from IMG to the unit circle going counter-clockwise — but that’s not possible since IMG goes clockwise. (The whole argument works similarly in reverse if IMG goes counter-clockwise.) QED.

It turns out the function x**i = cos(lnx(x)) + isin(ln(x)) gets around this problem because ln(x) is an infinitely-multi-valued function.

However since f(x) = x**i = cos(lnx(x)) + isin(ln(x)) has infinitely many branches, it’s still true that f(f(x)) = 1/x for all nonzero x, for *one* of those branches of the inner f and one of those branches of the outer f. I’ve had that explained to me once and could probably reproduce the steps, but I still can’t see intuitively why.

You wrote in your book that we all know the natural numbers are “real things” because we know intuitively that you can never get paradoxes and contradictions by adding and subtracting them. That’s certainly true for me for the natural numbers, but I’m still not convinced that imaginary exponents actually exist, because I still harbor lingering suspicions that these definitions could lead to a contradiction somewhere. I wonder if more experienced mathematicians have a stronger feeling in their bones that imaginary exponents are (no pun intended) “real”.

2. 2 2 ToddM

I always liked the form e^(i*pi) + 1 = 0
that way you get e, i, pi, 1, and 0 in a simple, beautiful relationship…

3. 3 3 Neil

Bennett,

Numbers do not exist, you just get used to them. After all, what is negative pi? A circle with a negative diameter and a positive circumference or vice versa?

4. 4 4 Bill Drissel

It’s all a matter of sequence. If you see “e to i pi equals minus one” first, you’re buffaloed. But if you expand the Taylor series for the left side, collect the odd and even terms, and recognize sin(pi), the equation comes to you as a result – not a mystery.

Regards,
Bill Drissel

5. 5 5 Bill Drissel

“if youâ€™re going to have banks, bank runs go with the territory.” Correction: if you’re going to have banks with fractional reserves and maintain the pretense (fraud) that the money is “in the bank” when it is actually lent for long-term mortgages, you’re going to have bank runs.

Regards,
Bill Drissel