As I Try To Make Wronski’s Formula For Pi Into Something I Like


Previously:

I remain fascinated with Józef Maria Hoëne-Wronski’s attempted definition of π. It had started out like this:

\pi = \frac{4\infty}{\sqrt{-1}}\left\{ \left(1 + \sqrt{-1}\right)^{\frac{1}{\infty}} -  \left(1 - \sqrt{-1}\right)^{\frac{1}{\infty}} \right\}

And I’d translated that into something that modern mathematicians would accept without flinching. That is to evaluate the limit of a function that looks like this:

\displaystyle \lim_{x \to \infty} f(x)

where

f(x) = -4 \imath x \left\{ \left(1 + \imath\right)^{\frac{1}{x}} -  \left(1 - \imath\right)^{\frac{1}{x}} \right\}

So. I don’t want to deal with that f(x) as it’s written. I can make it better. One thing that bothers me is seeing the complex number 1 + \imath raised to a power. I’d like to work with something simpler than that. And I can’t see that number without also noticing that I’m subtracting from it 1 - \imath raised to the same power. 1 + \imath and 1 - \imath are a “conjugate pair”. It’s usually nice to see those. It often hints at ways to make your expression simpler. That’s one of those patterns you pick up from doing a lot of problems as a mathematics major, and that then look like magic to the lay audience.

Here’s the first way I figure to make my life simpler. It’s in rewriting that 1 + \imath and 1 - \imath stuff so it’s simpler. It’ll be simpler by using exponentials. Shut up, it will too. I get there through Gauss, Descartes, and Euler.

At least I think it was Gauss who pointed out how you can match complex-valued numbers with points on the two-dimensional plane. On a sheet of graph paper, if you like. The number 1 + \imath matches to the point with x-coordinate 1, y-coordinate 1. The number 1 - \imath matches to the point with x-coordinate 1, y-coordinate -1. Yes, yes, this doesn’t sound like much of an insight Gauss had, but his work goes on. I’m leaving it off here because that’s all that I need for right now.

So these two numbers that offended me I can think of as points. They have Cartesian coordinates (1, 1) and (1, -1). But there’s never only one coordinate system for something. There may be only one that’s good for the problem you’re doing. I mean that makes the problem easier to study. But there are always infinitely many choices. For points on a flat surface like a piece of paper, and where the points don’t represent any particular physics problem, there’s two good choices. One is the Cartesian coordinates. In it you refer to points by an origin, an x-axis, and a y-axis. How far is the point from the origin in a direction parallel to the x-axis? (And in which direction? This gives us a positive or a negative number) How far is the point from the origin in a direction parallel to the y-axis? (And in which direction? Same positive or negative thing.)

The other good choice is polar coordinates. For that we need an origin and a positive x-axis. We refer to points by how far they are from the origin, heedless of direction. And then to get direction, what angle the line segment connecting the point with the origin makes with the positive x-axis. The first of these numbers, the distance, we normally label ‘r’ unless there’s compelling reason otherwise. The other we label ‘θ’. ‘r’ is always going to be a positive number or, possibly, zero. ‘θ’ might be any number, positive or negative. By convention, we measure angles so that positive numbers are counterclockwise from the x-axis. I don’t know why. I guess it seemed less weird for, say, the point with Cartesian coordinates (0, 1) to have a positive angle rather than a negative angle. That angle would be \frac{\pi}{2} , because mathematicians like radians more than degrees. They make other work easier.

So. The point 1 + \imath corresponds to the polar coordinates r = \sqrt{2} and \theta = \frac{\pi}{4} . The point 1 - \imath corresponds to the polar coordinates r = \sqrt{2} and \theta = -\frac{\pi}{4} . Yes, the θ coordinates being negative one times each other is common in conjugate pairs. Also, if you have doubts about my use of the word “the” before “polar coordinates”, well-spotted. If you’re not sure about that thing where ‘r’ is not negative, again, well-spotted. I intend to come back to that.

With the polar coordinates ‘r’ and ‘θ’ to describe a point I can go back to complex numbers. I can match the point to the complex number with the value given by r e^{\imath\theta} , where ‘e’ is that old 2.71828something number. Superficially, this looks like a big dumb waste of time. I had some problem with imaginary numbers raised to powers, so now, I’m rewriting things with a number raised to imaginary powers. Here’s why it isn’t dumb.

It’s easy to raise a number written like this to a power. r e^{\imath\theta} raised to the n-th power is going to be equal to r^n e^{\imath\theta \cdot n} . (Because (a \cdot b)^n = a^n \cdot b^n and we’re going to go ahead and assume this stays true if ‘b’ is a complex-valued number. It does, but you’re right to ask how we know that.) And this turns into raising a real-valued number to a power, which we know how to do. And it involves dividing a number by that power, which is also easy.

And we can get back to something that looks like 1 + \imath too. That is, something that’s a real number plus \imath times some real number. This is through one of the many Euler’s Formulas. The one that’s relevant here is that e^{\imath \phi} = \cos(\phi) + \imath \sin(\phi) for any real number ‘φ’. So, that’s true also for ‘θ’ times ‘n’. Or, looking to where everybody knows we’re going, also true for ‘θ’ divided by ‘x’.

OK, on to the people so anxious about all this. I talked about the angle made between the line segment that connects a point and the origin and the positive x-axis. “The” angle. “The”. If that wasn’t enough explanation of the problem, mention how your thinking’s done a 360 degree turn and you see it different now. In an empty room, if you happen to be in one. Your pedantic know-it-all friend is explaining it now. There’s an infinite number of angles that correspond to any given direction. They’re all separated by 360 degrees or, to a mathematician, 2π.

And more. What’s the difference between going out five units of distance in the direction of angle 0 and going out minus-five units of distance in the direction of angle -π? That is, between walking forward five paces while facing east and walking backward five paces while facing west? Yeah. So if we let ‘r’ be negative we’ve got twice as many infinitely many sets of coordinates for each point.

This complicates raising numbers to powers. θ times n might match with some point that’s very different from θ-plus-2-π times n. There might be a whole ring of powers. This seems … hard to work with, at least. But it’s, at heart, the same problem you get thinking about the square root of 4 and concluding it’s both plus 2 and minus 2. If you want “the” square root, you’d like it to be a single number. At least if you want to calculate anything from it. You have to pick out a preferred θ from the family of possible candidates.

For me, that’s whatever set of coordinates has ‘r’ that’s positive (or zero), and that has ‘θ’ between -π and π. Or between 0 and 2π. It could be any strip of numbers that’s 2π wide. Pick what makes sense for the problem you’re doing. It’s going to be the strip from -π to π. Perhaps the strip from 0 to 2π.

What this all amounts to is that I can turn this:

f(x) = -4 \imath x \left\{ \left(1 + \imath\right)^{\frac{1}{x}} -  \left(1 - \imath\right)^{\frac{1}{x}} \right\}

into this:

f(x) = -4 \imath x \left\{ \left(\sqrt{2} e^{\imath \frac{\pi}{4}}\right)^{\frac{1}{x}} -  \left(\sqrt{2} e^{-\imath \frac{\pi}{4}} \right)^{\frac{1}{x}} \right\}

without changing its meaning any. Raising a number to the one-over-x power looks different from raising it to the n power. But the work isn’t different. The function I wrote out up there is the same as this function:

f(x) = -4 \imath x \left\{ \sqrt{2}^{\frac{1}{x}} e^{\imath \frac{\pi}{4}\cdot\frac{1}{x}} - \sqrt{2}^{\frac{1}{x}} e^{-\imath \frac{\pi}{4}\cdot\frac{1}{x}} \right\}

I can’t look at that number, \sqrt{2}^{\frac{1}{x}} , sitting there, multiplied by two things added together, and leave that. (OK, subtracted, but same thing.) I want to something something distributive law something and that gets us here:

f(x) = -4 \imath x \sqrt{2}^{\frac{1}{x}} \left\{ e^{\imath \frac{\pi}{4}\cdot\frac{1}{x}} -  e^{- \imath \frac{\pi}{4}\cdot\frac{1}{x}} \right\}

Also, yeah, that square root of two raised to a power looks weird. I can turn that square root of two into “two to the one-half power”. That gets to this rewrite:

f(x) = -4 \imath x 2^{\frac{1}{2}\cdot \frac{1}{x}} \left\{ e^{\imath \frac{\pi}{4}\cdot\frac{1}{x}} -  e^{- \imath \frac{\pi}{4}\cdot\frac{1}{x}} \right\}

And then. Those parentheses. e raised to an imaginary number minus e raised to minus-one-times that same imaginary number. This is another one of those magic tricks that mathematicians know because they see it all the time. Part of what we know from Euler’s Formula, the one I waved at back when I was talking about coordinates, is this:

\sin\left(\phi\right) = \frac{e^{\imath \phi} - e^{-\imath \phi}}{2\imath }

That’s good for any real-valued φ. For example, it’s good for the number \frac{\pi}{4}\cdot\frac{1}{x} . And that means we can rewrite that function into something that, finally, actually looks a little bit simpler. It looks like this:

f(x) = -2 x 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)

And that’s the function whose limit I want to take at ∞. No, really.

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Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there.

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