# Wronski’s Formula For Pi: How Close We Came

Previously:

Józef Maria Hoëne-Wronski’s had an idea for a new, universal, culturally-independent definition of π. It was this formula that nobody went along with because they had looked at it: $\pi = \frac{4\infty}{\sqrt{-1}}\left\{ \left(1 + \sqrt{-1}\right)^{\frac{1}{\infty}} - \left(1 - \sqrt{-1}\right)^{\frac{1}{\infty}} \right\}$

I made some guesses about what he would want this to mean. And how we might put that in terms of modern, conventional mathematics. I describe those in the above links. In terms of limits of functions, I got this: $\displaystyle \lim_{x \to \infty} f(x) = \lim_{x \to \infty} -2 x 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$

The trouble is that limit took more work than I wanted to do to evaluate. If you try evaluating that ‘f(x)’ at ∞, you get an expression that looks like zero times ∞. This begs for the use of L’Hôpital’s Rule, which tells you how to find the limit for something that looks like zero divided by zero, or like ∞ divided by ∞. Do a little rewriting — replacing that first ‘x’ with ‘ $\frac{1}{1 / x}$ — and this ‘f(x)’ behaves like L’Hôpital’s Rule needs.

The trouble is, that’s a pain to evaluate. L’Hôpital’s Rule works on functions that look like one function divided by another function. It does this by calculating the derivative of the numerator function divided by the derivative of the denominator function. And I decided that was more work than I wanted to do.

Where trouble comes up is all those parts where $\frac{1}{x}$ turns up. The derivatives of functions with a lot of $\frac{1}{x}$ terms in them get more complicated than the original functions were. Is there a way to get rid of some or all of those?

And there is. Do a change of variables. Let me summon the variable ‘y’, whose value is exactly $\frac{1}{x}$. And then I’ll define a new function, ‘g(y)’, whose value is whatever ‘f’ would be at $\frac{1}{y}$. That is, and this is just a little bit of algebra: $g(y) = -2 \cdot \frac{1}{y} \cdot 2^{\frac{1}{2} y } \cdot \sin\left(\frac{\pi}{4} y\right)$

The limit of ‘f(x)’ for ‘x’ at ∞ should be the same number as the limit of ‘g(y)’ for ‘y’ at … you’d really like it to be zero. If ‘x’ is incredibly huge, then $\frac{1}{x}$ has to be incredibly small. But we can’t just swap the limit of ‘x’ at ∞ for the limit of ‘y’ at 0. The limit of a function at a point reflects the value of the function at a neighborhood around that point. If the point’s 0, this includes positive and negative numbers. But looking for the limit at ∞ gets at only positive numbers. You see the difference?

… For this particular problem it doesn’t matter. But it might. Mathematicians handle this by taking a “one-sided limit”, or a “directional limit”. The normal limit at 0 of ‘g(y)’ is based on what ‘g(y)’ looks like in a neighborhood of 0, positive and negative numbers. In the one-sided limit, we just look at a neighborhood of 0 that’s all values greater than 0, or less than 0. In this case, I want the neighborhood that’s all values greater than 0. And we write that by adding a little + in superscript to the limit. For the other side, the neighborhood less than 0, we add a little – in superscript. So I want to evalute: $\displaystyle \lim_{y \to 0^+} g(y) = \lim_{y \to 0^+} -2\cdot\frac{2^{\frac{1}{2}y} \cdot \sin\left(\frac{\pi}{4} y\right)}{y}$

Limits and L’Hôpital’s Rule and stuff work for one-sided limits the way they do for regular limits. So there’s that mercy. The first attempt at this limit, seeing what ‘g(y)’ is if ‘y’ happens to be 0, gives $-2 \cdot \frac{1 \cdot 0}{0}$. A zero divided by a zero is promising. That’s not defined, no, but it’s exactly the format that L’Hôpital’s Rule likes. The numerator is: $-2 \cdot 2^{\frac{1}{2}y} \sin\left(\frac{\pi}{4} y\right)$

And the denominator is: $y$

The first derivative of the denominator is blessedly easy: the derivative of y, with respect to y, is 1. The derivative of the numerator is a little harder. It demands the use of the Product Rule and the Chain Rule, just as last time. But these chains are easier.

The first derivative of the numerator is going to be: $-2 \cdot 2^{\frac{1}{2}y} \cdot \log(2) \cdot \frac{1}{2} \cdot \sin\left(\frac{\pi}{4} y\right) + -2 \cdot 2^{\frac{1}{2}y} \cdot \cos\left(\frac{\pi}{4} y\right) \cdot \frac{\pi}{4}$

Yeah, this is the simpler version of the thing I was trying to figure out last time. Because this is what’s left if I write the derivative of the numerator over the derivative of the denominator: $\displaystyle \lim_{y \to 0^+} \frac{ -2 \cdot 2^{\frac{1}{2}y} \cdot \log(2) \cdot \frac{1}{2} \cdot \sin\left(\frac{\pi}{4} y\right) + -2 \cdot 2^{\frac{1}{2}y} \cdot \cos\left(\frac{\pi}{4} y\right) \cdot \frac{\pi}{4} }{1}$

And now this is easy. Promise. There’s no expressions of ‘y’ divided by other expressions of ‘y’ or anything else tricky like that. There’s just a bunch of ordinary functions, all of them defined for when ‘y’ is zero. If this limit exists, it’s got to be equal to: $\displaystyle -2 \cdot 2^{\frac{1}{2} 0} \cdot \log(2) \cdot \frac{1}{2} \cdot \sin\left(\frac{\pi}{4} \cdot 0\right) + -2 \cdot 2^{\frac{1}{2} 0 } \cdot \cos\left(\frac{\pi}{4} \cdot 0\right) \cdot \frac{\pi}{4}$ $\frac{\pi}{4} \cdot 0$ is 0. And the sine of 0 is 0. The cosine of 0 is 1. So all this gets to be a lot simpler, really fast. $\displaystyle -2 \cdot 2^{0} \cdot \log(2) \cdot \frac{1}{2} \cdot 0 + -2 \cdot 2^{ 0 } \cdot 1 \cdot \frac{\pi}{4}$

And 20 is equal to 1. So the part to the left of the + sign there is all zero. What remains is: $\displaystyle 0 + -2 \cdot \frac{\pi}{4}$

And so, finally, we have it. Wronski’s formula, as best I make it out, is a function whose value is … $-\frac{\pi}{2}$

… So, what Wronski had been looking for, originally, was π. This is … oh, so very close to right. I mean, there’s π right there, it’s just multiplied by an unwanted $-\frac{1}{2}$. The question is, where’s the mistake? Was Wronski wrong to start with? Did I parse him wrongly? Is it possible that the book I copied Wronski’s formula from made a mistake?

Could be any of them. I’d particularly suspect I parsed him wrongly. I returned the library book I had got the original claim from, and I can’t find it again before this is set to publish. But I should check whether Wronski was thinking to find π, the ratio of the circumference to the diameter of a circle. Or might he have looked to find the ratio of the circumference to the radius of a circle? Either is an interesting number worth finding. We’ve settled on the circumference-over-diameter as valuable, likely for practical reasons. It’s much easier to measure the diameter than the radius of a thing. (Yes, I have read the Tau Manifesto. No, I am not impressed by it.) But if you know 2π, then you know π, or vice-versa.

The next question: yeah, but I turned up -½π. What am I talking about 2π for? And the answer there is, I’m not the first person to try working out Wronski’s stuff. You can try putting the expression, as best you parse it, into a tool like Mathematica and see what makes sense. Or you can read, for example, Quora commenters giving answers with way less exposition than I do. And I’m convinced: somewhere along the line I messed up. Not in an important way, but, essentially, doing something equivalent to divided by -2 when I should have multiplied by that.

I’ve spotted my mistake. I figure to come back around to explaining where it is and how I made it. ## Author: Joseph Nebus

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

Categories Math, Mathematics, Maths

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