On Wronski’s Definition of π

28 Nov 2017

On Wronski’s Definition of π

Joseph Nebus has recently written a couple of posts (here and here) in which he discusses an interesting attempt by Józef Maria Hoëne-Wronski to create a purely numerical definition of the mathematical constant π which is independent of the classical, geometric definition of “the ratio of the circumference of a circle to its diameter.” This has been a goal of many mathematicians, since the idea of π seems like it is more fundamental to mathematics than a definition based on circles would make it seem– as evidenced by the fact that it shows up in areas of mathematics which are seemingly unrelated to circles. Wronski’s idea, to this end, was the following formula:

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

At first glance, the formula seems inherently nonsensical. After all, $\infty$ is not a number, and therefore cannot be utilized in numerical operations in this way. However, one can get a sense of what Wronski may have intended by this equation. It appears that Wronski wanted to utilize $\infty$ to represent an infinite number, and modern mathematics actually gives us several tools for handling this sort of idea. One which might be of particular use, here, is Non-Standard Analysis with its infinite and infinitesimal Hyperreal numbers. In NSA, we have the ability to perform calculations with and upon infinite numbers perfectly consistently and reasonably.

First things first, let’s translate Wronski’s equation into a more modern form. Borrowing from the work Joseph Nebus already did in his second post on the subject, we can replace all the $\sqrt{-1}$ ‘s in the equation with $i$ ‘s, instead, in order to get:

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

Now, we can use our tools from NSA to find suitable substitutes for $\infty$ in the above equation. One immediate problem which a mathematician might notice is that replacing the three $\infty$ symbols with positive, infinite Hyperreal numbers $K$ , $M$ , and $N$ will lead to different solutions for the equation when one uses different values for $K$ , $M$ , and $N$ .

However, Wronski died well before Georg Cantor‘s brilliant work showing that there are different sizes of infinite sets was even published, let alone accepted by mainstream mathematicians. As such, it is very reasonable to assume that Wronski believed his $\infty$ symbol was referring to a single, specific quantity, rather than a range of possible infinite quantities. So, let’s replace all $\infty$ symbols with a single positive, infinite Hyperreal number, $N$ . This gives us:

$\pi = -4 i N \left\{\left(1+i\right)^{\frac{1}{N}} - \left(1 - i \right)^{\frac{1}{N}} \right\}$

Starting with the expression within the braces, we can explore to find something which may be a bit easier to work with. This takes a little bit of work, but we can show that:

$\left(1+i\right)^{\frac{1}{N}} - \left(1 - i \right)^{\frac{1}{N}}=\sqrt[{2N}]{2}\left(\sqrt[{4N}]{-1}-\frac{1}{\sqrt[{4N}]{-1}} \right)$

Let’s zoom in on our equation a little bit more, now. The expression $\sqrt[{4N}]{-1}$ is a Complex Hyperreal number which is infinitely close to the Real number, $1$ . As such, its reciprocal is also infinitely close to $1$ . Given this information, we know that the expression $\sqrt[{4N}]{-1}-\frac{1}{\sqrt[{4N}]{-1}}$ must simplify into some infinitesimal Complex Hyperreal number. Let’s call this number $\gamma + \delta i$ for Hyperreals $\gamma$ and $\delta$ .

Similarly, we know that $\sqrt[{2N}]{2}$ is a non-Complex Hyperreal number which is infinitely close to $1$ . Let’s call this number $1+\epsilon$ , where epsilon is some non-zero infinitesimal. Multiplying this by our earlier result yields $(1+\epsilon)(\gamma + \delta i)=\gamma + \gamma \epsilon + \delta i(1+\epsilon)$ . We can then take this expression and substitute it for the entire braced expression from our full equation:

$\pi= \left(-4iN\right) \left\{ \gamma + \gamma \epsilon + \delta i \left(1+\epsilon \right) \right\}$

This, in turn, can now be simplified to:

$\pi = 4N \left\{ \left( \delta + \delta \epsilon \right) - \left( \gamma + \gamma \epsilon \right) i \right\}$

We’re still far from anything which clearly resembles the π which we all know and love, but now we are getting to a place where we can really start to see some of the implications of Wronski’s definition. Notably, either $\gamma = 0$ , or else NSA seems to show that Wronski’s π is not a Real number. As such, it seems like Wronski’s definition is a failure if $\gamma \neq 0$ — presumably, Wronski was not attempting to redefine π out of the set of Real numbers!

However, it seems quite dubious that it would be the case that $\gamma =0$ . Looking back for a moment, we defined our $\gamma$ as the Real part of the expression $\sqrt[{4N}]{-1}-\frac{1}{\sqrt[{4N}]{-1}}$ . Let’s break this down a bit further, now. The term $\sqrt[{4N}]{-1}$ is a Complex Hyperreal number which is infinitely close to the Real number, $1$ ; let’s call it $1+\alpha+\beta i$ , for infinitesimal Hyperreals $\alpha$ and $\beta$ . I’ll spare my readers a few more convoluted formulae (feel free to work this out yourself!), but $\gamma=0$ if and only if $\alpha=0$ . However, it seems fairly clear that $\alpha \neq 0$

One of the properties of $\sqrt[x]{-1}$ for all Real $x$ is that it has a magnitude equal to 1. This means that for any Complex number $a+bi$ such that $a$ and $b$ are Real and that $a+bi=\sqrt[x]{-1}$ , it will be true that $a^2+b^2=1$ . The Transfer Principle of the Hyperreals allows us to extend this statement over the Hyperreal numbers, as well. Since the Complex Hyperreal which we were concerned with is $1+\alpha + \beta i$ , we therefore know that $(1+\alpha)^2 + \beta^2=1$ . Since $\beta$ is non-Complex, we know that its square must be positive or zero. Similarly, $(1+a)^2$ must be positive and greater-than-or-equal-to 1. As such, the only way for $(1+\alpha)^2 + \beta^2=1$ to be true is in the case that $\alpha = \beta = 0$ .

For this to be the case, then $\sqrt[{4N}]{-1}=1$ . In order for this to be true, it must be the case that $-1<N<1$ . However, this contradicts our initial definition for $N$ as a positive, infinite Hyperreal number.

Unfortunately, it seems that Wronski’s attempt to create a non-geometric definition for π simply does not work. That said, I’m still very curious about his thought process, here. What led him to this particular formulation, in the first place? Is it, perhaps, possible to salvage his work? Could there be some actual truth hidden underneath the apparent incoherence? It will certainly be fun to unravel this puzzle even further.

Posted in Mathematics, Philosophy and tagged Infinitesimal, Infinity, Joseph Nebus, mathematics, non-standard analysis, pi, Wronski

One thought on “On Wronski’s Definition of π”

Matthew McIrvin on 2018/01/02 at 2:52 pm said:

If you interpret N as a finite real number and then take the limit as N->infinity (the “safe” way to handle these things that we all learned in school)… you get 2pi. Close, but no cigar! Whether that corresponds in any way to Wronski’s reasoning, I don’t know…

(Unlike Nebus, I cheated by just throwing it at a CAS, namely Sagemath. It’s interesting to see that Sagemath doesn’t take it all the way to 2pi if you ask for the limit, but instead gives a complex logarithmic expression that you can simplify pretty easily by putting the 1+i and 1-i into polar form. If you instead just evaluate it numerically, you get something with a real part that looks like 2pi and a tiny imaginary part from rounding error. It’s amusingly reminiscent of what happens with the hyperreals.)

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