Treasures buried by pi in "the" Euler identity

An enlightening discussion about pi and tau.

Treasures buried by pi in "the" Euler identity

by Kodegadulo » Tue Jul 10, 2012 3:01 pm

There's even more to "Euler Identities" than meets the ... eye. ;) The Tau Manifesto made the point about how silly it is that there is all this hype over the equation

[tex]e^{i\pi} + 1 = 0[/tex]

otherwise known as "the" Euler Identity (so-called, as if there's only one). People marvel over how it combines the "five most important numbers in mathematics" and the operations of "exponentiation, multiplication, and addition" in one equation -- yet they do so without gaining any new insight or understanding of what the equation means. It amounts to nothing more than a piece of curious pop numerological mumbo-jumbo. But if we can just peel off the obscurity caused by [tex]\pi[/tex] there are interesting treasures we can dig up.

First off, looking at this "Euler Identity" is far less edifying than looking at what it's derived from, the Euler formula:

[tex]e^{i\theta} = \cos \theta + i \sin \theta[/tex]

This formula reveals that complex exponentiation is equivalent to a rotation around the unit circle, mapping an angle [tex]\theta[/tex] of polar coordinates to real and imaginary rectilinear coordinates via the circle functions sin and cos. In that light, plugging in interesting values for [tex]\theta[/tex] can be instructive:

[tex]\begin{tabular}{lll}
\underline{\operatorname{Identity}} & & \underline{\operatorname{Meaning}} \\
e^{i\tau} = 1 & & \operatorname{A rotation of a full turn is unity.} \\
e^{i\tau/2} = -1 & & \operatorname{A rotation of a half turn is negation. } \\
e^{i\tau/4} = i & & \operatorname{A rotation of a quarter turn is perpendicular.} \\
e^{i\tau \cdot k} = 1^k = 1 & & \operatorname{Any integer number of whole turns is unity.} \\
\end{tabular}[/tex]

I find each of these equations interesting and revealing. But for some reason the second of these is considered "ugly" because somehow division by two and negation are "inelegant" operations. But why should division and negation be disparaged as operations? Aren't they nearly as fundamental as multiplication and addition, and aren't they just as vital to mathematics? And why should 2 be disparaged? It's a rather important number in and of itself, being the first prime number and the only even prime. Yet it supposedly makes the equation more "beautiful" and "elegant" if we perform a bit of algebraic sleight-of-hand, burying the half in a [tex]\pi[/tex], and making a negative appear as if it's a positive. Hiding the dirty laundry as it were. But the only thing we accomplish in doing that is to mask the true importance of this identity:

What is the significance of a division when it appears within an exponentiation? In other words, what is the meaning of [tex]z^{1/n}[/tex]? The answer is:

[tex]z^{1/n} = \sqrt[n]{z}[/tex]

That means dividing an exponent by [tex]n[/tex] is the same as taking the [tex]n[/tex]-th root. But the Euler formula reveals that a complex exponentiation is equivalent to a rotation around the origin in the complex plane. So taking an [tex]n[/tex]-th root of a complex number is equivalent to dividing its rotation by [tex]n[/tex]. What are the consequences of that?

If we start with a full circle of rotation

[tex]e^{i\tau} = 1[/tex]

and divide the rotation in half we get

[tex]e^{i\tau\cdot 1/2} = -1 = 1^{1/2} = \sqrt[2]{1}[/tex]

In other words, this reveals that the square root of unity is negation. Or rather, the first square root of unity is negation. Because in fact if we take any number of whole turns

[tex]e^{i\tau \cdot k} = 1[/tex]

and divide their rotations in half

[tex]e^{i\tau \cdot k/2} = 1^{1/2}[/tex]

we see that the square roots of unity must include both

[tex]\begin{tabular}{lllll}
e^{i\tau \cdot 1/2} = -1 & & \operatorname{and} & & e^{i\tau \cdot 2/2} = 1 \\
\end{tabular}[/tex]

We can confirm that these are square roots by squaring them to get 1:

[tex]\begin{tabular}{lllll}
\left(e^{i\tau \cdot 1/2}\right)^2 = \left(-1\right)^2 = 1 & & \operatorname{and} & & \left(e^{i\tau \cdot 2/2}\right)^2 = \left(1\right)^2 = 1 \\
\end{tabular}[/tex]

But this means there is a very interesting bit of mathematical treasure buried unnoticed in the formula

[tex]e^{i\pi} + 1 = 0[/tex]

because we can substitute equivalent terms to yield

[tex]e^{i\tau \cdot 1/2} + e^{i\tau \cdot 2/2} = 0[/tex]

in other words, the sum of the square roots of unity (the multiplicative identity) is zero (the additive identity). Let's plot that on a unit circle:

Image

That makes sense: The negative unity counterbalances the positive unity, to average out to zero.

Hmm. What about the cube roots of unity? What would those be? And do they also add up to zero? Dividing up whole turns by 3 we get:

[tex]e^{i\tau \cdot k/3} = 1^{1/3}[/tex]

[tex]\begin{tabular}{lllll}
e^{i\tau \cdot 1/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i & & e^{i\tau \cdot 2/3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i & & e^{i\tau \cdot 3/3} = 1\\
\end{tabular}[/tex]

Let's plot them:

Image

Does this look familiar? That's right, those are the vertices of the equilateral triangle.

Let's confirm they're actually cube roots:

[tex]\left(e^{i\tau\cdot 1/3}\left)^3 = \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)^3 = \left(-\frac{1}{2}\right)^3 + 3\left(-\frac{1}{2}\right)^2\left(\frac{\sqrt{3}}{2}\right) + 3\left(-\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^3 = -\frac{1}{8} + \frac{3\sqrt{3}}{2} i + \frac{9}{8} - \frac{3\sqrt{3}}{2}i = \frac{8}{8} = 1[/tex]

[tex]\left(e^{i\tau\cdot 2/3}\left)^3 = \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)^3 = \left(-\frac{1}{2}\right)^3 + 3\left(-\frac{1}{2}\right)^2\left(-\frac{\sqrt{3}}{2}\right) + 3\left(-\frac{1}{2}\right)\left(-\frac{\sqr{3}}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^3 = -\frac{1}{8} - \frac{3\sqrt{3}}{2} i + \frac{9}{8} + \frac{3\sqrt{3}}{2}i = \frac{8}{8} = 1[/tex]

[tex]\left(e^{i\tau\cdot 3/3}\left)^3 = \left(1\right)^3 = 1[/tex]

Yes, that works. And what do they add up to?

[tex]e^{i\tau\cdot 1/3} + e^{i\tau\cdot 2/3} + e^{i\tau\cdot 3/3} = \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) + \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) + \left(1\right) = \left(-\frac{1}{2} - \frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}i - \frac{\sqrt{3}}{2}i\right) + 1 = -1 + 0 + 1 = 0[/tex]

They do add up to zero! But that makes sense too, because in the diagram, the two vectors on the left are mirror images of each other vertically, so they must cancel each out each other's imaginary component. And they each contribute a negative half in their real components, so they cancel out the unity on the right, to yield zero.

How about the fourth roots, where we divide up whole turns by 4? These are a bit easier:

[tex]e^{i\tau \cdot k/4} = 1^{1/4}[/tex]

[tex]\begin{tabular}{lllllll}
e^{i\tau \cdot 1/4} = i & & e^{i\tau \cdot 2/4} = -1 & & e^{i\tau \cdot 3/4} = -i & & e^{i\tau \cdot 4/4} = 1\\
\end{tabular}[/tex]

Image

And here we have the vertices of the square (disguised as its alter-ego, the diamond). They certainly look like they'll cancel each other out.

Let's confirm they're actually fourth roots:

[tex]\left( e^{i\tau \cdot 1/4}\right)^4 = \left(i\right)^4 = \left(-1\right)^2 = 1[/tex]
[tex]\left( e^{i\tau \cdot 2/4}\right)^4 = \left(-1\right)^4 = \left(1\right)^2 = 1[/tex]
[tex]\left( e^{i\tau \cdot 3/4}\right)^4 = \left(-i\right)^4 = \left(-1\right)^2 = 1[/tex]
[tex]\left( e^{i\tau \cdot 4/4}\right)^4 = \left(1\right)^4 = \left(1\right)^2 = 1[/tex]

And do they add up to zero?

[tex]e^{i\tau \cdot 1/4} + e^{i\tau \cdot 2/4} + e^{i\tau \cdot 3/4} + e^{i\tau \cdot 4/4} = \left( i\right) + \left(-1\right) + \left(-i\right) + \left(1\right) = \left(i - i\right) + \left(-1 + 1\right) = 0 + 0 = 0[/tex]

They do, as expected.

Are you beginning to see a pattern here? For every natural number [tex]n \ge 2[/tex] there are [tex]n[/tex] complex roots of unity. The first such [tex]n[/tex]-th root is defined as

[tex]\zeta_n = e^{i\tau_n}[/tex] where [tex]\tau_n = \frac{\tau}{n}[/tex]

In other words, it's positioned on the unit circle at the [tex]n[/tex]-th of the "candidate circle constants", which I identified as "special angles" in an earlier post, i.e., the angle that is an [tex]n[/tex]-th of a full turn from unity. And the entire set of [tex]n[/tex]-th roots are defined as the powers of this first [tex]n[/tex]-th root, up to and including unity itself:

[tex]{\zeta_n}^k = e^{i\tau_n\cdot k}[/tex] for all [tex]k \in \left[1, n \right][/tex]

These roots are spaced around the unit circle at multiples of the special angle, each position being an [tex]n[/tex]-th part of [tex]k[/tex] whole turns from unity, and each corresponding to a vertex of the regular [tex]n[/tex]-gon plotted on the unit circle. And the sum of these [tex]n[/tex]-th roots is zero:

[tex]\sum_{k=1}^{n} {\zeta_n}^k = 0[/tex]

You can try it out with the fifth roots and the vertices of the pentagon, using a calculator for those multiples of [tex]72^{\circ}[/tex] angles. Or a little more easily with the sixth roots at those sextant angles. But you'll see it works out for every [tex]n[/tex] you try. It makes sense geometrically: The [tex]n[/tex]th roots of unity are unit vectors centered at the origin and distributed evenly around the circle, so they must counterbalance each other to add up to that center point. But we can prove this rigorously by noting that [tex]{\zeta_n}^k[/tex] constitutes a geometric series. The formula for the sum of a geometric series [tex]a^k[/tex] is given as:

[tex]\sum_{k=0}^{n-1} a^k = \frac{a^n - 1}{a - 1}[/tex]

So if we substitute [tex]a = \zeta_n[/tex] and note that [tex]{\zeta_n}^n = {\zeta_n}^0 = 1[/tex]:

[tex]\sum_{k=1}^{n} {\zeta_n}^k = \sum_{k=0}^{n-1} {\zeta_n}^k = \frac{{\zeta_n}^n - 1}{{\zeta_n} - 1} = \frac{1 - 1}{{\zeta_n} - 1} = 0[/tex] Q.E.D.

Bottom line, ask yourself: Would any of this be any clearer or more "beautiful" or "elegant", if it were cast in terms of [tex]\pi[/tex]? Wouldn't it be incrementally more ugly, and therefore incrementally more obscure, and therefore incrementally harder for students to grasp, if it were encrusted with [tex]2\pi[/tex] everywhere? If you saw a [tex]\pi/3[/tex] in the Euler formula, would you immediately grasp that it was the rotation for a 6th-root of unity, and not a cube root? If you saw [tex]\pi \cdot 2/7[/tex] and [tex]\pi \cdot 4/7[/tex], would you immediately grasp they these were the first and second of the 7th-roots of unity, and not the second and fourth? If you are a confirmed [tex]\pi[/tex]-ist, are you really willing to go through all those mental gymnastics, just for the sake of [tex]\pi[/tex]?

We are dealing with complex numbers plotted on the unit circle. Isn't this just confirmation that the most fundamental constant associated with circles, and with radians as the ideal angular measure to use with circle functions such as [tex]\sin \theta[/tex], [tex]\cos \theta[/tex], and [tex]e^{i \theta}[/tex], is the number identified as [tex]\tau[/tex]? And isn't [tex]\pi[/tex], at best, just one of many possible numbers derivable from [tex]\tau[/tex]?

And given all this insight we can derive, doesn't "the" Euler Identity [tex]e^{i\pi} + 1 = 0[/tex] become more beautiful, not less, when we re-cast it as [tex]e^{i\tau/2} = -1[/tex]?
Kodegadulo
Kindergarten
 
Posts: 6
Joined: Sun Jun 10, 2012 3:04 am

Re: Treasures buried by pi in "the" Euler identity

by josephlindenberg » Sat Jul 21, 2012 6:31 pm

Excellent post! I'm glad to see more people using the sum-of-the-nth-roots-of-unity argument for tau in Euler's Identity, especially when it's expressed this well.

By the way, there's a great way to appeal to all those numerologists fawning over the traditional form of Euler's Identity. Show them this conjugate form of Euler's Identity using tau:

[tex]0\, =\, 1\, +\, e^{-i\cdot{\frac\tau{2}}}[/tex]

All four basic arithmetic operations appear exactly once, and in their standard order (+,−,*,/)
0,1,2 appear in numerical order
e,i,tau appear in alphabetical order
PI is the SEMICIRCLE constant, not the circle constant . . . http://sites.google.com/site/taubeforeitwascool
josephlindenberg
Elementary School
Elementary School
 
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Joined: Wed Jul 06, 2011 2:34 am


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