Posts tagged with reciprocal

Just playing with z² / z² + 2z + 2

$g(z)=\frac{z^2}{z^2+2z+2}$

on WolframAlpha. That’s Wikipedia’s example of a function with two poles (= two singularities = two infinities). Notice how “boring” line-only pictures are compared to the the 3-D ℂ→>ℝ picture of the mapping (the one with the poles=holes). That’s why mathematicians say ℂ uncovers more of “what’s really going on”.

As opposed to normal differentiability, ℂ-differentiability of a function implies:

• infinite descent into derivatives is possible (no chain of C¹ ⊂ C² ⊂ C³ ... Cω like usual)

• nice Green’s-theorem type shortcuts make many, many ways of doing something equivalent. (So you can take a complicated real-world situation and validly do easy computations to understand it, because a squibbledy path computes the same as a straight path.)


Pretty interesting to just change things around and see how the parts work.

• The roots of the denominator are 1+i and 1−i (of course the conjugate of a root is always a root since i and −i are indistinguishable)
• you can see how the denominator twists
• a fraction in ℂ space maps lines to circles, because lines and circles are turned inside out (they are just flips of each other: see also projective geometry)
• if you change the z^2/ to a z/ or a 1/ you can see that.
• then the Wikipedia picture shows the poles (infinities)

Complex ℂ→ℂ maps can be split into four parts: the input “real”⊎”imaginary”, and the output “real"⊎"imaginary”. Of course splitting them up like that hides the holistic truth of what’s going on, which comes from the perspective of a “twisted” plane where the elements z are mod z • exp(i • arg z).

ℂ→ℂ mappings mess with my head…and I like it.

I want to understand this connection better.

Although it sounds the most involved, arctan answers the simplest trigonometric question you could ask. “The full moon subtends seven degrees of my vision; so how big is the moon?” The arctan (with radius = distance from Earth to moon) is the answer.

But what does that simple operation have to do with:

1. continuous sum of
2. an inversion of
3. one more than
4. some continuous range of numbers
5. times themselves?

$\dpi{200} \bg_white \large \arctan = \mathtt{Int} \circ \mathtt{flip} \circ \mathtt{Peano} \circ \mathtt{square} \\ \\ \\ \text{where}\\ \!\!\!\!\!\!\! \bullet\ \mathtt{Peano}(x)\overset{\mathrm{def}}{=}x+1 \\ \bullet\ \mathtt{flip}(x)\overset{\mathrm{def}}{=} 1/x \\ \bullet\ \mathtt{square}(x)\overset{\mathrm{def}}{=} xx \\ \bullet\ \mathtt{Int}(x)\overset{\mathrm{def}}{=} \textstyle{\int x}$

That’s confusing.

Four years later…

This comment shows how to get the answer. (Pair with this on the derivative of logarithms.) A consequence of the chain rule (or perhaps another way to state it!) is that the derivative of ƒ⁻¹ at a point p is flip( derivative( ƒ )) evaluated at ƒ⁻¹(p), presupposing that all of the maps fit together right.

arctan is defined, for simplicity, on a circle of unit radius. (This just means if you’re looking at the moon then use units of “one moon-distance”.) It takes as an argument a ratio of sides and returns an angle θ. Since derivative ( tan ) = derivative (sin⁄cos) = 1⁄cos² = sec × sec (reasoned with calc 101), by combining that derivative ( tan ) with JavaMan’s perspective on the derivatives of inverse-functions we can argue that

Java Man’s idea gets us to look at the triangle

which, since it’s constrained to a circle by the equivalence-classing of triangles to be just the ones with a certain angle (see fibration), limits us to just one free parameter R (a ratio of opposite O to adjacent A side lengths of an equivalence-class of triangles). After following Java Man’s logic to see why derivative( tan ) = sec² implies derivative( inverse of tan ) = cos² [at the angle which is implied by tan R], we’re left knowing that A=the adjacent side implied by the cos R = cos O⁄A of the original O⁄A ratio we were given as the natural input space (ratios) which the tan function accepts. This isn’t enough because the answer needs to be in O⁄A terms to match the input. We have to do some Pythagorean jiu jitsu involving A = A⁄1 and 1=1²=A²+O² to get the answer into an O⁄A form (since that was the information we were given). Using the 1=A²+O² is using the natural delimitation of the inscribing circle to make the two A and O move together the way they should on the circle, by the way. The algebraic jiu jitsu then yields A = flip( 1+R² ), now using the proper input R=O⁄A.

The point of all this mangling was merely to match up cosine’s output with tangent’s input. Sheesh with all the symbols!

But that’s merely deriving the correct answer algebraically, with a bit of “why” from the comment’s perspective on inverse functions generally. What about my original question? Why does this sequence of mappings, if iterated, subtend the moon?