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).

a conformal map (angle-preserving map)

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?

image

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
image
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.

image

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?