arctan

4


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?

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?

Sábado 13, de Junio 2015.
Yomar Gonzalez ( Androidfast ) 


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