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?

4 notes

1. neljae reblogged this from isomorphismes and added:
Me too.
2. isomorphismes posted this