784 lines
23 KiB
Plaintext
784 lines
23 KiB
Plaintext
---
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title: "Further Notes on Algebraic Stereography"
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description: |
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How do you rotate in 2D and 3D without standard trigonometry?
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format:
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html:
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html-math-method: katex
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jupyter: python3
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date: "2021-10-10"
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date-modified: "2025-06-30"
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categories:
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- algebra
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- complex analysis
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- polar roses
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- generating functions
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---
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<style>
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.figure-img {
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max-width: 512px;
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object-fit: contain;
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height: 100%;
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}
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</style>
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```{python}
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#| echo: false
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from dataclasses import dataclass
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import matplotlib.pyplot as plt
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import sympy
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from sympy.abc import t
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from anim import SympyAnimationWrapper
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# rational circle function
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o = (1 + sympy.I*t) / (1 - sympy.I*t)
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# cache some values for plotting later
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def generate_os(amt):
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ret = sympy.sympify(1)
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for _ in range(amt):
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yield ret
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ret = (ret * o).expand().cancel().simplify()
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os = list(generate_os(11))
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cs = [sympy.re(i).simplify() for i in os]
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ss = [sympy.im(i).simplify() for i in os]
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```
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In my previous post, I discussed the stereographic projection of a circle as it pertains
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to complex numbers, as well as its applications in 2D and 3D rotation.
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In an effort to document more interesting facts about this mathematical object
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(of which scarce information is immediately available online),
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I will now elaborate on more of its properties.
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Chebyshev Polynomials
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---------------------
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[Previously](/posts/chebyshev/1), I derived the
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[Chebyshev polynomials](https://en.wikipedia.org/wiki/Chebyshev_polynomials)
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with the archetypal complex exponential.
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These polynomials express the sines and cosines of a multiple of an angle from
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the sine and cosine of the base angle.
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Where $T_n(t)$ are Chebyshev polynomials of the first kind and $U_n(t)$ are those of the second kind,
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$$
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\begin{gather*}
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\cos(n \theta) = T_n(\cos(\theta))
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\\
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\sin(n \theta) = U_{n - 1}(\cos(\theta)) \sin(\theta)
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\end{gather*}
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$$
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The complex exponential derivation begins by squaring and developing a second-order recurrence.
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$$
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\begin{align*}
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(e^{i\theta})^2 &= (\cos + i\sin)^2
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\\
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&= \cos^2 + 2i\cos \cdot \sin - \sin^2 + (0 = \cos^2 + \sin^2 - 1)
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\\
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&= 2\cos^2 + 2i\cos \cdot \sin - 1
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\\
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&= 2\cos \cdot (\cos + i\sin) - 1
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\\
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&= 2\cos(\theta)e^{i\theta} - 1
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\\
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(e^{i\theta})^{n+2} &= 2\cos(\theta)(e^{i\theta})^{n+1} - (e^{i\theta})^n
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\end{align*}
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$$
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This recurrence relation can then be used to obtain the Chebyshev polynomials, and hence,
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the expressions using sine and cosine above.
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Presented this way with such a simple derivation, it appears as though these relationships
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are inherently trigonometric.
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However, these polynomials actually have *nothing* to do with sine and cosine on their own.
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For one, [they appear in graph theory](/posts/chebyshev/2), and for two,
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the importance of the complex exponential is overstated.
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$e^{i\theta}$ really just specifies a point on the complex unit circle.
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This property is used on the second line to coax the equation into a quadratic in $e^{i\theta}$.
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This is also the *only* property upon which the recurrence depends; all else is algebraic manipulation.
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### Back to the Stereograph
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Knowing this, let's start over with the stereographic projection of the circle:
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$$
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o_1(t) = {1 + it \over 1 - it}
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= {1 - t^2 \over 1 + t^2} + i {2t \over 1 + t^2}
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= \text{c}_1 + i\text{s}_1
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$$
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The subscript "1" is because as *t* ranges over $(-\infty, \infty)$, the function loops once
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around the unit circle.
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Taking this to higher powers keeps points on the circle since all points on the circle
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have a norm of 1.
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It also makes more loops around the circle, which we can denote by larger subscripts:
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$$
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\begin{align*}
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o_n &= (o_1)^n
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= \left( {1 + it \over 1 - it} \right)^n
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\\
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\text{c}_n + i\text{s}_n
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&= (\text{c}_1 + i\text{s}_1)^n
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\end{align*}
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$$
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This mirrors raising the complex exponential to a power
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(which loops over the range $(-\pi, \pi)$ instead).
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The final line is analogous to de Moivre's formula, but in a form where everything is
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a ratio of polynomials in *t*.
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This means that the Chebyshev polynomials can be obtained directly from these rational expressions:
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$$
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\begin{align*}
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o_2 = (o_1)^2 &= (\text{c}_1 + i\text{s}_1)^2
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\\
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&= \text{c}_1^2 + 2i\text{c}_1\text{s}_1 - \text{s}_1^2
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+ (0 = \text{c}_1^2 + \text{s}_1^2 - 1)
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\\
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&= 2\text{c}_1^2 + 2i\text{c}_1\text{s}_1 - 1
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\\
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&= 2\text{c}_1(\text{c}_1 + i\text{s}_1) - 1
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\\
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&= 2\text{c}_1 o_1 - 1
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\\
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o_2 \cdot (o_1)^n &= 2\text{c}_1 o_1 \cdot (o_1)^n - (o_1)^n
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\\
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o_{n+2} &= 2\text{c}_1 o_{n+1} - o_n
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\end{align*}
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$$
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This matches the earlier recurrence relation with the complex exponential and therefore
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the recurrence relation of the Chebyshev polynomials.
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It also means that the the rational functions obey the same relationship as sine and cosine:
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$$
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\begin{matrix}
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\begin{gather*}
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\text{c}_n = T_n(\text{c}_1)
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\\
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\text{s}_n = U_{n-1}(\text{c}_1) \text{s}_1
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\end{gather*}
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& \text{where }
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\text{c}_1 = {1 - t^2 \over 1 + t^2}, &
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\text{s}_1 = {2t \over 1 + t^2}
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\end{matrix}
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$$
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Thus, the Chebyshev polynomials are tied to (coordinates on) circles,
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rather than explicitly to the trigonometric functions.
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It is a bit strange that these polynomials are in terms of rational functions, but no stranger
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than them being in terms of *ir*rational functions like sine and cosine.
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Calculus
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--------
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Since these functions behave similarly to sine and cosine, one might wonder about
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the nature of these expressions in the context of calculus.
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For comparison, the complex exponential (as it is a parallel construction) has a simple derivative[^1].
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Since the exponential function is its own derivative, the expression acquires
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an imaginary coefficient through the chain rule.
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[^1]: This is forgoing the fact that complex derivatives require more care than their real counterparts.
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It matters slightly less in this case since this function is complex-valued, but has a real parameter.
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$$
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\begin{align*}
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e^{it} &= \cos(t) + i\sin(t)
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\\
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{d \over dt} e^{it}
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&= {d \over dt} \cos(t) + {d \over dt} i\sin(t)
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\\
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i e^{it} &= -\sin(t) + i\cos(t)
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\\
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i[\cos(t) + i\sin(t)]
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&\stackrel{\checkmark}{=} -\sin(t) + i\cos(t)
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\end{align*}
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$$
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Meanwhile, the complex stereograph has derivative
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$$
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\begin{align*}
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{d \over dt} o_1(t) &= {d \over dt} {1 + it \over 1 - it}
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= {i(1 - it) + i(1 + it) \over (1 - it)^2}
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\\
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&= {2i \over (1 - it)^2}
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= {2i(1 + it)^2 \over (1 + t^2)^2}
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= {2i(1 - t^2 + 2it) \over (1 + t^2)^2}
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\\
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&= {-4t \over (1 + t^2)^2} + i {2(1 - t^2) \over (1 + t^2)^2}
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\\
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&= {-2 \over 1 + t^2}s_1 + i {2 \over 1 + t^2}c_1
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\\
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&= -(1 + c_1)s_1 + i(1 + c_1)c_1
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\\
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&= i(1 + c_1)o_1
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\end{align*}
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$$
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Just like the complex exponential, an imaginary coefficient falls out.
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However, the expression also accrues a $1 + c_1$ term, almost like an adjustment factor
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for its failure to be the complex exponential.
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Sine and cosine obey a simpler relationship with respect to the derivative,
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and thus need no adjustment.
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### Complex Analysis
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Since $o_n$ is a curve which loops around the unit circle *n* times, that possibly suits it
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to showing a simple result from complex analysis.
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Integrating along a contour which wraps around a sufficiently nice function's pole
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(i.e., where its magnitude grows without bound) yields a familiar value.
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This is easiest to see with $f(z) = 1 / z$:
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$$
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\oint_\Gamma {1 \over z} dz
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= \int_a^b {\gamma'(t) \over \gamma(t)} dt
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= 2\pi i
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$$
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In this example, Γ is a counterclockwise curve parametrized by γ which loops once around
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the pole at *z* = 0.
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More loops will scale this by a factor according to the number of loops.
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Normally this equality is demonstrated with the complex exponential, but will $o_1$ work just as well?
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If Γ is the unit circle, the integral is:
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$$
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\oint_\Gamma {1 \over z} dz
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= \int_{-\infty}^\infty {o_1'(t) \over o_1(t)} dt
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= \int_{-\infty}^\infty i(1 + c_1(t)) dt
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= 2i\int_{-\infty}^\infty {1 \over 1 + t^2} dt
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$$
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If one has studied their integral identities, the indefinite version of the final integral
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will be obvious as $\arctan(t)$, which has horizontal asymptotes of $\pi / 2$ and $-\pi / 2$.
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Therefore, the value of the integral is indeed $2\pi i$.
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If there are *n* loops, then naturally there are *n* of these $2\pi i$s.
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Since powers of *o* are more loops around the circle, the chain and power rules show:
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$$
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\begin{gather*}
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{d \over dt} (o_1)^n = n(o_1)^{n-1} {d \over dt} o_1
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\\[14pt]
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\oint_\Gamma {1 \over z} dz
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= \int_{-\infty}^\infty {n o_1(t)^{n-1} o_1'(t) \over o_1(t)^n} dt
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= n \int_{-\infty}^\infty {o_1'(t) \over o_1(t)} dt
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= 2 \pi i n
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\end{gather*}
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$$
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It is certainly possible to perform these contour integrals along straight lines;
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in fact, integrating along lines from 1 to *i* to -1 to -*i* deals with a
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similar integral involving arctangent.
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However, the best one can do to construct more loops with lines is to count each line
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multiple times, which isn't extraordinarily convincing.
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Perhaps the use of $\infty$ in the integral bounds is also unconvincing.
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The integral can be shifted back into the realm of plausibility by considering simpler bounds on $o_2$:
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$$
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\begin{align*}
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\oint_\Gamma {1 \over z} dz
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&= \int_{-1}^1 {2 o_1(t) o_1'(t) \over o_1(t)^2} dt
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\\
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&= 2 \int_{-1}^1 {o_1'(t) \over o_1(t)} dt
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\\
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&= 2(2i\arctan(1) - 2i\arctan(-1))
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\\
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&= 2\pi i
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\end{align*}
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$$
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This has an additional benefit: using the series form of $1 / (1 + t^2)$ and integrating,
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one obtains the series form of the arctangent.
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This series converges for $-1 \le t \le 1$, which happens to match the bounds of integration.
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The convergence of this series is fairly important, since it is tied to formulas for π,
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in particular [Leibniz's formula](https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80).
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Were one to integrate with the complex exponential, we would instead use the bounds $(0, 2\pi)$,
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since at this point a full loop has been made.
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But think to yourself -- how do you know the period of the complex exponential?
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How do you know that 2π radians is equivalent to 0 radians?
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The result using stereography relies on neither of these prior results and is directly pinned
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to a formula for π instead an apparent detour through the number *e*.
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Polar Curves
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------------
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Polar coordinates are useful for expressing for which the distance from the origin is
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a function of the angle with respect to the positive *x*-axis.
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They can also be readily converted to parametric forms:
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$$
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\begin{gather*}
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r(\theta) &\Longleftrightarrow&
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\begin{matrix}
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x(\theta) = r \cos(\theta) \\
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y(\theta) = r \sin(\theta)
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\end{matrix}
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\end{gather*}
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$$
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Polar curves frequently loop in on themselves, and so it is necessary to choose appropriate bounds
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for θ (usually as multiples of π) when plotting.
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Evidently, this is due to the use of sine and cosine in the above parametrization.
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Fortunately, $s_n$ and $c_n$ (as shown by the calculus above) have much simpler bounds.
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So what happens when one substitutes the rational functions in place of the trig ones?
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### Polar Roses
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[Polar roses](https://en.wikipedia.org/wiki/Rose_(mathematics)) are beautiful shapes which have
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a simple form when expressed in polar coordinates.
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$$
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r(\theta) = \cos \left( {p \over q} \cdot \theta \right)
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$$
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The ratio $p/q$ in least terms uniquely determines the shape of the curve.
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If you weren't reading this post, you might assume this curve is transcendental since it uses cosine,
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but you probably know better at this point.
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The Chebyshev examples above demonstrate the resemblance between $c_n$ and $\cos(n\theta)$.
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The subscript of $c$ is easiest to work with as an integer, so let $q = 1$.
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$$
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x(t) = c_p(t) c_1(t) \qquad y(t) = c_p(t) s_1(t)
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$$
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will plot a $p/1$ polar rose as t ranges over $(-\infty, \infty)$.
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```{python}
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#| echo: false
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#| output: false
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@dataclass
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class Rose:
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x: sympy.Expr
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y: sympy.Expr
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title: str | None = None
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@classmethod
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def from_rational(cls, p: int, q: int):
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return cls(
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cs[p]*cs[q],
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cs[p]*ss[q],
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title=f"$x = c_{{{p}}}c_{{{q}}}; y = c_{{{p}}}s_{{{q}}}$",
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)
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def animate_roses(filename: str, arguments: list[Rose], interval=500):
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with SympyAnimationWrapper(filename) as animate:
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@animate(list(range(len(arguments))), interval=interval)
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def ret(fr):
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plt.clf()
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argument = arguments[fr]
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sympy.plot_parametric(
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argument.x,
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argument.y,
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xlim=(-2,2),
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ylim=(-2,2),
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title=argument.title,
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backend="matplotlib",
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)
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ret.save() # type: ignore
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animate_roses(
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"polar_roses_1.mp4",
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[ Rose.from_rational(i, 1) for i in range(1, 11) ],
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)
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```
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::: {#fig-polar-roses-1}
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{{< video "./polar_roses_1.mp4" >}}
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p/1 polar roses as rational curves.
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Since *t* never reaches infinity, a bite appears to be taken out of the graphs near (-1, 0)."
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:::
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$q = 1$ happens to match the subscript *c* term of *x* and *s* term of *y*, so one might wonder
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whether the other polar curves can be obtained by allowing it to vary as well.
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And you'd be right.
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$$
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x(t) = c_p(t) c_q(t) \qquad y(t) = c_p(t) s_q(t)
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$$
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will plot a $p/q$ polar rose as t ranges over $(-\infty, \infty)$.
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```{python}
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#| echo: false
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#| output: false
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animate_roses(
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"polar_roses_2.mp4",
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[
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Rose.from_rational(i,j)
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for i,j in [(1,1),(1,2),(2,1),(3,1),(3,2),(2,3),(1,3),(1,4),(3,4),(4,3),(4,1)]
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],
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)
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||
```
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::: {#fig-polar-roses-2}
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{{< video "./polar_roses_2.mp4" >}}
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p/q polar roses as rational curves
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:::
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Just as with the prior calculus examples, doubling all subscripts of *c* and *s* will
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only require *t* to range over $(-1, 1)$, which removes the ugly bite mark.
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||
Perhaps it is also slightly less satisfying, since the fraction $p/q$ directly appears in the
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typical polar incarnation with cosine.
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On the other hand, it exposes an important property of these curves: they are all rational.
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This approach lends additional precision to a prospective pseudo-polar coordinate system.
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In the next few examples, I will be using the following notation for compactness:
|
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|
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$$
|
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\begin{gather*}
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R_n(t) = f(t) &\Longleftrightarrow&
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\begin{matrix}
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x(t) = f(t) c_n(t) \\
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y(t) = f(t) s_n(t)
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\end{matrix}
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\end{gather*}
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$$
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### Conic Sections
|
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The polar equation for a conic section (with a particular unit length occurring somewhere)
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in terms of its eccentricity $\varepsilon$ is:
|
||
|
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$$
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r(\theta) = {1 \over 1 - \varepsilon \cos(\theta)}
|
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$$
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Correspondingly, the rational polar form can be expressed as
|
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$$
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R_1(t) = {1 \over 1 - \varepsilon c_1}
|
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$$
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||
Since polynomial arithmetic is easier to work with than trigonometric identities,
|
||
it is a matter of pencil-and-paper algebra to recover the implicit form from a parametric one.
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#### Parabola ($|\varepsilon| = 1$)
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The conic section with the simplest implicit equation is the parabola.
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Since $c_n$ is a simple ratio of polynomials in *t*, it is much simpler to recover the implicit equation.
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For $\varepsilon = 1$,
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|
||
:::: {layout-ncol="2"}
|
||
```{python}
|
||
#| echo: false
|
||
#| fig-cap: $x = {c_1 \over 1 - c_1} \quad y = {s_1 \over 1 - c_1}$
|
||
|
||
sympy.plot_parametric(
|
||
1 / (1 - sympy.re(o)) * sympy.re(o),
|
||
1 / (1 - sympy.re(o)) * sympy.im(o),
|
||
xlim=(-5,5),
|
||
ylim=(-5,5),
|
||
)
|
||
```
|
||
|
||
::: {}
|
||
$$
|
||
\begin{align*}
|
||
1 - c_1 &= 1 - {1 - t^2 \over 1 + t^2}
|
||
= {2 t^2 \over 1 + t^2}
|
||
\\
|
||
y &= {s_1 \over 1 - c_1}
|
||
= {2t \over 1 + t^2} {1 + t^2 \over 2 t^2}
|
||
= {1 \over t}
|
||
\\
|
||
x &= {c_1 \over 1 - c_1}
|
||
= {1 - t^2 \over 1 + t^2} \cdot {1 + t^2 \over 2 t^2}
|
||
= {1 - t^2 \over 2t^2}
|
||
\\
|
||
&= {1 \over 2t^2} - {1 \over 2}
|
||
= {y^2 \over 2} - {1 \over 2}
|
||
\end{align*}
|
||
$$
|
||
:::
|
||
::::
|
||
|
||
*x* is a quadratic polynomial in *y*, so trivially the figure formed is a parabola.
|
||
Technically it is missing the point where $y = 0 ~ (t = \infty)$, and this is not a circumstance
|
||
where using a higher $c_n$ would help.
|
||
It is however, similar to the situation where we allow $o_1(\infty) = -1$, and an argument
|
||
can be made to waive away any concerns one might have.
|
||
|
||
|
||
#### Ellipse ($|\varepsilon| < 1$)
|
||
|
||
Ellipses are next.
|
||
The simplest fraction between zero and one is 1/2, so for $\varepsilon = 1/2$,
|
||
|
||
:::: {layout-ncol = "2"}
|
||
```{python}
|
||
#| echo: false
|
||
#| fig-cap: $x = {c_1 \over 1 - c_1 / 2} \quad y = {s_1 \over 1 - c_1 / 2}$
|
||
|
||
sympy.plot_parametric(
|
||
1 / (1 - (1/2)*sympy.re(o)) * sympy.re(o),
|
||
1 / (1 - (1/2)*sympy.re(o)) * sympy.im(o),
|
||
xlim=(-5,5),
|
||
ylim=(-5,5),
|
||
)
|
||
```
|
||
|
||
::: {}
|
||
$$
|
||
\begin{align*}
|
||
1 - {1 \over 2}c_1 &= 1 - {1 \over 2} \cdot {1 - t^2 \over 1 + t^2}
|
||
= {3 t^2 + 1 \over 2 + 2t^2}
|
||
\\
|
||
y &= {s_1 \over 1 - {1 \over 2}c_1}
|
||
= {4t \over 3t^2 + 1}
|
||
\\
|
||
x &= {c_1 \over 1 - {1 \over 2}c_1}
|
||
= {2 - 2t^2 \over 3t^2 + 1}
|
||
\end{align*}
|
||
$$
|
||
:::
|
||
::::
|
||
|
||
There isn't an obvious way to combine products of *x* and *y* into a single equation.
|
||
The general form of a conic section is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, so
|
||
we know that the implicit equation for the curve almost certainly involves $x^2$ and $y^2$.
|
||
|
||
$$
|
||
x^2 = {4 - 8t^2 + 4t^4 \over (3t^2 + 1)^2} \qquad
|
||
y^2 = {16t^2 \over (3t^2 + 1)^2}
|
||
$$
|
||
|
||
Squaring produces some $t^4$ terms which cannot exist outside of these terms and *xy*.
|
||
A linear combination of $x^2$ and $y^2$ never includes any cubic terms in the numerator
|
||
which would appear in *xy*, so $B = 0$.
|
||
Since all remaining terms are linear in *x* and *y*, their denominator must appear as a factor
|
||
in the numerator of $Ax^2 + Cy^2$, whatever *A* and *C* are.
|
||
|
||
Since the coefficient of $t^4$ in $x^2$ is 4, *A* must be multiple of 3.
|
||
Through trial and error, $A = 3, C = 4$ gives:
|
||
|
||
$$
|
||
\begin{align*}
|
||
3x^2 + 4y^2
|
||
&= {12 - 24t^2 + 12t^4 + 64t^2 \over (3t^2 + 1)^2}
|
||
\\
|
||
&= {12 - 40t^2 + 12t^4 \over (3t^2 + 1)^2}
|
||
\\
|
||
&= {(4t^2 + 12) (3t^2 + 1) \over (3t^2 + 1)^2}
|
||
\\
|
||
&= {4t^2 + 12 \over 3t^2 + 1}
|
||
\end{align*}
|
||
$$
|
||
|
||
Since the numerator of *y* has a *t*, this is clearly some combination of *x* and a constant.
|
||
By the previous line of thought, the constant term must be a multiple of 4, and picking the smallest
|
||
option finally results in the implicit form:
|
||
|
||
$$
|
||
\begin{align*}
|
||
4 &= {4(3t^2 + 1) \over 3t^2 + 1}
|
||
= {12t^2 + 4 \over 3t^2 + 1}
|
||
\\
|
||
{4t^2 + 12 \over 3t^2 + 1} - 4
|
||
&= {8 - 8t^2 \over 3t^2 + 1} = 4x
|
||
\\[14pt]
|
||
3x^2 + 4y^2 &= 4x + 4
|
||
\\
|
||
3x^2 + 4y^2 - 4x - 4 &= 0
|
||
\end{align*}
|
||
$$
|
||
|
||
Notably, the coefficients of *x* and *y* are 3 and 4.
|
||
Simultaneously, $o_1(\varepsilon) = o_1(1/2) = {3 \over 5} + i{4 \over 5}$.
|
||
This binds together three concepts: the simplest case of the Pythagorean theorem,
|
||
the 3-4-5 right triangle; the coefficients of the implicit form; and the role of eccentricity
|
||
with respect to stereography.
|
||
|
||
|
||
#### Hyperbola ($|\varepsilon| > 1$)
|
||
|
||
As evidenced by the bound on the eccentricity above, hyperbolae are in some way the inverses of ellipses.
|
||
Since $o_1(2)$ is a reflection of $o_1(1/2)$, you might think the implicit equation for
|
||
$\varepsilon = 2$ to be the same, but with a flipped sign or two.
|
||
Unfortunately, you'd be wrong.
|
||
|
||
:::: {layout-ncol="2"}
|
||
```{python}
|
||
#| echo: false
|
||
#| fig-cap: $x = {c_1 \over 1 - 2c_1} \quad y = {s_1 \over 1 - 2c_1}$
|
||
|
||
# Ignore discontinuity at sqrt(1/3)
|
||
disc = (1/3)**(1/2)
|
||
sympy.plot_parametric(
|
||
(
|
||
1 / (1 - 2*sympy.re(o)) * sympy.re(o),
|
||
1 / (1 - 2*sympy.re(o)) * sympy.im(o),
|
||
(t, -disc + 1e-5, disc - 1e-5)
|
||
),
|
||
(
|
||
1 / (1 - 2*sympy.re(o)) * sympy.re(o),
|
||
1 / (1 - 2*sympy.re(o)) * sympy.im(o),
|
||
(t, -5, -disc - 1e-5)
|
||
),
|
||
(
|
||
1 / (1 - 2*sympy.re(o)) * sympy.re(o),
|
||
1 / (1 - 2*sympy.re(o)) * sympy.im(o),
|
||
(t, disc + 1e-5, 5)
|
||
),
|
||
xlim=(-5,5),
|
||
ylim=(-5,5),
|
||
line_color = "blue",
|
||
)
|
||
```
|
||
|
||
::: {}
|
||
$$
|
||
\begin{gather*}
|
||
\begin{align*}
|
||
x &= {c_1 \over 1 - 2c_1} = {1 - t^2 \over 3t^2 - 1}
|
||
\\
|
||
y &= {s_1 \over 1 - 2c_1} = {2t \over 3t^2 - 1}
|
||
\\[14pt]
|
||
3x^2 - y^2 &= {3 - 6t^2 + 3t^4 - 4t^2 \over 3t^2 - 1}
|
||
\\
|
||
&= {(t^2 - 3)(3t^2 - 1) \over (3t^2 - 1)^2 }
|
||
\\
|
||
&= {t^2 - 3 \over 3t^2 - 1 }
|
||
= ... = -4x - 1
|
||
\end{align*}
|
||
\\[14pt]
|
||
3x^2 - y^2 + 4x + 1 = 0
|
||
\end{gather*}
|
||
$$
|
||
:::
|
||
::::
|
||
|
||
At the very least, the occurrences of 1 in the place of 4 have a simple explanation: 1 = 4 - 3.
|
||
|
||
|
||
### Archimedean Spiral
|
||
|
||
Arguably the simplest (non-circular) polar curve is $r(\theta) = \theta$, the unit
|
||
[Archimedean spiral](https://en.wikipedia.org/wiki/Archimedean_spiral).
|
||
Since the curve is defined by a constant turning, this is a natural application of the properties
|
||
of sine and cosine.
|
||
The closest equivalent in rational polar coordinates is $R_1(t) = t$.
|
||
But this can be converted to an implicit form:
|
||
|
||
$$
|
||
\begin{gather*}
|
||
x = tc_1 \qquad y = ts_1
|
||
\\[14pt]
|
||
x^2 + y^2 = t^2(c_1^2 + s_1^2) = t^2
|
||
\\
|
||
y = {2t^2 \over 1 + t^2} = {2(x^2 + y^2) \over 1 + (x^2 + y^2)}
|
||
\\[14pt]
|
||
(1 + x^2 + y^2)y = 2(x^2 + y^2)
|
||
\end{gather*}
|
||
$$
|
||
|
||
The curve produced by this equation is a
|
||
[right strophoid](https://mathworld.wolfram.com/RightStrophoid.html)
|
||
with a node at (0, 1) and asymptote $y = 2$.
|
||
This form suggests something interesting about this curve: it approximates the Archimedean spiral
|
||
(specifically the one with polar equation $r(\theta) = \theta/2$).
|
||
Indeed, the sequence of curves with parametrization $R_n(t) = 2nt$ approximate the (unit) spiral
|
||
for larger *n*, as can be seen in the following video.
|
||
|
||
```{python}
|
||
#| echo: false
|
||
#| output: false
|
||
|
||
with SympyAnimationWrapper("approximate_archimedes.mp4") as animate:
|
||
@animate(list(range(10)), interval=500)
|
||
def ret(fr):
|
||
plt.clf()
|
||
|
||
p = sympy.plot_parametric(
|
||
t*sympy.cos(t),
|
||
t*sympy.sin(t),
|
||
xlim=(-4,4),
|
||
ylim=(-4,4),
|
||
label="Archimedean Spiral",
|
||
backend="matplotlib",
|
||
show=False
|
||
)
|
||
i = fr + 1
|
||
p.extend(
|
||
sympy.plot_parametric(
|
||
2*i*t*cs[i],
|
||
2*i*t*ss[i],
|
||
(t, -5, 5),
|
||
line_color="black",
|
||
label=f"$R_{{{i}}}(t) = {2*i}t$",
|
||
backend="matplotlib",
|
||
show=False
|
||
)
|
||
)
|
||
p.show()
|
||
plt.legend()
|
||
|
||
ret.save() # type: ignore
|
||
```
|
||
|
||
::: {#fig-approx-archimedes}
|
||
{{< video ./approximate_archimedes.mp4 >}}
|
||
|
||
Approximations to the Archimedean spiral
|
||
:::
|
||
|
||
|
||
Since R necessarily defines a rational curve, the curves will never be equal,
|
||
just as any stretching of $c_n$ will never exactly become cosine.
|
||
|
||
|
||
Closing
|
||
-------
|
||
|
||
Sine, cosine, and the exponential function, are useful in a calculus setting precisely
|
||
because of their constant "velocity" around the circle.
|
||
Also, nearly every modern scientific calculator in the world features buttons
|
||
for trigonometric functions, so there seems to be no reason *not* to use them.
|
||
|
||
We can however be misled by their apparent omnipresence.
|
||
Stereographic projection has been around for *millennia*, and not every formula needs to be rewritten
|
||
in its language.
|
||
For example (and as previously mentioned), defining the Chebyshev polynomials really only requires
|
||
understanding the multiplication of two complex numbers whose norm cannot grow,
|
||
not trigonometry and dividing angles.
|
||
Many other instances of sine and cosine merely rely on a number (or ratio) of loops around a circle.
|
||
When velocity does not factor, it will obviously do to "stay rational".
|
||
|
||
One of my favorite things to plot as a kid were polar roses, so I was somewhat intrigued
|
||
to see that they are, in fact, rational curves.
|
||
On the other hand, their rationality follows immediately from the rationality of the circle
|
||
(which itself follows from the existence of Pythagorean triples).
|
||
If I were more experienced with manipulating Chebyshev polynomials or willing to set up a
|
||
linear system in (way too) many terms, I might have considered attempting to find
|
||
an implicit form for them as well.
|
||
|
||
Diagrams created with Sympy and Matplotlib.
|