708 lines
25 KiB
Plaintext
708 lines
25 KiB
Plaintext
---
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title: "Algebraic Rotations and 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: "2021-09-26"
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date-modified: "2025-06-28"
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categories:
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- geometry
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- symmetry
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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 IPython.display import display, Code
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import sympy
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from sympy.abc import t
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```
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Expressing rotation mathematically can be rather challenging, even in two dimensional space.
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Typically, the subject is approached from complicated-looking rotation matrices,
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with the occasional accompanying comment about complex numbers.
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The challenge only increases when examining three dimensional space,
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where everyone has different ideas about the best implementations.
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In what follows, I attempt to provide a derivation of 2D and 3D rotations based in intuition
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and without the use of trigonometry.
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Complex Numbers: 2D Rotations
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-----------------------------
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As points in the plane, multiplication between complex numbers is frequently analyzed
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as a combination of scaling (a ratio of two scales) and rotation (a point on the complex unit circle).
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However, the machinery used in these analyses usually expresses complex numbers in a polar form
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involving the complex exponential.
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This leverages prior knowledge of trigonometry and some calculus, and in fact obfuscates
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the algebraic properties of complex numbers.
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Neither concept is truly necessary to understand points on the complex unit circle,
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or complex number multiplication in general.
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### A review of complex multiplication
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A complex number $z = a + bi$ multiplies with another complex number
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$w = c + di$ in the obvious way: by distributing over addition.
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$$
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zw = (a + bi)(c + di) = ac + bci + adi + bdi^2 = (ac - bd) + i(ad + bc)
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$$
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*z* also has associated to it a conjugate $z^* = a - bi$.
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The product $zz^*$ is the *norm* $a^2 + b^2$, a positive real quantity whose square root
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is typically called the *magnitude* of the complex number.
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Taking the square root is frequently unnecessary, as many of the properties of
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this quantity do not depend on this normalization.
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One such property is that for two complex numbers *z* and *w*,
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the norm of the product is the product of the norm.
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$$
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\begin{align*}
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(zw)(zw)^* &= (ac - bd)^2 + (ad + bc)^2
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\\
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&= a^2c^2 - 2abcd + b^2d^2 + a^2d^2 + 2abcd + b^2c^2
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\\
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&= a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2
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= a^2(c^2 + d^2) + b^2(c^2 + d^2)
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\\
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&= (a^2 + b^2)(c^2 + d^2)
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\end{align*}
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$$
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Conjugation also distributes over multiplication, and complex numbers possess multiplicative inverses.
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Also, the norm of the multiplicative inverse is the inverse of the norm.
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$$
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\begin{gather*}
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z^*w^* = (a - bi)(c - di) = ac - bci - adi + bdi^2
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\\
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= (ac - bd) - i(ad + bc) = (zw)^{*}
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\\
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{1 \over z} = {z^{*} \over z z^{*}} = {a - bi \over a^2 + b^2}
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\\[14pt]
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\left({1 \over z}\right)
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\left({1 \over z}\right)^{*}
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= {1 \over zz^{*}} = {1 \over a^2 + b^2}
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\end{gather*}
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$$
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A final interesting note about complex multiplication is the product $z^{*} w$
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$$
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z^{*} w = (a - bi)(c + di) = ac - bci + adi - bdi^2 = (ac + bd) + i(ad - bc)
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$$
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The real part is the same as the dot product $(a, c) \cdot (b, d)$ and the imaginary part
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looks like the determinant of $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$.
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This shows the inherent value of complex arithmetic, as both of these quantities
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have important interpretations in vector algebra.
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### Onto the Circle
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If $zz^{*} = a^2 + b^2 = 1$, then *z* lies on the complex unit circle.
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Since the norm is 1, a general complex number *w* has the same norm as its product with *z*, *zw*.
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Specifically, if *w* is 1, multiplication by *z* can be seen as the rotation that maps 1 to *z*.
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But how do you describe a point on the unit circle?
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Simple.
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For any complex *w*, there are three other complex numbers that are guaranteed
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to share its norm: $-w, w^{*}$, and $-w^{*}$.
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Dividing any one of these numbers by another results in a complex number on the unit circle.
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Dividing *w* by -*w* is boring, since that just results in -1.
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However, division by $w^*$ turns out to be important:
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$$
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\begin{align*}
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z &= {w \over w^*} =
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\left( {c + di \over c - di} \right)
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\left( {c + di \over c + di} \right) = 1
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\\[8pt]
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&= {(c^2 - d^2) + (2cd)i \over c^2 + d^2}
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\end{align*}
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$$
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Dividing the numerator and denominator of this expression through by $c^2$ yields an expression
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in $t = d/c$, where t can be interpreted as the slope of the line joining 0 and *w*.
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Slopes range from $-\infty$ to $\infty$, which means that the variable *t* does as well.
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$$
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\begin{align*}
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{(c^2 - d^2) + (2cd)i \over c^2 + d^2}
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&= {1/c^2 \over 1/c^2} \cdot {(c^2 - d^2) + (2cd)i \over c^2 + d^2}
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\\
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&= {(1 - d^2/c^2) + (2d/c)i \over 1 + d^2/c^2}
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= {(1 - t^2) + (2t)i \over 1 + t^2}
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\\
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&= {1 - t^2 \over 1 + t^2} + i{2t \over 1 + t^2} = z(t)
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\end{align*}
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$$
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This gives us a point *z* on the unit circle.
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Its reciprocal is:
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$$
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{1 \over z} = {z^{*} \over zz^{*}} = z^{*}
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$$
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Applying the same operation to *z* as we did to *w* gives us $z / z^{*} = z^2$,
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or as an action on points, the rotation from 1 to $z^2$.
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This demonstrates the decomposition of the rotation into two parts.
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For a general complex number *w* which may not lie on the unit circle, multiplication by
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this number dilates the complex plane as well rotating 1 toward the direction of *w*.
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Then, division by $w^{*}$ undoes the dilation and performs the rotation again.
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General (i.e., non-unit) rotations *must* occur in pairs.
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### Two Halves, Twice Over
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Plugging in -1, 0, and 1 for *t* reveals some interesting behavior:
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$z(0) = 1 + 0i$, $z(1) = 0 + 1i$, and $z(-1) = 0 - 1i$.
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Assuming continuity, this shows that $t \in (-1, 1)$ plots the semicircle in the right half-plane.
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The other half of the circle comes from *t* with magnitude greater than 1,
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and the point where $z(t) = -1$ exists if we admit the extra point $t = \pm \infty$.
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Is it possible to get the other half into a nicer interval?
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No problem, just double the rotation.
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$$
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\begin{align*}
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z(t)^2 &= (\Re[z]^2 - \Im[z]^2) + i(2\Re[z]\Im[z])
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\\[8pt]
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&= {(1 - t^2)^2 - (2t)^2 \over (1 + t^2)^2} + i{2(1 - t^2)(2t) \over (1 + t^2)^2}
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\\[8pt]
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&= {1 - 6t^2 + t^4 \over 1 + 2t^2 + t^4}
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+ i {4t - 4t^3 \over 1 + 2t^2 + t^4}
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\end{align*}
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$$
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Now $z(-1)^2 = z(1)^2 = -1$ and the entire circle fits in the interval \[-1, 1\].
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This action of squaring *z* means that as *t* ranges from $-\infty$ to $\infty$, the point
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$z^2$ loops around the circle twice, since the unit rotation *z* is applied twice.
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### Plotting Transcendence
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Let's go on a brief tangent and compare these expressions to their
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transcendental trigonometric counterparts.
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Graphing the real and imaginary parts separately shows how much they resemble
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$\cos(\pi t)$ and $\sin(\pi t)$ around zero[^1].
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[^1]: An approximation of a function as the ratio of two polynomials is called a
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[Padé approximant](https://en.wikipedia.org/wiki/Pad%C3%A9_approximant).
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Specifically, the real part (which approximates cosine) has order \[4/4\] and the imaginary part
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(which approximates sine) has order \[3/4\].
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```{python}
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#| echo: false
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#| layout-ncol: 2
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z = (1 + sympy.I * t) / (1 - sympy.I * t)
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sympy.plot(
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sympy.re(z**2),
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sympy.im(z**2),
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(t, -2, 2),
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ylim=(-2, 2),
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label=[
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r"$\Re \left [ \left ( \frac{1 + it}{1 - it} \right )^2 \right ]$",
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r"$\Im \left [ \left ( \frac{1 + it}{1 - it} \right )^2 \right ]$",
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],
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legend=True,
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xlabel=False,
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ylabel=False,
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)
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sympy.plot(
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sympy.re(z**2) - sympy.cos(sympy.pi * t),
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sympy.im(z**2) - sympy.sin(sympy.pi * t),
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(t, -2, 2),
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ylim=(-2, 2),
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label=[
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r"Cosine error",
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r"Sine error",
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],
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legend=True,
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xlabel=False,
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ylabel=False,
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)
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```
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Ideally, the zeroes of the real part should occur at $t = \pm 1/2$, since $\cos(\pm \pi/2) = 0$.
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Instead, they occur at
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$$
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\begin{align*}
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0 &= 1 - 6t^2 + t^4
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= (t^2 - 2t - 1)(t^2 + 2t - 1)
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\\
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&\implies t = \pm {1 \over \delta_s} = \pm(\sqrt 2 - 1) \approx \pm 0.414
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\end{align*}
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$$
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With a bit of polynomial interpolation, this can be rectified.
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A cubic interpolation between the expected and actual values is given by:
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$$
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\begin{gather*}
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P(x) = ax^3 + bx^2 + cx + d
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\\
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\begin{pmatrix}
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1 & 0 & 0 & 0 \\
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1 & 1/2 & 1/4 & 1/8 \\
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1 & -1/2 & 1/4 & -1/8 \\
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1 & 1 & 1 & 1
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\end{pmatrix} \begin{pmatrix}
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d \\
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c \\
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b \\
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a
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\end{pmatrix} = \begin{pmatrix}
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0 \\
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\sqrt 2 - 1 \\
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-\sqrt 2 + 1 \\
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1
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\end{pmatrix}
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\\
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\implies \begin{pmatrix}
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d \\
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c \\
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b \\
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a
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\end{pmatrix} = \begin{pmatrix}
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0 \\
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-3 + 8\sqrt 2/3 \\
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0 \\
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4 - 8\sqrt 2/3
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\end{pmatrix}
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\implies P(x) \approx 0.229 x^3 + 0.771x
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\end{gather*}
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$$
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::: {}
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```{python}
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#| echo: false
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#| layout-ncol: 2
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interpolate = (-3 + 8 * sympy.sqrt(2)/3) * t + (4 - 8 * sympy.sqrt(2)/3) * t**3
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sympy.plot(
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sympy.re(z**2).subs(t, interpolate),
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sympy.im(z**2).subs(t, interpolate),
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(t, -2, 2),
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ylim=(-2, 2),
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label=[
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r"$\Re \left [ z(P(t))^2 \right ]$",
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r"$\Im \left [ z(P(t))^2 \right ]$",
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],
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legend=True,
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xlabel=False,
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ylabel=False,
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)
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sympy.plot(
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sympy.re(z**2).subs(t, interpolate) - sympy.cos(sympy.pi * t),
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sympy.im(z**2).subs(t, interpolate) - sympy.sin(sympy.pi * t),
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(t, -2, 2),
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ylim=(-2, 2),
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label=[
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r"Cosine error",
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r"Sine error",
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],
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legend=True,
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xlabel=False,
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ylabel=False,
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)
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```
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The error over the interval \[-1, 1\] in these approximations is around 1% (RMS).
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:::
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Notably, this interpolating polynomial is entirely odd, which gives it some symmetry about 0.
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Along with being a very good approximation, it has the feature that a rotation by an
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*n*^th^ of a turn is about $z(P(1/n))^2$.
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The approximation be improved further by taking *z* to higher powers and deriving a higher-order
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interpolating polynomial.
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It is impossible to shrink this error to 0 because the derivative of the imaginary part at
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$t = 1$ would be π.
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But the imaginary part is a rational expression, so its derivative is also a rational expression.
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Since π is transcendental, there must always be some error[^2].
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[^2]: Even with the interpolating polynomial, the quantity is a ratio of two algebraic numbers,
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which is also algebraic and not transcendental.
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Even though it is arguably easier to calculate, there probably isn't a definite benefit to this approximation.
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For example,
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- Other accurate approximants for sine and cosine can be calculated directly from their power series.
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- There are no inverse functions like `acos` or `asin` to complement these approximations.
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- This isn't that bad, since I have refrained from describing rotations with an angle
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(i.e., the quantity returned by these functions).
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Even so, the inverse functions have their use cases.
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- Trigonometric functions can be hardware-accelerated (at least in some FPUs).
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- If no such hardware exists, approximations of `sin` and `cos` can be calculated by software beforehand
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and stored in a lookup table (which is also probably involved at some stage of the FPU).
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Elaborating on the latter two points, using a lookup table means evaluation happens in roughly constant time.
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A best-case analysis of the above approximation, given some value *t* is...
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| Expression | Additions | Multiplications | Divisions |
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|-----------------------------------------------------------|-----------|-----------------|-----------|
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| $p = t \cdot (0.228763834 \cdot t \cdot t + 0.771236166)$ | 1 | 3 | |
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| $q = p \cdot p$ | | 1 | |
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| $r = 1 + q$ | 1 | | |
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| $c = { 1 - q \over r}$ | 1 | | 1 |
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| $s = { p + p \over r}$ | 1 | | 1 |
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| real = $c \cdot c - s \cdot s$ | 1 | 2 | |
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| imag = $c \cdot s + c \cdot s$ | 1 | 1 | |
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| Total | 6 | 7 | 2 |
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...or 15 FLOPs.
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On a more optimistic note, a Monte Carlo test of the above approximation on my computer yields
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promising results when compared with GCC's `libm` implementations of `sin`, `cos`, and `cexp`.
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```{python}
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#| echo: false
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for file in [
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"approx-results/results_math.txt",
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"approx-results/results_complex.txt",
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]:
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with open(file) as a:
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display(Code(a.read()))
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```
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For the source which I used to generate this output, see the repository linked at the bottom
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of this article.
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Even though results can be inaccurate, this exercise in the algebraic manipulation
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of complex numbers is fairly interesting since it requires no calculus to define,
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unlike sine, cosine, and their Padé approximants (and to a degree, π).
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From Circles to Spheres
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-----------------------
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The expression for complex points on the unit circle coincides with the stereographic projection of a circle.
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This method is achieved by selecting a point on the circle, fixing *t* along a line
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(such as the *y*-axis), and letting *t* range over all possible values.
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The equations of the line and circle appear in the diagram above.
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Through much algebra, expressions for *x* and *y* as functions of *t* can be formed.
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<!-- TODO: this is gross -->
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$$
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\begin{align*}
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y &= t(x + 1)
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\\
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1 &= x^2 + y^2 = x^2 + (tx + t)^2
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\\
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&= x^2 + t^2x^2 + 2t^2 x + t^2
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\\[14pt]
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{1 \over t^2 + 1} -\ {t^2 \over t^2 + 1}
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&= x^2 + {t^2 \over t^2 + 1} 2x
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\\
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{1 -\ t^2 \over t^2 + 1} + \left( {t^2 \over t^2 + 1} \right)^2
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&= \left(x + {t^2 \over t^2 + 1} \right)^2 + {t^2 \over t^2 + 1}
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\\
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{1 -\ t^4 \over (t^2 + 1)^2} + {t^4 \over (t^2 + 1)^2}
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&= \left(x + {t^2 \over t^2 + 1} \right)^2
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\\
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{1 \over (t^2 + 1)^2}
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&= \left(x + {t^2 \over t^2 + 1} \right)^2
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\\
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{1 \over t^2 + 1}
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&= x + {t^2 \over t^2 + 1}
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\\
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x &= {1 -\ t^2 \over 1 + t^2}
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\\
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y &= t(x + 1)
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= t\left( {1 -\ t^2 \over 1 + t^2} + {1 + t^2 \over 1 + t^2} \right)
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\\
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&= {2t \over 1 + t^2}
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\end{align*}
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$$
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These are exactly the same expressions which appear in the real and imaginary components
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of the ratio between $1 + ti$ and its conjugate.
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Compared to the algebra using complex numbers, this is quite a bit more work.
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One might ask whether, given a proper arithmetic setting, this can be extended to three dimensions.
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In other words, we want to find the projection of a sphere using some new number system
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analogously to the circle with respect to complex numbers.
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### Algebraic Projection
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The simplest thing to do is try it out.
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Let's say an extended number $h = 1 + si + tj$ has a conjugate of $h^{*} = 1 - si - tj$.
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Then, following our noses[^3]:
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[^3]: We don't know whether *i* and *j* form a
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[field](https://en.wikipedia.org/wiki/Field_%28mathematics%29)
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or [division ring](https://en.wikipedia.org/wiki/Division_ring)),
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so it's a bit of an assumption for division to be possible.
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We'll ignore that.
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$$
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\begin{gather*}
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{h \over h^{*}} = {1 + si + tj \over 1 - si - tj}
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= \left({1 + si + tj \over 1 - si - tj}\right)
|
|
\left({1 + si + tj \over 1 + si + tj}\right)
|
|
\\[8pt]
|
|
= {
|
|
1 + si + tj + si + s^2i^2 + stij + tj + stji + t^2j^2
|
|
\over 1 - si - tj + si - s^2 i^2 - stij + tj - stji - t^2j^2
|
|
}
|
|
\end{gather*}
|
|
$$
|
|
|
|
While there is some cancellation in the denominator, both products are rather messy.
|
|
To get more cancellation, we can add some nice algebraic properties between *i* and *j*.
|
|
If we let i and j be *anticommutative* (meaning that $ij = -ji$) then $stij$ cancels with $stji$.
|
|
So that the denominator is totally real (and therefore can be guaranteed to divide),
|
|
we can also assert that $i^2$ and $j^2$ are both real.
|
|
Then the expression becomes
|
|
|
|
$$
|
|
{1 + si + tj \over 1 - si - tj}
|
|
= {1 + s^2i^2 + t^2j^2 \over 1 - s^2 i^2 - t^2 j^2}
|
|
+ i{2s \over 1 - s^2 i^2 - t^2 j^2}
|
|
+ j{2t \over 1 - s^2 i^2 - t^2 j^2}
|
|
$$
|
|
|
|
If it is chosen that $i^2 = j^2 = -1$, this produces the correct equations parametrizing the unit sphere
|
|
(see [Wikipedia](https://en.wikipedia.org/wiki/Stereographic_projection#First_formulation)).
|
|
|
|
$$
|
|
{h \over h^{*}} = {1 - s^2 - t^2 \over 1 + s^2 + t^2}
|
|
+ i{2s \over 1 + s^2 + t^2}
|
|
+ j{2t \over 1 + s^2 + t^2}
|
|
$$
|
|
|
|
One can also check that this makes the squares of the real, *i*, and *j* components sum to 1.
|
|
|
|
If both *i* and *j* anticommute, then their product also anticommutes with both *i* and *j*.
|
|
|
|
$$
|
|
\textcolor{red}{ij}j = -j\textcolor{red}{ij}
|
|
~,~ i\textcolor{red}{ij} = -\textcolor{red}{ij}i
|
|
$$
|
|
|
|
Calling this product *k* and noticing that $k^2 = (ij)(ij) = -ijji = i^2 = -1$
|
|
completely characterizes the [quaternions](https://en.wikipedia.org/wiki/Quaternion).
|
|
|
|
|
|
### Other projections
|
|
|
|
Choosing different values for $i^2$ and $j^2$ yield different shapes than a sphere.
|
|
If you know a little group theory, you might know there are only two nonabelian
|
|
(noncommutative) groups of order 8:
|
|
|
|
- the [quaternion group](https://en.wikipedia.org/wiki/Quaternion_group)
|
|
- and the [
|
|
dihedral group of degree 4
|
|
](https://en.wikipedia.org/wiki/Examples_of_groups#dihedral_group_of_order_8).
|
|
|
|
In the latter group, *j* and *k* are both imaginary, but square to 1 (*i* still squares to -1)[^4].
|
|
|
|
[^4]: I'm being a bit careless with the meanings of "1" and "-1" here.
|
|
Properly, these are the group identity and another group element which commutes with all others.
|
|
|
|
Changing the sign of one (or both) of the imaginary squares in the expression $h / h^{*}$ above
|
|
switches the multiplicative structure from quaternions to the dihedral group.
|
|
In this group, picking $i^2 = -j^2 = -1$ parametrizes a hyperboloid of one sheet,
|
|
and picking $i^2 = j^2 = 1$ parametrizes a hyperboloid of two sheets.
|
|
|
|
|
|
Quaternions and Rotation
|
|
------------------------
|
|
|
|
We've already established that complex numbers are useful for describing 2D rotations.
|
|
They also elegantly describe the stereographic projection of a circle.
|
|
Consequently, since quaternions elegantly describe the stereographic projection of a sphere,
|
|
they are useful for 3D rotations.
|
|
|
|
Since the imaginary units *i*, *j*, and *k* are anticommutative, a general quaternion does not
|
|
commute with other quaternions like complex numbers do.
|
|
This embodies a difficulty with 3D rotations in general: unlike 2D rotations, they do not commute.
|
|
|
|
Only three of the four components in a quaternion are necessary to describe a point in 2D space.
|
|
The *ijk* subspace is ideal since the three imaginary axes are symmetric (i.e., they all square to -1).
|
|
Quaternions in this subspace are called *vectors*.
|
|
Another reason for using the *ijk* subspace comes from considering a point
|
|
$u = ai + bj + ck$ on the unit sphere ($a^2 + b^2 + c^2 = 1$).
|
|
Then the square of this point is:
|
|
|
|
$$
|
|
\begin{align*}
|
|
u^2 &= (ai + bj + ck)(ai + bj + ck)
|
|
\\
|
|
&= a^2i^2 + abij + acik + abji + b^2j^2 + bcjk + acki + bckj + c^2k^2
|
|
\\
|
|
&= (a^2 + b^2 + c^2)(-1) + ab(ij + ji) + ac(ik + ki) + bc(jk + kj)
|
|
\\
|
|
&= -1 + 0 + 0 + 0
|
|
\end{align*}
|
|
$$
|
|
|
|
This property means that *u* behaves similarly to a typical imaginary unit.
|
|
If we form pseudo-complex numbers of the form $a + b u$, then ${1 + tu \over 1 - tu} = \alpha + \beta u$
|
|
specifies *some kind* of rotation in terms of the parameter *t*.
|
|
As with complex numbers, the inverse rotation is the conjugate $1 - tu \over 1 + tu$.
|
|
|
|
Another useful feature of quaternion algebra involves symmetry transformations of the imaginary units.
|
|
If an imaginary unit (one of *i*, *j*, *k*) is left-multiplied by a quaternion $q$
|
|
and right-multiplied by its conjugate $q^{*}$, then the result is still imaginary.
|
|
In other words, if *p* is a vector, then $qpq^{*}$ is also a vector since
|
|
*i*, *j*, and *k* form a basis and multiplication distributes over addition.
|
|
I will not demonstrate this fact, as it requires a large amount of algebra.
|
|
|
|
These two features combine to characterize a transformation for a vector quaternion.
|
|
For a vector *p*, if $q_u(t)$ is a "rotation" as above for a point *u* on the unit sphere,
|
|
then another vector (the image of *p*) can be described by
|
|
|
|
$$
|
|
\begin{gather*}
|
|
qpq^* = qpq^{-1}
|
|
= \left({1 + tu \over 1 - tu}\right) p \left({1 - tu \over 1 + tu}\right)
|
|
= (\alpha + \beta u)p(\alpha - \beta u)
|
|
\\[4pt]
|
|
= (\alpha + \beta u)(\alpha p - \beta pu)
|
|
= \alpha^2 p - \alpha \beta pu + \alpha \beta up - \beta^2 upu
|
|
\end{gather*}
|
|
$$
|
|
|
|
|
|
### Testing a Transformation
|
|
|
|
3D space contains 2D subspaces.
|
|
If this transformation is a rotation, then a 2D subspace will be affected in a way
|
|
which can be described more simply by complex numbers.
|
|
For example, if $u = k$ and $p = xi + yj$, then this can be expanded as:
|
|
|
|
$$
|
|
\begin{align*}
|
|
qpq^{-1} &= (\alpha + \beta k)(xi + yj)(\alpha - \beta k)
|
|
\\[10pt]
|
|
&= \alpha^2(xi + yj) - \alpha \beta (xi + yj)k + \alpha \beta k(xi + yj) - \beta^2 k(xi + yj)k
|
|
\vphantom{\over}
|
|
\\[10pt]
|
|
&= \alpha^2(xi + yj) - \alpha \beta(-xj + yi) + \alpha \beta(xj - yi) - \beta^2(xi + yj)
|
|
\\[10pt]
|
|
&= (\alpha^2 - \beta^2)(xi + yj) + 2\alpha \beta(xj - yi)
|
|
\\[10pt]
|
|
&= [(\alpha^2 - \beta^2)x - 2\alpha \beta y]i + [(\alpha^2 - \beta^2)y + 2\alpha \beta x]j
|
|
\end{align*}
|
|
$$
|
|
|
|
If *p* is rewritten as a vector (in the linear algebra sense), then this final expression
|
|
can be rewritten as a linear transformation of *p*:
|
|
|
|
$$
|
|
\begin{align*}
|
|
\begin{pmatrix}
|
|
(\alpha^2 - \beta^2)x - 2 \alpha \beta y \\
|
|
(\alpha^2 - \beta^2)y + 2 \alpha \beta x
|
|
\end{pmatrix}
|
|
&= \begin{pmatrix}
|
|
\alpha^2 - \beta^2 & -2\alpha \beta \\
|
|
2\alpha \beta & \alpha^2 - \beta^2
|
|
\end{pmatrix}
|
|
\begin{pmatrix}
|
|
x \\
|
|
y
|
|
\end{pmatrix}
|
|
\\
|
|
&= \begin{pmatrix}
|
|
\alpha & -\beta \\
|
|
\beta & \alpha
|
|
\end{pmatrix}^2
|
|
\begin{pmatrix}
|
|
x \\
|
|
y
|
|
\end{pmatrix}
|
|
\end{align*}
|
|
$$
|
|
|
|
The complex number $\alpha + \beta i$ is isomorphic to the matrix on the second line[^5].
|
|
Since $\alpha$ and $\beta$ were chosen so to lie on "a" complex unit circle,
|
|
this means that the vector (*x*, *y*) is rotated by $\alpha + \beta i$ twice.
|
|
This also demonstrates that *u* specifies the axis of rotation, or equivalently,
|
|
the plane of a great circle.
|
|
This means that the "some kind of rotation" specified by *q* is a rotation around
|
|
the great circle normal to *u*.
|
|
|
|
[^5]: Observe this by comparing the entries of the square of the matrix and the square of the complex number.
|
|
|
|
The double rotation resembles the half-steps in the complex numbers,
|
|
where $1 + ti$ performs a dilation, ${1 \over 1 - ti}$ reverses it, and together they describe a rotation.
|
|
The form $qpq^{-1}$ is similar to this -- the $q$ on the left performs some transformation which is (partially)
|
|
undone by the $q^{-1}$ on the right.
|
|
In this case, since *q*'s conjugate is its inverse and quaternions do not commute, the only thing to do without
|
|
the transformation being trivial is to sandwich *p* between them.
|
|
|
|
As shown above, the product of a vector with itself should be totally real
|
|
and equal to the negative of the norm of *p*, the sum of the squares of its components.
|
|
The rotated *p* shares the same norm as *p*, so it should equal $p^2$.
|
|
Indeed, this is the case, as
|
|
|
|
$$
|
|
(qpq^{-1})^2 = (qpq^{-1})(qpq^{-1}) = q p p q^{-1} = p^2 q q^{-1} = p^2
|
|
$$
|
|
|
|
As a final remark, due to rotation through quaternions being doubled, the interpolating polynomial
|
|
used above to smooth out double rotations can also be used here.
|
|
That is, $q_u(P(t))pq_u^{-1}(P(t))$ rotates *p* by (approximately) the fraction of a turn specified
|
|
by *t* through the great circle which intersects the plane normal to *u*.
|
|
|
|
|
|
Closing
|
|
-------
|
|
|
|
It is difficult to find a treatment of rotations which does not hesitate
|
|
to use the complex exponential $e^{ix} = \cos(x) + i \sin(x)$.
|
|
The true criterion for rotations is simply that a point lies on the unit circle.
|
|
Perhaps this contributes to why understanding the 3D situation with quaternions is challenging for many.
|
|
Past the barrier to entry, I believe them to be rather intuitive.
|
|
I have outlined some futher benefits of this approach in this post.
|
|
|
|
As a disclaimer, this post was (at least subconsciously) inspired by
|
|
[this video](https://www.youtube.com/watch?v=d4EgbgTm0Bg) by 3blue1brown on YouTube
|
|
(though I had not seen the video in years before checking that I wasn't just plagiarizing it).
|
|
It *also* uses stereographic projections of the circle and sphere to describe rotations
|
|
in the complex plane and quaternion vector space.
|
|
However, I feel like it fails to provide an algebraic motivation for quaternions,
|
|
or even stereography in the first place.
|
|
Hopefully, my remarks on the algebraic approach can be used to augment the information in the video.
|
|
|
|
<!-- TODO: better repository link -->
|
|
Diagrams created with GeoGebra and Matplotlib.
|
|
Repository with approximations (as well as GeoGebra files) available [here](https://github.com/queue-miscreant/approx-trig).
|