diff --git a/posts/chebyshev/1/index.qmd b/posts/chebyshev/1/index.qmd new file mode 100644 index 0000000..bea551a --- /dev/null +++ b/posts/chebyshev/1/index.qmd @@ -0,0 +1,428 @@ +--- +format: + html: + html-math-method: katex +--- + + + + +Generating Polynomials, Part 1: Regular Constructibility +======================================================== + +[Recently](), I used coordinate-free geometry to derive the volumes of the Platonic solids, a problem which was very accessible to the ancient Greeks. On the other hand, they found certain problems regarding which figures can be constructed via compass and straightedge to be very difficult. For example, they struggled with problems like [doubling the cube](https://en.wikipedia.org/wiki/Doubling_the_cube) or [squaring the circle](https://en.wikipedia.org/wiki/Squaring_the_circle), which are known (through circa 19th century mathematics) to be impossible . However, before extending geometry with a third dimension or including the areas of circles in geometry, a simpler problem becomes apparent; namely, what kinds of regular polygons are constructible? + + +Regular Geometry and a Complex Series +------------------------------------- + +When constructing a regular polygon, one wants ratios relating a side, an inradius, and the circumradius. If one of these measurements is declared to be a unit, then only a single ratio is needed, since the other is available through the Pythagorean theorem. This is best exemplified by the following figure + +![]() + +In a convex polygon, the total central angle is always one full turn, or $2\pi$ radians. The central angle of a regular *n*-gon is ${2\pi \over n}$ radians, and the green angle above is half of it. This means that the ratio of half of the side and circumradius is $\sin(\pi / n)$ radians. But how can this quantity be derived? + +Let's turn this question on its head. Let $\theta = \pi / n$, meaning $n\theta = \pi$. Applying sine over this gives $\sin(n\theta) = \sin(\pi) = 0$. Therefore, constructing a polygon is equivalent to solving this equation. It then becomes a question of how to express $\sin(n\theta)$ (and $\cos(n\theta)$). + +Thanks to [Euler's formula](https://en.wikipedia.org/wiki/Euler%27s_formula) and [de Moivre's formula](https://en.wikipedia.org/wiki/De_Moivre%27s_formula), this can be phrased in terms of the complex exponential. + +$$ +\begin{align*} +e^{i\theta} &= \text{cis}(\theta) = \cos(\theta) + i\sin(\theta) +& \text{ Euler's formula} \\ +\text{cis}(n \theta) = e^{i(n\theta)} &= e^{(i\theta)n} = {(e^{i\theta})}^n = \text{cis}(\theta)^n \\ +\cos(n \theta) + i\sin(n \theta) &= (\cos(\theta) + i\sin(\theta))^n +& \text{ de Moivre's formula} +\end{align*} +$$ + +De Moivre's formula for $n = 2$ gives + +$$ +\begin{align*} +\text{cis}(\theta)^2 &= (\text{c} + i\text{s})^2 \\ +&= \text{c}^2 + 2i\text{cs} - \text{s}^2 + (0 = \text{c}^2 + \text{s}^2 - 1) \\ +&= 2\text{c}^2 + 2i\text{cs} - 1 \\ +&= 2\text{c}(\text{c} + i\text{s}) - 1 \\ +&= 2\cos(\theta)\text{cis}(\theta) - 1 \\ +\end{align*} +$$ + +This can easily be massaged into a recurrence relation. + +$$ +\begin{align*} +\text{cis}(\theta)^2 &= 2\cos(\theta)\text{cis}(\theta) - 1 \\ +\text{cis}(\theta)^{n+2} &= 2\cos(\theta)\text{cis}(\theta)^{n+1} - \text{cis}(\theta)^n \\ +\text{cis}((n+2)\theta) &= 2\cos(\theta)\text{cis}((n+1)\theta) - \text{cis}(n\theta) \\ +\end{align*} +$$ + +Through some fairly straightforward summatory manipulations, the sequence can be interpreted as the coefficients in a Taylor series, giving a [generating function](https://en.wikipedia.org/wiki/Generating_function). + +$$ +\begin{align*} +\sum_{n=0}^\infty \text{cis}((n+2)\theta)x^n +&= 2\cos(\theta) \sum_{n=0}^\infty \text{cis}((n+1)\theta) x^n + - \sum_{n=0}^\infty \text{cis}(n\theta) x^n \\ +{F(x; \text{cis}(\theta)) - 1 - x\text{cis}(\theta) \over x^2} +&= 2\cos(\theta) {F(x; \text{cis}(\theta)) - 1 \over x} + - F(x; \text{cis}(\theta)) \\ \\ +F - 1 - x\text{cis}(\theta) +&= 2\cos(\theta) x (F - 1) + - x^2 F \\ +F - 2\cos(\theta) x F + x^2 F +&= 1 + x(\text{cis}(\theta) - 2\cos(\theta)) \\ \\ +F(x; \text{cis}(\theta)) +&= {1 + x(\text{cis}(\theta) - 2\cos(\theta)) \over +1 - 2\cos(\theta)x + x^2} +\end{align*} +$$ + +Even though the generating function *F* is parametrized by the complex exponential of $\theta$, separating the real and imaginary components reveals an expression in terms of $\cos(\theta)$. This is useful because the real and imaginary parts of *F* correspond to $\cos(n\theta)$ and $\sin(n\theta)$, respectively. + +$$ +\begin{align*} +\Re[ F(x; \text{cis}(\theta)) ] &= {1 + x(\cos(\theta) - 2\cos(\theta)) \over +1 - 2\cos(\theta)x + x^2} \\ +&= {1 - x\cos(\theta) \over 1 - 2\cos(\theta)x + x^2} = A(x; \cos(\theta)) \\ +\Im[ F(x; \text{cis}(\theta)) ] +&= {x \sin(\theta) \over 1 - 2\cos(\theta)x + x^2} = B(x; \cos(\theta))\sin(\theta) +\end{align*} +$$ + +Extracting the coefficients of *x* in *A* and *B* yields an expression for $\cos(n\theta)$ and $\sin(n\theta)$ in terms of $\cos(\theta)$ (and in the latter case, a common factor of $\sin(\theta)$). If $\cos(\theta)$ in *A* and *B* is replaced with the parameter *z*, then all of the trigonometric functions are removed from the equation, and it becomes evident that this relationship is polynomial. This can actually be observed as early as the recurrence relation. + +$$ +\begin{align*} +\text{cis}(\theta)^{n+2} &= 2\cos(\theta)\text{cis}(\theta)^{n+1} - \text{cis}(\theta)^n \\ +a_{n+2} &= 2 z a_{n+1} - a_n \\ +\Re[ a_0 ] &= 1,~~ \Im[ a_0 ] = 0 \\ +\Re[ a_1 ] &= z,~~ \Im[ a_1 ] = 1 \cdot \sin(\theta) +\end{align*} +$$ + +In other words, *A* starts off with $1, z, ...$, and *B* starts off with $0, 1, ...$. These polynomials are [*Chebyshev polynomials*](https://en.wikipedia.org/wiki/Chebyshev_polynomial) *of the first (A) and second (B) kind*. Actually, the polynomials of the second kind are typically not offset (the x in the numerator of *B* is omitted). However, this form makes explicit which polygon (remember? we were working with those) we are considering, and will be important later. For future reference, the first few polynomials of the second kind (at $z / 2$) are: + +| n | $[x^n]B(x; z / 2) = U_{n -\ 1}(z / 2)$ | Factored | +|----|-----------------------------------------|---------------------------------------------------| +| 0 | 0 | 0 | +| 1 | 1 | 1 | +| 2 | $z$ | $z$ | +| 3 | $z^2 -\ 1$ | $(z -\ 1)(z + 1)$ | +| 4 | $z^3 -\ 2z$ | $z(z^2 -\ 2)$ | +| 5 | $z^4 -\ 3z^2 + 1$ | $(z^2 -\ z -\ 1)(z^2 + z -\ 1)$ | +| 6 | $z^5 -\ 4 z^3 + 3z$ | $z(z^2 -\ 1)(z^2 -\ 3)$ | +| 7 | $z^6 -\ 5 z^4 + 6 z^2 -\ 1$ | $(z^3 -\ z^2 -\ 2 z + 1)(z^3 + z^2 -\ 2 z -\ 1)$ | +| 8 | $z^7 -\ 6 z^5 + 10 z^3 -\ 4 z$ | $z(z^2 -\ 2)(z^4 -\ 4 z^2 + 2)$ | +| 9 | $z^8 -\ 7 z^6 + 15 z^4 -\ 10 z^2 + 1$ | $(z^2 -\ 1)(z^3 -\ 3 z -\ 1)(z^3 -\ 3 z + 1)$ | +| 10 | $z^9 -\ 8 z^7 + 21 z^5 -\ 20 z^3 + 5 z$ | $z(z^2 -\ z -\ 1)(z^2 + z -\ 1)(z^4 -\ 5z^2 + 5)$ | + +[OEIS A049310](http://oeis.org/A049310) + +Evaluating the polynomials at $z / 2$ cancels the 2 in the denominator (and recurrence), making these expressions much simpler. This evaluation can also be interpreted intuitively by studying the the previous diagram. Since the side length was bisected by the inradius, the side length in terms of a unit circumradius is $2\sin ({\pi / n} )$. To compensate for this doubling, the Chebyshev polynomial must be evaluated at half its normal argument. + + +### Back on the Plane + +The constructibility criterion is deeply connected to the Chebyshev polynomials. In compass and straightedge constructions, one only has access to linear forms (lines) and quadratic forms (circles). This means that a figure is constructible if and only if the root can be achieved through normal arithmetic (which is linear) and square roots (which are quadratic). + +Let's look at a regular pentagon. The relevant polynomial is + +$$ +[x^5]B \left( x; {z \over 2} \right) += z^4 - 3z^2 + 1 += (z^2 - z - 1) (z^2 + z - 1) +$$ + +As a reminder, when $z = 2\cos( \pi / 5 )$, the polynomial evaluates to 0. Also, either factor is the other evaluated at -*z*. Thus, one of the factors in terms is the minimal polynomial of $2\cos(\pi / 5 )$. The former is correct, since $2\cos( \pi / 5 ) = \phi$, the golden ratio. + +An example of where constructability fails is for $2\cos( \pi / 7 )$. + +$$ +\begin{align*} +[x^7]B \left( x; {z \over 2} \right) +&= z^6 - 5 z^4 + 6 z^2 - 1 \\ +&= ( z^3 - z^2 - 2 z + 1 ) (z^3 + z^2 - 2 z - 1 ) +\end{align*} +$$ + +Whichever is the minimal polynomial (the former), it is a cubic, and constructing a regular heptagon is equivalent to solving it for *z*. But there are no (nondegenerate) cubics that one can produce via compass and straightedge, and the construction fails. + +One might think the same of $2\cos(\pi /10 )$ + +$$ +\begin{align*} +[x^{10}]B \left( x; {z \over 2} \right) +&= z^9 - 8 z^7 + 21 z^5 - 20 z^3 + 5 z \\ +&= z ( z^2 - z - 1 )( z^2 + z - 1 )( z^4 - 5 z^2 + 5 ) +\end{align*} +$$ + +This expression also contains the polynomials for $2\cos( \pi / 5 )$. This is because a regular decagon would contain two disjoint regular pentagons, produced by connecting every other vertex. + +![]() + +The polynomial which actually corresponds to $2\cos( \pi / 10 )$ is the quartic, which seems to suggest that it will require a fourth root and somehow decagons will not be constructible. However, it can be solved by completing the square, and we can breathe a sigh of relief. + +$$ +\begin{align*} +z^4 - 5z^2 &= -5 \\ +z^4 - 5z^2 + (5/2)^2 &= -5 + (5/2)^2 \\ +( z^2 - 5/2)^2 &= {25 - 20 \over 4} \\ +( z^2 - 5/2) &= {\sqrt 5 \over 2} \\ +z^2 &= {5 \over 2} + {\sqrt 5 \over 2} \\ +z &= \sqrt{ {5 + \sqrt 5 \over 2} } +\end{align*} +$$ + + +The Triangle behind Regular Polygons +------------------------------------ + +Preferring *z* to be halved in $B(x; z/2)$ makes something else more evident. Observe these four rows of the Chebyshev polynomials + +| n | $[x^n]B(x; z / 2)$ | k | *m* = 2*k* + 1 | *n* - *m* | $[z^{n -\ m}][x^n]B(x; z / 2)$ | +|----|-----------------------------|---|----------------|-----------|--------------------------------| +| 4 | $z^3 -\ 2z$ | 0 | 1 | 3 | 1 | +| 5 | $z^4 -\ 3z^2 + 1$ | 1 | 3 | 2 | -3 | +| 6 | $z^5 -\ 4 z^3 + 3z$ | 2 | 5 | 1 | 3 | +| 7 | $z^6 -\ 5 z^4 + 6 z^2 -\ 1$ | 3 | 7 | 0 | -1 | + +The last column looks like an alternating row of Pascal's triangle, and can be expressed as ${n -\ k -\ 1 \choose k}(-1)^k$. This resemblance can be made more apparent by listing the coefficients of the polynomials in a table. + + +```{python} +#| echo: false + +from IPython.display import Markdown +from tabulate import tabulate +from math import comb + +rainbow = [ + "", + "red", + "orange", + "yellow", + "aqua", + "green", + "cyan", + "blue", + "purple" + "", + "", +] + +entry = lambda x, color: f"{x}" + +Markdown(tabulate( + [ + [ + n, + *[" " for nm in range(1, 11 - n)], + *[ + 0 if k % 2 == 1 + else entry( + comb(n - (k // 2) - 1, nm - (k // 2)) * (-1)**(k // 2), + n - (k // 2) - 1, + ) + for nm, k in zip(range(n), range(10)) + ] + ] + for n in range(1, 11) + ], + headers=[ + "n", + *[f"$z^{nm}$" for nm in reversed(range(2, 10))], + "$z$", + "$1$", + ], + numalign="right", + stralign="right", +)) +``` + +Though they alternate in sign, the rows of Pascal's triangle appear along diagonals, which I have marked in rainbow. Meanwhile, alternating versions of the naturals (1, 2, 3, 4...), the triangular numbers (1, 3, 6, 10...), the tetrahedral numbers (1, 4, 10, 20...), etc. are present along the columns, albeit spaced out by 0's. + +The relationship of the Chebyshev polynomials to the triangle is easier to see if the coefficient extraction of $B(x; z / 2)$ is reversed. + +$$ +\begin{align*} +B(x; z / 2) &= {x \over 1 - zx + x^2} += {x \over 1 + x^2 - zx} += {x \over 1 + x^2} \cdot +{1 \over {1 + x^2 \over 1 + x^2} - z{x \over 1 + x^2}} \\ +[z^n]B(x; z / 2) &= {x \over 1 + x^2} [z^n] {1 \over 1 - z{x \over 1 + x^2}} += {x \over 1 + x^2} \left( {x \over 1 + x^2} \right)^n \\ +&= \left( {x \over 1 + x^2} \right)^{n+1} += x^{n+1} (1 + x^2)^{-n - 1} \\ +&= x^{n+1} \sum_{k=0}^\infty {-n - 1 \choose k}(x^2)^k +\end{align*} +$$ + +While the use of the binomial theorem in $1 + x^2$ is more than enough to justify the appearance of Pascal's triangle (along with explaining the 0's), I will press onward until it becomes excruciatingly obvious. + +$$ +\begin{align*} +{(-n - 1)_k} &= (-n - 1)(-n - 2) \cdots (-n - k) \\ +&= (-1)^k (n + k)(n + k - 1) \cdots (n + 1) \\ +&= (-1)^k (n + k)_k \\ +\implies {-n - 1 \choose k} +&= {n + k \choose k}(-1)^k \\ \\ +[z^n]B(x; z / 2) &= x^{n+1} \sum_{k=0}^\infty {n + k \choose k} (-1)^k x^{2k} \\ +\end{align*} +$$ + +Squinting hard enough, the binomial coefficient is similar to the earlier which gave the third row of Pascal's triangle. If k is fixed, then this expression actually generates the antidiagonal entries of the coefficient table, which are the columns with uniform sign. The alternation instead occurs between antidiagonals (one is all positive, the next is 0's, the next is all negative, etc.). The initial $x^{n+1}$ lags these sequences so that they reproduce the triangle. + + +### Imagined Transmutation + +The generating function of the Chebyshev polynomials resembles other two term recurrences. This resemblance can be made explicit with a simple algebraic manipulation. + +$$ +\begin{align*} +B(ix; -iz / 2) &= {1 \over 1 -\ (-i z)(ix) + (ix)^2} += {1 \over 1 -\ (-i^2) z x + (i^2)(x^2)} \\ +&= {1 \over 1 -\ z x -\ x^2} +\end{align*} +$$ + +The poles of this expression are the reciprocals and additive inverses of the metallic means. For example, if $z = 1$, then the denominator is $1 -\ x -\ x^2$, which has $1 / \phi$ as a root. Equally well, it is also a generating function of the Fibonacci numbers. The same can be said for the silver ratio and Pell numbers, and so on for higher $z$. + +In terms of the Chebyshev polynomials, this removes the alternation in the coefficients of $U_n$, and restores Pascal's triangle to its nonalternating form. Related to the previous point, it is possible to find the Fibonacci numbers (Pell numbers, etc.) in Pascal's triangle, which you can read more about [here](http://users.dimi.uniud.it/~giacomo.dellariccia/Glossary/Pascal/Koshy2011.pdf). + + +Manipulating the Series +----------------------- + +### Total Degrees + +Look back to the table of $U_{n -\ 1}(z / 2)$. When I brought up decagons, I pointed out their relationship to pentagons as an explanation for why $U_{5 -\ 1}(z / 2)$ appears as a factor. Conveniently, $U_{2 -\ 1}(z / 2)$ is also a factor, and 10 is an even number. This pattern is present throughout the table; $n = 6$ contains factors for $n = 2 \text{ and } 3$ and the prime numbers have no smaller factors. If this observation is legitimate, call the factor polynomials $f_n(z)$ and denote $p_n(z) = U_{n -\ 1}( z / 2 )$. + +It can also be observed that a factor is either symmetric ($g(z) = g(-z)$), or is the product of another polynomial and its reflection, potentially negated. For example, $p_9(z) = f_3(z) \cdot g_9(z) \cdot -g_9(-z)$, where $g_9(z) = z^3 -\ 3z -\ 1$. These reflections are necessary for $n = 3, 5, 7, 9$, strongly implying that it occurs on the odd terms. + +In other words, if $f_n$ is the new polynomial introduced by $p_n$ of, then denote its conditional factorization $g_n$. + +$$ +f_n(z) = \begin{cases} +g_n(z) & n \text{ is even} \\ +g_n(z)g_n(-z) & \text{$n$ is odd and ${\deg(f_n) \over 2}$ is even } \\ +-g_n(z)g_n(-z) & \text{otherwise} +\end{cases} +$$ + +The final case seems to be tied to [OEIS A004614](http://oeis.org/A004614), which are primes of the form $4k+3$ and products thereof. + +Without resorting to any advanced techniques, the degrees of $f_n$ are not too difficult to work out. The degree of $p_n(z)$ is $n -\ 1$, which is also the degree of $f_n(z)$ if *n* is prime. If *n* is composite, then the degree of $f_n(z)$ is $n -\ 1$ minus the degrees of the divisors of $n -\ 1$. This leaves behind how many numbers less than *n* are coprime to *n*. Therefore $\deg(f_n) = \varphi(n)$, the Euler totient function. + +The totient function can be used to examine the parity of *n*. If *n* is odd, it is coprime to 2 and all even numbers. The introduced factor of 2 to 2*n* removes the evens from the totient, but this is compensated by the addition of the odd multiples of old numbers coprime to *n* and new primes. This means that $\varphi(2n) = \varphi(n)$ for odd *n* ($n \neq 1$). + +The same argument can be used for even *n*: there are as many odd numbers from 0 to *n* as there are from *n* to 2*n*, and there are an equal number of numbers coprime to 2*n* in either interval. Therefore, $\varphi(2n) = 2\varphi(n)$ for even *n*. + +This collapses the cases of the conditional factorization of $f_n$ into one, and the degrees of $g_n$ are + +$$ +\begin{align*} \deg( g_n(z) ) &= \begin{cases} +\deg( f_n(z) ) = \varphi(n) & n \text{ is even} & \implies \varphi(2n) / 2 \\ +\deg( f_n(z) ) / 2= \varphi(n) / 2 & n \text{ is odd} & \implies \varphi(2n) / 2 \\ +\end{cases} \\ +&= \varphi(2n) / 2 +\end{align*} +$$ + +Though they were present in the earlier Chebyshev table, the $g_n$ themselves are presented again, along with the expression for their degree + + +| n | $\varphi(2n)/2$ | $g_n(z)$ | Coefficient List, Rising Powers | +|----|-----------------------------------------|------------------------|-----------------------------------------| +| 2 | 1 | $z$ | [0, 1] | +| 3 | 1 | $z -\ 1$ | [-1, 1] | +| 4 | 2 | $z^2 -\ 2$ | [-2, 0, 1] | +| 5 | 2 | $z^2 -\ z -\ 1$ | [-1, -1, 1] | +| 6 | 2 | $z^2 -\ 3$ | [-2, 0, 1] | +| 7 | 3 | $z^3 -\ z^2 -\ 2z + 1$ | [1, -2, -1, 1] | +| 8 | 4 | $z^4 -\ 4z^2 + 2$ | [2, 0, -4, 0, 1] | +| 9 | 3 | $z^3 -\ 3z + 1$ | [1, -3, 0, 1] | +| 10 | 4 | $z^4 -\ 5z^2 + 5$ | [5, 0, -5, 0, 1] | +| | [OEIS A055034](http://oeis.org/A055034) | | [OEIS A187360](http://oeis.org/A187360) | + + +### Attempts to Factor + +The relationship between $p_n$ and the intermediate $f_d$, where *d* is a divisor of *n*, can be made explicit by a [Moebius inversion](https://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula). + +$$ +\begin{align*} +p_n(z) &= \prod_{d|n} f_n(z) \\ +\log( p_n(z) ) &= \log \left( \prod_{d|n} f_d(z) \right) += \sum_{d|n} \log( f_d(z) ) \\ +\log( f_n(z) ) &= \sum_{d|n} {\mu \left({n \over d}\right)} +\log( p_d(z) ) \\ +f_n(z) &= \prod_{d|n} p_d(z)^{\mu \left({n \over d}\right)} \\ \\ +f_6(z) = g_6(z) &= p_6(z)^{\mu(1)} +p_3(z)^{\mu(2)} +p_2(z)^{\mu(3)} \\ +&= {p_6(z) \over p_3(z) p_2(z)} \\ +\end{align*} +$$ + +Mobius inversion by means of [Dirichlet generating functions](https://en.wikipedia.org/wiki/Dirichlet_series#Formal_Dirichlet_series) is algebraically clear. But moving from $f_n$ to $g_n$ is not easy. $\varphi(n)$ can also be easily expressed by in the Dirichlet world (it is $\zeta(s -\ 1) / \zeta(s)$, where $\zeta$ is the Riemann zeta function), which is useful for knowing when to negate a term. However, the move which doubles the argument of $\varphi$ is difficult to perform in terms of a Dirichlet series, and leaves me stuck. + +Even reformatting as a [Lambert series](https://en.wikipedia.org/wiki/Lambert_series) (another kind amenable to Moebius inversion) doesn't provide any intuition, since the product relationship between the polynomials necessitates a logarithm. If you're in the mood for awful-looking math, it is + +$$ +\begin{align*} +\log( p_n(z) ) &= \sum_{d|n} \log( f_d(z) ) \\ +\sum_{n = 1}^\infty \log( p_n ) x^n +&= \sum_{n = 1}^\infty \sum_{d|n} \log( f_d ) x^n \\ +&= \sum_{k = 1}^\infty \sum_{m = 1}^\infty \log( f_m ) x^{m k} \\ +&= \sum_{m = 1}^\infty \log( f_m ) \sum_{k = 1}^\infty (x^m)^k \\ +&= \sum_{m = 1}^\infty \log( f_m ) {x^m \over 1 - x^m} \\ +\prod_{n = 1}^\infty p_n(z)^{(x^n)} &= \prod_{n = 1}^\infty f_n(z)^{\left({x^n \over 1 - x^n}\right)} +\end{align*} +$$ + +Either way, the number-theoretic properties of this sequence are very difficult to ascertain without advanced techniques. If research has been done, it is not easily available in the OEIS. + + +Closing +------- + +My initial jumping off point for writing this article was completely different. However, in the process of writing, its share of the article shrank and shrank until its introduction was only vaguely related to what preceded it. But alas, the introduction via geometric constructions flows better coming off my [post about the Platonic solids](). As well, it reads better if I rely less on "if you search for this sequence of numbers" and more on how to interpret the definition. Consider reading [the follow-up]() to this post if you're interested in another way one can obtain the Chebyshev polynomials. + +Diagrams created with GeoGebra. + +Update: I have since rederived the Chebyshev polynomials without the complex exponential, which you can read about in [this post](). diff --git a/posts/chebyshev/2/index.qmd b/posts/chebyshev/2/index.qmd new file mode 100644 index 0000000..23568f2 --- /dev/null +++ b/posts/chebyshev/2/index.qmd @@ -0,0 +1,327 @@ +--- +format: + html: + html-math-method: katex +--- + + +Generating Polynomials, Part 2: Ghostly Chains +============================================== + +In the [previous post](), I tied the geometry regular polygons to a sequence of polynomials though some clever algebraic manipulation. But let's deign to ask a very basic question: what is a polygon? + +Fundamentally, it is just a collection of vertices and edges. Each vertex is connected to its two neighbors by two edges. Only examining these figures by their connectedness is precisely the kind of thing *graph theory* deals with. Graph can seem like a strange name, as it has no relation to the familiar graphs on a Cartesian plane one may be familiar with. Either way, it is a worthy subject of study, and will be the focus of this article. + + +Loops without Distance +---------------------- + +For polygons in a Euclidean setting, the position of points matters, as well as which points connect to which. A rectangle is different from a trapezoid or a kite. However, topologically, all of these are quadrilateral graphs and cannot be distinguished; they are but a simple notion of 4 points in a loop. + +![]() + +From a graph theory perspective, the vertices are sometimes called nodes. In these graphs, each edge has no direction associated to it, so the graph is called *undirected*. Additionally, these graphs are *planar* since the nodes can be arranged so that no nodes cross another, despite the appearance of the lower-right figure. + +If the graph is a simple loop, it is called a [*cycle graph*](https://en.wikipedia.org/wiki/Cycle_graph), denoted $C_n$, where n is the number of nodes. In a cycle graph, all nodes and all edges are identical to each of the others. Therefore, the best geometric interpretation is a shape which is + +- Regular, so that each edge and each angle (vertex) are of equal measure +- Convex, so that no edge meets another without creating a vertex (or node) + +In other words, $C_3$ is analogous to an equilateral triangle, $C_4$ is analogous to a square, and so on. + + +### Encoding Graphs + +There are two primary ways to store information about a graph. The first is by labelling each node (for example, with integers), then recording the edges as a list of pairs of connected nodes. In the case of an undirected graph, these are unordered pairs. While such a list is convenient, it doesn't convey a lot of information about the graph besides the number of edges. + +Alternatively, these pairs can also be interpreted as addresses in a matrix, called an *adjacency matrix*. + +$$ +\begin{align*} +C_3 := \begin{matrix} [(0, 1),\\ (1, 2),\\ (2, 0)] \end{matrix} &\cong +\begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} \\ \\ +C_4 := \begin{matrix} [(0, 1),\\ (1, 2),\\ (2, 3),\\ (3, 0)] \end{matrix} & \cong +\begin{pmatrix} 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \end{pmatrix} \\ \\ +C_5 := \begin{matrix} [(0, 1),\\ (1, 2),\\ (2, 3),\\ (3, 4),\\ (4, 0)] \end{matrix} &\cong +\begin{pmatrix} 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ +0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 \end{pmatrix} \\ +\end{align*} +$$ + +Each adjacency matrix is square, where each column and row refer to a specific node. An entry is 1 when the nodes corresponding to the row and column of its address are joined by an edge (and zero otherwise). For undirected graphs, these matrices are symmetric, since it is possible to traverse an edge in either direction. + +Swapping the labels on two nodes is as simple as exchanging two rows and two columns. Just one of these swaps would flip the sign of the determinant of the adjacency matrix. However, since they occur in pairs, the determinant is invariant of the labelling (equally, a graph invariant). + + +Prismatic Recurrence +-------------------- + +The determinant of a matrix is also the product of its eigenvalues, which are another matrix invariant. The set of eigenvalues is also called its *spectrum*, and the study of the spectra of graphs is called [*spectral graph theory*](https://en.wikipedia.org/wiki/Spectral_graph_theory), which is among the most mystifying names in math to read for the first time. + +Eigenvalues are the roots of the characteristic polynomial of a matrix. The matrix $C_5$ is sufficiently large enough to generalize to $C_n$, and its characteristic polynomial by [Laplace expansion](https://en.wikipedia.org/wiki/Laplace_expansion) is: + +$$ +\begin{gather*} +Ax = \lambda x \implies (\lambda I - A)x = 0 \\ \\ +c_5(\lambda) = |\lambda I - C_5| = \left | +\begin{matrix} \lambda & -1 & 0 & 0 & -1 \\ -1 & \lambda & -1 & 0 & 0 \\ +0 & -1 & \lambda & -1 & 0 \\ 0 & 0 & -1 & \lambda & -1 \\ -1 & 0 & 0 & -1 & \lambda \end{matrix} +\right | \\ += \lambda m_{1,1} ++ \overbrace{(-1)}^\text{entry}\overbrace{(-1)^{1 + 2 \ }}^\text{sign} m_{1, 2} ++ \overbrace{(-1)}^\text{entry}\overbrace{(-1)^{1 + 5 \ }}^\text{sign} m_{1, 5} +\end{gather*} +$$ + +Note that every occurrence of "5" generalizes to higher *n*. The first minor is easily expressed in terms of *another* matrix's characteristic polynomial. + +$$ +m_{1, 1} = \left | +\begin{matrix} \lambda & -1 & 0 & 0 \\ -1 & \lambda & -1 & 0 \\ +0 & -1 & \lambda & -1 \\ 0 & 0 & -1 & \lambda \end{matrix} +\right | = |\lambda I - P_4| = p_{5-1}(\lambda) +$$ + +We will come to the meaning of the $P_n$ in a moment. Meanwhile, the second and third minors require another expansion, but one that (thankfully) quickly terminates. + +$$ +\begin{matrix} +m_{1, 2} =& \left | +\begin{matrix}-1 & -1 & 0 & 0 \\ 0 & \lambda & -1 & 0 \\ +0 & -1 & \lambda & -1 \\ -1 & 0 & -1 & \lambda \end{matrix} +\right | &=& (-1) \left | +\begin{matrix}\lambda & -1 & 0 \\ -1 & \lambda & -1 \\ 0 & -1 & \lambda \end{matrix} +\right | &+& (-1)(-1)^{1 + 4} \left | +\begin{matrix}-1 & 0 & 0 \\\lambda & -1 & 0 \\ -1 & \lambda & -1 \end{matrix} +\right | \\ +&&=& (-1)|\lambda I - P_3| &+& (-1)\overbrace{(-1)^{5}(-1)^{5 - 2}}^{\text{even, even when $\scriptsize n \neq 5$}} \\ +&&=& ((-1)p_{5 - 2}(\lambda) &+& (-1)) \\ +&&=& -(p_{5 - 2}(\lambda) &+& 1) \\ +\\ +m_{1, 5} =& \left | +\begin{matrix}-1 & \lambda & -1 & 0 \\ 0 & -1 & \lambda & -1 \\ +0 & 0 & -1 & \lambda \\ -1 & 0 & 0 & -1 \end{matrix} +\right | &=& (-1) \left | +\begin{matrix}-1 & \lambda & -1 \\ 0 & -1 & \lambda \\ 0 & 0 & -1 \end{matrix} +\right | &+& (-1)(-1)^{5 - 2} \left | +\begin{matrix}\lambda & -1 & 0 \\ -1 & \lambda & -1 \\ +0 & -1 & \lambda \end{matrix} +\right | \\ +&&=& (-1)(-1)^{5-2} &+& (-1)(-1)^{5 - 2}|\lambda I - P_3| \\ +&&=& (-1)^{5-1}((-1)(-1) &+& (-1)(-1)p_{5 - 2}(\lambda)) \\ +&&=& (-1)^{5-1}(1 &+& p_{5 - 2}(\lambda)) +\end{matrix} +$$ + +All together, this produces a characteristic polynomial in terms of the polynomials $p_n$: + +$$ +\begin{align*} +&&c_5(\lambda) &= \lambda p_{5 - 1} ++ (-1)(p_{5 - 2} + 1) ++ (-1)\overbrace{(-1)^{5 - 1} (-1)^{5 - 1}}^{\text{even, even when $\scriptsize n \neq 5$}}(p_{5 - 2} + 1) \\ +&&&= \lambda p_{5 - 1} +- (p_{5 - 2} + 1) +- (p_{5 - 2} + 1) \\ +&&&= \lambda p_{5 - 1} +- 2(p_{5 - 2} + 1) \\ +&&\implies c_n(\lambda) &= \lambda p_{n - 1}(\lambda) +- 2(p_{n - 2}(\lambda) + 1) \\ +\end{align*} +$$ + +This resembles a recurrence relation, which is great, but it is meaningless without knowing $p_n$. + + +Powerful Chains +--------------- + +The various $P_n$ are in fact the adjacency matrices of a path on *n* nodes. + +![]() + +$$ +\begin{align*} +P_2 &:= \begin{matrix} [(0, 1)] \end{matrix} \cong +\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ +P_3 &:= \begin{matrix} [(0, 1),\\ (1, 2)] \end{matrix} \cong +\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \\ +P_4 &:= \begin{matrix} [(0, 1),\\ (1, 2),\\ (2, 3)] \end{matrix} \cong +\begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} \\ \\ +\end{align*} +$$ + +Without the entries in the corners of the matrix, the characteristic polynomials of $P_n$ are much easier to solve for. + +$$ +\begin{gather*} +p_4(\lambda) = |\lambda I - P_4| = \left | +\begin{matrix} \lambda & -1 & 0 & 0 \\ -1 & \lambda & -1 & 0 \\ +0 & -1 & \lambda & -1 \\ 0 & 0 & -1 & \lambda \end{matrix} +\right | \\ \\ += \lambda \left | +\begin{matrix} \lambda & -1 & 0 \\ -1 & \lambda & -1 \\ 0 & -1 & \lambda \end{matrix} +\right | + (-1)(-1)^{1+2} \left | +\begin{matrix} -1 & -1 & 0 \\ 0 & \lambda & -1 \\ 0 & -1 & \lambda \end{matrix} +\right | \\ \\ += \lambda |\lambda I - P_3| + \left ( (-1) \left | +\begin{matrix} \lambda & -1 \\ -1 & \lambda \end{matrix} +\right | + (-1)(-1) \left | +\begin{matrix} 0 & -1 \\ 0 & \lambda \end{matrix} +\right | \right) \\ \\ += \lambda |\lambda I - P_3| - |\lambda I - P_2| \\ += \lambda p_{4 - 1}(\lambda) - p_{4 - 2}(\lambda) \\ +\implies p_{n}(\lambda) = \lambda p_{n - 1}(\lambda) - p_{n - 2}(\lambda) +\end{gather*} +$$ + +While the earlier equation for $c_n$ in terms of $p_n$ reminded of a recurrence relation, actually is one (if perhaps eerily familiar). + +Since the recurrence has order 2, it requires two initial terms: $p_0$ and $p_1$. The graph corresponding to $p_1$ is a single node, not connected to anything. Therefore, its adjacency matrix is a 1x1 matrix with 0 as its only entry, and its characteristic polynomial is $\lambda$. By the recurrence, $p_2 = \lambda p_1 -\ p_0 = \lambda^2 -\ p_0$. Equating terms with the characteristic polynomial of $P_2$, it is obvious that + +$$ +|\lambda I - P_2| = \begin{pmatrix}\lambda & -1 \\ -1 & \lambda \end{pmatrix} += \lambda^2 - 1 = \lambda p_1 - p_0 \\ +\implies p_0 = 1 +$$ + +which makes sense, since $p_0$ should have degree zero. Therefore, the sequence of polynomials $p_n(\lambda)$ is: + +$$ +\begin{gather*} +p_0(\lambda) =& && 1 \\ +p_1(\lambda) =& && \lambda \\ +p_2(\lambda) =& \lambda \lambda - 1 &=& \lambda^2 - 1 \\ +p_3(\lambda) =& \lambda (\lambda^2 - 1) - \lambda &=& \lambda(\lambda^2 - 2) \\ +p_4(\lambda) =& \lambda (\lambda(\lambda^2 - 2)) - (\lambda^2 - 1) +&=& \lambda^4 - 3\lambda^2 + 1 \\ +\vdots&\vdots&&\vdots +\end{gather*} +$$ + +But wait, we've seen these before (if you read the previous post, that is). These are just the Chebyshev polynomials of the second kind, evaluated at $\lambda / 2$. Indeed, their recurrence relations are identical, and the characteristic polynomial of $P_n$ is $U_n(\lambda / 2)$. Effectively, this connects an n-path to a regular $n+1$-gon. + +Since the generating function of $U_n$ is known, the generating function for the $c_n$ which prompted this is also easily determined. For ease of use, let + +$$ +P(x; \lambda) = {B(x; \lambda / 2) \over x} = {1 \over 1 - \lambda x +\ x^2} +$$ + +Discarding the initial $c_0$ and $c_1$ by setting them to 0, the generating function is + +$$ +\begin{align*} +c_{n+2}(\lambda) &= \lambda p_{n+1}(\lambda) - 2(p_n(\lambda) + 1) \\ \\ +{C(x; \lambda) - c_0(\lambda) - x c_1(\lambda) \over x^2} +&= \lambda \left( {P(x; \lambda) - 1 \over x} \right) +- 2\left( P(x; \lambda) + {1 \over 1 - x} \right) \\ +C &= x \lambda (P - 1) -\ +2x^2\left( P + {1 \over 1 - x} \right) \\ +C{(1 - x) \over P} &= x \lambda \left(1 - {1 \over P} \right)(1 - x) -\ +2x^2\left( (1 - x) + {1 \over P} \right) \\ +&= x^4 \lambda - 2 x^4 - x^3 \lambda^2 + x^3 \lambda ++ 2 x^3 + x^2 \lambda^2 - 4 x^2 \\ +&= x^2 (\lambda - 2) (x^2 - \lambda x - x + \lambda + 2) \\ \\ +C(x; \lambda) &= x^2 (\lambda - 2) +{(x^2 - (\lambda + 1) x + \lambda + 2) +\over (1 - x)(1 - \lambda x + x^2)} +\end{align*} +$$ + +While the numerator is considerably more complicated than the one for P, the factor $\lambda -\ 2$ drops out of the entire series, pleasantly informing that 2 is an eigenvalue of all $C_n$. + + +Some other Exceptional Graphs +----------------------------- + +And now for something completely different (and which prompted me to research this topic in the first place). Another simple family of graphs are [*trees*](https://en.wikipedia.org/wiki/Tree_%28graph_theory%29). They are in some sense opposite to the cycle graphs, since they contain no cycles. + + +### Platonic Trees + +Of trees, three small ones (which are not also paths) stand out. They are based on the [Schläfli symbols](https://en.wikipedia.org/wiki/Schl%C3%A4fli_symbol) of the Platonic solids. These expressions convey which regular polygon is present, and how many of them are found at every vertex. + +- Tetrahedron: {3, 3} + - Equilateral triangles, three around a vertex +- Octahedron: {3, 4} + - Equilateral triangles, four around a vertex +- Icosahedron: {3, 5} + - Equilateral triangles, five around a vertex +- Cube: {4, 3} + - Squares, three around a vertex +- Dodecahdedron: {5, 3} + - Regular pentagons, three around a vertex + +Each edge in a Platonic solid connects two faces, which gives a third parameter of 2. From each parameter, create a $n-1$-path, and join the three to a central node. Intuitively, this is because we have shown that $n-1$-paths are related to regular *n*-gons, which show up in the pairs of dual Platonic solids. + +::: {} +![]() +Adjacency matrices of T as a heatmap. Purple is 0, yellow is 1. +::: + +Dispensing with the Laplace expansion and finding a recurrence relation, the characteristic polynomials for each of the three trees is + +$$ +\begin{align*} +|\lambda I - T_{3,3}| = t_{3,3}(\lambda) +&= \lambda^{6} - 5 \lambda^{4} + 5 \lambda^{2} - 1 \\ +&= (\lambda - 1) (\lambda + 1) (\lambda^{4} - 4 \lambda^{2} + 1) \\ +|\lambda I - T_{3,4}| = t_{3,4}(\lambda) +&= \lambda^{7} - 6 \lambda^{5} + 9 \lambda^{3} - 3 \lambda \\ +&= \lambda (\lambda^{6} - 6 \lambda^{4} + 9 \lambda^{2} - 3) \\ +|\lambda I - T_{3,5}| = t_{3,5}(\lambda) +&= \lambda^{8} - 7 \lambda^{6} + 14 \lambda^{4} - 8 \lambda^{2} + 1 +\end{align*} +$$ + +Searching the OEIS for the coefficients of $t_{3,5}$ returns sequence [A228786](http://oeis.org/A228786), which informs that it is the minimal polynomial of $2\sin( \pi/15 )$. This sequence also informs that, where $m_n$ is the minimal polynomial of $2\sin( \pi / n )$: + +- $t_{3,3} = m_6(\lambda) \cdot -m_6(-\lambda) \cdot m_{12}(\lambda)$ +- $t_{3,4} = m_1(\lambda) \cdot m_9(\lambda)$ + +Perhaps this is not surprising, given how $2\cos(\pi / n)$ also appear in the spectra of graphs. However, I am perplexed by the apparent relationship of tetrahedra to dodecagons, octahedra to enneagons, and icosahedra to 15-gons. It is also strange that 9 is *not* related to the smallest Platonic solid, and is sandwiched between 12 and 15. + + +### Wooden Tiles + +Platonic solids can be seen as regular tilings of the sphere. Schläfli symbols also exist for the regular tilings of the plane (which can also tile the torus, another topologically significant object). + +- Triangular tiling: {3, 6} + - Equilateral triangles; 6 around a vertex +- Square tiling: {4, 4} + - Squares; 4 around a vertex +- Hexagonal tiling: {6, 3} + - Regular hexagons; 3 around a vertex + +::: {} +![]() +Adjacency matrices as a heatmap +::: + +I would prefer *not* to wrestle the adjacency matrices into a general recurrence relation, but if one would like to go down this route, try using the Laplace expansion of the southmost tip of the "island" in the $T_{3,n}$ graphs as an example. + +The characteristic polynomials of *these* graphs are + +$$ +\begin{align*} +|\lambda I - T_{4,4}| = t_{4,4}(\lambda) +&= \lambda^{8} - 7 \lambda^{6} + 14 \lambda^{4} - 8 \lambda^{2} \\ +&= (\lambda - 1) (\lambda + 1) (\lambda^{4} - 4 \lambda^{2} + 1) \\ +&= m_6(\lambda) \cdot -m_6(-\lambda) \cdot m_8(\lambda) \\ +|\lambda I - T_{3,6}| = t_{3,6}(\lambda) +&= \lambda^{9} - 8 \lambda^{7} + 20 \lambda^{5} - 17 \lambda^{3} + 4 \lambda \\ +&= \lambda (\lambda - 1)(\lambda + 1) (\lambda - 2)(\lambda + 2) +(\lambda^{2} - \lambda - 1) (\lambda^{2} + \lambda - 1) \\ +&= m_1(\lambda) \cdot m_6(\lambda) \cdot -m_6(-\lambda) +\cdot m_2(\lambda) \cdot -m_2(-\lambda) +\cdot m_{10}(\lambda) \cdot m_{10}(-\lambda) \\ +\end{align*} +$$ + +Where the $m_n$ come from the sine polynomial series. To me, this implies that where chains are connected to cosines of fractions of a half-turn, these trefoil trees are connected to sines of fractions of turns (but you'd have to derive a general rule to be sure). + + +Closing +------- + +Regardless of whether chains or polygons are more fundamental, it is certainly interesting that they are just an algebraic stone's (a *calculus*'s?) toss away from one another. Perhaps Euler skipped such stones from the bridges of Koenigsberg which inspired him to initiate graph theory. diff --git a/posts/chebyshev/extra/index.qmd b/posts/chebyshev/extra/index.qmd new file mode 100644 index 0000000..8b22533 --- /dev/null +++ b/posts/chebyshev/extra/index.qmd @@ -0,0 +1,305 @@ +--- +format: + html: + html-math-method: katex +--- + + +Generating Polynomials Extra: Legendary +======================================= + +In the [previous]() [two]() posts, I made clear the ties between the Chebyshev polynomials and polygonal constructibility and graph theory, the benefits of having an infinite family of polynomials are manifold. In this post, I will construct another family with different properties, but one that is still related to the Chebyshev polynomials. + +The format of this post will be slightly different: it will have a less concrete goal and be geared more toward following mathematical intuition. + + +Orthogonality +------------- + +Another very basic question we can ask is: what exactly is a polynomial? Formally (or by viewing it as a container for data), it is no more than a list of numbers with the following operations: + +- Addition between two polynomials, implying +- Scalar multiplication of a polynomial +- Multiplication between two polynomials +- Evaluation at a number + +By "number", I mean an object of the same type as the coefficients, which can be a natural, an integer, a rational, etc. The first two properties above and existence of the zero polynomial imply that polynomials can be considered a [vector space](https://en.wikipedia.org/wiki/Vector_space). + +Another thing they have in common with typical Cartesian vectors is that a dot product can be defined on them. Again from a data perspective, a dot product must be calculated from information in *both* polynomials and have a value whose type is the same as the coefficients of the polynomial. Naively, this could be done by using the already-defined multiplication on two polynomials, then evaluating the product at some value. + +In actuality, this is done by *integrating* the product of two polynomials. Sometimes a third expression is included in the product, termed a *weight function*, but the simplest weight function is 1. Similarly, the simplest interval is from -1 to 1, as this can be used to exploit symmetry for odd integrands. + +$$ +\begin{gather} +P(z) \cdot_w Q(z) = \int_a^b P(z)Q(z)w(z)dz \\ +P(z) \cdot Q(z) = \int_{-1}^1 P(z)Q(z)dz +\end{gather} +$$ + +If the product of *P* and *Q* is an odd polynomial, then their dot product is trivially 0, and they are orthogonal. For example, if $P(z) = 1$ and $Q(z) = z$, then the integrand is simply *z*; this polynomial is odd, so $1$ and $z$ are orthogonal. The inner products of $1$ and $z$ with themselves (or norms, by analogy with typical vectors) are: + +$$ +\begin{gather} +1 \cdot 1 = \int_{-1}^1 1^2 dz = \left.{z^{\vphantom{0}}}\right]_{-1}^1 = 1 -\ (-1) = 2 \\ +z \cdot z = \int_{-1}^1 z^2 dz = \left.{z^3 \over 3}\right]_{-1}^1 = {1 \over 3} -\ \left(-{1 \over 3}\right) = {2 \over 3} +\end{gather} +$$ + +What if we were to add a third polynomial into the mix, which is orthogonal to both $1$ and $z$, has degree 2, and whose norm is, say, 2/5? To start, just use a generic quadratic polynomial $p_2(z) = a z^2 + b z + c$. Then enforce all three constraints: + +$$ +\begin{align*} +1 \cdot p_2 &= \int_{-1}^1 (a z^2 + b z + c) dz += \left[ {a z^3 \over 3} + c z \right]_{-1}^1 += {2 \over 3}a + 2c = 0 \\ +z \cdot p_2 &= \int_{-1}^1 (a z^3 + b z^2 + c z) dz += \left[ {b z^3 \over 3} \right]_{-1}^1 += {2 \over 3}b = 0 \\ +p_2 \cdot p_2 &= \int_{-1}^1 (a z^2 + c)^2 dz += \int_{-1}^1 (a^2 z^4 + 2ac z^2 + c^2) dz = {2 \over 5} +\end{align*} +$$ + +Already it can be seen that *b* is 0, which implies the new polynomial is completely even, like $1$ which preceded it. The final integral, though not challenging, requires a fair number of steps to resemble a constraint on *a* and *c* which solves the system. + +$$ +\begin{align*} +\int_{-1}^1 (a^2 z^4 + 2ac z^2 + c^2) dz +=&~ \left[ a^2 {z^5 \over 5} + ( 2ac ) {z^3 \over 3} + c^2 z \right]_{-1}^1 \\ +=&~ {2a^2 \over 5} + {4ac \over 3} + 2c^2 += {2a^2 \over 5} + {2ac \over 3} + {2ac \over 3} + 2c^2 \\ +=&~ {2 \over 15} ( 3a^2 + 5ac ) + c \left( {2a \over 3} + 2c \right) \\ +=&~ {2 \over 15} ( 3(-3c)^2 + 5(-3c)c ) + c (0) += {2 \over 15}(27c^2 -\ 15c^2) \\ +=&~ {2 \over 5} (4c^2) = {2 \over 5} \implies c = \pm {1 \over 2} \\ +\implies& {2a \over 3} \pm 1 = 0 \qquad a = \mp {3 \over 2} +\end{align*} +$$ + +Preferring the leading coefficient a to be positive, this yields the polynomial $p_2(z) = {1 \over 2}(3z^2 -\ 1)$. We might have sought a different constraint which requires less algebra to evaluate, for example by forcing $p_2(1) = 1$ (though this turns out to be equivalent to the one already given). + +This family of polynomials is called the [Legendre polynomials](https://en.wikipedia.org/wiki/Legendre_polynomials), named after Adrien-Marie Legendre. I bring up the man behind them in particular because only one surviving portrait of him exists, and it is a fairly humorous caricature. + +::: {} +![]() +Public domain image retrieved from Wikimedia +::: + +An entire family of mutually orthogonal polynomials has its uses. When doing numeric computation involving integrals, expressing a (scaled and shifted) general polynomial in the orthogonal basis not only simplifies the math, but can prevent errors from accumulating. They also have applications in electrical engineering. + +Even with the alternative, it becomes clear that extending the family further will require more free coefficients, more integrals, and an overall bigger system. How did Legendre do it? Is there a way to bypass so much math? + + +Presupposition and Trial and Error +---------------------------------- + +Of course there is. What do we know about this problem? We're dealing with a sequence of polynomials, where + +- A polynomial's degree uniquely identifies it in the sequence, +- Any two polynomials which are not the same are orthogonal, +- Any polynomial's norm is $2/(2n + 1)$, where *n* is the degree of the polynomial +- The first two terms are $1$ and $z$ + +In the same manner as with the Chebyshev polynomials, considering a sequence of polynomials is easiest if given as an expression whose Taylor series coefficients are the requisite polynomials. As a reminder, and for the sake of consistency with the previous post, the polynomial variable has been *z*, and the series variable will be *x*. + +We don't know what this expression is, and we only know a few terms of the series + +$$ +\begin{align*} +F(x; z) &= \sum_{n=0}^\infty p_n(z) x^n \\ +&= p_0(z) + p_1(z)x + p_2(z)x^2 + ... \\ +&= 1 + zx + {1 \over 2}(3z^2 -\ 1) x^2 + ... +\end{align*} +$$ + +We want to exploit the norm and orthogonality condition, which are obtained by integrating the product of any two coefficients. As a single object *F*, it is very easy to get every possible product of two terms -- it is simply given by $F^2$. Despite the apparent difficulty in distributing over an infinite number of terms, the result can be expressed somewhat simply. + +$$ +\begin{align*} +F(x; z)^2 &= \left( \sum_{n=0}^\infty p_n x^n \right)^2 += (p_0 + p_1 x + p_2 x^2 + ...)(p_0 + p_1 x + p_2 x^2 + ...) \\ +&= p_0^2 + (2p_0 p_1) x + (2p_0 p_2 + p_1^2) x^2 + (2p_0 p_3 + 2p_1 p_2) x^3 + ... \\ +&= \sum_{\substack{n = 0 \\ r + s = n}}^\infty p_r p_s x^n += \sum_{n = 0}^\infty p_n^2 x^{2n} ++ \sum_{\substack{n = 0 \\ r + s = n \\ r \neq s}}^\infty p_r p_s x^n +\end{align*} +$$ + +This expression can be integrated over *z* on the boundary defined by the inner product. The integral of the sum is the sum of the integrals since x as a series variable is totally independent of *z* (as well as [a separate kind of object](https://en.wikipedia.org/wiki/Indeterminate_%28variable%29)). + +$$ +\begin{align*} +\int_{-1}^1 F(x; z)^2 dz +&= \int_{-1}^1 \left( \sum_{n = 0}^\infty p_n^2 x^{2n} \right) dz ++ \int_{-1}^1 \left( \sum_{ +\substack{n = 0 \\ r + s = n \\ r \neq s}}^\infty p_r p_s x^n \right) \\ +&= \sum_{n = 0}^\infty \left( \int_{-1}^1 p_n^2 dz \right) x^{2n} ++ \sum_{\substack{n = 0 \\ r + s = n \\ r \neq s}}^\infty +\left( \int_{-1}^1 p_r p_s dz \right) x^n \\ +&= \sum_{n = 0}^\infty {2 \over 2n + 1} x^{2n} +\end{align*} +$$ + +When multiplied by *x*, this sum very much looks like the integral of another expression, + +$$ +\begin{align*} +x \int_{-1}^1 F(x; z)^2 dz +&= \sum_{n = 0}^\infty {2 \over 2n + 1} x^{2n+1} += \int \left( \sum_{n = 0}^\infty 2 x^{2n} \right) dx \\ +&= \int {2dx \over 1 -\ x^2} = \int { \left( {A \over 1 + x} + {B \over 1 -\ x} \right) } \\ +&= A\ln(1+x) -\ B\ln(1 -\ x) \\ +2 &= A(1 -\ x) + B(1 + x) +\implies \stackrel{x = -1}{A = 1}, \quad \stackrel{x = 1}{B = 1} \\ +\end{align*} +$$ + +As a reminder, the goal of these manipulations is to find a closed expression which resembles the evaluation of integral on the LHS with respect to *z*. Fortunately, the partial fraction decomposition is not only simple, but resembles an evaluated integral. + +Before continuing, it is imperative that *x* be divided out of both sides. This implies that the coefficient of *z* in the expression of the logarithm is a multiple of *x*. For example: + +$$ +\begin{align*} +\int_{-1}^1 F^2 dz +&= {1 \over x}( \ln(1+x) -\ \ln(1 -\ x) ) \\ +&= {1 \over x} \left[ \ln(1 +\ zx)^{\vphantom{0}} \right]_{z = -1}^1 \\ +\partial_z \ln(1 + zx) &= {x \over 1 + zx} \\ +\implies F^2 &\stackrel{?}{=} {1 \over 1 + zx} +\quad F \stackrel{?}{=} {1 \over \sqrt{1 + zx} } +\end{align*} +$$ + +But this expression is clearly wrong. It can be interpreted as a power series in *zx*, rather than just in *x*. This series can also come from a binomial expansion in $1 + zx$, and its terms compared with the ones we already know. The second Legendre polynomial is ${1 \over 2}(3z^2 -\ 1)$, but it is impossible that this expression will be generated, since it contains a term other than a multiple of $z^2$. + +Using logarithm identities, an alternate expression can be obtained + +$$ +\begin{align*} +\int_{-1}^1 F^2 dz +&= {1 \over x}( \ln(1+x) -\ \ln(1 -\ x) ) \\ +&= {1 \over 2x}( \ln((1+x)^2) -\ \ln((1 -\ x)^2) )\\ +&= {1 \over 2x}( \ln(1+ 2x + x^2) -\ {1 \over 2}\ln(1 -\ 2x + x^2) ) \\ +&= {1 \over 2x} \left[ \ln(1 + 2zx + x^2) \right]_{z = -1}^1 \\ +\partial_z \ln(1 + 2zx + x^2) &= {2x \over 1 + 2zx + x^2} \\ +\implies F^2 &\stackrel{?}{=} {1 \over 1 + 2zx + x^2} +\quad F \stackrel{?}{=} {1 \over \sqrt{1 + 2zx + x^2} } +\end{align*} +$$ + +The justification for this move is that we start with two polynomials, $1$ and $z$, so the expression must be quadratic in *x* to fit with the two initial values. Examining the first term of the series (i.e., the coefficient of *x*) by taking its derivative, this expression is only *almost* correct. + +$$ +\begin{align*} +\left. \left( \partial_x (1 + 2zx + x^2)^{-1/2} \right) \right|_{x = 0} &= +\left. -{1 \over 2} (1 + 2zx + x^2)^{-3/2} (2z + 2x) \right|_{x = 0} \\ +&= (1 + 0 + 0)^{-3/2} (-z -\ 0) \\ +&= -z +\end{align*} +$$ + +This has the opposite sign of $z$, the first term in the series. However, the series does correctly generate the zeroth term 1. Thus, by replacing *z* with -*z* in the expression of *F*, one obtains a new expression, which can be integrated to verify its validity. + +$$ +\begin{align*} +F &\stackrel{?}{=} {1 \over \sqrt{1 -\ 2zx + x^2} } \\ +\int_{-1}^1 {dz \over 1 -\ 2zx + x^2} +&= -{1 \over 2x}\left[ \ln( 1 -\ 2zx + x^2 ) \right]_{z=-1}^{1} \\ +&= -{1 \over 2x}( \ln( 1 -\ 2x + x^2 ) -\ \ln( 1 + 2x + x^2 ) ) \\ +&= -{1 \over 2x}( \ln( (1 -\ x)^2 ) -\ \ln( (1 + x)^2 ) ) \\ +&= {1 \over x}( \ln( 1 + x ) -\ \ln( 1 -\ x ) ) +\end{align*} +$$ + +Which is the exact same as the previous constraint given by the orthogonality condition. These previous hurdles are in part due to the non-uniqueness of the square root, but also because reversing integration over symmetric bounds is challenging. + +This final expression is correct, and the generating function for the Legendre polynomials is + +$$ +F(x; z) = {1 \over \sqrt{1 -\ 2zx + x^2} } +$$ + +Isn't this form somewhat familiar? It is simply the square root of the expression which generates the Chebyshev polynomials of the second kind (without an offset). + +$$ +F(x; z) = \sqrt{B(x; z) \over x} +$$ + +The generating function approach is (at least according to Wikipedia) how Legendre approached this problem. From the math above, it certainly has a degree of elegance to it, albeit one which requires some apparent (but justified) leaps in logic. + + +Recurrence +---------- + +With a closed expression finally known, is it possible to obtain a recurrence? It is, but doing so is no easier than the previous steps. The result and process for doing so are outlined on Wikipedia, but I will do so explicitly here. Begin by differentiating *F* with respect to *x*, which shifts the series and multiplies each polynomial by its place in the sequence (i.e., its degree) + +$$ +\begin{align*} +F(x; z) &= \sum_{n=0}^\infty p_n(z)x^n += {1 \over \sqrt{1 -\ 2zx + x^2} } \\ +\partial_x F(x; z) &= \sum_{n=1}^\infty n p_n(z)x^{n-1} += {z -\ x \over ( 1 -\ 2zx + x^2 )^{3/2} } += {z -\ x \over 1 -\ 2zx + x^2 } F(x; z) \\ +(z -\ x) \sum_{n=0}^\infty p_n x^n +&= (1 -\ 2zx + x^2) \sum_{n=1}^\infty n p_n x^{n-1} +\end{align*} +$$ + +With the sums re-exposed, the most difficult part is aligning like terms. The most natural range of summation starts at 1 (like the RHS), but the power of *x* is simply $x^n$ (like the LHS). Therefore, distributing over each sum and realigning indices yields: + +$$ +\text{LHS} \left \{ \begin{align*} +z\sum_{n=0}^\infty p_n x^n +&= z p_0 + z \sum_{n=1}^\infty p_n x^n +\\ +x\sum_{n=0}^\infty p_n x^n +&= \sum_{n=0}^\infty p_n x^{n+1} += \sum_{n=1}^\infty p_{n-1} x^n +\end{align*} \right . +\\ +\text{RHS} \left \{ \begin{align*} +\sum_{n=1}^\infty n p_n x^{n-1} +&= \sum_{n=0}^\infty (n + 1) p_{n + 1} x^n += p_1 + \sum_{n=1}^\infty (n + 1) p_{n + 1} x^n +\\ +2zx\sum_{n=1}^\infty n p_n x^{n-1} +&= 2z\sum_{n=1}^\infty n p_n x^n \\ +x^2\sum_{n=1}^\infty n p_n x^{n-1} +&= \sum_{n=1}^\infty n p_n x^{n+1} += \sum_{n=0}^\infty n p_n x^{n+1} += \sum_{n=1}^\infty (n -\ 1) p_{n-1} x^n +\end{align*}\right. +$$ + +Now the only problem which remains are the elements which are *not* in a sum, $zp_0$ and $p_1$. But these are both just *z*, and they are on opposite sides of the equation, so they cancel out. The remaining sums can be dropped entirely, and an explicit recurrence is formed. + +$$ +\begin{align*} +(z -\ x) \sum_{n=0}^\infty p_n x^n +&= (1 -\ 2zx + x^2) \sum_{n=1}^\infty n p_n x^{n-1} \\ +z \sum_{n=1}^\infty p_n x^n -\ \sum_{n=1}^\infty p_{n-1} x^n +&= \sum_{n=1}^\infty (n + 1) p_{n + 1} x^n -\ 2z\sum_{n=1}^\infty n p_n x^n ++ \sum_{n=1}^\infty (n -\ 1) p_{n-1} x^n \\ +z p_n -\ p_{n-1} &= +(n + 1) p_{n + 1} -\ 2z n p_n + (n -\ 1) p_{n-1} \\ +-(n + 1) p_{n + 1} &= +-z p_n (n + 1) -\ 2z n p_n + (n -\ 1) p_{n-1} + p_{n-1} \\ +(n + 1) p_{n + 1} &= (2n + 1)z p_n -\ n p_{n-1} +\end{align*} +$$ + +This form is called *Bonnet’s recursion formula*. Trying out this formula with $1$ and $z$ indeed produces the correct next term. + +$$ +\begin{align*} +(n + 1) p_{n + 1} &= (2n + 1)z p_n -\ n p_{n-1} \\ +(1 + 1) p_{1 + 1} &= (2(1) + 1)z p_1 -\ 1 p_{1-1} \\ +2 p_2 &= 3z(z) -\ p_0 \\ +p_2 &= {1 \over 2}(3z^2 -\ 1) +\end{align*} +$$ + +The recursion formula also implies something interesting: like the Chebyshev polynomials of the second kind, all polynomials are either totally even or totally odd. + +*** + +With the Chebyshev polynomials, the goal was to manipulate a recurrence relation into a generating function. However, the Legendre polynomials require the opposite; one first starts by assuming they have the generating function. Only through clever manipulations and experimentation can an expression be recovered, and from the expression, a recurrence. diff --git a/posts/misc/platonic_volume/index.qmd b/posts/misc/platonic_volume/index.qmd new file mode 100644 index 0000000..922412e --- /dev/null +++ b/posts/misc/platonic_volume/index.qmd @@ -0,0 +1,371 @@ +--- +format: + html: + html-math-method: katex +--- + + +On the Volume of the Platonic Solids +==================================== + + +The Platonic solids have been known for millennia. They bear the name of Plato, who spoke of them in his dialogue *Timaeus*. He describes their "construction" (sans the dodecahedron) from the most basic "isosceles and scalene" triangles, or in modern parlance, the "45-45-90 and 30-60-90" triangles. However, the construction was not mathematical, and to my knowledge, each solid was first rigorously described from principles in Book XIII of Euclid's Elements. + +In my teenage years, I recall viewing articles on the solids with their volumes so proudly displayed along with their surface area. While the latter quantity may be troublesome in the case of the dodecahedron (as the geometry of regular pentagons is not widely taught), it is easy for any student of trigonometry to calculate the surface area of the solids made of equilateral triangles, and easy for any child who knows of squares to do so for the cube. On the other hand, the volume is somewhat mystical. + +There is only one free variable in a Platonic solid, its edge length, which means that their volumes are parametrized by this value alone. To be a true unitless value, a volume must be in a ratio with another volume. The cube has the simplest expression for its volume; it is simply the side length cubed. Therefore in the following post, I will derive the ratio of each solid's volume to the volume of the cube formed by any side. + +This post will calculate the volume without using any trigonometric functions (sine, cosine tangent), opting instead for a more compass-and-straightedge approach. For this reason, *square* of the volume will be calculated initially to better cooperate with the Pythagorean theorem. Additionally, except for the octahedron, the edge length of every solid will be 2 to simplify the bisection of edges. This happens to coincide with Plato's description; two 30-60-90 triangles were used to make an equilateral triangle, meaning that its edge length was twice the "unit" length: the shortest side of the 30-60-90 triangle. + + +A Recap of Geometry +------------------- + +For those with only a vague recollection (or perhaps none at all) of geometry, this section is intended as a refresher. + + +### Planar Geometry + +There are [many centers of a triangle](https://faculty.evansville.edu/ck6/encyclopedia/etc.html), but there are two of primary interest: + +- The *circumcenter* is the center of the circle containing every vertex, meaning that it is equidistant from every vertex. + - It is constructed by finding the intersection of the edges' perpendicular bisectors. + - The distance from a vertex to the circumcenter is called the *circumradius* (*c*). +- The *incenter* is equidistant from every edge, meaning that the length of the perpendicular segment connecting an edge and the incenter is the same for all edges. + - It is constructed by finding the intersection of the lines which bisect each angle. + - The perpendicular distance from an edge to the incenter is called the *inradius* (*a*). + +::: {} +![](index_files/triangle centers.png) + +Constructing the circumcenter and incenter. Angle bisectors in blue, perpendicular bisectors in red, in- and circumradii in green. +::: + +The inradius is special because it is also an altitude for a triangle formed by the inradius and an edge of the larger triangle. This means that the area of the larger triangle is the sum of these smaller triangles. + +![](index_files/incenter area.png) + +$$ +\begin{align*} +A &= \left ({e_1 a \over 2} + {e_2 a \over 2} + {e_3 a \over 2} \right) = +\left ({a \over 2} \right ) (e_1 + e_2 + e_3) \\ +&= {Pa \over 2} +\end{align*} +$$ + +This gives an expression for the area. For an equilateral triangle, these two centers coincide. This is because the perpendicular bisectors of the edges *are* the angle bisectors. In fact, the bisection of an angle involves constructing a rhombus, which is made up of two isosceles triangles (of which the equilateral triangle is a special case). In this case, the inradius is also called the *apothem*, and the difference between it and the circumradius is immediately apparent and called the *sagitta* (*s*). + +![](index_files/equilateral center.png) + +This idea of incenters and circumcenters can be extended to other 2D figures such as the square and regular pentagon. For a square, the center is simply the intersection of the diagonals (i.e., the diagonals' common midpoint). The pentagon is trickier, and will be discussed later. Regardless, the expression for the area ${Pa \over 2}$ still works, since the polygon can be triangulated through the center in a similar way. + +![](index_files/square pentagon circles.png) + + +### Cubes, Prisms, and Pyramids + +Now we speak of 3D geometry. The volume of a prism is equal to the height times the area of the base, where the "height" is orthogonal to the plane of the base. Pyramids with the same height and base have one-third this area. + +$$ +V_\text{prism} = Bh,~~ V_\text{pyramid} = {Bh \over 3} +$$ + +This volume formula can be made more intuitive by considering the cube. The pyramid formed by one of the faces and an edge perpendicular to it will contain one square and two half-squares, or two squares in total. Therefore three pyramids are needed to recreate all six faces of the cube. + +For a slightly more detailed explanation, consider a point inside the face on top of the cube. Its (perpendicular) distance from one edge is *x* and its distance to an edge adjacent to that is *y*. Connecting all other bases to this point produces five pyramids, whose bases all have the same area. Designate these pyramids as "bottom", "left", "right", "front", and "back", where left and right correspond to *x* and front and back correspond to *y*. + +![](index_files/cube pyramids.png) + +$$ +\begin{align*} +V_\text{cube} &= Bh = V_\text{bottom} + V_\text{left} + V_\text{right} ++ V_\text{front} + V_\text{back} \\ +&= rBh + rBx + rB(h-x) + rBy + rB(h-y) \\ +&= rBh + rBh + rBh \implies 1 = 3r \\ +r &= {1 \over 3} +\end{align*} +$$ + +This can be generalized to a pyramid based on any prism, where the top point lies in the plane of one of the bases. However, this is beyond the scope of this post. + + +Simple Solids: the Octahedron and the Tetrahedron +------------------------------------------------- + +While "simple" is a bit of a misnomer, their volumes are easiest to appreciate, since they do not need regular pentagons. + + +### Octahedron + +The octahedron can be thought of as two square pyramids joined end-on-end, with uniform edge length throughout. Since the base is a square, its center is equidistant from the vertices of the base. An alternative, congruent square can be noticed by the symmetry of the octahedron, meaning the center is also equidistant from the top of the square pyramid, and that the segment connecting the two is an altitude of the pyramid. + +::: {} +![](index_files/octahedron squares.png) + +Primary square in blue, secondary square in red. Diagonals of both squares shown. +::: + +The length of this altitude is simply half of the diagonal of the square. Therefore, the volume of an octahedron (calculated using edge length 1) is: + +$$ +\begin{align*} +B^2 &= (1^2)^2 \\ +(2h)^2 &= 4h^2 = 1^2 + 1^2 = 2 \\ +V_\text{sq.pyr.}^2 &= {B^2 h^2 \over 3^2} = +{1 \cdot {2 / 4} \over 3^2} \\ +(1^3 \cdot V_\text{oct})^2 &= (2V_\text{sq.pyr})^2 = 4V_\text{sq.pyr}^2 = +4 \cdot {2 / 4 \over 3^2} = {2 \over 3^2} \\ \\ +V_\text{oct} &= {\sqrt{2} \over 3} +\end{align*} +$$ + + +### Tetrahedron + +The tetrahedron is itself a pyramid. First, the (square of the) area of the base of an equilateral triangle must be known. As a reminder, this and all future solids will have edge length 2. + +:::: {layout-ncol="2"} +::: {.column width="49%"} +![](index_files/equilateral triangle area.png) +::: + +::: {.column width="49%"} +$$ +\begin{align*} +d_\text{altitude}^2 &= +\textcolor{orange}{2}^2 -\ \textcolor{green}{1}^2 = 3 \\ +B^2 &= \left ( {2 \cdot d_\text{altitude} \over 2} \right )^2 = 3 +\end{align*} +$$ +::: +:::: + +Next, bisect the tetrahedron through one edge and the altitudes of two faces. The altitudes form the legs of an isosceles triangle, so bisecting the angle where they meet (perpendicularly) bisects the remaining edge. + +:::: {layout-ncol="2"} +::: {.column width="49%"} +![](index_files/tetrahedron bisect.png) +::: + +::: {.column width="49%"} +$$ +\begin{align*} +\textcolor{blue}{d_\text{length}}^2 &= +d_\text{altitude}^2 -\ \textcolor{green}{1}^2 = 2 \\ +(2A_\text{center})^2 &= (2d_\text{length})^2 = (\textcolor{red}{h} d_\text{altitude})^2 \\ +&= 4 \cdot 2 = 3h^2 \\ +h^2 &= 8 / 3 +\end{align*} +$$ +::: +:::: + +Since *h* is known, we can calculate the volume. Note that the volume is multiplied by the cube of the side length to produce the correct ratio to a unit cube. + +$$ +\begin{align*} +({2^3 \cdot V_\text{tet}})^2 &= {B^2 h^2 \over 3^2} = {3 \cdot (8/3) \over 3^2} \\ +V_\text{tet}^2 &= {8 \over 2^6 \cdot 3^2} = {1 \over 2^3 \cdot 3^2} = {1 \over 2 \cdot 6^2} \\ +V_\text{tet} &= \sqrt{1 \over 6^2 \cdot 2} = {1 \over 6\sqrt 2} +\end{align*} +$$ + + +### Returning to 2D: Regular Pentagons + +Both of the icosahedron and dodecahedron contain regular pentagons. Thus, it is necessary to examine them in detail. + +The regular pentagon has five diagonals, which form a pentagram. Since all angles in a regular pentagon are equal, the trapezoid formed by three consecutive edges and one diagonal is isosceles. This means the diagonal is parallel to one of the edges, which applies to all diagonals by symmetry. Since the diagonal and side are parallel, this means that any two adjacent edges form a parallelogram with part of the diagonal. More specifically, it is a rhombus and those "parts of diagonals" have length equal to the side. + +::: {} +![](index_files/pentagram rhombus.png) + +Left: Pentagram in regular pentagon; Middle: Isosceles trapezoid, with parallel lines marked in blue; Right: Rhombus in regular pentagon +::: + +Bisect the pentagon vertically and let the length of half of the diagonal of a pentagon be *d*, half the length of the other diagonal of a rhombus be *h*, and the remaining height of the pentagon be *g*. + +![](index_files/pentagon measurements.png) + +$$ +\begin{align*} +\textcolor{orange}{d}^2 + \textcolor{red}{h}^2 &= \textcolor{blue}{2}^2 \\ +2\textcolor{darkblue}{A_\text{blue}} &= 2\textcolor{green}{g} = h(\textcolor{magenta}{d -\ (2 -\ d)}) = h (2d -\ 2) \\ +\implies g &= {h(2d -\ 2) \over 2} = h(d -\ 1) +\end{align*} +$$ + +Notice that the center of a pentagram contains a regular pentagon. This means that the ratio of its height to the side is equal to the ratio of the larger pentagon's height to its side. This is enough information to deduce *d*: + +\begin{align*} +{\textcolor{red}{h} \over 2(\textcolor{brown}{2 -\ d})} &= +{2\textcolor{red}{h} + \textcolor{green}{g} \over \textcolor{blue}{2}} = + +{2h + h(d-1) \over 2} = {h(1 + d) \over 2 } \\ +2h &= 2h(1 + d)(2 -\ d) \\ +1 &= (1 + d)(2 -\ d) = 2 -\ d + 2d -\ d^2 \\ +0 &= d^2 -\ d -\ 1 +\end{align*} + +This is the minimal polynomial of the golden ratio $\phi$; it is half the length of the diagonal, so the ratio of a diagonal to a side is also $\phi$. To make calculations easier, some conversions will be made to base $\phi$, or phinary. If you are not familiar already with phinary, I have already written at length about it [here](). Finally, the apothem *a* and height *l* can be calculated by similar triangles. + +::: {} +![](index_files/pentagon apothem.png) +::: + +$$ +\begin{align*} +\textcolor{blue}{c \over a} &= \textcolor{brown}{2 \over \phi},~ +a^2 + 1^2 = c^2 \implies 1 = c^2 -\ a^2 = (c + a)(c -\ a) \\ +l &= c + a = {2a \over \phi} + a = a{2 + \phi \over \phi} = a{12_\phi \over 10_\phi} = +a{2\bar{1}0_\phi \over 10_\phi} = a(2\bar{1}_\phi) \\ \\ +s &= c -\ a = {2a \over \phi} -\ a = a{2 -\ \phi \over \phi} = a{\bar{1}2_\phi \over 10_\phi} = +a{2\bar{3}0_\phi \over 10_\phi} = a(2\bar{3}_\phi) \\ \\ \\ +1 &= ls = a^2(2\bar{1}_\phi)(2\bar{3}_\phi) = a^2(4\bar{8}3_\phi) = +a^2(\bar{4}7_\phi) = a^2(3.\bar{4}_\phi) \\ +a^2 &= {1 \over 3.\bar{4}_\phi} \cdot {43_\phi \over 43_\phi} = +{43_\phi \over [12]\bar{7}.[\bar{12}]_\phi} = {3 + 4\phi \over 5} \\ \\ +\implies l^2 &= a^2(2\bar{1}_\phi)^2 = {3 + 4\phi \over 5} \cdot (4\bar{4}1_\phi) += {3 + 4\phi \over 5} \cdot 5 = 3 + 4\phi +\end{align*} +$$ + +The last few steps in solving for $a^2$ are somewhat tricky. The conjugate of $\phi$ is $-{1 \over \phi}$. Since the digit in the $\phi^{-1}$ place value is negative, its conjugate has a positive value in the $\phi$ place value; i.e., $3.\bar{4}_\phi^* = 43_\phi$. Multiplying a quadratic root by its conjugate produces an integer value, which means that the scary quantity $[12]\bar{7}.[\bar{12}]_\phi$ resolves cleanly to 5. + +The division can also be done explicitly in phinary: + +$$ +\begin{align*} +{1 \over 3.\bar{4}_\phi} &= {1 \over 0.\bar{1}3_\phi} = +{500_\phi \over 5 (\bar{1}3_\phi)} = +{233_\phi + (0 = \textcolor{red}{4\bar{4}}\bar{4}0_\phi = +\textcolor{red}{26\bar{2}}\bar{4}0_\phi) \over 5 (\bar{1}3_\phi)} \\ +&= {\bar{2}60\bar{1}3_\phi \over 5 (\bar{1}3_\phi)} = +{2001_\phi \over 5} = +{221_\phi \over 5} = +{43_\phi \over 5} = +{3 + 4\phi \over 5} +\end{align*} +$$ + +Finally, with the length and apothem of a regular pentagon in tow, the geometry of the final two solids may be explored. + + +The Remaining Solids +-------------------- + +The icosahedron and dodecahedron are easiest to dissect as many pyramids joined to a single center. This is reminiscent of the area formula which uses the triangulation of a polygon through the incenter. + +The altitude (*h*) of any one pyramid is the radius of the *insphere* of the solid, which is tangent to the plane of every face. Similarly, the *circumsphere* (circumradius, *r*) contains all vertices, and the *midsphere* (midradius, $\rho$) is tangent to every edge. These will become important shortly. + + +### The Icosahedron: an Antiprism in Profile + +The icosahedron may also be thought of as two pentagonal pyramids connected to either base of a pentagonal *antiprism*. An antiprism is a figure similar to a prism, but with the one of the bases twisted relative to the other and with (equilateral) triangles joining them. + +:::: {layout-ncol="2"} +::: {.column width="49%"} +![](index_files/icosahedron with antiprism.png) + +Icosahedron with pentagonal antiprism in blue +::: + +::: {.column width="49%"} +![](index_files/equilateral triangles.png) + +Construction showing $2a = c$ +::: +:::: + +A segment connecting the centers of two antipodal faces is a diameter of the insphere. The altitude of one of these faces will be cut into circumradius and inradius. By similar triangles, the circumradius of an equilateral triangle is exactly twice the length of the inradius. This means the inradius is 1/3 of the altitude, or 1/9 of the square of the altitude. With edge length 2, the square of the altitude is 3, so the square of the inradius is ${3 \over 9} = {1 \over 3}$ . + +The pentagonal antiprism may be bisected bisected along the plane containing the altitudes of two triangles opposite one another. This forms a parallelogram with side lengths of the altitude of an equilateral triangle and height of a pentagon. + +![](index_files/antiprism measures.png) + +$$ +\begin{align*} +(\textcolor{green}{2h})^2 &= \textcolor{red}{l}^2 -\ {1 \over 3} = 3 + 4\phi -\ {1 \over 3} = +{3(3 + 4\phi) -\ 1 \over 3} \\ +h^2 &= {8 + 12\phi \over 3 \cdot 4} = {2 + 3\phi \over 3} = {32_\phi \over 3} \\ +32_\phi &= 210_\phi = 1100_\phi = 10000_\phi = \phi^4 \\ \\ +(2^3 \cdot V_\text{ico})^2 &= \left ( 20 \cdot {Bh \over 3} \right )^2 = +{20^2 B^2 h^2 \over 3^2} = {5^2 \cdot 4^2 \cdot 3 \cdot {\phi^4 \over 3} \over 3^2} = +{5^2 \cdot 2^4 \cdot \phi^4 \over 3^2} \\ +V^2 &= {5^2 \cdot 2^4 \cdot \phi^4 \over 2^6 \cdot 3^2} = +{5^2 \cdot \phi^4 \over 2^2 \cdot 3^2} \\ +V &= {5 \phi^2 \over 6} +\end{align*} +$$ + + +### The Dodecahedron + +The dodecahedron is a bit trickier. It belongs to a class of polyhedra known as *truncated trapezohedra*. However, the bisection trick from the icosahedron still works. Begin by bisecting the solid through antipodal altitudes. This produces an oblong hexagon made up of four pentagon heights and two edges. + +![](index_files/dodecahedron with hemisphere.png) + +The segment connecting the antipodal midpoints bisects the hexagon into two (isosceles) trapezoids, and is a diameter of the midsphere. Additionally, it is parallel to the two edges. A second midradius is perpendicular to this one, bisecting the trapezoid. + +![](index_files/dodecahedron measures.png) + +The inradius is the altitude of a triangle formed by the length of a pentagon (its base), a midradius, and a circumradius. However, the altitude with respect to the midradius is another midradius. This means that the height can be found by equating areas and completing the square. + +$$ +\begin{align*} +a^2 + \textcolor{green}{h}^2 &= \textcolor{blue}{\rho}^2,~~ +\textcolor{orange}{l}\textcolor{green}{h} = +\textcolor{blue}{\rho \rho} \implies \textcolor{orange}{l}^2\textcolor{green}{h}^2 = +(\textcolor{orange}{(2\bar{1}_\phi) a})^2 h^2 = 5a^2h^2 = \textcolor{blue}{\rho}^4 \\ +5a^2h^2 &= (a^2 + h^2)^2 = a^4 + 2a^2h^2 + h^4 \\ \\ +0 &= a^4 -\ 3a^2h^2 + h^4 = (h^2 -\ x)^2 + y = h^4 -\ 2xh^2 + x^2 + y \\ +-2x &= -3a^2 \implies x = {3a^2 \over 2},~~ +x^2 + y = {9a^4 \over 4} + y = a^4 \\ +y &= {4a^4 \over 4} -\ {9a^4 \over 4} = -{5a^4 \over 4} = +-{(2\bar{1}_\phi)^2 a^4 \over 4} = -\left ( {(2\bar{1}_\phi) a^2 \over 2} \right )^2 +\\ \\ +(h^2 -\ {3a^2 \over 2})^2 &= -y = \left ( {(2\bar{1}_\phi) a^2 \over 2} \right )^2 \\ +h^2 -\ {3a^2 \over 2} &= {(2\bar{1}_\phi) a^2 \over 2} \\ +h^2 &= {(2\bar{1}_\phi) a^2 \over 2} + {3a^2 \over 2} = {(22_\phi) a^2 \over 2} = +(11_\phi) a^2 \\ +&= {(11_\phi)(43_\phi) \over 5} = {473_\phi \over 5} = +{[11]7_\phi \over 5} = {7 + 11\phi \over 5} +\end{align*} +$$ + +With the square of the height known, all that is left to do is find the volume. + +$$ +\begin{align*} +B^2 &= \left( {Pa \over 2} \right)^2 = (5a)^2 = 25a^2 = 5(43_\phi) \\ +5h^2 &= [11]7_\phi = 740_\phi = 4300_\phi = (43_\phi)(100_\phi) \\ +(2^3 \cdot V_\text{dodec})^2 &= \left (12 \cdot {Bh \over 3} \right )^2 = +4^2 B^2 h^2 = 2^4 \cdot 5(43_\phi) \cdot {(43_\phi)(100_\phi) \over 5} \\ +V^2 &= {2^4 \cdot (43_\phi)^2 \cdot (100_\phi) \over 2^6} = +{(43_\phi)^2(10_\phi)^2 \over 2^2} \\ +V &= {(43_\phi)(10_\phi) \over 2} = {(430_\phi) \over 2} = {(74_\phi) \over 2} = +{4 + 7\phi \over 2} +\end{align*} +$$ + + +Closing +------- + +Since each of these volumes has been calculated algebraically, there have been no approximate decimal forms. Ordered by size, the volumes of each of the solids are: + +| Solid | Volume | Approximation | Length of Side with Unit Volume | +|--------------|-----------------------|-----------------|---------------------------------| +| Tetrahedron | ${1 \over 6\sqrt 2}$ | 0.1178511302... | 2.039648903... | +| Octahedron | ${\sqrt{2} \over 3}$ | 0.4714045208... | 1.284898293... | +| Cube | $1^3$ | 1 | 1 | +| Icosahedron | ${5\phi^2 \over 6}$ | 2.181694991... | 0.7710253465... | +| Dodecahedron | ${4 + 7\phi \over 2}$ | 7.663118961... |0.5072220724... | + +The dodecahedron being so much larger than the icosahedron surprised me, to be honest. When one glances at a set of dice (as one does), it seems like the d20 is larger than the d12, albeit with smaller edges. However, at least in my set, the edges of the d20 are in fact about 1.5 times as long as those of the d12, implying their volumes are roughly equal. + +*** + +I tried to use as much coordinate-free geometry as I could in producing these diagrams. GeoGebra lacks a tool for producing Platonic solids other than cubes and tetrahedra, so I ended up using approximations for the octahedron and icosahedron diagrams. On the other hand, the hexagon I described in the dodecahedron is of such importance to its construction that I ended up constructing it from scratch. I am rather proud of this because I did so without looking up someone else's. After having written this post, I feel much more competent with compass-and-straightedge constructions. + +All diagrams made with GeoGebra.