tweak language in chebyshev.1 and stereo.2

posts/math/chebyshev/1/index.qmd

@@ -259,18 +259,18 @@ $$
   = (z^2 - z - 1) (z^2 + z - 1)
 $$
 
-According to how we derived this series, when $z = 2\cos(\theta)$, the roots of this polynomial
+According to how we derived this polynomial, when $z = 2\cos(\theta)$, the roots
   correspond to when $\sin(5\theta) / \sin(\theta) = 0$.
 This relation itself is true when $\theta = \pi / 5$, since $\sin(5 \pi / 5) = 0$.
 
 One of the factors must therefore be the minimal polynomial of $2\cos(\pi / 5 )$.
-The former happens to be correct correct, since $2\cos( \pi / 5 ) = \varphi$, the golden ratio.
+The former happens to be the correct choice, since $2\cos( \pi / 5 ) = \varphi$, the golden ratio.
 Note that the second factor is the first evaluated at -*z*.
 
 
 #### Heptagons
 
-An example of where constructability fails is for $2\cos( \pi / 7 )$.
+An example where constructability fails is for $2\cos( \pi / 7 )$.
 
 $$
 \begin{align*}
@@ -289,7 +289,8 @@ But there are no (nondegenerate) cubics that one can produce via compass and str
 
 #### Decagons
 
-One might think the same of $2\cos(\pi /10 )$
+One might think the same of $2\cos(\pi /10 )$.
+The relevant polynomial is:
 
 $$
 \begin{align*}
@@ -300,16 +301,17 @@ $$
 \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.
+This expression also contains the polynomials for $2\cos( \pi / 5 )$,
+  since connecting every other vertex of a regular decagon produces
+  two disjoint regular pentagons.
 
 ![
    
 ](./decagon_divisible.png)
 
 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 are not constructible.
+  which seems to suggest that it will require a fourth root and imply that decagons
+  are somehow not constructible.
 However, it can be solved by completing the square...
 
 $$
@@ -496,12 +498,12 @@ $$
 If $z = 1$, these two generating functions are equal.
 The same can be said for $z = 2$ with the generating function of the Pell numbers,
   and so on for higher recurrences (corresponding to metallic means) for higher integral *z*.
+In fact, 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).
 
 In terms of the Chebyshev polynomials, this series manipulation removes the alternation in
   the coefficients of $U_n$, restoring 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
@@ -674,6 +676,7 @@ Though they were present in the earlier Chebyshev table,
 
 ```{python}
 #| echo: false
+#| label: tbl-cosinepolynomials
 #| classes: plain
 
 def poly_to_rising_power_list(poly, var):

posts/math/stereo/2/index.qmd

@@ -190,7 +190,7 @@ For comparison, the complex exponential (as it is a parallel construction) has a
 Since the exponential function is its own derivative, the expression acquires
   an imaginary coefficient through the chain rule.
 
-[^1]: This is forgoing the fact that complex derivatives require more care than their real counterparts.
+[^1]: This is forgoing the extra care that complex derivatives require beyond their real counterparts.
     It matters slightly less in this case since this function is complex-valued, but has a real parameter.
 
 $$
@@ -249,12 +249,12 @@ $$
   = 2\pi i
 $$
 
-In this example, Γ is a counterclockwise curve parametrized by γ which loops once around
+In this example, *Γ* is a counterclockwise curve parametrized by *γ* which loops once around
   the pole at *z* = 0.
 More loops will scale this by a factor according to the number of loops.
 
 Normally this equality is demonstrated with the complex exponential, but will $o_1$ work just as well?
-If Γ is the unit circle, the integral is:
+If *Γ* is the unit circle, the integral is:
 
 $$
 \oint_\Gamma {1 \over z} dz
@@ -281,9 +281,9 @@ $$
 \end{gather*}
 $$
 
-It is certainly possible to perform these contour integrals along straight lines;
-  in fact, integrating along lines from 1 to *i* to -1 to -*i* deals with a
-  similar integral involving arctangent.
+It is certainly possible to perform these contour integrals along straight lines
+  in the complex plane; in fact, making *Γ* a diamond-shaped contour from
+  1 to *i* to -1 to -*i* produces a similar integral involving arctangent.
 However, the best one can do to construct more loops with lines is to count each line
   multiple times, which isn't extraordinarily convincing.
 
@@ -304,7 +304,7 @@ $$
 $$
 
 This has an additional benefit: using the series form of $1 / (1 + t^2)$ and integrating,
-  one obtains the series form of the arctangent.
+  one obtains [the series form of arctangent](https://en.wikipedia.org/wiki/Arctangent_series).
 This series converges for $-1 \le t \le 1$, which happens to match the bounds of integration.
 The convergence of this series is fairly important, since it is tied to formulas for π,
   in particular [Leibniz's formula](https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80).