add drafts from wordpress

drafts/alg_homo/index.qmd

@@ -0,0 +1,55 @@
+---
+format:
+  html:
+    html-math-method: katex
+---
+
+
+Viewing Homology Geometrically
+==============================
+
+Delta sets for the Klein bottle, torus, and projective plane.
+
+Define the delta maps in the natural way.
+
+$$
+0  \overset{\partial_3}{\longrightarrow}
+C_2 \overset{\partial_2}{\longrightarrow}
+C_1 \overset{\partial_1}{\longrightarrow}
+C_0
+$$
+
+Note that multiplying d1 by d2
+
+$$
+0  \overset{\delta_3}{\longleftarrow}
+C_2 \overset{\delta_2}{\longleftarrow}
+C_1 \overset{\delta_1}{\longleftarrow}
+C_0
+$$
+
+Fortunately, defining these maps is easy; we can just transpose the matrices by letting $\delta_\bullet = \partial_\bullet {}^T$.
+
+For the projective plane, we end up modding out the base space. But what does this actually look like?
+
+$$
+\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \left/
+\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}
+\right.
+$$
+
+On the left is the integer lattice and on the right is the sublattice generated by (1, -1) and (1, 1). These can be overlaid to show that there are obviously twice as many points in the former as there are the latter. When we perform the action of "modding out", all of the points in the sublattice go to 0. So in the quotient space, we only have two types of points: ones which were on the sublattice (0) and points which were not (1). But this describes the exact same thing as $Z_2$, which happens to be the homology group under discussion.
+
+The Klein bottle is a little more complicated. In this case, we end up modding a rank 3 space out by a rank 2 space.
+
+$$
+\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \left/
+\begin{pmatrix} 1 & -1 \\ 1 & 1 \\ -1 & 1 \end{pmatrix}
+\right.
+$$
+
+The free component is the line normal to the plane described by the column space. But this is precisely the kernel of the map.
+
+The torsion subgroup turns out to be identical to the one we already discussed for the Klein bottle, only we're in a plane that's slightly skewed with respect to the integer lattice.
+
+![]()

drafts/pseudo_taylor/index.qmd

@@ -0,0 +1,195 @@
+---
+format:
+  html:
+    html-math-method: katex
+---
+
+
+Notes on Pseudo-Taylor Series
+=============================
+
+Whilst examining type algebra, I found that the geometric series, a cornerstone of calculus, could be derived by analyzing its definition as a rewrite rule. That is, $(1 - X)^{-1}$ is expressed similarly to an exponential object; when it is written as $-1 \rightarrow (1 - X)$, it can be interpreted thus:
+
+$$
+(1 - X)^{-1} = -1 ~\rightarrow~ 1 - X \\ \\
+\begin{matrix}
+-1
+& \rightarrow&
+1 - X
+\\&&
+1 + (-1) \times X
+& \rightarrow &
+1 +(1 - X)X
+\\  & & & &
+1 + X + (- 1) \times X^2
+& \rightarrow &
+1 + X + (1-X)X^2
+\\ &&&&&& ...
+\end{matrix}
+$$
+
+Which, in the limit, becomes the geometric series
+
+$$
+1 + X + X^2 + X^3 + X^4 + ... = \sum_{n=0}^\infty X^n
+$$
+
+I would like dedicate a bit more of my time to examining this apparent correspondence between Taylor series and rewrite rules, which as far as I can tell, appears to be new territory.
+
+
+Why is -1 Special?
+------------------
+
+One of my first questions upon seeing this was why -1 is special. In type theory, integers can represent the number of values which occupy a type. Naturally (if you'll pardon the pun), this only makes sense for the positives and zero. But what is it that distinguishes -1 from any other negative number?
+
+Naively, we can examine what happens when examining $-2 \rightarrow 1 - 2X$:
+
+$$
+\textcolor{blue}{-2} \rightarrow 1 - 2X \\
+\begin{matrix}
+\textcolor{blue}{-2}
+& \stackrel{?}{\rightarrow} &
+1 - 2X
+\\&&
+1 + (\textcolor{blue}{-2}) \times X
+& \stackrel{?}{\rightarrow} &
+1 +(1 - 2X)X
+\\  & & & &
+1 + X + (\textcolor{blue}{-2}) \times X^2
+& \stackrel{?}{\rightarrow} &
+1 + X + (1-2X)X^2
+\\ &&&&&& ...
+\end{matrix}
+$$
+
+But we know that this arrowed expression should evaluate to $(1-2X)^{-2}$
+
+$$
+\textcolor{blue}{-2} \rightarrow 1 - 2X  =
+(1 - 2X)^{-2} =
+(1 - 4X + 4X^2)^{-1} =
+\textcolor{red}{-1} \rightarrow 1 - 4X + 4X^2 \\ \ \\
+\begin{matrix}
+\textcolor{red}{-1}
+& \rightarrow &
+1 - 4X + 4X^2
+\\&&
+1 + (\textcolor{red}{-1}) \times 4X + 4X^2
+& \rightarrow &
+1 +(1 - 4X + 4X^2) \times 4X + 4X^2
+\\  & & & &
+1 + 4X - 12X^2  + 16X^3
+\\  & & & &
+1 + 4X + (\textcolor{red}{-1}) \times 12X^2  + 16X^3
+& \rightarrow & ...
+\end{matrix}
+$$
+
+Continuing the rewrite rule gives coefficients of 12, 32, 80, 192... for $X^2$ onward. And indeed, the [Taylor series for the reciprocal](https://www.wolframalpha.com/input?i=taylor+series+%281+-+4X+%2B+4X%5E2%29%5E-1) of $1 - 4X + 4X^2$ is
+
+This gives us a bit more information: -1 is special because it factors out of all other negatives. From there, we can evaluate the remaining positive exponent by the binomial theorem.
+
+
+Ordering
+--------
+
+What happens when there are multiple negatives in the denominator? For example, the expression $(1 - X -X^2)^{-1}$ has the Fibonacci numbers as its Taylor series coefficients. If we try replacing all the -1s at once, we don't even come close past the first numbers:
+
+$$
+{1 \over 1 - X -X^2} = \textcolor{red}{-1} \rightarrow 1 - X - X^2 \\ \ \\
+\begin{matrix}
+\textcolor{red}{-1}
+& \rightarrow &
+1 - X - X^2
+\\&&
+1 + (\textcolor{red}{-1}) \times X + (\textcolor{red}{-1}) \times X^2
+& \rightarrow &
+1 + (1 - X - X^2 )X + (1 - X - X^2 )X^2
+\\  & & & &
+1 + X - X^2 - X^3 + X^2 - X^3 - X^4
+\\  & & & &
+1 + X - 2X^3  - X^4
+\\  & & & &
+1 + X + (\textcolor{red}{-1}) \times 2X^3  + (\textcolor{red}{-1}) \times X^4
+\\  & & & & ...
+\end{matrix}
+$$
+
+The [katex]X^2[/katex] terms cancel after the first iterated replacement. This is clearly not right; the Fibonacci number following two 1's is 2.
+
+In fact, all is well if we replace less aggressively.
+
+$$
+\begin{matrix}
+\textcolor{red}{-1}
+& \rightarrow &
+1 - X - X^2
+\\&&
+1 + (\textcolor{red}{-1}) \times X -X^2
+& \rightarrow &
+1 + (1 - X - X^2 ) \times X - X^2
+\\  & & & &
+1 + X - X^2 - X^3 -X^2
+\\  & & & &
+1 + X - 2X^3  - X^3
+\\  & &
+1 + X+  (1- X - X^2)\times2X^2 - X^3
+& \leftarrow &
+1 + X + (\textcolor{red}{-1}) \times 2X^2 - X^3
+\\  & &
+1 + X + 2X^2 - 2X^3 - 2X^4 - X^3
+\\  & &
+1 + X + 2X^2 - 3X^3 - 2X^4
+\\  & &
+1 + X + 2X^2 + (\textcolor{red}{-1})\times 3X^3 - 2X^4
+& \rightarrow & ...
+\end{matrix} \\ \ \\
+{1 \over 1 - X - X^2} = 1 + X + 2X^2 + 3X^3 + 5X^4 + 8X^5 + ...
+$$
+
+This gives us another way of obtaining the generating function for the Fibonacci numbers.
+
+$$
+{1 \over 1 - X -X^2} = \textcolor{red}{-1} \rightarrow 1 - X - X^2 \\ \ \\
+\begin{matrix}
+\textcolor{red}{-1}
+& \rightarrow &
+F_0 + F_1 X + F_2 X^2 + ... - F_{n-1} X^{n-1} - F_{n-2} X^n
+\\ & &
+... + (-1) \times  F_{n-1} X^{n-1} - F_{n-2} X^n
+\\ & &  \downarrow
+\\ & &
+... + (1 - X - X^2) \times  F_{n-1} X^{n-1} - F_{n-2} X^n
+\\ & &
+... + F_{n-1} X^{n-1} - F_{n-1} X^n - F_{n-1} X^{n+1} - F_{n-2} X^n
+\\ & &
+... + F_{n-1} X^{n-1} - (F_{n-1} + F_{n-2}) X^n - F_{n-1} X^{n+1}
+\\ & &
+... + F_{n-1} X^{n-1} - F_n X^n - F_{n-1} X^{n+1}
+\end{matrix}
+$$
+
+
+Complex Roots
+-------------
+
+Negatives?
+
+Must have a pole?
+
+
+Continued Fractions
+-------------------
+
+I am the most interested in generalizing rewrite rule-style types. The recursive equation for a forest is actually similar to the continued fraction for the Catalan numbers:
+
+$$
+F(X) = {1 \over 1 - X \times F(X)} = {1 \over 1 - X \times {1 \over 1 - X \times F(X)}} =...= \cfrac{1}{1 - \cfrac{X}{1- \cfrac{X}{1- \cfrac{X}{_\ddots} }}}
+\\
+= 1 + X\times F(X) + X^2 \times F(X)^2 +X^3 \times F(X)^3 + ... \\
+= 1 + X + 2X^2 +5X^3 + 14X^4 + 42X^5+...
+$$
+
+Successively replacing F(X) with itself will produce the desired series.
+
+Access to -1 and exponentials implies access to fractions and all surds.