help I have to give a talk

summer-notes-evan/0xparc-summer-2024-notes.typ

@@ -13,3 +13,4 @@
 #include "src/0803-research-workshop.typ"
 #include "src/0812-math-seminar.typ"
 #include "src/0813-math-seminar.typ"
+#include "src/0813-evan-talk.typ"

summer-notes-evan/src/0813-evan-talk.typ

@@ -0,0 +1,120 @@
+#import "@local/evan:1.0.0":*
+
+= Practice run for Evan's talk
+
+== Overview
+
+Brian mentioned that some of the engineers were curious why polynomials keep on reappearing
+over and over, and asked me to give a talk on this.
+So this is going to be a math _philosophy_ talk.
+
+I don't know if I'm able to give _the_ reason why polynomials, because that's above my pay-grade,
+so I'm content to try to explain _one_ reason why polynomials appear.
+The overall assertion is the following pseudo-theorem.
+
+#proposition[
+  Whenever you are trying to do linear algebra,
+  the "best" choice of coefficients are given by polynomials.
+]
+
+What do I mean by this?
+Let me give a concrete example by talking about secret sharing.
+
+== Secret-sharing
+
+Let's say we want to do a $2$-of-$N$ secret sharing scheme,
+which means we give each of $N$ people a secret share
+such that any two of them can reconstruct the secret, but no single one of them.
+If we denote the secret by, say, an ordered pair of integers $(x,y)$,
+then there is a pretty easy way to do this:
+
+- Send person 1 the value of $x+y$
+- Send person 2 the value of $x+2y$
+- ...
+- Send person $N$ the value of $x+N y$.
+
+And then you can solve a two-variable system of equations.
+In fact, no person can even recover one of $x$ or $y$,
+so in practice often $x$ is the secret and $y$ is just a random blinding factor.
+
+What if you want to do $3$-of-$N$?
+If you want to keep buying the linear algebra idea,
+then the general principle is that we want to choose some functions $f$, $g$, $h$ such that
+
+- Send person 1 the value of $f(1) x + g(1) y + h(1) z$
+- Send person 2 the value of $f(2) x + g(2) y + h(2) z$
+- ...
+- Send person $N$ the value of $f(N) x + g(N) y + h(N) z$.
+
+In the two-variable example, we just used $f(n) = 1$ and $g(n) = n$.
+For three functions, what would we pick?
+Well, let's first make a wish-list of properties that we would want to satisfy:
+
+1. For every three indices $1 <= a < b < c <= N$, we want there to be a unique $(x,y,z)$
+  solution to the linear algebra task
+  $
+    f(a) x + g(a) y + h(a) z &= 0 \
+    f(b) x + g(b) y + h(b) z &= 0 \
+    f(c) x + g(c) y + h(c) z &= 0.
+  $
+  This is equivalent to
+  $ det M_(a,b,c) != 0 " where "
+  M_(a,b,c) = mat(f(a), g(a), h(a); f(b), g(b), h(b); f(c), g(c), h(c)). $
+2. Ideally, for practical reasons, we'd like $f$, $g$, $h$ to not grow too quickly
+  and be pretty easy to calculate.
+3. Ideally, for practical reasons, we'd like inverting $M_(a,b,c)^(-1)$ to be
+  easy to implement.
+
+Now, the thing I want to communicate is that the first property is "easy" to satisfy,
+in the sense that I pull random numbers out of a hat,
+the chance that the determinant happens to be zero is vanishingly small.
+The issue with random garbage is that the second and third properties become harder to check,
+and anyway we want a deterministic promise that $det M_(a,b,c) != 0$, not just a "probably".
+
+== Vandermonde determinant
+
+Up until now I've made no mention of polynomials yet, and that's on purpose,
+because my thesis is that even in pure linear-algebra contexts
+there are good reasons to want polynomials.
+Now we'll bring them in.
+
+If you know how Shamir's secret-sharing works already, you know what the functions
+$f$, $g$, $h$ should be: it uses $f(n) = 1$, $g(n) = n$, $h(n) = n^2$.
+The relevant math theorem is this:
+#theorem[Vandermonde determinant][
+  $
+    det M_(a,b,c) = det mat(1, a, a^2; 1, b, b^2; 1, c, c^2)
+    = -(a-b)(a-c)(b-c)
+  $
+  The obvious generalization to more than three variables is also valid, e.g.
+  $
+    det mat(1, a, a^2, a^3; 1, b, b^2, b^3; 1, c, c^2, c^3; 1, d, d^2, d^3)
+    = (a-b)(a-c)(a-d)(b-c)(b-d)(c-d).
+  $
+  although I never remember which $plus.minus$ sign is on the front.
+]
+This theorem provides an affirmative answer to the first criteria,
+and gives a reason to believe this is "pretty good" for the second criteria.
+The point is that $M_(a,b,c)$ is _always_ going to equal zero if any of the variables are equal.
+That means we expect $det M_(a,b,c)$ will always at least have to be "divisible" by
+$(a-b)(a-c)(b-c)$, depending on what you mean by "divisible".
+The fact that the Vandermonde determinant has no additional garbage factors,
+just a single $plus.minus$ sign, means we have a reason to believe
+that we can't get a more reasonably slowly growing $f$, $g$, $h$.
+
+== Lagrange interpolation
+
+What about the last criteria?
+Let's specialize again to three-variables,
+and in fact let's specialize to $a=2$, $b=3$, $c=5$ so we have fewer symbols.
+That means we'd like to "rapidly" solve equations of the form
+$
+  x + 2y + 4z &= lambda_2 \
+  x + 3y + 9z &= lambda_3 \
+  x + 5y + 25z &= lambda_5.
+$
+Linear algebra means that it suffices to do the case where one of the $lambda_*$'s is zero,
+
+- TODO: switch order so leading coeffs first
+- TODO: mention FTA is a _linear algebra_ fact
+- TODO: evaluate this