#import "@local/handout:0.1.0": *
#import "@preview/cetz:0.4.2"

#show: handout.with(
  title: [Warm-Up: Snakes!],
  by: "Mark",
  page_numbers: false,
)

#box(place(box(width: 100%, height: 8in, align(horizon, image(
  "7.svg",
  width: 360mm,
)))))

#problem()
Evaluate this integral. \
All integrals are of the form $integral_a^b 1 #h(1mm) d x$.

#if_solutions(pagebreak())

#solution[
  This integral is drawn recursively with the following code: \

  #v(2mm)

  #let snake(depth, x, y) = {
    if depth == 0 {
      math.attach(math.integral, br: x, tr: y)
    } else {
      let bot = snake(depth - 1, x, "1")
      let top = snake(depth - 1, "0", y)
      snake(depth - 1, bot, top)
    }
  }

  ```typst
  #let snake(depth, x, y) = {
    if depth == 0 {
      $integral_#x ^#y$
    } else {
      let bot = snake(depth - 1, x, "1")
      let top = snake(depth - 1, "0", y)
      snake(depth - 1, bot, top)
    }
  }
  ```

  #v(5mm)

  For example:

  $ snake\(0, x, y) = #snake(0, "x", "y") $

  #v(2mm)

  $ snake\(1, x, y) = #snake(1, "x", "y") $

  #v(2mm)

  $ snake\(2, x, y) = #snake(2, "x", "y") $

  In other words, we want $f_7 (0, 1)$, where...
  - $f_0(x, y) := integral_x^y$
  - $f_(n+1)(x, y) := f_n [f_n (x, 1), f_n (0, y)]$

  #v(5mm)
  Expanding a few iterations, we find:
  - $f_0(x, y) = integral_x^y = y-x$
  - $f_1(x, y) = f_0 [f_0 (x, 1), f_0 (0, y)] = y+x-1$
  - $f_2(x, y) = f_1 [f_1 (x, 1), f_1 (0, y)] = y+x-2$

  #v(5mm)

  We can use induction to show that this pattern continues. \
  If $g_n = y+x+c$, then $g_(n+1) = y+x+(3c+1)$.


  #v(5mm)
  Finally, use this recursion to find that
  $f_0, f_1, ..., f_7 = 1, 0, -1, -4, -13, -40, -121, -364$

  One can also find an explicit formula for $g_n$:

  $
    f_(n+1) = g_n & = x + y + 3^n c + 3^0 + 3^1 + ... + 3^n \
                  & = x + y + 3^n c + sum_(i=0)^n 3^i \
                  & = x + y + 3^n c + (3^(n+1) + 1)/2
  $
]