90 lines
1.9 KiB
Typst
90 lines
1.9 KiB
Typst
#import "@local/handout:0.1.0": *
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#import "@preview/cetz:0.4.2"
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#show: handout.with(
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title: [Warm-Up: Snakes!],
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by: "Mark",
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page_numbers: false,
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)
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#box(place(box(width: 100%, height: 8in, align(horizon, image(
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"7.svg",
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width: 360mm,
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)))))
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#problem()
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Evaluate this integral. \
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All integrals are of the form $integral_a^b 1 #h(1mm) d x$.
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#if_solutions(pagebreak())
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#solution[
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This integral is drawn recursively with the following code: \
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#v(2mm)
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#let snake(depth, x, y) = {
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if depth == 0 {
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math.attach(math.integral, br: x, tr: y)
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} else {
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let bot = snake(depth - 1, x, "1")
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let top = snake(depth - 1, "0", y)
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snake(depth - 1, bot, top)
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}
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}
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```typst
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#let snake(depth, x, y) = {
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if depth == 0 {
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$integral_#x ^#y$
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} else {
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let bot = snake(depth - 1, x, "1")
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let top = snake(depth - 1, "0", y)
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snake(depth - 1, bot, top)
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}
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}
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```
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#v(5mm)
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For example:
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$ snake\(0, x, y) = #snake(0, "x", "y") $
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#v(2mm)
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$ snake\(1, x, y) = #snake(1, "x", "y") $
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#v(2mm)
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$ snake\(2, x, y) = #snake(2, "x", "y") $
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In other words, we want $f_7 (0, 1)$, where...
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- $f_0(x, y) := integral_x^y$
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- $f_(n+1)(x, y) := f_n [f_n (x, 1), f_n (0, y)]$
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#v(5mm)
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Expanding a few iterations, we find:
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- $f_0(x, y) = integral_x^y = y-x$
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- $f_1(x, y) = f_0 [f_0 (x, 1), f_0 (0, y)] = y+x-1$
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- $f_2(x, y) = f_1 [f_1 (x, 1), f_1 (0, y)] = y+x-2$
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#v(5mm)
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We can use induction to show that this pattern continues. \
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If $g_n = y+x+c$, then $g_(n+1) = y+x+(3c+1)$.
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#v(5mm)
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Finally, use this recursion to find that
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$f_0, f_1, ..., f_7 = 1, 0, -1, -4, -13, -40, -121, -364$
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One can also find an explicit formula for $g_n$:
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$
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f_(n+1) = g_n & = x + y + 3^n c + 3^0 + 3^1 + ... + 3^n \
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& = x + y + 3^n c + sum_(i=0)^n 3^i \
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& = x + y + 3^n c + (3^(n+1) + 1)/2
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$
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]
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