484 lines
10 KiB
Typst
484 lines
10 KiB
Typst
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#import "../handout.typ": *
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#import "../macros.typ": *
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#import "@preview/cetz:0.3.1"
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= Tropical Polynomials
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#definition()
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A _polynomial_ is an expression formed by adding and multiplying numbers and a variable $x$. \
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Every polynomial can be written as
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#align(
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center,
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box(
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inset: 3mm,
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$
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c_0 + c_1 x + c_2 x^2 + ... + c_n x^n
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$,
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),
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)
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for some nonnegative integer $n$ and coefficients $c_0, c_1, ..., c_n$. \
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The _degree_ of a polynomial is the largest $n$ for which $c_n$ is nonzero.
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#theorem()
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The _fundamental theorem of algebra_ states that any non-constant polynomial with real coefficients
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can be written as a product of polynomials of degree 1 or 2 with real coefficients.
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#v(2mm)
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For example, the polynomial $-160 - 64x - 2x^2 + 17x^3 + 8x^4 + x^5$ \
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can be written as $(x^2 + 2x+5)(x-2)(x+4)(x+4)$
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#v(2mm)
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A similar theorem exists for polynomials with complex coefficients. \
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These coefficients may be found using the _roots_ of this polynomial. \
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As you already know, there are formulas that determine the roots of quadratic, cubic, and quartic #note([(degree 2, 3, and 4)]) polynomials. There are no formulas for the roots of polynomials with larger degrees---in this case, we usually rely on appropriate roots found by computers.
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#v(2mm)
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In this section, we will analyze tropical polynomials:
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- Is there a fundamental theorem of tropical algebra?
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- Is there a tropical quadratic formula? How about a cubic formula?
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- Is it difficult to find the roots of tropical polynomials with large degrees?
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#definition()
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A _tropical_ polynomial is a polynomial that uses tropical addition and multiplication. \
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In other words, it is an expression of the form
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#align(
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center,
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box(
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inset: 3mm,
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$
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c_0 #tp (c_1 #tm x) #tp (c_2 #tm x^2) #tp ... #tp (c_n #tm x^n)
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$,
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),
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)
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where all exponents represent repeated tropical multiplication.
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#pagebreak() // MARK: page
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#problem()
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Draw a graph of the tropical polynomial $f(x) = x^2 #tp 1x #tp 4$. \
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#hint([$1x$ is not equal to $x$.])
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#notsolution(graphgrid(none))
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#solution([
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$f(x) = min(2x , 1+x, 4)$, which looks like:
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#graphgrid({
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import cetz.draw: *
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let step = 0.75
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line(
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(0, 0),
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(4 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 1 * step),
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(7 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 4 * step),
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(8 * step, 4 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line((0, 0), (1 * step, 2 * step), (3 * step, 4 * step), (7.5 * step, 4 * step), stroke: 1mm + oblue)
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})
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])
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#problem()
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Now, factor $f(x) = x^2 #tp 1x #tp 4$ into two polynomials with degree 1. \
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In other words, find $r$ and $s$ so that
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#align(
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center,
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box(
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inset: 3mm,
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$
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x^2 #tp 1x #tp 4 = (x #tp r)(x #tp s)
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$,
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),
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)
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#note([Naturally, we will call $r$ and $s$ the _roots_ of $f$.])
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#solution([
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Because $(x #tp r)(x #tp s) = x^2 #tp (r #tp s)x #tp s r$, we must have $r #tp s = 1$ and $r #tm s = 4$. \
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In standard notation, we need $min(r, s) = 1$ and $r + s = 4$, so we take $r = 1$ and $s = 3$:
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#v(2mm)
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$
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f(x) = x^2 #tp 1x #tp 4 = (x #tp 1)(x #tp 3)
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$
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])
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#v(1fr)
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#problem()
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Can you see the roots of this polynomial in the graph? \
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#hint([Yes, you can. What "features" do the roots correspond to?])
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#solution([The roots are the corners of the graph.])
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#v(0.5fr)
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#pagebreak() // MARK: page
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#problem()
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Graph $f(x) = -2x^2 #tp x #tp 8$. \
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#hint([Use half scale. 1 box = 2 units.])
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#notsolution(graphgrid(none))
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#solution([
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#graphgrid({
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import cetz.draw: *
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let step = 0.75
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line(
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(0, 0),
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(8 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0.5 * step, 0),
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(4 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 4 * step),
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(8 * step, 4 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line((0.5 * step, 0), (1 * step, 1 * step), (4 * step, 4 * step), (7.5 * step, 4 * step), stroke: 1mm + oblue)
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})
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])
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#problem()
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Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$.
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#solution([
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We (tropically) factor out $-2$ to get
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#align(
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center,
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box(
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inset: 3mm,
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$f(x) = -2(x^2 #tp 2x #tp 10)$,
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),
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)
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by the same process as the previous problem, we get
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#align(
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center,
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box(
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inset: 3mm,
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$f(x) = -2(x #tp 2)(x #tp 8)$,
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),
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)
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])
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#v(1fr)
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#problem()
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Can you see the roots $r$ and $s$ in the graph? \
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How are the roots related to the coefficients of $f$? \
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#hint([look at consecutive coefficients: $0 - (-2) = 2$])
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#solution([
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The roots are the differences between consecutive coefficients of $f$:
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- $0-(-2) = 2$
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- $8 - 0 = 8$
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])
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#v(0.5fr)
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#problem()
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Find a tropical polynomial that has roots $4$ and $5$ \
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and always produces $7$ for sufficiently large inputs.
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#solution([
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We are looking for $f(x) = a x^2 #tp b x #tp c$. \
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Since $f(#sym.infinity) = 7$, we know that $c = 7$. \
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Using the pattern from the previous problem, we'll subtract $5$ from $c$ to get $b = 2$, \
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and $4$ from $b$ to get $a = -2$.
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And so, $f(x) = -2x^2 #tp 2x #tp 7$
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#v(2mm)
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Subtracting roots in the opposite order does not work.
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])
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#v(1fr)
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#pagebreak() // MARK: page
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#problem()
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Graph $f(x) = 1x^2 #tp 3x #tp 5$.
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#notsolution(graphgrid(none))
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#solution([
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The graphs of all three terms intersect at the same point:
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#graphgrid({
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import cetz.draw: *
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let step = 0.75
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line(
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(0, 1 * step),
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(3.5 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 4 * step),
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(8 * step, 4 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 3 * step),
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(5 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line((0, 1 * step), (2 * step, 5 * step), (7.5 * step, 5 * step), stroke: 1mm + oblue)
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})
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])
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#problem()
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Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$.
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#solution(
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align(
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center,
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box(
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inset: 3mm,
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$f(x) = 1x^2 #tp 3 x #tp 5 = 1(x #tp 2)^2$,
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),
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),
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)
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#v(1fr)
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#problem()
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How is this graph different from the previous two? \
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How is this polynomial's factorization different from the previous two? \
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How are the roots of $f$ related to its coefficients?
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#solution([
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The factorization contains the same term twice. \
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Also note that the differences between consecutive coefficients of $f$ are both two.
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])
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#v(0.5fr)
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#pagebreak() // MARK: page
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#problem()
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Graph $f(x) = 2x^2 #tp 4x #tp 4$.
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#notsolution(graphgrid(none))
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#solution(
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graphgrid({
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import cetz.draw: *
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let step = 0.75
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line(
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(0, 2 * step),
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(3 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 4 * step),
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(5 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 4 * step),
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(8 * step, 4 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line((0, 2 * step), (1 * step, 4 * step), (7.5 * step, 4 * step), stroke: 1mm + oblue)
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}),
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)
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#problem()
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Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$, or show that one does not exist.
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#solution([
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We can factor out a 2 to get $f(x) = 2(x^2 #tp 2x #tp 2)$,
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but $x^2 #tp 2x #tp 2$ does not factor. \
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There are no $a$ and $b$ with minimum 2 and sum 2.
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])
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#v(1fr)
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#problem()
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Find a polynomial that has the same graph as $f$, but can be factored.
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#solution([
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$
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2x^2 #tp 3x #tp 4 = 2(x #tp 1)^2
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$
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])
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#v(1fr)
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#pagebreak() // MARK: page
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#theorem()
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The _fundamental thorem of tropical algebra_ states that for every tropical polynomial $f$, \
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there exists a _unique_ tropical polynomial $accent(f, macron)$ with the same graph that can be factored \
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into linear factors.
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#v(2mm)
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Whenever we say "the roots of $f$", we really mean "the roots of $accent(f, macron)$." \
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Also, $f$ and $accent(f, macron)$ might be the same polynomial.
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#problem()
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If $f(x) = a x^2 #tp b x #tp c$, then $accent(f, macron)(x) = a x^2 #tp B x #tp c$ for some $B$. \
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Find a formula for $B$ in terms of $a$, $b$, and $c$. \
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#hint([there are two cases to consider.])
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#solution([
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If we want to factor $a(x^2 #tp (b-a)x #tp (c-a))$, we need to find $r$ and $s$ so that
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- $min(r,s) = b-a$, and
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- $r + s = c - a$
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#v(2mm)
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This is possible if and only if $2(b-a) <= c-a$, \
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or equivalently if $b <= (a+c) #sym.div 2$
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#v(8mm)
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*Case 1:* If $b <= (a + c #sym.div) 2$, then $accent(f, macron) = f$ and $b = B$.
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#v(2mm)
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*Case 2:* If $b > (a + c #sym.div) 2$, then
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$
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accent(f, macron)(x)
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&= a x^2 #tp ((a+c)/2)x #tp c \
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&= a(x #tp (c-a)/2)^2
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$
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has the same graph as $f$, and thus $B = (a+c) #sym.div 2$
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#v(8mm)
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We can combine these results as follows:
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$
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B = min(b, (a+c)/2)
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$
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])
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#v(1fr)
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#problem()
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Find a tropical quadratic formula in terms of $a$, $b$, and $c$ \
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for the roots $x$ of a tropical polynomial $f(x) = a x^2 #tp b x #tp c$. \
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#hint([
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again, there are two cases. \
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Remember that "roots of $f$" means "roots of $accent(f, macron)$".
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])
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#solution([
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*Case 1:* If $b <= (a+c) #sym.div 2$, then $accent(f, macron) = f$ has roots $b-a$ and $c-b$, so
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$
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accent(f, macron)(x) = a(x #tp (b-a))(x #tp (c-b))
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$
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#v(8mm)
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*Case 2:* If $b > (a+c) #sym.div 2$, then $accent(f, macron)$ has root $(c-a) #sym.div$ with multiplicity 2, so
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$
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accent(f, macron)(x) = a(x #tp (c-a)/2)^2
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$
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#v(8mm)
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It is interesting to note that the condition $2b < a+ c$ for there to be two distinct roots becomes $b^2 > a c$ in tropical notation. This is reminiscent of the discriminant condition for standard polynomials!
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])
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#v(1fr)
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