Section 3 solutions

Advanced/Tropical Polynomials/parts/02 cubic.typ

@@ -10,7 +10,33 @@ Consider the polynomial $f(x) = x^3 #tp x^2 #tp 3x #tp 6$. \
 - use this graph to find the roots of $f$
 - write (and expand) a product of linear factors with the same graph as $f$.
 
-#graphgrid(none)
+#notsolution(graphgrid(none))
+
+#solution([
+  - Roots are 1, 2, and 3.
+  - $accent(f, macron)(x) = x^3 #tp 1x^2 #tp 3x #tp 6 = (x #tp 1)(x #tp 2)(x #tp 3)$
+
+
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 0), (2.66 * step, 8 * step))
+    dotline((0, 1 * step), (3.5 * step, 8 * step))
+    dotline((0, 3 * step), (5 * step, 8 * step))
+    dotline((0, 6 * step), (8 * step, 6 * step))
+
+    line(
+      (0, 0),
+      (1 * step, 3 * step),
+      (2 * step, 5 * step),
+      (3 * step, 6 * step),
+      (7.5 * step, 6 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
+
 
 
 #v(1fr)
@@ -22,7 +48,30 @@ Consider the polynomial $f(x) = x^3 #tp x^2 #tp 6x #tp 6$. \
 - use this graph to find the roots of $f$
 - write (and expand) a product of linear factors with the same graph as $f$.
 
-#graphgrid(none)
+#notsolution(graphgrid(none))
+
+#solution([
+  - Roots are 1, 2.5, and 2.5.
+  - $accent(f, macron)(x) = x^3 #tp 1x^2 #tp 3.5x #tp 6 = (x #tp 1)(x #tp 2.5)^2$
+
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 0), (2.66 * step, 8 * step))
+    dotline((0, 1 * step), (3.5 * step, 8 * step))
+    dotline((0, 6 * step), (2 * step, 8 * step))
+    dotline((0, 6 * step), (8 * step, 6 * step))
+
+    line(
+      (0, 0),
+      (1 * step, 3 * step),
+      (2.5 * step, 6 * step),
+      (7.5 * step, 6 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
 
 
 #v(1fr)
@@ -33,7 +82,29 @@ Consider the polynomial $f(x) = x^3 #tp 6x^2 #tp 6x #tp 6$. \
 - use this graph to find the roots of $f$
 - write (and expand) a product of linear factors with the same graph as $f$.
 
-#graphgrid(none)
+#notsolution(graphgrid(none))
+
+#solution([
+  - Roots are 2, 2, and 2.
+  - $accent(f, macron)(x) = x^3 #tp 2x^2 #tp 4x #tp 6 = (x #tp 2)^3$
+
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 0), (2.66 * step, 8 * step))
+    dotline((0, 6 * step), (1 * step, 8 * step))
+    dotline((0, 6 * step), (2 * step, 8 * step))
+    dotline((0, 6 * step), (8 * step, 6 * step))
+
+    line(
+      (0, 0),
+      (2 * step, 6 * step),
+      (7.5 * step, 6 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
 
 
 #v(1fr)
@@ -44,6 +115,16 @@ Consider the polynomial $f(x) = x^3 #tp 6x^2 #tp 6x #tp 6$. \
 If $f(x) = a x^3 #tp b x^2 #tp c x #tp d$, then $accent(f, macron)(x) = a x^3 #tp B x^2 #tp C x #tp d$ for some $B$ and $C$. \
 Using the last three problems, find formulas for $B$ and $C$ in terms of $a$, $b$, $c$, and $d$.
 
+#solution([
+
+  $
+    B = min(b, (a+c)/2, (2a+d)/2)
+  $
+  $
+    C = min(c, (b+d)/2, (a+2d)/2)
+  $
+])
+
 #v(1fr)
 #pagebreak() // MARK: page
 
@@ -60,6 +141,13 @@ What are the roots of the following polynomial?
   ),
 )
 
+#solution([
+  We have
+  $
+    accent(f, macron)(x) = 3x^6 #tp 2x^5 #tp 1x^4 #tp x^3 #tp 1x^2 #tp 3x #tp 5
+  $
+  which has roots $-1$, $-1$, $-1$, $1$, $2$, $2$
+])
 
 
 #v(1fr)
@@ -77,8 +165,34 @@ $
 
 #v(2mm)
 
-Find a formula for each $C_i$ in terms of $c_0, c_1, ..., c_n$. \
-Then, find formulas for the roots $r_1, r_2, ..., r_n$.
+Find a formula for each $C_i$ in terms of $c_0, c_1, ..., c_n$.
+
+#solution([
+  $
+    A_j
+    &= min_(l<=j<k)( (a_l - a_k) / (k-l) (k-j) + a_k ) \
+    &= min_(l<=j<k)( a_l (k-j) / (k-l) + a_k (j-l) / (k-l) )
+  $
+
+  #v(2mm)
+
+  which is a weighted average of some $a_l$ and $a_k$, with $l<=j<k$
+])
+
+#v(1fr)
+
+
+#problem()
+With the same setup as the previous problem, \
+find formulas for the roots $r_1, r_2, ..., r_n$.
+
+#solution([
+  The roots are the differences between consecutive coefficients of $accent(f, macron)$:
+  $
+    r_i = A_i - A_(i-1)
+  $
+  where we set $A_n = a_n$ and $A_0 = a_0$.
+])
 
 #v(1fr)
 
@@ -86,4 +200,13 @@ Then, find formulas for the roots $r_1, r_2, ..., r_n$.
 Can you find a geometric interpretation of these formulas \
 in terms of the points $(-i, c_i)$ for $0 <= i <= n$?
 
+#solution([
+  The inequality #note([(for $l <= k < k$)])
+  $
+    A_j <= (a_l - a_k) / (k-l) (k-j)+a_k
+  $
+  states that the point $(-j,A_j)$ must lie on or below the line segment between the points $(-k, a_k)$ and $(-l, a_l)$.
+  This makes it easy to find the $A_j$ using a graph of the points $(-i, a_i)$ for $0 <= i <=n$.
+])
+
 #v(0.5fr)