commit allerede lavet opgaver

Diskret Mat/Binomial formula/Exercises.typ

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+Section 6.4: 7, *9*, *11*, 17, *21*, 23, *29, 33, 35, 37, 39*
+
+= Exercise 7
+What is the coefficient of $x⁹ "in" (2-x)^(19)$
+= Exercise 9
+What is the coefficient of $x^101y^99 "in the expansion of" (2x-3y)^200$
+
+*Answer:*
+
+$ -200!/(101! dot 99!)dot 2^101 dot 3^99 = -2^101 dot 3^99 dot mat(200;99) $
+= Exercise 11
+Use the binomial theorem to expand $(3x⁴-2y³)⁵$ into a sum of terms of the form $c x^a y^b$, where $c$ is a real number and $a$ and $b$ are nonnegative integers
+
+*Answer:*
+
+$ sum^5_(k=0)mat(5;k)(3x^4)^(5-k) (-2y^3)^(k) =\
+243x^20-81dot x^16dot 2y^3 dot 5 + 27 dot x^12dot 4y^6dot 10 - 9x^8dot 8y^9dot 10 + 3x^4 dot 16y^12dot 5 -32y^15 =\
+243x^20-810x^16y^3+1080x^12y^6-720x^8y^9+240x^4y^12-32y^15 $
+= Exercise 17
+What is the row of Pascal's triangle containing the binomial coefficients $mat(9;k),0<=k<=9$
+
+ 
+= Exercise 21
+Show that if $n$ and $k$ are integers with $1<=k<=n$, then $mat(n;k)<=n^k / 2^(k-1)$
+
+*Answer:*
+
+We know that $mat(n;k)$ is $n!/((n-k)!k!) = (n(n-1)(n-2)dots n-k+1)/(k(k-1)(k-2)(k-3)dots 2) <= (n dot n dot n dot dots)/(2 dot 2 dot 2)$
+= Exercise 23
+Prove Pascal's identity, using the formula for $mat(n;r)$
+= Exercise 29
+Let $n$ be a positive integer. Show that $ mat(2n;n+1)+mat(2n;n) = mat(2n+2;n+1) / 2 $
+
+*Answer:*
+
+We know:
+
+$mat(2n;n+1)+mat(2n;n)=mat(2n+1;n+1)$
+
+$mat(2n+1;n+1)=1/2dot (mat(2n+1;n+1)mat(2n+1;n+1))=mat(2n+2;n+1)/2$
+= Exercise 33
+Give a combinatorial proof that $sum_(k=1)^n k mat(n;k)=n 2^(n-1)$.
+_Hint: Count in two ways the number of ways to select a committee and to then select a leader of the committee._
+
+*Answer:*
+
+To select a committee, you have $2^(n-1)$ choices (n-1 because when you have n=1, then the choices are $2^0=1$) and then you select one from the `n` people in the committee.
+
+
+= Exercise 35
+Show that a nonempty set has the same number of subsets with an odd number of elements as it does subsets with an even number of elements
+= Exercise 37
+In this exercise we will count the number of paths in the $x y$ plane between the origin $(0,0)$ and point $(m,n)$, where $m$ and $n$ are nonnegative integers, such that each path is made up of a series of steps, where each step is a move unit to the right or a move unit upward. (No moves to the left or downward are allowed.) Two such paths from $(0,0)" to "(5,3)$ are illustrated here.
+#image("Exercise 6.4-37.png")
+a) Show that each path of the type described can be represented by a bit string consisting of $m$ 0s and $n$ 1s, where a 0 represents a move one unit to the right and a 1 represents a move one unit upward.
+
+b) Conclude from part (a) that there are $mat(m+n;n)$ paths of the desired type
+= Exercise 39
+Use Exercise 37 to prove Theorem 4. _Hint: Count the number of paths with $n$ steps of the type described in Exercise 37. Every such path must end at one of the points $(n-k,k)$ for $k=0,1,2, dots, n$._
+
+Theorem 4: Let $n$ and $r$ be nonnegative integers with $r<=n$. Then: $mat(n+1;r+1)=sum^n_(j=r) mat(j;r)$

Diskret Mat/Binomial formula/main.typ

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+#import "@preview/cetz:0.4.2"
+Section 6.4
+
+#title[Binomial coefficients, formula and identities]
+= Binomial coefficient
+From last week:
+
+$C(n, k) = $ the number of k-combinations from an n-set. Or the number of ways to select k elements from an n-set. Or the number of k-subsets of an n-set. Formula: $n!/(k!-(n-k)!)=mat(n;k)$ for $0<=k<=n$ 
+
+
+== Pascal's triangle
+#align(center, grid(columns: 2, column-gutter: 1em, cetz.canvas({
+  import cetz.draw: *
+
+  let n = 6
+
+  // calculate the triangle
+  let pascal = ()
+  for row in range(n + 1) {
+    let row-data = ()
+    for col in range(row + 1) {
+      let value = if col == 0 or col == row {
+        1
+      } else {
+        let prev = pascal.at(row - 1)
+        prev.at(col - 1) + prev.at(col)
+      }
+      row-data.push(value)
+    }
+    pascal.push(row-data)
+  }
+
+  // draw lines
+  for (row-idx, row) in pascal.enumerate() {
+    if row-idx < n {
+      let row-len = row.len()
+      let y = n - row-idx
+      for (col-idx, val) in row.enumerate() {
+        let x = col-idx - row-len / 2 + 0.5
+
+        let next-row-len = row-len + 1
+        let left-child-x = col-idx - next-row-len / 2 + 0.5
+        let right-child-x = (col-idx + 1) - next-row-len / 2 + 0.5
+        let child-y = y - 1
+
+        line((x, y / 1.5), (left-child-x, child-y / 1.5), stroke: gray)
+        line((x, y / 1.5), (right-child-x, child-y / 1.5), stroke: gray)
+      }
+    }
+  }
+
+  // draw values
+  for (row-idx, row) in pascal.enumerate() {
+    let row-len = row.len()
+    let y = (n - row-idx) / 1.5
+    for (col-idx, val) in row.enumerate() {
+      let x = col-idx - row-len / 2 + 0.5
+      circle((x, y), radius: 0.25, fill: white, stroke: none, name: "c-" + str(row-idx) + "-" + str(col-idx))
+      content((x, y), $binom(#str(row-idx), #str(col-idx))$)
+    }
+  }
+
+  // draw n
+  for i in range(n + 1) {
+    content((rel: (-2, 0), to: ("c-" + str(i) + "-0", "-|", "c-" + str(n - 1) + "-0")), [n=#i])
+  }
+}), cetz.canvas({
+  import cetz.draw: *
+
+  let n = 6
+
+  // calculate the triangle
+  let pascal = ()
+  for row in range(n + 1) {
+    let row-data = ()
+    for col in range(row + 1) {
+      let value = if col == 0 or col == row {
+        1
+      } else {
+        let prev = pascal.at(row - 1)
+        prev.at(col - 1) + prev.at(col)
+      }
+      row-data.push(value)
+    }
+    pascal.push(row-data)
+  }
+
+  // draw lines
+  for (row-idx, row) in pascal.enumerate() {
+    if row-idx < n {
+      let row-len = row.len()
+      let y = n - row-idx
+      for (col-idx, val) in row.enumerate() {
+        let x = col-idx - row-len / 2 + 0.5
+
+        let next-row-len = row-len + 1
+        let left-child-x = col-idx - next-row-len / 2 + 0.5
+        let right-child-x = (col-idx + 1) - next-row-len / 2 + 0.5
+        let child-y = y - 1
+
+        line((x, y / 1.5), (left-child-x, child-y / 1.5), stroke: gray)
+        line((x, y / 1.5), (right-child-x, child-y / 1.5), stroke: gray)
+      }
+    }
+  }
+
+  // draw values
+  for (row-idx, row) in pascal.enumerate() {
+    let row-len = row.len()
+    let y = (n - row-idx) / 1.5
+    for (col-idx, val) in row.enumerate() {
+      let x = col-idx - row-len / 2 + 0.5
+      circle((x, y), radius: 0.25, fill: white, stroke: none, name: "c-" + str(row-idx) + "-" + str(col-idx))
+      content((x, y), str(val))
+    }
+  }
+})))
+
+#figure(image("Pascals triangle-2.png"),caption: [Another form of Pascals triangle. It is mirrored])
+
+Binomial identity:
+
+$ mat(n;k) = mat(n;n-k) $
+You can do analytic proof (mathematic proof using the formulas) or the combinatorial proof (count it in one way to get first result, then in a different way to get the other result).
+
+*Combinatorial proof for $mat(n;k)=mat(n;n-k)$:* You could say that for n people, you ask k people to walk out the door, or you could ask `n-k` people to stay in the room.
+
+In Pascal's triangle, adding the rows: 
+$ mat(n;0)+mat(n;1)+mat(n;2)+dots+(n;n)=sum^n_(k=0)mat(n;k)=2^k $
+
+= Binomial identity
+For $mat(n;k)$ you can also write $mat(n-1;k)+mat(n-1;k-1)$. It is called Pascal's identity.
+#image("pascals identity.png")
+
+= Binomial formula
+$ (x+y)¹=x¹+y¹\
+(x+y)²=x²+2x y+y²\
+(x+y)³=x³ + 3x²y + 3x y³ + y³\
+(x+y)⁴=x⁴+4x³y+6x²y²+4x y³+y⁴
+$
+Note for each $x^(a)y^(b)$, you have $(x+y)^(a+b)$
+
+You also have that the coefficients (for $(x+y)⁴$ you have 1,4,6,4,1) they follow Pascal's triangle
+
+== The formula
+$ (x+y)^n&=mat(n;0)x^n+mat(n;1)x^(n-1) y + mat(n;2)x^(n-2)y²+dots+mat(n;n-1)x y^(n-1)+mat(n;n)y^n\
+&=sum^n_(k=0) mat(n;k)x^(n-k)y^k $
+
+$ (x+y)^n=sum^n_(k=0)mat(n;k)x^k y^(n-k)$
+
+== Proof by induction with respect to `n`
+*Basis step:* Put $n=1$ and prove that the formular is true:
+$ (x+y)¹&=mat(1;0)dot x⁰ dot y¹ + mat(1;1)dot x¹ dot y⁰\
+ &=y+x $
+ *Induction step:* Assume true for `n-1`, then prove that it's then true for `n`:
+ $ (x+y)^(n-1)=sum^(n-1)_(k=0)mat(n-1;k)x^k y^(n-1-k) $
+ $  
+ (x+y)^n&=(x+y)dot (x+y)^(n-1) \
+ &= (x+y)dot sum^(n-1)_(k=0)mat(n-1;k)x^k y^(n-1-k)\
+ &=sum^(n-1)_(k=0)mat(n-1;k)x^(k+1)y^(n-1-k)+sum^(n-1)_(k=0)mat(n-1;k)x^k y^(n-k)\
+ &= x^n+ sum^(n-2)_(k=0)mat(n-1;k)x^(k+1)y^(n-1-k) + sum^(n-1)_(k=1)mat(n-1;k)x^k y^(n-k)+y^n\
+ &"Replace k+1 with k"\
+ &= x^n+ sum^(n-1)_(k=1)mat(n-1;k-1)x^(k)y^(n-k) + sum^(n-1)_(k=1)mat(n-1;k)x^k y^(n-k)+y^n\
+ & "Note from pascals identity:" mat(n-1;k-1) + mat(n-1;k) = mat(n;k)\
+ $
+
+ *Combinatorial Proof:*
+  
+  Consider expanding $(x+y)^n = (x+y)(x+y)dots.c(x+y)$ (n factors).
+  
+  To get a term $x^(n-k)y^k$, we must:
+  - Choose $k$ of the $n$ factors to contribute a $y$ (the rest contribute $x$)
+  - There are $binom(n,k)$ ways to do this
+  
+  Therefore, the coefficient of $x^(n-k)y^k$ is $binom(n,k)$.
+
+= Van der Mande
+ $mat(m+n;r)=sum^r_(k=0)mat(m;r-k) mat(n;k)$