11.6 Three principles

1. Theorem (De Moivre): Let \(\theta\) be a real number, \(n \in \mathbb{N}\). Then
   \(\left( \cos(\theta) + i \sin(\theta) \right)^n = \cos(n \theta) + i \sin(n \theta)\).

   Proof:

   1. Let's see that \(n = 1\) satisfies:
      \(\left( \cos(\theta) + i \sin(\theta) \right)^1 = \cos(1 \cdot \theta) + i \sin(1 \cdot \theta)\).

   2. Suppose that \(\left( \cos(\theta) + i \sin(\theta) \right)^n = \cos(n \theta) + i \sin(n \theta)\).

      We want to show that \(\left( \cos(\theta) + i \sin(\theta) \right)^{n+1} = \cos((n+1) \theta) + i \sin((n+1) \theta)\).

      Now,
      \(\left( \cos(\theta) + i \sin(\theta) \right)^{n+1} = \left( \cos(\theta) + i \sin(\theta) \right)^n \cdot \left( \cos(\theta) + i \sin(\theta) \right)\).

      Using the induction hypothesis, this becomes
      \(\left( \cos(n \theta) + i \sin(n \theta) \right) \cdot \left( \cos(\theta) + i \sin(\theta) \right)\).
      \(\Rightarrow\cos(n\theta)\cos(\theta)+i\cos(n\theta)\sin(\theta)+i\sin(n\theta)\cos(\theta)-\sin(n\theta)\sin(\theta)\).

      \(\Rightarrow\left\lbrack\cos(n\theta)\cos(\theta)-\sin(n\theta)\sin(\theta)\right\rbrack+i\left\lbrack\sin(n\theta)\cos(\theta)+\cos(n\theta)\sin(\theta)\right\rbrack\).

      By the angle addition formulas, we get \(\cos(n\theta+1\theta)+i\sin(n\theta+1\theta)=\cos((n+1) \theta) + i \sin((n+1) \theta)\)

   Since the base case \(n=1\) satisfies the property, and if \(n\) satisfies it, then \(n+1\) does as well, we conclude by induction that this property is satisfied for all \(n \in \mathbb{N}\).

2. Evaluate or simplify:

   a) \(\frac{8!}{3! \cdot 5!}\)

   Expanding factorials, we get \(\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 \cdot 2 \cdot 3 \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5)} = \frac{7 \cdot 8}{1} = 56\)

   b) \(\frac{(n+3)!}{(n+1)!}\)

   Expanding the factorials, we have \(\frac{(n+3)(n+2)(n+1)!}{(n+1)!} = (n+3)(n+2)\)

   c) \(\sum_{k=1}^4 \left( k^2 - (k-1)^2 \right)\)

   Expanding each term \(= (1^2 - 0^2) + (2^2 - 1^2) + (3^2 - 2^2) + (4^2 - 3^2)=4^2\)

3. Prove that \(2^n \leq n!\) for all \(n \geq 4\).

   Let's prove this by induction:

   1. First, let's see that it satisfies for \(n = 4\):

      \(16 = 2 \cdot 2 \cdot 2 \cdot 2 \leq 1 \cdot 2 \cdot 3 \cdot 4 = 24\), so \(2^4 \leq 4!\) ✅ 2. Suppose that it satisfies for some \(n \geq 4\), i.e., \(2^n \leq n!\). We need to show that it also satisfies for \(n+1\).

      We have:
      ​\(2^{n+1} = 2 \cdot 2^n \leq n! \cdot 2 \leq n! \cdot (n+1) = (n+1)!\)

      The first inequality holds by the induction hypothesis, and the second inequality holds because \(2 \leq n+1\) for \(n \geq 4\).

   Therefore, by induction, \(2^n \leq n!\) for all \(n \geq 4\).

4. Prove that the sum of the angles of a convex polygon with \(n\) vertices is \((n-2)\pi\).

   I know that for a polygon of 3 vertices, this is true.

   Suppose it is true for all convex polygons of \(n\) vertices. Let's see that it is also true for \(n+1\).

   I will enumerate the vertices like \(V_1, V_2, V_3, \ldots, V_n, V_{n+1}\) clockwise. I will draw a line from \(V_1\) to \(V_3\).

   Then \(\triangle V_1 V_2 V_3\) forms a triangle.

   And \(V_{n+1}, V_1, V_3, \ldots, V_n\) forms a convex polygon of \(n\) vertices.

   Then by inductive step, I can say that the sum of the angles of \(V_{n+1}, V_1, V_4, \ldots, V_3\) is \((n-2)\pi\).

   And I also know that the sum of the angles of \(\triangle V_1 V_2 V_3\) is \(\pi\).

   I also know that the sum of the angles of my initial polygon is equal to the sum of the angles of \(\angle V_1,\angle V_2,\ldots,\angle V_{n+1}\) and \(\angle V_2,V_1,V_3\).

   So the sum is \(\pi+(n-2)\pi=(\left(n+1)-2\right)\pi\).

   Since it satisfies for \(n=3\) and also if it satisfies for \(n\), then it satisfies for \(n+1\). Then it satisfies for \(n \geq 3\)

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