Day 11, Mat385

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Announcements
- Your quiz on Thursday will cover induction.
- ChatGPT is a lousy poet:
  Here was my "haiku Valentine" (that I might have sent to my wife, but didn't...:)
  (What is a haiku?)
  As Confucius implied, "If names be not correct, affairs cannot be carried on to success."
- Your quiz: ouch! Let's go over this, very slowly.... We want to prove that \[ P \rightarrow Q \] First of all, remember what Confucius implied:
  - Contraposition: \[ Q' \rightarrow P' \]
  - Contradiction: \[ P \land Q' \rightarrow 0 \]
  - It's also true that "If logical rules be not followed, affairs cannot be carried on to success."
    - Pay attention to scope.
    - Pay attention to parentheses.
    - Rules of Negation: \[ \left[\left(\forall x\right)P(x)\right]' \leftrightarrow \left(\exists x\right)P(x)' \] \[ \left[\left(\exists x\right)P(x)\right]' \leftrightarrow \left(\forall x\right)P(x)' \]
    - So the theorem was \[ \left(\forall y\right) [Q(x,y) \rightarrow P(x)] \longrightarrow \left[\left(\exists y\right) Q(x,y) \rightarrow P(x)\right] \]
      1. Connor - contraposition
      2. Kaleab/Isaac -- contradiction
  - Key
  - I'm going to add another "quiz drop" (I'll drop your lowest three quizzes before computing your quiz average).
    But I'm going to put a problem like this on the test....
Last time:
- We continued with recursion: Section 3.1 (Recursive Definitions), while working on highlight worksheet 3.1
Today: We'll
1. wrap up recursion: Section 3.1 (Recursive Definitions)
2. Then we'll hit Section 3.2 (Recurrence Relations), while working on highlight worksheet 3.2
  Note: we won't spend time on closed form solutions of second-order recurrence with constant coefficients (like Fibonaccis). But it is possible (the method described in the text seems ad hoc to me -- and not very enlightening).
  \[ Fib(n) = \frac{1}{\sqrt{5}} \left( \left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n \right) \]
  \[ \gamma = \frac{1+\sqrt{5}}{2} \approx 1.618 \] is called the "golden ratio", while \[ \frac{1-\sqrt{5}}{2} = 1-\gamma = \frac{-1}{\gamma} \approx -0.618 \] It turns out that \[ Fib(n) \approx \frac{1}{\sqrt{5}}\gamma^n \] So well, that by simply rounding this quantity you get all the right values! \[ Fib(n) = Round\left(\frac{1}{\sqrt{5}}\gamma^n\right) \] This shows that the Fibonacci numbers grow exponentially fast, with base $\gamma$.
  This is called a "closed-form solution" for the Fibonacci number sequence.
Links:
- Recursive computation of Fibonacci numbers (in lisp) (Here's a nice YouTube about the recursive definition of Fibonaccis)

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