Day 11, Mat385

Last time

Next time

Announcements
- Bit of a nerve-wracking day: let's hope that we can cool things down in Ukraine....
- The Zooms continue
  A menu of our past zooms...
- Your 2.2 homework is graded: problems 22, 27, 50
  Generally, the worst problem is writing $P(k)=$.
  $P(k)$ is an assersion. It has a truth value, but not a numeric value. So we don't let it equal anything (except another assertion). Use a colon: $P(k):$.
  Write, for example, \[ P(k): 1\cdot 1! + 2\cdot 2! + \ldots + k\cdot k! = (k+1)! - 1 \]
  $P(k)$ is the assertion that the sum on the left is equal to the difference on the right. It is not a number. It is not a quantity. It is predicate wff, which may or may not be true.
  But I fear that I'm shouting into the wind here. Those who want to please the grader (and avoid saying nonsense) -- let them hear.
  It also may be wise for some of you to state $P(k+1)$. That's your target, using the first principle: show that $P(k) \rightarrow P(k+1)$.
  - 22: straightforward; the biggest trouble is algebra. Some folks are a little disturbed by factorials.
  - 27: straightforward; again, it's the algebra. Some have trouble with exponentials.
  - 50: this requires a clever rethinking; but the primary objective is to use $P(k)$, so we have to figure out how to rewrite \[ 13^{k+1}-6^{k+1} \] so that we can use the fact that $13^{k}-6^{k}$ is divisible by 7.
Last time:
- We discussed the exam, and looked over the key.
- We started 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 that we don'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 pretty ad hoc, however -- not 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 \[ \frac{1-\sqrt{5}}{2} = 1-\gamma = \frac{-1}{\gamma} \approx -0.618 \] As I show in some lisp code below, \[ Fib(n) \approx \frac{1}{\sqrt{5}}\gamma^n \] So well, that by simply rounding this quantity you get all the right values! This shows that the Fibonacci numbers grow exponentially fast, with base $\gamma$.
Links:
- Recursive computation of Fibonacci numbers (in lisp)
- Recursive computation of rows from Pascal's triangle (in lisp)
- "String exponention" (in lisp)
- Here is the recursive selection sort algorithm, in lisp:
- Here is a recursive binary search algorithm in lisp:
- Closed-form computation of Fibonacci numbers (in lisp)
- Here is a recursive bubbleSort algorithm in lisp
  
  (coded up following our author's algorithm, in section 3.3, p. 214, as closely as possible).

Website maintained by Andy Long. Comments appreciated.