### Problem Set #1

#### Due: Fri, Sep 11th

1. In each of the following situations, indicate whether f = O(g), or f = Omega(g), or both (in which case f = Theta(g)).
(a) f(n) = n1.01, g(n) = n (lg n)2
(b) f(n) = lg n, g(n) = ln n
(c) f(n) = nn, g(n) = n!

2. The text spends a lot of time talking about fib(200). Compute this number exactly. (Hint: python has arbitrary-precision arithmetic... you can learn enough python in about 15 mins to solve this problem.)

3. Look up the "golden ratio" online. Call this number φ. Using your favorite multi-precision calculator (python works again), compute φ200/sqrt(5) rounded to the nearest integer. Compare the efficiency of this method to the method of problem 2.

4. As we said in class, if lim n → ∞ f(n)/g(n) is constant, then f(n) = O(g(n)). Show that the converse is false. In other words, give some f(n) and g(n) where f(n) = O(g(n)) but the limit of f(n)/g(n) isn't constant. (Hint: choose an f(n) where the limit doesn't exist. Remember to use functions from Z+ to R+ only.)

5. I mentioned in class that n.01 grows faster than lg n and that the proof takes the limit, requiring l'Hopital's rule. Let's use a simpler approach: set n=2100k and simplify. What's the smallest integer k such that n.01 is larger than lg n? How do we know that n.01 will remain larger forever after?