1. Chapter 2 - Mathematical Review Functions
Factorial - x!=1*2*3*…..*n
Permutation - a rearrangement of a sequence
Boolean variable - takes on 2 values
TRUE, FALSE
0, 1
Floor - round down x
Ceiling - round up x
Modulus - mod - Remainder
x mod y - remainder when divide x by y
10 mod 3 is 1

2. Logarithms logby=x  bx=y  blogby=y log mn = log m + log n
XAXB=XA+B
log m/n = log m - log n
XA/XB=XA-B
log nr = r log n
(XA)B=XAB
logan=logbn/logba
2n+2n=2*2n=2n+1

3. Recursion A recursive algorithm calls itself.
Must have a base case to stop recursion
Calls must pass a “smaller problem” to the function. This will cause the recursion to eventually stop.
Factorial example:
int fact(int n) {
if (n<=1) return 1;
else return n*fact(n-1) }
Looks a bit like induction
Better to use for’s and while’s when appropriate.

4. Summations Adding up a list of values f(1)+f(2)+f(3)+…+f(n)
if f(i)=i then sum is n(n+1)/2
if f(i)=i2 then sum is n(n+1)(2n+1)/6
if f(i)=2i then sum is 2n+1–1
if f(i)=Ai then sum is (An+1 –1)/(A-1)

5. Recurrence Relations
Many of our operations (functions) will be recursive.
We will want to analyze the running time for a problem of size n.
Recursion defines this in terms of some work plus the amount of time to solve a smaller problem (using the same algorithm).
A mathematical representation of this time will look like:
T(n) = T(n-1) + 1
Need the time for the base case
T(1) = 0

6. Recurrence Relation Solving
Many recurrence relations are solved using the brute force method. Expand it; see the pattern; solve it.
For example:
T(n) = T(n-1)+1 = T(n-1)+1+1 = T(1)+(n-1) = n-1
Another example:
T(n) = T(n-1)+n; T(1)=1
T(n) = T(n-1)+n = T(n-2)+(n-1)+n = T(1)+2+…n = …+n = n(n+1)/2

7. Proof Techniques Proof by Contradiction Proof by induction
Find a counterexample to prove theorem is false
Assume the opposite of the conclusion is true. Use steps to prove a contradiction (for example 1=2), then the original theorem must be true.
Proof by induction
Show the theorem holds for a base case (n=0 or n=1)
Assume the theorem holds for n=k-1
Prove the theorem is true for n=k
Example: show …+n=n(n+1)/2
for n=1 1=1(2)/2=1
assume …+n-1=(n-1)(n)/2; show 1+2+…+n=n(n+1)/2
(n-1)(n)/2+n=[(n-1)(n)+2n]/2=[n(n-1+2)]/2=n(n+1)/2