Mathematics Review

Mathematics Review • Exponents • Logarithms • Series • Modular arithmetic • Proofs

Manipulation of exponents • When the operation involves multiplication, add the exponents algebraically • When the operation involves division, subtract the divisor exponent from the numerator exponent. • When the operation involves powers or roots, multiply the exponent by the power number or divide the exponent by the power number, respectively. • Others

Logarithms • In computer science, all logarithms are to the base 2 unless specified otherwise. • DEFINITION: XA = B if and only if logXB = A

THEOREM 1.1 Proof: Let X = logCB, Y = logCA, Z = logAB. By the definition of logarithms, CX = B, CY = A, and AZ = B Combining these three equalities yields CX = B = AZ = (CY)Z = CYZ Therefore, X = YZ, which implies Z = X/Y, proving the theorem.

AB = 2X2y = 2x+y 2Z = AB • THEOREM 1.2 logAB = logA + logB; A, B > 0 Proof: Let X = logA, Y = logB, Z = logAB. By the definition of logarithms, 2X = A, 2Y = B, and 2Z = AB Combining these three equalities yields 2X2y = 2x+y= AB = 2Z Therefore, X+Y = Z, proving the theorem.

Some other useful formulas: log (A/B) = log A – log B log (AB) = B log A log X < X for all X > 0

Series • Two usual types • Geometric series: • Such that for a constant g, n=1, 2, … • Mathematical progressions: • Such that for some constant d , n=1, 2, …

Let S = , That is, S = A0 + A1 + A2 + … + AN-1 + AN AS = A1 + A2 + … + AN-1 + AN +AN+1 AS – S = AN+1 – 1 Therefore, If 0 < A < 1, then • Geometric series • A common one: • Its companion: How to compute it? Why ?

For 0 < A < 1 and N tends to , • Another sum that occurs frequently We write S = and multiply by 2, obtaining 2S = Subtracting these two equations yields S = Thus, S = 2.

How to compute ? • Arithmetic series 1 2 3 4 5 … N-2 N-1 N Therefore, (N+1)*(N/2) = N+1 N+1 N+1 N/2 pairs

Harmonic series For positive integers N , the N th harmonic number is HN= = • Other formulas

Modular Arithmetic • A is congruentto B modulo N, written A B (mod N), if N divides A-B. • Intuitively, this means that the remainder is the same when either A or B is divided by N. • Ex’s, • Check the following • 20 ? 6 (mod 2) • 25 ? 10 (mod 3) • 20 ? 6 (mod 4) • 100 ? 23 (mod 6)

Since A B (mod N), based on the definition, there exists an integer k such that A-B = k N. Now, AD-BD = (A-B)D = k ND So, N divides AD-BD. Therefore, AD BD (mod N) (A+C) – (B+C) = A – B Since A B (mod N), from the definition, N divides A-B. Thus, N divides (A+C)-(B+C) Therefore, A+C B+C (mod N) • If A B (mod N), then A+C B+C (mod N) AD BD (mod N)

Proofs • Two most common ways of proving statements in data structure analysis: proof by inductionand proof by contradiction. • The best way of proving that a theorem is false is by exhibiting a counterexample.

Proof by induction (two steps) • Base case: Establish that a theorem is true for some small value(s). This step is almost always trivial. • Induction step: • An inductive hypothesis is assumed (which means that the theorem is assumed to be true for all cases up to some limit k) • Using this assumption, the theorem is then shown to be true for the next value, which is typically k+1.

Example 1: Fibonacci numbers Fi < (5/3)i, for i >=1 Fibonacci numbers: F0 = 1, F1 = 1, F2 = 2, …, Fi = Fi-1 + Fi-2 Base case: verify that the theorem is true for the trivial cases. F1 = 1 < 5/3 F2 = 2 < 25/9 Inductive hypothesis: We assume that the theorem is true for i = 1, 2, …, k. To prove the theorem, we need to show that Fk+1 < (5/3)k+1 We have Fk+1 = Fk + Fk-1 < (5/3)k+ (5/3)k-1 (using inductive hypothesis) = (3/5) (5/3)k+1 + (3/5)2(5/3)k+1 = (3/5) (5/3)k+1 + (9/25)(5/3)k+1 = (3/5 + 9/25) (5/3)k+1 = (24/25) (5/3)k+1 < (5/3)k+1 which proves the theorem.

= (N+2)(2N+3) • Example 2: THEOREM 1.3 If N >= 1, then Proof: The proof is by induction. Base case: Obviously, the theorem is true when N = 1 Inductive hypothesis: Assume that the theorem is true for 1 <= k <= N. We need to establish that, under this assumption, the theorem is true for N+1. We have Applying the inductive hypothesis, we obtain Thus,

Proof by contradiction • Strategies: • Assume that the theorem is false • Show that this assumption implies that some know property is false, which indicates the original assumption was wrong. • Example 1: The number of primes is infinite. A positive integer number is a prime if and only if only 1 and itselfdivide the number. Proof: We assume that there is a finite number of primes, so that there is some largest prime Pk. Let P1, P2, …, Pk be all the primes in order and consider N = P1P2 … Pk + 1 Clearly, N > Pk, so by assumption N is not prime. However, none of P1, P2, …, Pk divides N exactly. This is a contradiction, because every number is either prime or a product of primes. Hence, the original assumption is false, which implies that the theorem is true.

Example 1 • is not a rational number. • Note: a rational number can be represented by a irreducible fraction of two integers • Proof. By contradiction • (Who can do this?)

Proof by counterexample • Proof by counterexample is usually used to prove that a theorem is false. • Constructing a counterexample is not as easy as it seems • Example 1: The statement Fibonacci number Fk <= k2 is false. • Proof: F11 = 144 > 112. Therefore, the statement is false.

A Brief Introduction to Recursion • A function is recursive if itself is used in its definition. • A recursive function must have a base (base cases) and a general relation which reduces a general case to simple case, and eventually to the base (or base cases). • Ex’s?

A good way to understand recursion is through building a recursion tree • The good points for recursion are • To write elegant codes • Easier analysis of the algorithm performance • The bad points are • Time consuming (Why?) • Space consuming (Why?)

Four Basic Rules of Recursion • Base cases: You must always have some base cases, which can be solved without recursion. • Making progress: For cases that are to be solved recursively, the recursive call must be always be to a case that makes progress toward a base case. • Design rule. Assume that all the recursive calls work. • Compound interest rule. Never duplicate work by solving the same instance of a problem in separate recursive calls.

Mathematics Review