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Mathematics Review • • • • • Exponents Logarithms Series Modular arithmetic Proofs • Manipulation of exponents – When the operation involves multiplication, add the exponents algebraically X A X B X A B – – When the operation involves division, subtract the divisor exponent from the numerator exponent. XA X A B B X When the operation involves powers or roots, multiply the exponent by the power number or divide the exponent by the power number, respectively. X A B – X AB B XA X Others X N X N 2 X N X 2N 2 N 2 N 2 N 1 A B 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 log A B log C B ; log C A Proof: Let X = logCB, Y = logCA, Z = logAB. A, B, C 0, A 1 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. – 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. y AB = 2X2 = 2x+y 2Z = AB – 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: a1 , a2 ,, an , an1 , • Such that an 1 g for a constant g, n=1, 2, … an – Mathematical progressions: a1 , a2 ,, an , an1 , • Such that an1 an d for some constant d , n=1, 2, … – Geometric series N • A common one: 2 • Its companion: A N 1 1 A A 1 i 0 i 2 N 1 1 i 0 N i How to compute it? N Let S = ,A i i 0 That is, S = A0 + A1 + A2 + … + AN-1 + AN AS = A1 + A2 + … + AN-1 + AN +AN+1 AS – S = AN+1 – 1 N A N 1 1 Therefore, i A A 1 i 0 If 0 < A < 1, then N Ai i 0 Why ? 1 1 A A N 1 1 1 A N 1 A A 1 1 A i 0 N i 0 A 1 0 A N 1 1 0 1 A N 1 1 1 A N 1 1 A 1 A 1 A i 0 N i • For 0 < A < 1 and N tends to , Ai i 0 1 1 A • Another sum that occurs frequently We write 1 2 3 4 5 S = 2 3 4 5 ... 2 2 2 2 2 and multiply by 2, obtaining 2 2 2S = 1 3 4 5 6 ... 2 2 23 2 4 25 Subtracting these two equations yields S = 1 1 12 13 14 15 ... 2 2 2 2 2 Thus, S = 2. i i i 1 2 – Arithmetic series N ( N 1) N 2 i 2 2 i 1 N 1 2 3 4 5 … N-2 N-1 N N+1 N+1 N+1 N/2 pairs Therefore, (N+1)*(N/2) = N N ( N 1) 2 How to compute (a kd ) ? k 1 – Harmonic series For positive integers N , the N th harmonic number is HN = 1 1 1 1 1 ... 2 3 4 N N 1 log e N i i 1 = – Other formulas N ( N 1)( 2 N 1) N 3 i 6 3 i 1 N 2 N k 1 i k 1 i 1 N k k 1 • Modular Arithmetic – A is congruent to 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) – If A B (mod N), then A+C B+C (mod N) 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) 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) • Proofs – Two most common ways of proving statements in data structure analysis: proof by induction and 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. – Example 2: N N ( N 1)( 2 N 1) 2 i THEOREM 1.3 If N >= 1, then 6 i 1 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. N 1 N We have i 2 i 2 ( N 1) 2 i 1 i 1 Applying the inductive hypothesis, we obtain N 1 i i 1 2 N ( N 1)( 2 N 1) ( N 1) 2 6 N (2 N 1) ( N 1) ( N 1) 6 2 2N 7N 6 ( N 1) 6 ( N 1)( N 2)( 2 N 3) 6 Thus, N 1 i i 1 2 ( N 1)[( N 1) 1][ 2( N 1) 1] 6 = (N+2)(2N+3) – 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 itself divide 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 • 2 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.