CmSc 250 Intro to Algorithms. Mathematical Review. 1. Basic Algebra. (a + b) 2 = a 2 + 2ab + b 2 (a - b) 2 = a 2-2ab + b 2 a 2 - b 2 = (a + b)(a - b)

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1 CmSc 250 Intro to Algorithms Mathematical Review 1. Basic Algebra (a + b) 2 = a 2 + 2ab + b 2 (a - b) 2 = a 2-2ab + b 2 a 2 - b 2 = (a + b)(a - b) a/x + b/y = (ay + bx)/xy 2. Exponents X n = XXX..X, n times X 0 = 1 by definition X a X b = X (a+b) X a / X b = X (a-b) Show that: X -n = 1 / X n (X a ) b = X ab 3. Logarithms log a X is defined for a > 0, X > 0 Definition: log a X = Y a Y = X E.G: log 2 8 = 3; 2 3 = 8 log a 1 = 0 because a 0 = 1 logx means log 2 X lgx means log 10 X lnx means log e X, where e is the natural number log a (XY) = log a X + log a Y log a (X/Y) = log a X - log a Y log a (X n ) = nlog a X 1

2 4. Series Σ 2 i = 2 (N+1) - 1 can be proved by induction Σ A i = (A (N+1) - 1)/(A - 1) 5. Limits of functions lim(n) =, n lim(n a ) =, n, a > 0 lim(1/n) = 0, n lim(1/(n a ) = 0, n, a > 0 lim(log(n)) =, n lim(a n ) =, n, a > 0 lim(f(x) + g(x)) = lim(f(x)) + lim(g(x)) lim(f(x) * g(x)) = lim(f(x)) * lim(g(x)) lim(f(x)/g(x)) = lim(f(x))/lim(g(x)) lim(f(x)/g(x)) = lim(f'(x)/g'(x)) (L'Hopital's rule, applied when both limits are 0, or both are infinity) Examples: lim(n/ n 2 ) = 0, n, lim(n 2 / n) =, n lim(n 2 / n 3 ) = 0, n, lim(n 3 / n 2 ) =, n lim(n/((n+1)/2) = 2. Derivatives: (log a n)' = (1/n) log a e (a n )' = (a n )ln(a) 2

3 Exercise: Consider the following functions: n 2, 2 n, nlogn, n 6 + n 2, n 4, n 6 - n 3, 4 n, n, n n Find the limit of the quotient of any two functions 6. Recursive definitions Basic idea: To define objects, processes and properties in terms of simpler objects, simpler processes or properties of simpler objects/processes. A recursive definition consists of: 1. Terminating rule - defining the object explicitly. 2. Recursive rules - defining the object in terms of a simpler object. Examples of recursive definitions: 6. Proofs 1. Factorials f(n) = n! 1. f(0) = 1 i.e. 0! = 1 2. f(n) = n * f(n-1) i.e. n! = n * (n-1)! 2. Fibonacci numbers F(1) = 1 F(2) = 1 F(k+1) = F(k) + F(k-1) 1, 1, 2, 3, 5, 8, Proof by Induction We use proof by induction when our claim concerns a sequence of cases, which can be numbered. An example is the equality S(N) = Σ 2 i = 2 (N+1) - 1 The cases here depend on N: 3

4 1. N = 0: S(0) = N = 1: S(1) = N = 2: S(2) = N = k: S(k) = k. Steps of proof by induction: EXAMPLE: 1. Inductive base: Show that the claim is true for the smallest case, usually n = 0 or n = Proof under inductive hypothesis: a. Assume that the claim is true for some k b. Prove that the claim is true for k+1 Prove by induction that S(N) = Σ 2 i = 2 (N+1) - 1 for any N 0 1. Inductive base: Show that the formula is true for N = 0 When substituting N=0 in the formula, we get S(0) = Σ 2 i = 2 (0+1) - 1 = 1 i=0 to 0 Is this true? Solution: By definition of S, S(0) = 2 0 = 1 i=0 to 0 By the formula we have: 2 (0+1) - 1 = = 1 Right sides are equal, hence the formula is true for N = 0 S(0) = Σ 2 i = 2 (0+1) - 1 = 2-1 = 1 4

5 2. Proof under inductive hypothesis: Assume that the formula is true for some k and prove that under this assumption the formula is true for k+1 We assume that S(k) is equal to 2 (k+1) - 1, i.e. i=0 to k S(k) = Σ 2 i = 2 (k+1) - 1 Now we shall show that S(k+1) = Σ 2 i = 2 (k+2) - 1 i=0 to k+1 We use the defining property of this particular sequence of sums: S(k+1) = S(k) + 2 (k+1) and we substitute S(k) with its equal expression from our assumption. Thus we have: S(k+1) = S(k) + 2 (k+1) = = 2 (k+1) (k+1) = = 2. 2 (k+1) - 1 = = 2 1+(k+1) - 1 = = 2 (k+2) - 1 This completes our proof : We conclude that S(N) = Σ 2 i = 2 (N+1) - 1 for any N Proof by counterexample Used when we want to prove that a statement is false. Types of statements: a claim that refers to all members of a class. EXAMPLE: The statement "all odd numbers are prime" is false. A counterexample is the number 9: it is odd but it is not prime. 5

6 6.3. Proof by contradiction Assume that the statement is false, i.e. its negation is true. Show that the assumption implies that some known property is false - this would be the contradiction. EXAMPLE: Prove that there is infinite number of primes. Proof: 1. Assume that the statement is false, i.e. the number of primes is finite. 2. Consequences of that assumption: There is a prime number that is the largest one - P k Let P 1, P 2,.. P k are all the prime numbers, ordered. Consider the number N = P 1 *P 2 *.P k + 1 ( N is the product of all prime numbers plus one) N is greater than P k, and following our assumption N is not prime. All non-prime numbers are equally divided by some prime number (This is a known property) Hence N should be equally divided by at least one of P 1 through P k. 3. Contradiction: We can see that none of the prime numbers P 1 through P k divides N equally - there is a remainder of 1. This is the contradiction. Hence our assumption that the number of primes is finite, is false. Hence the number of primes is infinite Proof by contraposition Based on the equivalence P Q ~Q ~P In order to prove P Q, we prove ~Q ~P Example: Prove that if the square of an integer is odd, then the integer is odd. Proof: We will prove that if an integer is even, then its square even. We can do this using direct proof. Let m be an even integer. This means that m can be represented as a multiple of 2: m = 2k m 2 = (2k) 2 = 4k 2 = 2.(2k 2 ) Therefore m 2 is even. 6