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Chapter 4 Sequences and Mathematical Induction 4.1 Sequences Sequences • The main mathematical structure used to study repeated processes is the sequence. • The main mathematical tool used to verify conjectures about patterns governing the arrangement of terms in sequences is mathematical induction. Example • Ancestor counting with a sequence – two parents, four grandparents, eight greatgrandparents, etc. – Number of ancestors can be represented as 2position – Example: 23 = 8 (great grandparents), therefore parents removed three generations are great grandparents for which you have a total of 8. Sequences • Sequence is a set of elements written in a row as illustrated on prior slide. (NOTE: a sequence can be written differently) • Each element of the sequence is a term. • Example – am, am+1, am+2, am+3, …, an – terms a sub m, a sub m+1, a sub m+2, etc. – m is subscript of initial term – n is subscript of final term Example • Finding terms of a sequence given explicit formulas – ak = k/(k+1) for all integers k ≥ 1 – bi = (i-1)/i for all integers i ≥ 2a a1 = 1/(1+1) = ½ b2 = (2-1)/2 = ½ a2 = 2/(2+1) = 2/3 b3 = (3-1)/3 = 2/3 a3 = 3/(3+1) = 3/4 b4 = (4-1)/4 = 3/4 a4 = 4/5 b5 = 4/5 – the sequences a and b have the same terms and hence, are identical Example • Alternating Sequence – cj = (-1)j for all integers j≥0 c0 = (-1)0 = 1 c1 = (-1)1 = -1 c2 = (-1)2 = 1 c3 = (-1)3 = -1 c4 = (-1)4 = 1 … – sequence has bound values for the term. – term ∈ {-1, 1} Example • Find an explicit formula to fit given initial terms – sequence = 1, -1/4, 1/9, -1/16, 1/25, -1/36, … – What can we observe about this sequence? • alternate in sign • numerator is always 1 • denominator is a square 1/12 -1/22 1/32 -1/42 1/52 -1/62 a1 a2 a3 a4 a5 a6 – ak = ±1 / k2 (from the previous example we know how to create oscillating sign sequence, odd negative and even positive. – ak = (-1)k+1 / k2 Summation Notation • Summation notation is used to create a compact form for summation sequences governed by a formula. n a k 1 k a1 a2 a3 a4 ... an • the sequence is governed by k which has lower limit (1) and a upper limit of n. • This sequence is finite because it is bounded on the lower and upper limits. Example • Computing summations 5 a k a1 a2 a3 a4 a5 k 1 – a1 = -2, a2 = -1, a3 = 0, a4 = 1, and a5 =2. 5 a k k 1 2 1 0 1 2 0 Example • Computing summation from sum form. 5 k 2 k 1 5 k k 1 2 12 22 32 4 2 52 55 Example • Changing from Summation Notation to Expanded form n (1) i i 1 i0 n (1)i (1) 0 (1)1 (1)2 (1)n i 1 0 1 11 2 1 ... n 1 i0 Example • Changing from expanded to summation form. • Find a close form for the following: 1 2 3 n 1 ... n n 1 n 2 2n 1 2 3 n 1 k 1 ... n n 1 n 2 n n k 0 n k n Separating Off a Final Term • A final term can be removed from the summation form as follows. n 1 n a a k k m k an k m • Example of use: Rewrite the following separating the final term 1 n 1 1 1 k2 k2 n2 k 1 k 1 n Example • Combining final term n 1 2 k 2n k 0 n 1 2 k 0 n k 2n 2k k 0 Telescoping Sum • Telescoping sum can be evaluated to a closed form. 1 1 (k 1) k 1 k k 1 k(k 1) k(k 1) n 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ... 1 k(k 1) k k 1 1 2 2 3 3 4 n 1 n n n 1 n 1 k 1 k 1 n Product Notation 5 a k a1a2 a3 a4 a5 k 1 n 1 ak ak an k 1 k 1 n Recursive form Example • Compute the following products: 5 k 1* 2 * 3* 4 * 5 120 k 1 1 k 1 1 k 1 11 2 k 1 Factorial • Factorial is for each positive integer n, the quantity n factorial denoted n! is defined to be the product of all the integers from 1 to n: – n! = n * (n-1) *…*3*2*1 • Zero factorial denoted 0! is equal to 1. Example • Computing Factorials 8! 8 * 7! 8 7! 7! 5! 5 * 4 * 3! 5 * 4 2!*3! 2!*3! 2! Properties of Summations and Products • Theorem 4.1.1 – If am, am+1, am+2, … and bm, bm+1, bm+2, … are sequences of real numbers and c is any real number, then the following equations hold for any integer n≥m: n 1. n n a b a k k m k k m k bk k m n n k m k m n 2. c * ak c * ak n n 3. ak * bk ak * bk k m k m k m Examples • Let ak = k +1 and bk = k – 1 for all integers k n a n k 2 k m b k k m n n k m k m k 1 2 k 1 n k 1 2 k 1 k m n 3k 1 k m Examples • Let ak = k +1 and bk = k – 1 for all integers k n n n n ak bk (k 1) (k 1) k m k m k m k m n (k 1)(k 1) k k m 2 1 Transforming a Sum by Change of Variable • Transform the following by changing the variable. 6 – summation: k 1 1 k 0 – change of variable: j = k+1 – Solution: • compute the new limits: – lower: j=k+1, j=0+1=1 – upper: j=k+1, j=6+1=7 7 7 1 1 ( j 1) 1 j j 1 j 1