S5_Unit_1_Outcome_4 - www.mathsrevision.com Higher Unit 1...

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Unformatted text preview: www.mathsrevision.com Higher Unit 1 Higher Unit 1 www.mathsrevision.com www.mathsrevision.com Recurrence Relations Grow and Decay Linear Recurrence Relation Divergence / Convergence / Limits Applications Find a formula Exam Type Questions Hig h e r O utc o m e 4 www.mathsrevision.com Recurrence Relations Sequences A 5 9 13 17 . B 3 6 12 24 . C 2 3 5 8 13 .. D 17 23 41 77 137 E 2 3 5 7 11 In the above sequences some have obvious patterns while others dont however this does not mean that a pattern doesnt exist. Higher Outcome 4 www.mathsrevision.com Notation Suppose we write the term of a sequence as u 1 , u 2 , u 3 , .., u n-1 , u n , u n+1 , ... where u 1 is the 1 st term, u 2 is the 2 nd term etc. and u n is the n th term ( n being any whole number.) The terms of a sequence can then be defined in two ways Recurrence Relations Higher Outcome 4 www.mathsrevision.com Either Using a formula for the nth term, u n in terms of the value n Or By expressing each term using the previous term(s) in the sequence. This is called a Recurrence Relation Now reconsider the sequences at the start Recurrence Relations Higher Outcome 4 www.mathsrevision.com Recurrence Relation : u n+1 = u n + 4 with u 1 = 5 Formula : u n = 4n + 1 A 5 9 13 17 . So u 100 = 4 X 100 + 1 = 401 u 2 = u 1 + 4 = 5 + 4 = 9 u 3 = u 2 + 4 = 9 + 4 = 13 Recurrence Relations Higher Outcome 4 www.mathsrevision.com Recurrence Relation : u n+1 = 2u n with u 1 = 3. B 3 6 12 24 Formula : u n = 3 X 2 n-1 So u 10 = 3 X 2 9 = 3 X 512 = 1536 u 2 = 2u 1 = 2 X 3 = 6, u 3 = 2u 2 = 2 X 6 = 12, etc Recurrence Relations Higher Outcome 4 www.mathsrevision.com C 2 3 5 8 13 .. No formula this time but we have a special type of recurrence relation called a FIBONACCI SEQUENCE . Here u 1 = 2 , u 2 = 3 then we have u 3 = u 2 + u 1 = 3 + 2 = 5 , u 4 = u 3 + u 2 = 5 + 3 = 8 In general u n+2 = u n+1 + u n ie apart from 1st two, each term is the sum of the two previous terms. Recurrence Relations Higher Outcome 4 www.mathsrevision.com D 17 23 41 77 137 This sequence doesnt have a recurrence relation but the terms can be found using the formula u n = n 3 - n + 17 Quite a tricky formula but it does work ......
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This note was uploaded on 02/13/2012 for the course MAT 205 math 205 taught by Professor Google during the Spring '10 term at University of Phoenix.

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S5_Unit_1_Outcome_4 - www.mathsrevision.com Higher Unit 1...

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