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Seq_Ser_Gro

# Seq_Ser_Gro - Â 1 Sequences Consider the function f with...

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Unformatted text preview: Â§ 1. Sequences Consider the function f with domain the set of nonnegative integers, IN = { , 1 , 2 , . . . } , which is defined as follows: i) f (0) = 6, ii) f ( n ) = 2 f ( n- 1)- 5 for each positive integer n . This is an example of a recursive definition of a function, meaning that first a finite set of the values are given, in this case just the value at 0, and then the remaining values are defined in terms of one or more previous values. We can use this definition to compute as many values of f as we wish by starting with the value f (0) = 6 given in part (i), and then using part (ii) repeatedly. At the beginning we obtain f (1) = 2 f (0)- 5 = 2 Â· 6- 5 = 7 , f (2) = 2 f (1)- 5 = 2 Â· 7- 5 = 9 , f (3) = 2 f (2)- 5 = 2 Â· 9- 5 = 13 , f (4) = 2 f (3)- 5 = 2 Â· 13- 5 = 21 . Continuing in this way would generate a list of values of f which begins with 6 , 7 , 9 , 13 , 21 , 37 , 69 , 133 , 261 , 517 , . . . . The entire infinite list of values in the proper order would give an alternate definition of the function f . Such an ordered list is an example of an infinite sequence . The locations of the individual terms of the sequence are usually indicated by using the members of IN as subscripts, as in a , a 1 , a 2 , a 3 , . . . , a n , . . . . The entire sequence is then often denoted by ( a n ) or ( a n ) n â‰¥ or ( a n ) n = âˆž n =0 . By an infinite sequence we always mean just such an ordered list. Thus each func- tion with domain IN determines an infinite sequence of its values and each sequence 1 Â§ 1. Sequences a , a 1 , a 2 , . . . , a n , . . . defines a function f on IN, where f ( n ) = a n for integers n â‰¥ 0. (In some cases we may begin a sequence with a 1 instead of a , corresponding to a function on the positive integers, or even start with some a m , m > 1, corresponding to a function on { m, m + 1 , m + 2 , . . . } .) Example 1.1. In dealing with sequences we will usually want some way to compute the n th term , a n , (actually the ( n + 1) st if we start with a , but this is what people call a n ) from some general formula. For example, consider the sequence ( b n ) which begins with b = 1, b 1 = 2, b 2 = 4, b 3 = 8, and which is defined recursively by i) b = 1 ii) b n = 2 b n- 1 for n â‰¥ 1. Here it is easy to see that the n th term is given by b n = 2 n for each integer n â‰¥ 0. To prove this it is enough to check that this formula is correct for 0 since b = 1 = 2 , and that if it is correct for any integer n- 1 so that b n- 1 = 2 n- 1 , then it is also true for the next integer n since b n = 2 b n- 1 = 2 Â· 2 n- 1 = 2 n . (This is the essence of a proof by induction .) If you write out a few more terms of the sequence ( b n ) with b n = 2 n , you may note a similarity to the sequence of values of the first example: ( b n ) 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, . . ....
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Seq_Ser_Gro - Â 1 Sequences Consider the function f with...

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