Unformatted text preview: Math 17A
Kouba Recursions, Sequences, Fixed Points, and Limits EXAMPLE : The following recursions and inital values determine a sequence. Find an for n=1,2,3,4,5. 1.) an+1 = 2a,, + 3,a0 = —1 a1=2a0+3=2(—1)+3=1,
a2=2a1+3=2(1)+3=5,
a3=2a2+3=2(5)+3=13,
a4=2a3+3:2(13)+3=29,
a5=2a4+3=2(29)+3:61. a1=2a0+3=2(— 3)+3 (3)+3
a322a2+3=2(~—3)+3
(3)+3 (3)+3 a5=2a4+3=2— Hence lim an 2 oo (DNE) . TL—>OO Hence lim an 2 —3 .
”"900 DEFINITION : Let an+1 = ﬂan), a0 = L, for n = 1, 2, 3, 4,   . be a recursion and initial
value which determines a sequence. The initial value L is called a ﬁxed point for the
recursion if all successive values of an are equal to L, i.e., if L = f (L) NOTE: I.) The number —3 is a ﬁxed point for the previous example.
II.) The initial value is sometimes critical in determining if the sequence converges or diverges. III.) A ﬁxed point represents a potential limit for the sequence generated by the recursion and its initial value. IV.) Every limit of an associated sequence is a ﬁxed point for the recursion. EXAMPLE : Find all ﬁxed points for each recursion. 1) an+1 = (1/2)an — (3/4)
2
an — 1 2) an+1 3.) an+1 ...
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 Winter '08
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