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... Find the first 4 terms and the 9th term of the sequence whose general term is given by an = 4(2)n. Solution: We have an = 4(2)n, so a1 = 4(2)1 = 8 a2 = 4(2)2 = 16 a3 = 4(2)3 = 32 a4 = 4(2)4 = 64 a9 = 4(2)9 = 2048 The power (2)n causes the sign of the terms to alternate between positive ...
... Find the first 4 terms and the 9th term of the sequence whose general term is given by an = 4(2)n. Solution: We have an = 4(2)n, so a1 = 4(2)1 = 8 a2 = 4(2)2 = 16 a3 = 4(2)3 = 32 a4 = 4(2)4 = 64 a9 = 4(2)9 = 2048 The power (2)n causes the sign of the terms to alternate between positive ...
Sequences
... Arithmetic Sequences • When you want to find a large sequence, this process is long and there is great room for error. • To find the 20th, 45th, etc. term use the following formula: an = a1 + (n - 1)d ...
... Arithmetic Sequences • When you want to find a large sequence, this process is long and there is great room for error. • To find the 20th, 45th, etc. term use the following formula: an = a1 + (n - 1)d ...
Chapter 3
... Big-Omega and Big-Theta Nation • Definition 2: Let f and g be functions from the set of integers or the set of real numbers to the set of real numbers. • We say that f(x) is Ω(g(x)) if there are positive constants C and k such that |f(x)|≥ C|g(x)| Whenever x > k. ( this is read as “f(x) is big-Omeg ...
... Big-Omega and Big-Theta Nation • Definition 2: Let f and g be functions from the set of integers or the set of real numbers to the set of real numbers. • We say that f(x) is Ω(g(x)) if there are positive constants C and k such that |f(x)|≥ C|g(x)| Whenever x > k. ( this is read as “f(x) is big-Omeg ...
Chapter 7 Functions
... In 1890, Giuseppe Peano discovered a continuous function from [0, 1] onto the unit square [0,1] ×[0,1]. However, this function cannot be one-to-one. The following function is due to David Hilbert, only the 1st 6 iterations are shown, the final function is the limit of these iterations. ...
... In 1890, Giuseppe Peano discovered a continuous function from [0, 1] onto the unit square [0,1] ×[0,1]. However, this function cannot be one-to-one. The following function is due to David Hilbert, only the 1st 6 iterations are shown, the final function is the limit of these iterations. ...
The quadratic recurrence for matchings of the 2-by
... The upper left entry of P_n has F_n terms, each of which has coefficient +1. To understand F_{-1}, etc., we look at P_{-1}, etc., where P_2 = M_1 M_2 P_1 = M_1 = P_2 (M_2)^{-1} P_0 = P_1 (M_1)^{-1} = I, P_{-1} = P_0 (M_0)^{-1} = (M_0)^{-1}, P_{-2} = P_{-1} (M_{-1}}^{-1} = (M_0)^{-1} (M_{-1})^{-1}, ...
... The upper left entry of P_n has F_n terms, each of which has coefficient +1. To understand F_{-1}, etc., we look at P_{-1}, etc., where P_2 = M_1 M_2 P_1 = M_1 = P_2 (M_2)^{-1} P_0 = P_1 (M_1)^{-1} = I, P_{-1} = P_0 (M_0)^{-1} = (M_0)^{-1}, P_{-2} = P_{-1} (M_{-1}}^{-1} = (M_0)^{-1} (M_{-1})^{-1}, ...
