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Expressing Sequences Explicitly By: Matt Connor Fall 2013 • Pure Math •Analysis •Calculus and Real Analysis •Sequences • Sequence- A list of numbers or objects in a specific order • • Finite Sequence- contains a finite number of terms • • 1,3,5,7,9,..... 2,4,6,8 Infinite Sequence- contains an infinite number of terms • 2,4,8,16, ........ • Arithmetic Sequence- add or subtract a constant to get from one term to the next • 88, 77, 66, 55,....... • Geometric Sequence- multiply or divide by a common ratio to get from one term to the next • 6, 12, 24, 48,........ • Recursive Formula- formula for a sequence that relates the previous term(s) to find the new one. • ex: An = A(n-1)+ 4 • Explicit Formula- formula that finds any term in the sequence without knowing any other terms. • ex: An = 1+ 2(n-1) • all you need to know is n Arithmetic Sequences • General Forms • Recursive formula • • An = A(n-1) + d Explicit formula • An = A1 + d(n-1) Geometric Sequences • General Forms • Recursive formula • • An = r(An-1) Explicit formula • An = A1 (rn-1) • What about sequences that are not arithmetic or geometric? • This means they do not have a common constant or ratio • These • The are commonly called Fibonacci-type difficult thing about these is finding an explicit formula Fibonacci Sequence Explicit Formula • Now we will go through deriving an explicit formula for the Fibonacci Sequence • We know the relational formula is • • An = An−1 + An−2 We guess an explicit formula of the form An =Cxn and plug it in to the relational equation and get • n Cx = n−1 Cx + n−2 Cx • n−1 Cx n Cx n−2 Cx this = + will always simplify to an equation with the same coefficients as the relational equation, • 2 x =x+1 • Then we collect the terms on one side to use the quadratic formula. • x2 −x−1=0 • The quadratic formula gives us x=(1/2)(1±√5) • Therefore: An= n B((1/2)(1+√5)) • Next + n C((1/2)(1-√5)) we use the first two Fibonacci numbers to find two equations representing B and C • A0=1 and A1=1 • This gives us two equations for B and C • B+C=1 and • B(1/2)(1+√5) • Then + C(1/2)(1-√5)=1 we simplify the second equation we have • (B + C) + (B - C)√5 = 2 and since our first equation tells us that B+C=1 we can replace that. •1 + (B-C)√5 = 2 • We then further simplify this to get the second of our two equations • B+C=1 and B-C=1/√5 • If we add these two equations and simplify we can then solve for B • B= (√5+1)/(2√5) • And then insert the value of B to find the value of C • C=(√5-1)/(2√5) • One More Step!! • If we replace the B and C in our equation for An An= • This is Binet’s formula, an explicit formula for th finding the n Fibonacci number. • As you have seen finding an explicit formula th for the n term in a Fibonacci-type sequence is much more difficult. •. . . . . but they are possible to find! • Resources • http://www.kenston.k12.oh.us/khs/academics/math/AA _11-3A_geometric_sequences_explicit.pdf • http://www.kenston.k12.oh.us/khs/academics/math/AA _11-2A_arithmetic_sequences.pdf • http://www.geom.uiuc.edu/~demo5337/s97b/fibonacci. html • http://faculty.mansfield.edu/hiseri/MA1115/1115L30.pdf