1. Expressing Sequences Explicitly By: Matt Connor Fall 2013

2. Pure Math Analysis Calculus and Real Analysis Sequences

3. Sequence- A list of numbers or objects in a specific order 1,3,5,7,9,..... Finite Sequence- contains a finite number of terms 2,4,6,8 Infinite Sequence- contains an infinite number of terms 2,4,8,16,........

4. 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,........

5. Recursive Formula- formula for a sequence that relates the previous term(s) to find the new one. ex: A n = A (n-1) + 4 Explicit Formula- formula that finds any term in the sequence without knowing any other terms. ex: A n = 1+ 2(n-1) all you need to know is n

6. General Forms Recursive formula A n = A (n-1) + d Explicit formula A n = A 1 + d(n-1) Arithmetic Sequences

7. Geometric Sequences General Forms Recursive formula A n = r(A n-1 ) Explicit formula A n = A 1 (r n-1 )

8. What about sequences that are not arithmetic or geometric? This means they do not have a common constant or ratio These are commonly called Fibonacci-type The difficult thing about these is finding an explicit formula

9. Now we will go through deriving an explicit formula for the Fibonacci Sequence We know the relational formula is A n = A n−1 + A n−2 We guess an explicit formula of the form A n =Cx n and plug it in to the relational equation and get Cx n = Cx n−1 + Cx n−2 Fibonacci Sequence Explicit Formula

10. Cx n = Cx n−1 + Cx n−2 this will always simplify to an equation with the same coefficients as the relational equation, x 2 = x + 1 Then we collect the terms on one side to use the quadratic formula. x 2 −x−1=0

11. The quadratic formula gives us x=(1/2)(1±√5) Therefore: A n = B((1/2)(1+√5)) n + C((1/2)(1-√5)) n Next we use the first two Fibonacci numbers to find two equations representing B and C A 0 =1 and A 1 =1

12. This gives us two equations for B and C B+C=1 and B(1/2)(1+√5) + C(1/2)(1-√5)=1 Then 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.

13. 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)

14. And then insert the value of B to find the value of C C=(√5-1)/(2√5) One More Step!!

15. If we replace the B and C in our equation for A n This is Binet’s formula, an explicit formula for finding the n th Fibonacci number. An=An=

16. As you have seen finding an explicit formula for the n th term in a Fibonacci-type sequence is much more difficult...... but they are possible to find!

17. 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 _11-3A_geometric_sequences_explicit.pdf http://www.kenston.k12.oh.us/khs/academics/math/AA _11-2A_arithmetic_sequences.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://www.geom.uiuc.edu/~demo5337/s97b/fibonacci. html http://faculty.mansfield.edu/hiseri/MA1115/1115L30.pdf