Expressing Sequences ExplicitlyBy: Matt Connor

Fall 2013

•Pure Math

•Analysis

•Calculus and Real Analysis

•Sequences

• 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, ........

•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

• General Forms• Recursive formula

• An = A(n-1) + d• Explicit formula

• An = A1 + d(n-1)

Arithmetic Sequences

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 are commonly called Fibonacci-type

•The difficult thing about these is finding an 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

• Cxn = Cxn−1 + Cxn−2

Fibonacci Sequence Explicit Formula

•Cxn = Cxn−1 + Cxn−2 this will always simplify to an equation with the same coefficients as the relational equation,

•x2 = 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= 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

•A0=1 and A1=1

•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.

•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

•This is Binet’s formula, an explicit formula for finding the nth Fibonacci number.

An=

•As you have seen finding an explicit formula for the nth 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/m

ath/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