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 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: 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

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

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

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

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

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 A n This is Binet’s formula, an explicit formula for finding the n th Fibonacci number. An=An=

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!

Resources _11-3A_geometric_sequences_explicit.pdf _11-3A_geometric_sequences_explicit.pdf _11-2A_arithmetic_sequences.pdf _11-2A_arithmetic_sequences.pdf html html