Matthew

Report
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

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