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Finite Differences
By Daniela G. Ambrosano
Finite Differences
 Finite
differences is a method to find the
polynomial that could generate a sequence,
assuming the sequence can be generated by a
polynomial.
Finding Successive Differences
15
Find successive difference between terms of sequence;
the difference sequence.
Term #n
f(n)
1
5
2
15  15 - 5 = 10
3
33  33 - 15 = 18
4
59  59 – 33 =26
5
93  93 – 59 = 34

Finding differences, continued…
Continue finding differences between the terms of
difference sequences until the difference sequence is a
constant.
Term#n
f(n)
1
5
2
15  15 - 5 = 10
3
33  33 - 15 = 18  18 - 10 = 8
4
59  59 - 33 = 26  26 - 18 = 8
5
93  93 - 59 = 34  34 - 26 = 8

Continuing on…
 The
second difference in the sequence is constant
so we need a 2nd degree polynomial.
 General Form:
an  bn  c
2
Why?
 The
number of difference sequences is the power
of the polynomial that generated the original
sequence.
 This is because it takes that many derivatives to get
to a constant.
Generating a system (n+1) equations in
(n+1) variables
 The
variables in (n+1) are the coefficients of the
general polynomial.
 To generate a system (n+1) equations in (n+1)
variables, we subtract the various term numbers to
equal the corresponding term values.
For n = 1
an  bn  c  a(1 )  b(1)  c  5
2
2
Polynomials
2
 For n = 1 a(1 )  b(1)  c  5
 For
n=2
 For n = 3
a(2 )  b(2)  c  15
2
a(3 )  b(3)  c  33
2
Solve the System of Equations
 Solve
the system of equations to find the
coefficients/variables in two ways:
*substitution
*elimination
Eliminating C
the 2nd equation for n=2 and subtract from it
the 1st equation for n=1
 Take
4a  2b  c  15
(a  b  c  5)
3a  b  10
take the 3rd equation for n=3 and subtract
from it the 2nd equation for n=2 9a  3b  c  33
 Now
(4a  2b  c  15)
5a  b  18
Eliminate b and Solve for a
 Now
we use the new equations 3a + b = 10 and
5a + b = 18, subtract them and then solve for a
5a  b  18
(3a  b  10)
2a  8
a4
Substituting
 Now
we can substitute a = 4 in 3a + b = 10 to find
the value of b 3(4)  b  10
b  2
 Then
substitute the values of a = 4 and b = -2 into
the 1st equation a + b + c = 5 to find the value of c
4  (2)  c  5
c3
What is f(n)?
 Now
that we know the values of a, b, and c we can
substitute them into the original 2nd degree
polynomial that will generate the difference
sequences.
f (n)  4n  (2)n  3
2
Check Polynomial
 Plug
in a value for n to check if the polynomial
generated is correct
Ex: n = 2
f (2)  4(2 )  (2)  3
2
 16  (4)  3
 15
Original Sequence Table
Term # n
1
2
3
4
5
f(n)
5
15
33
59
93
The End