CS 70
Discrete Mathematics and Probability Theory
Spring 2010
Alistair Sinclair
Note 7
Polynomials
Recall from your high school math that a
polynomial
in a single variable is of the form
p
(
x
) =
a
d
x
d
+
a
d

1
x
d

1
+
...
+
a
0
. Here the
variable x
and the
coefficients a
i
are usually real numbers. For example,
p
(
x
) =
5
x
3
+
2
x
+
1, is a polynomial of
degree d
=
3. Its coefficients are
a
3
=
5,
a
2
=
0,
a
1
=
2, and
a
0
=
1.
Polynomials have some remarkably simple, elegant and powerful properties, which we will explore in this
note.
First, a definition: we say that
a
is a
root
of the polynomial
p
(
x
)
if
p
(
a
) =
0. For example, the degree
2 polynomial
p
(
x
) =
x
2

4 has two roots, namely 2 and

2, since
p
(
2
) =
p
(

2
) =
0.
If we plot the
polynomial
p
(
x
)
in the
x

y
plane, then the roots of the polynomial are just the places where the curve crosses
the
x
axis:
We now state two fundamental properties of polynomials that we will prove in due course.
Property 1:
A nonzero polynomial of degree
d
has at most
d
roots.
Property 2:
Given
d
+
1 pairs
(
x
1
,
y
1
)
,...,
(
x
d
+
1
,
y
d
+
1
)
, with all the
x
i
distinct, there is a unique polynomial
p
(
x
)
of degree (at most)
d
such that
p
(
x
i
) =
y
i
for 1
≤
i
≤
d
+
1.
Let us consider what these two properties say in the case that
d
=
1. A graph of a linear (degree 1) polynomial
y
=
a
1
x
+
a
0
is a line. Property 1 says that if a line is not the
x
axis (i.e. if the polynomial is not
y
=
0), then
it can intersect the
x
axis in at most one point.
CS 70, Spring 2010, Note 7
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Property 2 says that two points uniquely determine a line.
Polynomial Interpolation
Property 2 says that two points uniquely determine a degree 1 polynomial (a line), three points uniquely
determine a degree 2 polynomial, four points uniquely determine a degree 3 polynomial, and so on. Given
d
+
1 pairs
(
x
1
,
y
1
)
,...,
(
x
d
+
1
,
y
d
+
1
)
, how do we determine the polynomial
p
(
x
) =
a
d
x
d
+
...
+
a
1
x
+
a
0
such
that
p
(
x
i
) =
y
i
for
i
=
1 to
d
+
1? We will give two different efficient algorithms for reconstructing the
coefficients
a
0
,...,
a
d
, and therefore the polynomial
p
(
x
)
.
In the first method, we write a system of
d
+
1 linear equations in
d
+
1 variables: the coefficients of the
polynomial
a
0
,...,
a
d
. The
i

th
equation is:
a
d
x
d
i
+
a
d

1
x
d

1
i
+
...
+
a
0
=
y
i
.
Since
x
i
and
y
i
are constants, this is a linear equation in the
d
+
1 unknowns
a
0
,...,
a
d
. Now solving these
equations gives the coefficients of the polynomial
p
(
x
)
. For example, given the 3 pairs
(

1
,
2
)
,
(
0
,
1
)
, and
(
2
,
5
)
, we will construct the degree 2 polynomial
p
(
x
)
which goes through these points. The first equation
says
a
2
(

1
)
2
+
a
1
(

1
)+
a
0
=
2. Simplifying, we get
a
2

a
1
+
a
0
=
2. Applying the same technique to the
second and third equations, we get the following system of equations:
a
2

a
1
+
a
0
=
2
a
0
=
1
4
a
2
+
2
a
1
+
a
0
=
5
Substituting for
a
0
and multiplying the first equation by 2 we get:
2
a
2

2
a
1
=
2
4
a
2
+
2
a
1
=
4
Then, adding down we find that 6
a
2
=
6, so
a
2
=
1, and plugging back in we find that
a
1
=
0. Thus, we
have determined the polynomial
p
(
x
) =
x
2
+
1. To justify this method more carefully, we must show that
the equations always have a solution and that it is unique. This involves showing that a certain determinant
is nonzero. We will leave that as an exercise, and turn to the second method.
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