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18.014
Problem
Set
2
Solutions
Total:
24
points
Problem
1:
Let
f
(
x
) =
k
n
=0
c
k
x
k
be
a
polynomial
of
degree
n
.
(d)
If
f
(
x
)
=
0
for
n
+
1
distinct
real
values
of
x
,
then
every
coeﬃcient
c
k
of
f
is
zero
and
f
(
x
)
=
0
for
all
real
x
.
(e)
Let
g
(
x
) =
m
b
k
x
k
be
a
polynomial
of
degree
m
where
m
≥
n
.
If
g
(
x
) =
f
(
x
)
k
=0
for
m
+
1
distinct
real
values
of
x
,
then
m
=
n
,
b
k
=
c
k
for
each
k
,
and
g
(
x
) =
f
(
x
)
for
all
real
x
.
Solution
(4
points)
We
prove
(
d
)
by
induction
on
n
.
Since
the
statement
is
true
for
all
integers
n
≥
0,
our
base
case
is
n
=
0.
If
f
is
a
polynomial
of
degree
0,
then
f
=
c
is
a
constant.
If
f
has
n
+
1
=
1
real
roots,
then
f
(
x
)
=
0
for
some
x
;
hence,
c
=
0
and
f
(
x
)
=
0
for
all
x
.
Assume
the
statement
is
true
for
all
polynomials
of
degree
n
;
we
prove
the
statement
for
a
polynomial
f
of
degree
n
+
1.
By
hypothesis,
f
has
n
+
2
distinct
real
roots,
{
a
1
, . . . , a
n
+2
}
.
Using
part
(
c
)
of
this
problem
(which
we
did
together
in
recitation),
f
(
x
) = (
x
−
a
n
+2
)
f
n
(
x
)
where
f
n
is
a
polynomial
of
degree
n
.
But,
the
roots
{
a
1
, . . . , a
n
+2
}
are
distinct;
hence,
a
i
−
a
n
+2
=
0
if
i < n
+
2
and
f
n
(
a
i
)
=
0
for
i
= 1
, . . . , n
+
1.
Thus,
by
the
in
duction
hypothesis,
f
n
=
0
because
f
n
is
a
polynomial
of
degree
n
with
n
+1
distinct
real
roots,
{
a
l
, . . . , a
n
+1
}
.
Since
f
n
=
0,
we
conclude
f
(
x
) = (
x
−
a
n
+2
)
f
n
(
x
)
=
0
for
every
real
x
.
Moreover,
since
every
coeﬃcient
of
f
n
is
zero,
every
coeﬃcient
of
f
is
zero.
This
is
what
we
wanted
to
show.
(e).
If
g
(
x
) =
f
(
x
)
for
m
+
1
distinct
values
of
x
,
then
g
(
x
)
−
f
(
x
)
has
m
+ 1
distinct
real
roots.
Moreover,
since
deg
f
=
n
,
deg
g
=
m
,
and
m
≥
n
,
we
observe
deg(
g
−
f
)
≤
m
.
Thus,
by
part
(d),
g
(
x
)
−
f
(
x
)
=
0
and
every
coeﬃcient
of
g
(
x
)
−
f
(
x
)
is
zero.
This
implies
g
(
x
) =
f
(
x
)
for
all
real
x
;
moreover,
since
the
coeﬃcients
of
g
(
x
)
−
f
(
x
)
are
b
k
−
a
k
,
it
implies
b
k
=
a
k
for
all
k
and
m
=
deg
g
=
deg
f
=
n
.
Problem
2:
Let
A
=
{
1
,
2
,
3
,
4
,
5
}
,
and
let
M
denote
the
set
of
all
subsets
of
A
.
For
each
set
S
in
M
,
let
n
(
S
)
defnote
the
number
of
elements
of
S
.
If
S
=
{
1
,
2
,
3
,
4
}
1
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�
and
T
=
{
3
,
4
,
5
}
,
compute
n
(
S
∪
T
),
n
(
S
∩
T
),
n
(
S
−
T
),
and
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 Fall '10
 ChristineBreiner
 Calculus, Derivative, Trigraph, Natural number, distinct real roots

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