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Exercises 6.1
25.
Clearly, it is suﬃcient to prove (9) just for two functions,
y
1
and
y
2
. Using the linear property
of diFerentiation, we have
L
[
y
1
+
y
2
]=[
y
1
+
y
2
]
(
n
)
+
p
1
[
y
1
+
y
2
]
(
n
−
1)
+
···
+
p
n
[
y
1
+
y
2
]
=
h
y
(
n
)
1
+
y
(
n
)
2
i
+
p
1
h
y
(
n
−
1)
1
+
y
(
n
−
1)
2
i
+
+
p
n
[
y
1
+
y
2
]
=
h
y
(
n
)
1
+
p
1
y
(
n
−
1)
1
+
+
p
n
y
1
i
+
h
y
(
n
)
2
+
p
1
y
(
n
−
1)
2
+
+
p
n
y
2
i
=
L
[
y
1
]+
L
[
y
1
]
.
Next, we verify (10).
L
[
cy
cy
]
(
n
)
+
p
1
[
cy
]
(
n
−
1)
+
+
p
n
[
cy
]=
cy
(
n
)
+
p
1
cy
(
n
−
1)
+
+
p
n
cy
=
c
±
y
(
n
)
+
p
1
y
(
n
−
1)
+
+
p
n
y
²
=
cL
[
y
]
.
27.
A linear combination
c
0
+
c
1
x
+
c
2
x
2
+
+
c
n
x
n
of the functions from the given set is a polynomial of degree at most
n
and so, by the funda
mental theorem of algebra, it cannot have more than
n
zeros unless it is the zero polynomial,
i.e., it has all zero coeﬃcients. Thus, if this linear combination vanishes on a whole interval
(
a, b
), then it follows that
c
0
=
c
1
=
c
2
=
...
=
c
n
= 0. Therefore, the set of functions
{
1
,x,x
2
,...,x
n
}
is linearly independent on any interval (
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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