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The Chinese University of Hong Kong *
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Course
3230
Subject
Mathematics
Date
Nov 24, 2024
Type
Pages
4
Uploaded by MinisterGorilla8297
MATH3230 Numerical Analysis
Tutorial 1
1
Recall:
The theorems and tools below will be frequently used in this course.
Theorem 1
(Polynomials Theorem)
.
A polynomial of degree
n
has exactly
n
roots, and a polynomial of degree
n
which vanishes at
(
n
+ 1)
points is identically zero.
Theorem 2
(Rolle’s theorem)
.
Let
f
(
x
)
be continuous on
[
a, b
]
and differentiable in
(
a, b
)
, and
f
(
a
) =
f
(
b
)
, then
there exists
ζ
∈
(
a, b
)
such that
f
′
(
ζ
) = 0
.
Theorem 3
(Mean-value theorem)
.
Let
f
(
x
)
be continuous on
[
a, b
]
and its derivative
f
′
(
x
)
be continuous in
(
a, b
)
,
then there exists
ζ
∈
(
a, b
)
such that
f
(
b
) =
f
(
a
) + (
b
−
a
)
f
′
(
ζ
)
,
i.e.,
f
′
(
ζ
) =
f
(
b
)
−
f
(
a
)
b
−
a
.
Theorem 4
(Intermediate value theorem)
.
Let
f
(
x
)
be continuous on
[
a, b
]
, then for any real number
g
that lies
between
f
(
a
)
and
f
(
b
)
there exists
ζ
∈
[
a, b
]
such that
f
(
ζ
) =
g
.
Theorem 5
(Taylor’s theorem in one dimension)
.
Let
f
(
x
)
be continuously differentiable on
[
a, b
]
up to order
n
−
1
and its
n
th derivative
f
(
n
)
(
x
)
be continuous in
(
a, b
)
. Then for any two points
x
0
, x
∈
(
a, b
)
,
f
(
x
) =
f
(
x
0
) +
f
′
(
x
0
)(
x
−
x
0
) +
f
′′
(
x
0
)
2!
(
x
−
x
0
)
2
+
· · ·
+
f
(
n
)
(
x
0
)
n
!
(
x
−
x
0
)
n
+
R
n
(
x
)
where
R
n
(
x
)
is the remainder of Taylor’s series and can be given by
R
n
(
x
) =
1
n
!
Z
x
x
0
f
(
n
+1)
(
t
)(
x
−
t
)
n
dt
or
R
n
(
x
) =
f
(
n
+1)
(
ζ
)
(
n
+ 1)!
(
x
−
x
0
)
n
+1
for some
ζ
lying between
x
0
and
x
.
Theorem 6
(Mean-value theorem for integrals)
.
Let
f
(
x
)
and
g
(
x
)
be continuous on
[
a, b
]
, and
f
(
x
)
does not
change signs in
[
a, b
]
. Then there exists
ζ
∈
(
a, b
)
such that
Z
b
a
f
(
x
)
g
(
x
)
dx
=
g
(
ζ
)
Z
b
a
f
(
x
)
dx
Theorem 7
(Taylor’s theorem in higher dimensions)
.
Let
f
(
x
) :
R
m
→
R
be sufficiently smooth in
Ω
⊂
R
m
. Then
for any two points
x
0
,
x
∈
Ω
, we have
f
(
x
) =
f
(
x
0
) +
Df
(
x
0
)
T
(
x
−
x
0
) +
1
2!
(
x
−
x
0
)
T
D
2
f
(
x
0
)(
x
−
x
0
) +
· · ·
where
Df
(
x
)
is the gradient of
f
(
x
)
and
D
2
f
(
x
)
is the Hessian matrix of
f
(
x
)
.
Theorem 8
(Fundamental theorem of calculus)
.
Let
F
be differentiable in an open set
Ω
⊂
R
n
and
x
∗
∈
Ω
. Then
for all
x
∈
Ω
such that the line segment connecting
x
∗
and
x
lines inside
Ω
, we have
F
(
x
)
−
F
(
x
∗
) =
Z
1
0
F
′
(
x
∗
+
t
(
x
−
x
∗
))(
x
−
x
∗
)
dt
.
1
2
Exercises:
Please do the star problem (*) in tutorial class first and finish the rest after class.
1.* Let
p, q
:
R
→
R
be two polynomials of degree
n
. Suppose
p
and
q
coincide on a distinct set of
n
+ 1
real
numbers
{
x
i
}
n
i
=0
, i.e.
p
(
x
i
) =
q
(
x
i
)
for all
i
= 0
,
1
,
2
,
· · ·
, n.
Using the polynomials theorem, show that
p
(
x
) =
q
(
x
)
for all
x
∈
R
.
Proof.
Let
f
=
p
−
q
, then
f
is a polynomial of at most degree
n
. Then we have
f
(
x
i
) = 0
for all
i
= 0
,
1
,
2
..., n
.
From polynomials theorem, we know this polynomial
f
of degree
n
that vanishes at
n
+ 1
points is a constant
function of zero, so
f
= 0
and
p
=
q
.
2.
(a)* Determine whether we are allowed to use Rolle’s Theorem to guarantee the existence of some
c
in
(
a, b
)
such that
f
′
(
c
) = 0
. If not, explain why not.
i.
f
(
x
) = cos(
x
)
, for
x
∈
[0
,
2
π
]
;
ii.
f
(
x
) =
|
x
−
2
|
, for
x
∈
[0
,
4]
;
iii.
f
(
x
) = 1
/x
2
, for
x
∈
[
−
1
,
1]
.
(b)
For
f
∈
C
n
(
R
)
and
f
(
x
0
) =
f
(
x
1
) =
· · ·
=
f
(
x
n
) = 0
for
x
0
< x
1
<
· · ·
< x
n
, show that there exists
f
n
(
ζ
) = 0
for some
ζ
∈
(
x
0
, x
n
)
.
Solution.
(a) We recall the condition of applying Rolle’s theorem is
f
(
x
)
must be continuous in
[
a, b
]
, differ-
entiable in
(
a, b
)
, and
f
(
a
) =
f
(
b
)
.
i. Since
f
(
x
)
is continuous in
[0
,
2
π
]
, differentiable in
(0
,
2
π
)
, and
f
(0) =
f
(2
π
) = 1
, there exists
c
∈
(0
,
2
π
)
such that
f
′
(
c
) = 0
.
ii. Since
f
(
x
)
is not differentiable at
x
= 2
, Rolle’s theorem does not apply.
iii. Since
f
(
x
)
is undefined at
x
= 0
, Rolle’s theorem does not apply.
(b) By Rolle’s theorem, there exist
x
1
j
∈
(
x
j
, x
j
+1
)
, for
j
= 0
,
· · ·
, n
−
1
such that
f
′
(
x
1
j
) = 0
for
j
= 0
,
· · ·
, n
−
1
,
and
x
1
0
< x
1
1
<
· · ·
< x
1
n
−
1
.
Then by using Rolle’s Theorem sequentially on
f
k
(
x
)
for
k
= 1
,
· · ·
, n
−
1
, we finally have there exists
ζ
∈
(
x
0
, x
n
)
such that
f
n
(
ζ
) = 0
.
3.
(a)* Prove the mean value theorem using Rolle’s Theorem.
(b)
Let
f, g
: [
a, b
]
→
R
be continuous on
[
a, b
]
and differentiable on
(
a, b
)
. Assume
g
′
(
x
)
>
0
for all
x
∈
(
a, b
)
.
Using the mean value theorem, show that there exists
c
∈
(
a, b
)
such that
f
′
(
c
)
g
′
(
c
)
=
f
(
b
)
−
f
(
a
)
g
(
b
)
−
g
(
a
)
.
Solution.
(a) Let
f
(
x
)
be continuous on
[
a, b
]
and
f
′
(
x
)
be continuous in
(
a, b
)
, consider
F
(
x
) :=
f
(
x
)
−
f
(
a
)
−
f
(
b
)
−
f
(
a
)
b
−
a
(
x
−
a
)
.
Then we have
F
(
a
) =
F
(
b
) = 0
. By Rolle’s theorem, there exists
ζ
∈
(
a, b
)
such that
F
′
(
ζ
) =
f
′
(
ζ
)
−
f
(
b
)
−
f
(
a
)
b
−
a
= 0
,
which implies
f
(
b
) =
f
(
a
) + (
b
−
a
)
f
′
(
ζ
)
.
2
(b) Consider
G
(
x
) :=
g
(
x
)(
f
(
b
)
−
f
(
a
))
−
f
(
x
)(
g
(
b
)
−
g
(
a
))
, we have
G
(
a
) =
G
(
b
)
. Using the mean value
theorem, there exists
c
∈
(
a, b
)
such that
G
′
(
c
) = 0
.
This leads to
g
′
(
c
)(
f
(
b
)
−
f
(
a
))
−
f
′
(
c
)(
g
(
b
)
−
g
(
a
)) = 0
.
Since
g
′
(
x
)
>
0
in
(
a, b
)
, we have
g
(
b
)
−
g
(
a
)
>
0
, g
′
(
c
)
>
0
. Therefore we can derive from the equation
above that
f
′
(
c
)
g
′
(
c
)
=
f
(
b
)
−
f
(
a
)
g
(
b
)
−
g
(
a
)
.
4.
Using the intermediate value theorem to show the following.
(a)* Show that the equation
cos(
x
) = 2
x
has a solution in the interval
[0
, π/
2]
.
(b)
Let
f
: [0
,
1]
→
R
be continuous such that
f
(0) =
f
(1)
. Show that there exists
c
∈
[0
,
1
/
2]
such that
f
(
c
) =
f
(
c
+ 1
/
2)
.
Solution.
(a) Let
f
(
x
) = cos(
x
)
−
2
x
, since
f
(
x
)
is continuous and
f
(0) = 1
>
0
,
f
(
π/
2) = 0
−
π <
0
, there
exists
ζ
∈
(0
, π/
2)
such that
f
(
ζ
) = 0
. Hence, there is a solution
ζ
to
cos(
x
) = 2
x
in
[0
, π/
2]
.
(b) Consider
g
(
x
) =
f
(
x
)
−
f
(
x
+ 1
/
2)
, since
g
(
x
)
is continuous and
g
(0) =
f
(0)
−
f
(1
/
2)
,
g
(1
/
2) =
f
(1
/
2)
−
f
(1) =
−
g
(0)
, there exists
c
∈
[0
,
1
/
2]
such that
g
(
c
) =
f
(
c
)
−
f
(
c
+ 1
/
2) = 0
, i.e.,
f
(
c
) =
f
(
c
+ 1
/
2)
. In
particular,
c
= 0
or
1
/
2
when
f
(0) =
f
(1
/
2) =
f
(1)
.
5.
(a)* Let
F
(
x, y
) = (
xy
)
2
+ sin(
x
+
y
) + 1
, given
F
(0
,
0) = 1
, using the fundamental theorem of calculus to
compute
F
(1
,
1)
by evaluating a line integral. Please specify all your steps clearly.
(b)
Using Taylor’s theorem to prove
1 +
x < e
x
is valid for all real numbers
x
except
0
.
Solution.
(a) By definition, we have
∂F
(
x, y
)
∂x
= 2
xy
2
+ cos(
x
+
y
)
,
∂F
(
x, y
)
∂y
= 2
x
2
y
+ cos(
x
+
y
)
.
Then by the fundamental theorem of calculus, we have
F
(1
,
1) =
F
(0
,
0) +
Z
1
0
2
t
3
+ cos(2
t
)
2
t
3
+ cos(2
t
)
T
1
1
dt
= 1 +
Z
1
0
(4
t
3
+ 2 cos(2
t
))
dt
= 2 + sin(2)
.
(b) Let
f
(
x
) =
e
x
−
(1 +
x
)
, by the Taylor’s theorem, we have
f
(
x
) =
f
(0) +
xf
′
(0) +
f
2
(
ζ
)
2
x
2
,
for some
ζ
∈
(0
, x
)
when
x >
0
and
ζ
∈
(
x,
0)
when
x <
0
. Then as
f
(0) = 0
,
f
′
(
x
) =
e
x
−
1
where
f
′
(0) = 0
, and
f
′′
(
x
) =
e
x
>
0
for all
x
, we have
f
(
x
) =
f
2
(
ζ
)
2
x
2
that is larger than
0
for all
x
except
0
which further implies
1 +
x < e
x
for all
x
except
0
.
6.
Let
R
n
(
x
) =
1
n
!
R
x
x
0
f
(
n
+1)
(
t
)(
x
−
t
)
n
dt
.
3
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(a)* Using the mean value theorem for integrals, show
R
n
(
x
)
can also be written as
R
n
(
x
) =
f
(
n
+1)
(
ζ
)
(
n
+ 1)!
(
x
−
x
0
)
n
+1
.
(b)
Using integration by parts, show
R
n
=
−
1
n
!
f
(
n
)
(
x
0
)(
x
−
x
0
)
n
+
R
n
−
1
.
(c)
Applying the integration by parts repeatedly, show
f
(
x
) =
f
(
x
0
) +
n
X
k
=1
1
k
!
f
(
k
)
(
x
0
)(
x
−
x
0
)
k
+
R
n
(
x
)
.
Solution.
(a) By the mean value theorem for integrals there exists
ζ
∈
(
x
0
, x
)
such that
R
n
(
x
) =
f
n
+1
(
ζ
)
n
!
Z
x
x
0
(
x
−
t
)
n
dt
=
−
f
n
+1
(
ζ
)
(
n
+ 1)!
(
x
−
t
)
n
+1
x
x
0
=
f
(
n
+1)
(
ζ
)
(
n
+ 1)!
(
x
−
x
0
)
n
+1
.
(b) Recall the formula for integration by parts
Z
udv
=
uv
−
Z
vdu ,
and we apply it to the integral
R
n
with
u
=
(
x
−
t
)
n
n
!
,
v
=
f
(
n
)
(
t
)
,
then we obtain
R
n
(
x
) =
1
n
!
h
f
(
n
)
(
t
)(
x
−
t
)
n
i
t
=
x
t
=
x
0
+
Z
x
x
0
f
(
n
)
(
t
)
(
x
−
t
)
n
−
1
(
n
−
1)!
dt
=
−
1
n
!
f
(
n
)
(
x
0
)(
x
−
x
0
)
n
+
R
n
−
1
(
x
)
.
(c) If the integrating parts is applied repeatedly on
R
j
for
j
=
n
−
1
, n
−
2
,
· · ·
,
1
, we eventually obtain
R
n
(
x
) =
−
n
X
n
=1
1
k
!
f
(
k
)
(
x
0
)(
x
−
x
0
)
k
+
R
0
(
x
)
.
Since
R
0
(
x
) =
Z
x
x
0
f
′
(
t
)
dt
=
f
(
x
)
−
f
(
x
0
)
,
we have
f
(
x
) =
f
(
x
0
) +
n
X
k
=1
1
k
!
f
(
k
)
(
x
0
)(
x
−
x
0
)
k
+
R
n
(
x
)
.
4