SOLUTION OF HW5
MINGFENG ZHAO
October 15, 2011
1. [Page 44, Problem 34] Use induction to prove the binomial theorem:
n
(a + b) =
n X
n
k
k=0
ak bn−k .
Then use the theorem to derive the formulas:
n X
n
k
k=0
n
Proof. Claim I: (a + b) =
= 2n ,
n
X
(−1)k
k=0
n X
n
k
k=0
and
n
= 0, if n > 0.
k
ak bn−k .
When n = 1, we have
(a + b)1 = a + b =
1
1
a+
b.
0
1
So the Claim I is true for n = 1. Now assume n = k, the Claim I is true for n = m, that is,
m X
m k m−k
=
a b
.
k
m
(a + b)
k=0
Look at n = m + 1. We have
(a + b)m+1
=
=
(a + b)m · (a + b)
m X
m
k
k=0
=
m X
m
k=0
=
m X
k=0
=
k
a b
· (a − b)
ak bm−k · a +
By the assumption for n = m
m X
m k m−k
a b
·b
k
k=0
m m k+1 m−k X m k m+1−k
a
b
+
a b
k
k
m+1
X
i=1
!
k m−k
k=0
m X
m
m k m+1−k
ai bm−i+1 +
a b
i−1
k
k=0
1
2
MINGFENG ZHAO
=
m+1
X
i=1
m+1
= a
m X
m
m i m+1−i
i m+1−i
ab
+
ab
i−1
i
i=0
m m X
X
m
m i m+1−i
i m+1−i
+
ab
+
ab
+ bm+1
i
−
1
i
i=1
i=1
= am+1 +
m X
m
m i m+1−i
ai bm+1−i +
ab
+ bm+1
i
−
1
i
i=1
= am+1 +
m X
m
m
+
· ai bm+1−i + bm+1
i
−
1
i
i=1
m+1
= a
+
m X
m+1
i=1
=
m+1
X
i=0
i
· ai bm+1−i + bm+1
By the Problem 3, in Page 44
m+1
· ai bm+1−i .
i
That is, the Claim is true for n = m + 1. By induction, we can conclude that for all n ∈ N, we
have
(a + b)n =
n X
n
k=0
k
ak bn−k .
By taking a = b = 1, we get
(1 + 1)n =
n X
n
k=0
k
1k 1n−k .
That is, we have
n X
n
k=0
k
= 2n .
By taking a = −1, b = 1, we get
(−1 = 1)n =
n X
n
k
k=0
(−1)k 1n−k .
That is, we have
n
k = 0 (−1)
= 0.
k
n
k
SOLUTION OF HW5
3
2. [Page 84, Problem 25] Let f be a function whose domain contains −x whenever it contains x. We
say that f is an even function if f (−x) = f (x) and an odd function if f (−x) = −f (x) for all x in the
domain off. If f is integrable on [0, b], prove that
Z
b
b
Z
a. If f is even, then
f (x) dx = 2
f (x) dx.
Z−b
b
0
b. If f is even, then
f (x) dx = 0.
−b
Proof. a. Since f is even, then f (x) = f (−x). Then
b
Z
0
Z
f (x) dx
b
Z
=
f (x) dx +
−b
f (x) dx
−b
0
Z
f (−x) dx +
−b
f (x) dx
Since f (x) = f (−x)
f (x) dx
By the Theorem 1.19, in Page 81
0
0
Z
= −
b
Z
f (x) dx +
b
0
b
=
b
Z
f (x) dx +
0
f (x) dx
By the Theorem 1.17, in page 80
0
b
Z
=
b
Z
=
Z
By the Theorem 1.17, in page 80
0
2
f (x) dx.
0
Therefore, we get
Z
b
Z
b
f (x) dx = 2
−b
f (x) dx.
0
b. Since f is odd, then f (x) = −f (−x). Then
Z
b
Z
f (x) dx
0
=
−b
b
Z
f (x) dx +
f (x) dx
−b
Z
0
−b
0
Z
By the Theorem 1.16
b
Z
f (x) dx +
b
f (x) dx
By the Theorem 1.19, in Page 81
0
Z
= −
b
Z
f (x) dx +
0
0.
f (x) dx
0
0
=
Since f (x) = −f (−x)
b
Z
f (−x) dx +
−b
Z
f (x) dx
0
= −
=
b
Z
−f (−x) dx +
=
By the Theorem 1.17, in page 80
0
b
f (x) dx
0
By the Theorem 1.17, in page 80
4
MINGFENG ZHAO
Therefore, we have
b
Z
f (x) dx = 0.
−b
3. [Page 84, Problem 28] Show that
b
Z
Z
c−a
f (c − x) dx =
f (x) dx.
a
c−b
Proof. Let g(x) = f (c − x), then
b
Z
b
Z
f (c − x) dx
=
g(x)
a
a
Z −b
1
·
g(−x) dx
−1 −a
Z −b
= −
g(−x) dx
=
By the Theorem 1.19
−a
Z
−b
= −
f (c + x) dx
−a
−a
Z
=
f (c + x) dx
By the Theorem 1.17.
−b
Let h(x) = f (c + x), then
Z
b
Z
f (c − x) dx
−a
=
f (c + x) dx
−b
a
Z
−a
=
h(x) dx
−b
Z
c−a
h(x − c) dx
=
By the Theorem 1.18
c−b
Z
c−a
f (c + x − c) dx
=
c−b
Z
c−a
=
f (x) dx.
c−b
Therefore, we conclude that
Z
b
Z
c−a
f (c − x) dx =
a
f (x) dx.
c−b
SOLUTION OF HW5
5
4. [Page 93, the Theorem 2.2] For any a, b > 0, and ∈ N, then we have
Z
1
b
x
1
b1+ n − a1+ n
dx =
.
1 + n1
1
n
a
1
Proof. Let f (x) = x n for all x ≥ 0, and g(y) = y n for all y ≥ 0.
For any a > 0, we let
a
Z
a
Z
S(a) =
0
1
x n dx.
f (x) dx =
0
Then S(a) denotes the area between the graph of f and x-axis over the interval [0, a]. For any
c > 0, we let
Z
T (c) =
c
Z
c
g(y) dy =
0
y n dy.
0
Then T (c) denotes the area between the graph of g and y-axis over the interval [0, c]. Also we can
find that g is the inverse function of f on [0, ∞. And we know that
1
f (a) = a n .
1
1
1
So S(a) + T a n = a · a n = a1+ n . But we know
1
T a
1
n
Z
=
an
y n dy
0
=
1 n+1
1
· an
n+1
=
1
1
· a1+ n .
n+1
By the Theorem 1.15, in Page 80
Hence, we get
S(a)
=
1
1
a1+ n − T a n
=
a1+ n −
=
1
n
· a1+ n .
n+1
1
1
1
· a1+ n
n+1
6
MINGFENG ZHAO
That is, we get
Z
a
1
x n dx =
0
1
n
· a1+ n .
n+1
Now for any a, b > 0, by the additive property of the integral, the Theorem 1.17, in Page 80, we
can get
Z
b
x
1
n
Z
dx
x
=
a
b
1
n
Z
dx −
0
a
1
x n dx
0
=
1
1
n
n
· b1+ n −
· a1+ n
n+1
n+1
=
h
i
1
1
n
· b1+ n − a1+ n
n+1
=
b1+ n − a1+ n
.
1 + n1
1
1
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu