SOLUTION OF HW4
MINGFENG ZHAO
October 01, 2011
1. [Page 44, Problem 3] Prove that
n+1
k
=
n
n
+
.
k−1
k
Proof. In fact, we have
n
n
+
=
k−1
k
n!
n!
+
(k − 1)!(n − k + 1)! k!(n − k)!
=
n! · k
n! · (n − k + 1)
+
k · (k − 1)!(n − k + 1)! k!(n − k)! · (n − k + 1)
=
k · n!
(n + 1 − k) · n!
+
k!(n + 1 − k)!
k!(n + 1 − k)!
=
k · n! + (n + 1 − k) · n!
k!(n + 1 − k)!
=
[k + (n + 1 − k)] · n!
k!(n + 1 − k)!
=
(n + 1) · n!
k!(n + 1 − k)!
(n + 1)!
k!(n + 1 − k)!
n+1
=
.
k
=
2. [Page 45, Problem 13] a. Use the binomial theorem to prove that for n a positive integer we have
1
1+
n
n
=1+
n
X
k=1
"
#
k−1
1 Y
r
·
1−
.
k! r=0
n
b. If n > 1, use part a and Exercise 11, in Page 45 to deduce the inequality
2<
1+
1
n
n
<1+
n
X
1
< 3.
k!
k=1
1
2
MINGFENG ZHAO
Proof. a. By the binomial theorem, the Exercise 4, in Page 44, we know that
1+
1
n
n X
n
n
=
k
k=0
1n−k ·
k
1
n
n X
n 1
k nk
=
k=0
n X
n 1
1+
k nk
=
k=1
=
1+
n
X
k=1
=
1+
n!
1
k!(n − k)! nk
n
X
1
n!
1
·
·
k! (n − k)! nk
k=1
=
1+
n
X
1 n(n − 1) · · · (n − k + 1) · (n − k) · · · 1 · 1 1
·
· k
k!
(n − k)(n − k − 1) · · · 2 · 1
n
k=1
=
1+
n
X
1 n(n − 1) · · · (n − k + 1)
·
k!
nk
k=1
=
1+
n
X
1 (n − 1)(n − 2) · · · (n − k + 1)
·
k!
nk−1
k=1
n
X
n−k+1
1 n−1 n−2
·
·
···
1+
k!
n
n
n
=
k=1
=
1+
n
k−1
X
1 Y n−r
·
k! r=1 n
k=1
=
1+
n
X
k=1
"
#
k−1
1 Y
r
·
.
1−
k! r=0
n
b. If n > 1, then we know that
1+
1
n
n
=
n X
n
k=0
≥
2 X
n
k=0
=
=
=
k
k
1n−k ·
n−k
1
k
1
n
By the binomial theorem, the Exercise 4, in Page 44
k
1
·
n
2 X
n 1
k nk
k=0
n
1
n
1
1+
· +
· 2
1
n
2
n
1+n·
1
n(n − 1) 1
+
· 2
n
2
n
SOLUTION OF HW4
>
1+1
=
2.
3
On the other hand, by the result of part a, we can get
n
1
1+
n
=
1+
n
X
"
k=1
<
1+
k−1
1 Y
r
·
1−
k! r=0
n
#
n
X
1
k!
k=1
=
1+1+
n
X
1
k!
k=2
=
2+
n
X
k=2
≤
2+
n
X
k=2
=
2+
1
k(k − 1) · · · 2 · 1
1
k(k − 1)
n X
k=2
=
2+1−
<
3.
1
1
−
k−1 k
1
n
By the Exercise 2, in Page 40
Therefore, we get
2<
1
1+
n
n
<1+
n
X
1
< 3.
k!
k=1
Z
3. [Page 70, Problem 6] a. If n is a positive integer, prove that
n
[t]2 dt =
0
b. If f (x) =
Rx
0
2
[t] dt for x ≥ 0, draw the graph offover the interval [0, 3].
c. Find all x > 0 for which
Rx
0
[t]2 dt = 2(x − 1).
Proof. a. Since n is a positive integer, then we have
Z
0
n
[t]2 dt
=
n−1
X Z k+1
k=0
k
[t]2 dt
n(n − 1)(2n − 1)
.
6
4
MINGFENG ZHAO
=
n−1
X Z k−1
k=0
=
n−1
X
k2
n−1
X
By the definition of [t], and k ≤ t < k − 1
k+1
Z
dt
k
k=0
=
k 2 dt
k
k2
k=1
=
n(n − 1)(2n − 1)
,
6
By the Exercise 6, in Page 40.
. b. For x ≥ 0, then for any 0 ≤ t < x, we know that [t] ≥ 0, so
Rx
0
[t]2 dt is an increasing function on
[0, ∞). We have the following cases:
(1) If 0 ≤ x ≤ 1, we know that for all 0 ≤ t < x ≤ 1, so [t] = 0, which implies that
Z
x
[t]2 dt = 0.
0
(2) If 1 < x ≤ 2, then
Z
x
[t]2 dt =
1
Z
0
[t]2 dt +
0
Z
x
[t]2 dt = 0 +
1
Z
x
dt = x − 1.
1
(3) If 2 < x ≤ 3, we have
Z
x
[t]2 dt =
Z
0
2
[t]2 dt +
x
Z
0
[t]2 dt = 2 − 1 +
2
Z
x
22 dt = 1 + 4(x − 2) = 4x − 7.
2
In a summary, we get




0,
if 0 ≤ x ≤ 1




Z x

[t]2 dt =
x − 1, if 1 < x ≤ 2

0






 4x − 7, if 2 < x ≤ 3.
Therefore, the graph of f (x) =
Rx
0
[t]2 dt on [0, 3] should be:
SOLUTION OF HW4
5
c. For any x > 0, then x − 1 < [x] ≤ x, so we know that
Z
x
Z
[t]2 dt =
0
[x]
[t]2 dt +
0
Z
x
[t]2 dt
[x]
[x]([x] − 1)(2[x] − 1)
+
6
=
Z
x
[x]2 dt
By the result of the part a
[x]
[x]([x] − 1)(2[x] − 1)
+ [x]2 (x − [x]).
6
=
For 2(x − 1), we have
2([x] − 1) ≤ 2(x − 1) < 2([x] + 1 − 1) = 2[x].
We have the following cases:
(1) If x ≥ 4, we know that
Z
x
[t]2 dt ≥
0
[x]([x] − 1)(2[x] − 1)
+ [x]2 (x − [x])
6
≥
[x]([x] − 1)(2[x] − 1)
6
≥
[x](4 − 1)(2 × 4 − 1)
6
=
21[x]
6
>
2[x]
>
2(x − 1).
6
MINGFENG ZHAO
(2) If 3 ≤ x < 4, then we have
x
Z
2
Z
Z
2
x
32 dt = 5 + 9(x − 3) = 9x − 22.
3
3
0
x
Z
2
[t] dt = 4 × 3 − 7 +
[t] dt +
[t] dt =
0
3
If 9x − 22 = 2(x − 1), that is, 7x = 20, then x =
(3) If 0 ≤ x ≤ 1, then we have
(4) If 1 < x ≤ 2, then
Rx
(5) If 2 < x ≤ 3, then
Rx
0
0
Rx
0
20
7
< 3, which is impossible.
[t]2 dt = 0. If 2(x − 1) = 0, then x = 1, which is possible.
[t]2 dt = x − 1. If x − 1 = 2(x − 1), which is impossible.
[t]2 dt = 4x − 7. If 4x − 7 = 2(x − 1), that is, 2x = 5, then x = 52 , which
is possible.
In a summary, only solutions of
Rx
0
[t]2 dt = 2(x − 1) with x > 0 are:
x = 1,
and x =
5
.
2
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu