SOLUTION OF HW10
MINGFENG ZHAO
December 04, 2011
Z
1. [The Problem 8, in Page 155] Assume f is continuous on [a, b]. Assume also that
b
f (x)g(x) dx = 0
a
for every function g that is continuous on [a, b]. Prove that f (x) ≡ 0 for all x ∈ [a, b].
Z
Proof. Since
b
f (x)g(x) dx = 0 for every function g that is continuous on [a, b], and f is continuous
a
on [0, 1], so when we take g(x) ≡ f (x) for all x ∈ [0, 1], then we can get
b
Z
Z
f (x) · f (x) dx =
a
b
[f (x)]2 dx = 0.
a
Since f is continuous on [a, b], then f 2 is also continuous on [a, b], and [f (x)]2 ≥ 0 for all x ∈ [a, b].
Z b
But we know that
[f (x)]2 dx =. By the result of the Problem, in Page, we know that [f (x)]2 = 0
a
at all continuity point of f 2 on [a, b]. But f 2 is continuous at every point in [a, b], so we get [f (x)]2 =
for all x ∈ [a, b], which implies that f (x) ≡ 0 for all x ∈ [a, b].
2. [The Problem 38, in Page 168] Give the formula
1 + x + x2 + · · · + xn =
xn+1 − 1
.
x−1
Determine, by differentiation, formulas for the following sums:
a. 1 + 2x + 3x2 + · · · + nxn−1 .
b. 12 x + 22 x2 + 32 x3 + · · · + n2 xn .
Proof. a. In fact, we have
1 + 2x + 3x2 + · · · + nxn−1
=
n
X
k=1
1
kxk−1
2
MINGFENG ZHAO
=
n
X
(xk )0
k=1
n
X
=
!0
k
x
k=1
=
xn+1 − 1
x−1
0
=
(n + 1)xn · (x − 1) − (xn+1 − 1)
(x − 1)2
=
(n + 1)xn+1 − (n + 1)xn − xn+1 + 1
(x − 1)2
=
nxn+1 − (n + 1)xn + 1
.
(x − 1)2
So we get
1 + 2x + 3x2 + · · · + nxn−1 =
nxn+1 − (n + 1)xn + 1
.
(x − 1)2
b. In fact, we have
12 x + 22 x2 + 32 x3 + · · · + n2 xn
=
n
X
k 2 xk
k=1
= x·
n
X
k 2 xk−1
k=1
= x·
n
X
(kxk )0
k=1
n
X
= x·
!0
kx
k
k=1
= x·
= x·
nxn+1 − (n + 1)xn + 1
(x − 1)2
0
[n(n + 1)xn − n(n + 1)xn−1 ](x − 1)2 − [nxn+1 − (n + 1)xn + 1] · 2(x − 1)
(x − 1)4
=
x
· [[n(n + 1)xn − n(n + 1)xn−1 ](x − 1) − 2[nxn+1 − 2(n + 1)xn + 1]]
(x − 1)3
=
x
· [(n2 + n)xn+1 − (n2 + n)xn − (n2 + n)xn + (n2 + n)xn−1 − 2nxn+1
(x − 1)3
+(2n + 2)xn − 2]
SOLUTION OF HW10
=
3
x
· [(n2 − n)xn+1 − (2n2 − 2)xn − (n2 + n)xn−1 − 2]
(x − 1)3
3. [The Problem 9, in Page 173] A function f is defined as follows:




 x2 ,
if x ≤ c
f (x) =



 ax + b, if x > c.
Find values of a and b in terms of c such that f 0 (c) exists.
Proof. Since f 0 (c) exists, then f is continuous at x = c, so we have
f (c+) = lim f (x) = lim [ax + b] = ac + b.
x&c
x&c
And
f (c−) = lim f (x) = lim x2 = c2 .
x%c
x&c
Since f is continuous at x = c, then f (c+) = f (c−), that is, ac + b = c2 . Since f 0 (c) exists, then
f 0 (c+) = lim
f (c + h) − f (c)
= a.
h
f 0 (c−) = lim
f (c + h) − f (c)
= 2c.
h
h&0
And
h%0
So we have a = 2c. Since ac + b = c2 , then b = −c2 . In a summary, we get a = 2c, and b = −c2 . Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu