M ATH 140 B - HW 7 S OLUTIONS
Problem 1 (WR Ch 8 #1). Define
(
f (x) =
e −1/x
2
0
(x 6= 0),
(x = 0).
Prove that f has derivatives of all orders at x = 0, and that f (n) (0) = 0 for n = 1, 2, 3, . . ..
Solution.
2
Claim 1. For any rational function R(x), limx→0 R(x)e −1/x = 0.
Let R(x) =
p(x)
q(x)
for polynomials p and q. Let m be the smallest power of x in q. Then
by dividing the top and bottom of the fraction
p(x)
q(x)
by x m , we have that the bottom does not
vanish, and then the claim follows from the linearity of limits and Theorem 8.6(f), which says
limx→∞ x n e −x = 0 for any n ∈ Z.
2
Claim 2. For x 6= 0, f (n) (x) = R n (x)e −1/x for some rational function R n (x).
³ ´
2
We prove this by induction. For n = 1, we have f 0 (x) = e −1/x ddx −1
=
x2
satisfies the requirement. Now assume for our induction hypothesis that f
2 −1/x 2
e
, which
x3
p(x) −1/x 2
(n)
.
(x) = q(x) e
Then by the product rule and the quotient rule:
µ
¶
¶ µ
p(x) 2 −1/x 2
q(x)p 0 (x) − q 0 (x)p(x) −1/x 2
(n+1)
f
(x) =
e
e
+
q(x) x 3
q 2 (x)
µ
¶
2p(x)
q(x)p 0 (x) − q 0 (x)p(x) −1/x 2
= 3
+
e
x q(x)
q 2 (x)
µ
¶
2p(x)q(x) + x 3 q(x)p 0 (x) − x 3 q 0 (x)p(x) −1/x 2
=
e
x 3 q 2 (x)
which satisfies the claim.
Now we solve the problem by induction using the two claims. For n = 1,
2
f 0 (0) = lim
h→0
f (h) − f (0)
e −1/h − 0
= lim
= 0.
h→0
h
h
Assuming that f (n) (0) = 0, we then have
2
f (n+1) (0) = lim
h→0
f (n) (h) − f (n) (0)
R n (h)e −1/h − 0
= lim
= 0.
h→0
h
h
1
Problem 2 (WR Ch 8 #2). Let a i j be the number in the i th row and j th column of the
array
−1
0
0
0
···
1
2
1
4
1
8
−1
0
0
···
1
2
1
4
−1
0
···
1
2
−1
..
.
···
..
.
..
.
..
.
..
.
so that
ai j


 0
=
−1

 j −i
2
(i < j ),
(i = j ),
(i > j ).
Prove that
XX
i
XX
a i j = −2,
j
j
a i j = 0.
i
Solution. Summing over columns first, we have
X
for any j ,
a i j = −1 +
∞
X
n=1
i
1
2n
= −1 + 1 = 0
j
Summing over rows first, using the identity
N
P
n=0
X
j
ai j
i
e − (1 + x)1/x
.
x→0
x
lim
n→∞
XX
i
lim
n
[n 1/n − 1].
log n
Solution.
2
X
0 = 0.
j
1−x N +1
1−x , we have
=⇒
Problem 3 (WR Ch 8 #5).
(b)
xn =
ai j =
1
1 − 2i
1
−1
= −1 +
− 1 = −1 +
= i −1
1
n
2
2
1
−
n=1
2
iX
−1
(a)
XX
=⇒
j
ai j =
X −1
i
2i −1
= −2.
(a)
1
log(1+x)
e − (1 + x)1/x
e −e x
lim
= lim
x→0
x→0
x
x
1
−e x
= lim
³
log(1+x) −1
x2
1
log(1 + x) + x(x+1)
´
1
¶
1
−
log(1
+ x)
+
= lim −(1 + x)1/x
x→0
x2
x(x + 1)
µ
¶
−(1 + x) log(1 + x) + x
1/x
= lim −(1 + x)
lim
x→0
x→0
x 2 (x + 1)
Ã
!
− log(1 + x) − 1+x
1+x + 1
= −e lim
x→0
3x 2 + 2x
¶
µ
log(1 + x)
= e lim
x→0 3x 2 + 2x
à 1 !
x→0
µ
= e lim
x→0
1+x
6x + 2
e
= .
2
(b) Since
1
n
log n → 0 as n → ∞, we have
·
¸
1
n
n
log n
1/n
n
[n
− 1] = lim
e
−1
lim
n→∞ log n
n→∞ log n
1
= lim
en
log n
1
n
n→∞
−1
log n
h
e −1
= lim
h→0
h
µ
¶¯
d x ¯¯
=
e ¯
dx
x=0
= e0
= 1.
Problem 4 (WR Ch 8 #6). Suppose f (x) f (y) = f (x + y) for all real x and y.
(a) Assuming that f is differentiable and not zero, prove that
f (x) = e c x
where c is a constant.
(b) Prove the same thing, assuming only that f is continuous.
3
Solution. We simply prove part (b). Assuming f is not identically zero, there exists some x
such that f (x) 6= 0. Then
f (0) f (x) = f (0 + x) = f (x)
f (0) = 1.
=⇒
By continuity at 0 this means there is a neighborhood (−², ²) (for ² > 0) on which f is greater
than 0. This means that for any n, m ∈ Z with n >
1
²
we have
1
1
1 m
f (m
n ) = f ( n ) · . . . · f ( n ) = [ f ( n )] > 0,
|
{z
}
m times
so f is positive on a dense set of R and thus is positive everywhere. Next, we let c = log f (1),
and let g (x) = f (x) − e cx , which is a continuous function. We will prove that g vanishes on
a dense set, and thus by continuity must vanish everywhere. For any nonzero n ∈ Z, by the
uniqueness of the positive nth root, we have
³ c ´n
f ( n1 )n = f (1) = e c = e n
c
f ( n1 ) = e n
=⇒
=⇒
Now for any m ∈ N and nonzero n ∈ Z
m
g(m
n ) = f ( n )−e
m
cn
¡
¢m ³ c ´m
= f ( n1 ) − e n
= 0.
4
g ( n1 ) = 0.