Math 194, problem set #6, SOLUTIONS
1. Evaluate
lim
n→∞
n
X
k=1
1
√
.
2
k + n2
(Larson 6.8.5)
Solution:
lim
n→∞
n
X
k=1
n
1X
1
1
√
p
,
= lim
2
2
n→∞ n
k +n
(k/n)2 + 1
k=1
and this, by the definition of the Riemann integral, equals
Z
0
2. Let f (x) =
x, then
Pn
k=1
1
1
√
√
dx
√
= ln |x + x2 + 1|
= ln(1 + 2).
x2 + 1
x=0
ak sin(kx) with ai ∈ R, n ≥ 1. Prove that if f (x) ≤ | sin(x)| for every
n
X
kak ≤ 1.
k=1
(Putnam 1967)
Solution: Since f (−x) = −f (x) and | sin(−x)| = | sin(x)|, we have |f (x)| ≤ | sin(x)|.
Consider
|f 0 (0)| = limx→0P
|f (x)/x| ≤ limx→0
| sin(x)/x| = 1. Also, since f 0 (x) =
P
Pn
n
n
0
k=1 kak | ≤ 1 as desired.
k=1 kak . Thus |
k=1 kak cos(kx), f (0) =
3. Let f (x) be a continuous function on [0, 1] such that f (0) = f (1) = 0 and 2f (x)+f (y) =
3f ( 2x+y
) for all x, y ∈ [0, 1]. Prove that f (x) = 0 for all x ∈ [0, 1].
(Vietnamese
3
Mathematical Olympiad, 1999)
Solution: Suppose for some a ∈ (0, 1), f (a) 6= 0. If f (a) > 0, f must attain its
maximum M at some c ∈ (0, 1). If c ∈ (0, 1/3), notice 2f (0) + f (3c) = f (3c) ≤ M <
3M = 3f (c) = 3f ( 2·0+3c
) a contradiction. If c ∈ (2/3, 1), notice 2f (1) + f (3c − 2) =
3
f (3c − 2) ≤ M < 3M = 3f (c) = 3f ( 2·1+3c−2
) a contradiction. If c ∈ [1/3, 2/3],
3
2· 1 (3c−1)+1
notice 2f ( 12 (3c − 1)) + f (1) = 2f ( 12 (3c − 1)) ≤ 2M < 3M = 3f (c) = 3f ( 2 3
)
a contradiction. Thus f (a) < 0, but then the same argument with f replaced by −f
shows that this is also a contradiction. Thus f (x) = 0 for all x ∈ [0, 1].
4. Suppose f : R → R is a continuous function such that |f (x) − f (y)| ≥ |x − y| for every
x, y ∈ R. Show that the range of f is R, i.e., for every c ∈ R there is an x such that
f (x) = c.
(De Souza & Silva, Berkeley Problems in Mathematics)
Solution: Suppose f (a) > c for every a ∈ R. Let d = f (0) − c, and notice that
|f (0) − f (d)| ≥ d and |f (0) − f (−d)| ≥ d by our hypothesis. Since c is a lower bound
for the values of f , we must have f (d) − f (0) ≥ d and f (0) − f (−d) ≤ −d (otherwise,
f (d) or f (−d) would have to be ≤ c). Since g(x) = f (x) − f (x − d) is positive at x = d
and negative at x = 0, by the Intermediate Value Theorem, there is some e ∈ (0, d)
such that g(e) = 0. Thus |f (e) − f (e − d)| = 0 6≥ d, a contradiction. If we replace f
by −f and c by −c, this argument also shows that we cannot have f (a) < c for every
a ∈ R. Thus, there must be some a, b ∈ R such that f (a) < c < f (b), and so c is in
the range of f by the Intermediate Value Theorem.
5. Suppose that f : R → R is a continuous function, and define
Z x
g(x) = f (x)
f (t)dt.
0
Prove that if g is a nonincreasing function, then f (x) = 0 for every x.
(Romanian Olympiad 1978)
2
Rx
Rx
Solution: Consider h(x) = 0 f (t)dt . Since h0 (x) = 2f (x) 0 f (t)dt = 2g(x), we
have that h0 (x) is nonincreasing. Notice that h(0) = 0 and h0 (0) = 0. Since h0 (x) is
nonincreasing, we have that h0 (x) ≤ 0 when x ≥ 0. Thus h(x) ≤ 0 when x ≥ 0, but
since h(x) is a square, we must have h(x) = 0 when x ≥ 0, and thus f (x) = 0 when
x ≥ 0. Similarly, we have that h0 (x) ≥ 0 when x ≤ 0, and so h(x) ≤ 0 when x ≤ 0,
which implies h(x) = 0 when x ≤ 0, and thus f (x) = 0 when x ≤ 0.
6. Let f : [0, 1] → R be a function with a continuous derivative, such that f (0) = 0 and
0 < f 0 (x) ≤ 1 for every x. Show that
Z
2
1
f (x)dx
0
Z
≥
1
(f (x))3 dx.
0
(Putnam 1973)
Solution: See Larson 7.4.4.
7. Suppose f and g are n-times continuously differentiable functions in a neighborhood
of a point a, such that f (a) = g(a), f 0 (a) = g 0 (a), . . . , f (n−1) (a) = g (n−1) (a), and
f (n) (a) 6= g (n) (a). Evaluate
ef (x) − eg(x)
.
lim
x→a f (x) − g(x)
(N. Georgescu-Roegen)
Solution: The jth derivative of eh(x) can be written as a function of eh(x) , h(x), and the
first j derivatives of h(x). Thus the expression in question is an indeterminate form,
and will remain indeterminate through n − 1 applications of L’Hôpital’s Rule by our
hypotheses. After the nth application, almost all terms in the numerator cancel upon
taking the limit, and we have
f (n) (x)ef (x) − g (n) (x)eg(x)
[f (n) (a) − g (n) (a)]ef (a)
ef (x) − eg(x)
= lim
=
= ef (a) .
x→a
x→a f (x) − g(x)
f (n) (x) − g (n) (x)
f (n) (a) − g (n) (a)
lim
8. Let n > 1 be a positive integer, and f : [a, b] → R a continuous function, n-times
differentiable on (a, b). Prove that if the graph of f has n + 1 collinear points, then
there is a point c ∈ (a, b) such that f (n) (c) = 0.
(G. Sireţchi, Mathematics Gazette, Bucharest)
Solution: Let g be a linear function whose graph intersects n + 1 points of the graph
of f . Then f − g has at least n + 1 roots, and so (f − g)(n) = f (n) − g (n) = f (n) has at
least one root.
9. Suppose x1 , x2 , . . . , xn ∈ R. Find the real number(s) a that minimize the expression
|a − x1 | + |a − x2 | + · · · + |a − xn |.
(Andreescu & Gelca)
Solution: Denote f (a) = |a − x1 | + |a − x2 | + · · · + |a − xn |. Without loss of generality,
assume x1 ≤ x2 ≤ · · · ≤ xn . For a ∈ (xj , xj+1 ), (a − xi ) will be positive for exactly
j values of i, and thus the slope of f is j − (n − j) = 2j − n. If n is odd, we then
see that f is decreasing on (−∞, x(n+1)/2 ) and increasing on (x(n+1)/2 , ∞), and so f is
minimized at a = x(n+1)/2 . If n is even, we then see that f is decreasing on (−∞, xn/2 ),
increasing on (x(n+2)/2 , ∞), and horizontal on [xn/2 , x(n+2)/2 ], and so f is minimized for
a ∈ [xn/2 , x(n+2)/2 ].
10. Prove that for every natural number n ≥ 2 and every x ∈ [−1, 1],
(1 + x)n + (1 − x)n ≤ 2n .
(Exercises and Problems in Algebra, Bucharest 1983)
Solution: We proceed by induction on n. For n = 1, the result holds. Assume the
result holds for all k ≤ n, and consider
(1 + x)k+1 + (1 − x)k+1 = (1 + x)(1 + x)k + (1 − x)(1 − x)k
= (1 + x)k + (1 − x)k + x[(1 + x)k − (1 − x)k ]
≤ 2k + x[2k − 2(1 − x)k ] ≤ 2k+1 ,
and so our result holds for all n ≥ 1.