SOLUTION OF HW11
MINGFENG ZHAO
December 07, 2011
1. [The Problem 7, in Page 186] Let f be a polynomial. A real number α is said to be a zero of f of
multiplicity m if f (x) = (x − α)m g(x), where g(α) 6= 0.
a. If f has r zeros in an interval [a, b], prove that f 0 has at least r − 1 zeros, and in general, when
k < r, the k-th derivative f (k) has at least r − k zeros in [a, b].
b. If the k-th derivative f (k) has exactly r zeros in [a, b], what can you conclude about the number
of zeros of f in [a, b]?
Proof. a. Assume f is a polynomial and fixed
Claim I: If α is a zero of f with multiplicity m ≥ 1, then α is a zeros of f 0 with multiplicity m − 1 ≥ 0.
If m = 1, then there exist some polynomial g such that f (x) = (x − α)g(x), where g(α) 6= 0.
Differentiating it, we get
f 0 (x) = g(x) + (x − a)g 0 (x).
So f 0 (α) = g(α) 6= 0, that is, α is not a zero of f 0 , so the Claim I is true.
If m ≥ 2, then there exist some polynomial g such that f (x) = (x − α)m g(x), where g(α) 6= 0.
Differentiating it, we get
f 0 (x) = m(x − α)m−1 g(x) + (x − α)m g 0 (x) = (x − α)m−1 [mg(x) + (x − α)g 0 (x)].
Let h(x) = mg(x) + (x − α)g 0 (x), then h(α) = mg(α) 6= 0, so α is a zero of f 0 with multiplicity
m − 1.
Claim II: If f has r zeros in an interval [a, b], then f 0 has at least r − 1 zeros in [a, b].
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2
MINGFENG ZHAO
Assume f has r zeros in [a, b], that is, there exists some {αi }ni=1 ⊂ [a, b] such that a ≤ α1 < α2 <
· · · < αn ≤ b, and for each αi , αi is a zero of f with multiplicity mi ≥ 1, and
n
X
mi = r. So f 0 has
i=1
zeros α1 , · · · , αn with multiplicity m1 − 1, · · · , mn − 1, respectively.
For each αi , 1 ≤ i ≤ n, since αi is a zero of f with multiplicity mi ≥ 1, then by the Claim I, we
know that αi is a zero of f 0 with multiplicity mi − 1 ≥ 0.
On the other hand, for each 1 ≤ i ≤ n − 1, since αi and αi+1 are zeros of f , that is, f (αi ) =
f (αi+1 ) = 0, then by the Rolle’s Theorem, we know that there exists some βi ∈ (αi , αi+1 ) such that
f 0 (βi ) = 0. So f 0 has zeros β1 , · · · , βn with multiplicity at least 1.
In a summary, we know that f 0 at least has
n
X
i=1
(mi − 1) + (n − 1) =
n
X
mi − n + (n − 1) = r − 1
i=1
many zeros in [a, b].
Claim III: If f has r zeros in an interval [a, b], for any 1 ≤ k < r, then f k has at least r − k zeros in
[a, b].
If k = 1, by the Claim I, we know that the Claim III is true. Now assume when k = m < r,
and the Claim III is true, that is, f m has at least r − m zeros in [a, b]. If m + 1 = r, we are done.
If m + 1 < r, since f m has r − m zeros in [a, b], by the Claim I, we know that f m+1 has at least
r − m − 1 = r − (m + 1) zeros in [a, b], that is, the Claim III is true for k = m + 1. By induction, we
can conclude that for any 1 ≤ k < r, then f k has at least r − k zeros in [a, b].
b. If the k-th derivative f (k) has exactly r zeros in [a, b], we can not say f at least has how many
zeros in [a, b], but we can say that f at most has k + r zeros in [a, b]. Since if f has more than k + r
zeros in [a, b], say f at least has k + m + 1 zeros in [a, b]. Then by the result of part a, we know that
f k at least has k + r + 1 − k = r + 1 zeros in [a, b], which contradicts with the assumption that the
k-th derivative f (k) has exactly r zeros in [a, b].
For example, k = 1, and r = 2, consider f (x) = x3 + 10 in [−1, 1], then f 0 (x) = 3x2 in [−1, 1], then
0 is a zero in [−1, 1] of f 0 with multiplicity 2, but f has no solutions in [−1, 1], since f (x) = x3 + 10 ≥
(−1)3 + 10 = 7 > 0 for all x ∈ [−1, 1].
SOLUTION OF HW11
3
2. [The Problem 9, in Page 187] A function f , continuous on [a, b], has a second derivative f 00
everywhere on the open interval (a, b). The line segment joining (a, f (a)) and (b, f (b)) intersects the
graph of f at a third point (c, f (c)), where a < c < b. Prove that f 00 (t) = 0 for at least one point t in
(a, b).
Proof. Since the line segment joining (a, f (a)) and (b, f (b)) intersects the graph of f at a third point
(c, f (c)), where a < c < b, then we have
f (b) − f (c)
f (c) − f (a)
=
.
b−c
c−a
By the Mean Value Theorem, there exists some d ∈ (a, c), and e ∈ (c, b) such that
f (c) − f (a) = f 0 (d)(c − a),
and f (b) − f (c) = f 0 (e)(b − c).
That is,
f (c) − f (a)
= f 0 (d),
c−a
and
f (b) − f (c)
= f 0 (e).
b−c
So we get f 0 (d) = f 0 (e). Notice that a < d < c < e < b, by the Rolle’s Theorem, we know that
there exists some t ∈ (d, e) ⊂ (a, b) such that f 00 (t) = 0.
3. [The Problem 19, in Page 209] Give a function g, continuous everywhere, such that g(1) = 5 and
Z 1
Z
1 x
g(t) dt = 2. Let f (x) =
(x − t)2 g(t) dt. Prove that
2 0
0
0
Z
x
Z
g(t) dt −
f (x) = x
0
x
tg(t) dt.
0
Then compute f 00 (1) and f 000 (1).
Proof. Since f (x) =
1
2
Z
x
(x − t)2 g(t) dt for all x ∈ R, and g is continuous on R, then we know that
0
f is differentiable on R, and
f 0 (x) =
1
1
(x − x)2 g(x) +
2
2
Z
x
2(x − t)g(t) dt,
0
for all x ∈ R.
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MINGFENG ZHAO
That is,
f 0 (x) =
x
Z
x
Z
g(t) dt −
0
tg(t) dt,
0
Z
for all x ∈ R.
0
x
x
Z
tg(t) dt are differentiable on R, which implies
g(t) dt and
Since g is continuous on R, then x
x
Z
(x − t)g(t) dt = x
0
0
that f 0 is differentiable on R, and
x
Z
00
Z
0
In particular, f 00 (1) =
x
g(t) dt + xg(x) − xg(x) =
f (x) =
for all x ∈ R.
g(t) dt,
0
R1
0
g(t) dt = 2. Since g is continuous on R, then
Rx
0
g(t) dt is differentiable
on R, which implies that f 00 (x) is differentiable on R, and
f 000 (x) = g(x),
for all x ∈ R.
In particular, f 000 (1) = g(1) = 5.
4. [The Problem 35, in Page 238] A function f , continuous on the positive real axis, has the property
that
Z
xy
Z
f (t) dt = y
1
x
y
Z
f (t) dt + x
f (t) dt,
1
for all x > 0 and all y > 0.
1
If f (1) = 3, compute f (x) for each x > 0.
Proof. For any y > 0 and fix, we define
Z
gy (x) =
xy
Z
f (t) dt,
and hy (x) = y
1
x
Z
f (t) dt + x
1
y
f (t) dt,
for all x > 0.
1
Then we know that gy (x) ≡ hy (x) for all x > 0. Since f is continuous on (0, ∞), then both gy and
hy are differentiable on (0, ∞), and
0
(gy ) (x) = yf (xy),
and
y
Z
0
f (t) dt,
(hy ) (x) = yf (x) +
for all x > 0.
1
Since gy (x) ≡ hy (x) for all x > 0, then (gy )0 (x) ≡ (hy )0 (x) for all x > 0, that is,
Z
yf (xy) = yf (x) +
y
f (t) dt,
1
for all x > 0.
SOLUTION OF HW11
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In particular, taking x = 1, then
y
Z
yf (y) = yf (1) +
Z
f (t) dt = 3y +
1
y
f (t) dt.
1
By the arbitrary of y > 0, so
y
Z
yf (y) = 3y +
f (t) dt,
for all y > 0.
1
That is,
Ry
f (y) = 3 +
Since
Ry
1
f (t) dt
,
y
1
for all y > 0.
f (t) dt and y are differentiable functions on (0, ∞), then we know that f is also differen-
tiable on (0, ∞), which implies that yf (y) is differentiable on (0, ∞). Since
y
Z
yf (y) = 3y +
f (t) dt,
for all y > 0.
f (y) + yf 0 (y) = 3 + f (y),
for all y > 0.
1
Differentiating both sides, so
That is,
f 0 (y) =
3
,
y
for all y > 0.
For any x > 0, integrating both sides along [1, x], then
x
Z
f 0 (y) dy =
1
Z
1
x
3
dy,
y
for all x > 0.
That is,
f (x) − f (1) = 3 ln x,
So we have f (x) = 3 + 3 ln x for all x > 0.
for all x > 0.
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MINGFENG ZHAO
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu