Prove that the sequence an 3 2 n−1 has a finite limit or tends to infinity.. Then an+1> 32an is equivalent to bn+1> bn, thus the sequence bn is strictly increasing.. Each increasing sequ
Trang 110 International Mathematical Competition for University Students
Cluj-Napoca, July 2003
Day 1
1 (a) Let a1, a2, be a sequence of real numbers such that a1 = 1 and an+1> 3
2an for all n Prove that the sequence
an
3 2
n−1
has a finite limit or tends to infinity (10 points)
(b) Prove that for all α > 1 there exists a sequence a1, a2, with the same properties such that
lim an
3 2
n−1 = α
(10 points)
Solution (a) Let bn= an
3 2
n−1 Then an+1> 32an is equivalent to bn+1> bn, thus the sequence
(bn) is strictly increasing Each increasing sequence has a finite limit or tends to infinity
(b) For all α > 1 there exists a sequence 1 = b1< b2 < which converges to α Choosing
an= 32n−1bn, we obtain the required sequence (an)
2 Let a1, a2 , a51be non-zero elements of a field We simultaneously replace each element with the sum of the 50 remaining ones In this way we get a sequence b1 , b51 If this new sequence is
a permutation of the original one, what can be the characteristic of the field? (The characteristic
of a field is p, if p is the smallest positive integer such that x + x + + x
p
= 0 for any element x
of the field If there exists no such p, the characteristic is 0.) (20 points)
Solution Let S = a1+ a2+ · · · + a51 Then b1+ b2+ · · · + b51 = 50S Since b1, b2, · · · , b51 is a permutation of a1, a2, · · · , a51, we get 50S = S, so 49S = 0 Assume that the characteristic of the field is not equal to 7 Then 49S = 0 implies that S = 0 Therefore bi = −ai for i = 1, 2, · · · , 51
On the other hand, bi= aϕ(i), where ϕ ∈ S51 Therefore, if the characteristic is not 2, the sequence
a1, a2, · · · , a51can be partitioned into pairs {ai, aϕ(i)} of additive inverses But this is impossible, since 51 is an odd number It follows that the characteristic of the field is 7 or 2
The characteristic can be either 2 or 7 For the case of 7, x1 = = x51 = 1 is a possible choice For the case of 2, any elements can be chosen such that S = 0, since then bi= −ai= ai
3 Let A be an n × n real matrix such that 3A3= A2+ A + I (I is the identity matrix) Show that the sequence Ak converges to an idempotent matrix (A matrix B is called idempotent if
B2= B.) (20 points)
Solution The minimal polynomial of A is a divisor of 3x3− x2− x − 1 This polynomial has three different roots This implies that A is diagonalizable: A = C−1DC where D is a diagonal matrix The eigenvalues of the matrices A and D are all roots of polynomial 3x3− x2− x − 1 One of the three roots is 1, the remaining two roots have smaller absolute value than 1 Hence, the diagonal elements of Dk, which are the kth powers of the eigenvalues, tend to either 0 or 1 and the limit
M = lim Dk is idempotent Then lim Ak= C−1M C is idempotent as well
4 Determine the set of all pairs (a, b) of positive integers for which the set of positive integers can be decomposed into two sets A and B such that a · A = b · B (20 points)
Solution Clearly a and b must be different since A and B are disjoint
Trang 210th International Mathematical Competition for University Students
Cluj-Napoca, July 2003
Day 2
1 Let A and B be n × n real matrices such that AB + A + B = 0 Prove that AB = BA Solution Since (A + I)(B + I) = AB + A + B + I = I (I is the identity matrix), matrices
A + I and B + I are inverses of each other Then (A + I)(B + I) = (B + I)(A + I) and
AB + BA.
2 Evaluate the limit
lim x→0+
Z 2x
x
sinmt
tn dt (m, n ∈ N).
Solution We use the fact that sin t
t is decreasing in the interval (0, π) and limt→0+0
sin t
t = 1. For all x ∈ (0,π2) and t ∈ [x, 2x] we have sin 2x
2 x <
sin t
t < 1, thus
sin 2x 2x
mZ 2x
x
tm
tn <
Z 2x
x
sinmt
tn dt <
Z 2x
x
tm
tndt,
Z 2x
x
tm
tndt = xm−n+1
Z 2
1
um−ndu.
The factor sin 2x
2x
m tends to 1 If m − n + 1 < 0, the limit of xm−n+1 is infinity; if
m − n + 1 > 0 then 0 If m − n + 1 = 0 then xm−n+1R2
1 um−ndu = ln 2 Hence,
lim x→0+0
2x Z
x
sinmt
tn dt =
ln 2, n − m = 1 +∞, n − m > 1.
.
3 Let A be a closed subset of Rn and let B be the set of all those points b ∈ Rn for which there exists exactly one point a0 ∈ A such that
|a0− b| = inf
a∈A|a − b|.
Prove that B is dense in Rn; that is, the closure of B is Rn.
Solution Let b0 ∈ A (otherwise b / 0 ∈ A ⊂ B), % = inf
a∈A|a − b0| The intersection of the ball
of radius % + 1 with centre b0 with set A is compact and there exists a0 ∈ A: |a0− b0| = %.
Trang 3Denote by Br(a) = {x ∈ Rn : |x − a| ≤ r} and ∂Br(a) = {x ∈ Rn : |x − a| = r} the ball and the sphere of center a and radius r, respectively.
If a0 is not the unique nearest point then for any point a on the open line segment (a0, b0)
we have B|a−a 0 |(a) ⊂ B%(b0) and ∂B|a−a 0 |(a) T
∂B%(b0) = {a0}, therefore (a0, b0) ⊂ B and
b0 is an accumulation point of set B.
4 Find all positive integers n for which there exists a family F of three-element subsets
of S = {1, 2, , n} satisfying the following two conditions:
(i) for any two different elements a, b ∈ S, there exists exactly one A ∈ F containing both a, b;
(ii) if a, b, c, x, y, z are elements of S such that if {a, b, x}, {a, c, y}, {b, c, z} ∈ F , then {x, y, z} ∈ F
Solution The condition (i) of the problem allows us to define a (well-defined) operation
∗ on the set S given by
a ∗ b = c if and only if {a, b, c} ∈ F, where a 6= b.
We note that this operation is still not defined completely (we need to define a ∗ a), but nevertheless let us investigate its features At first, due to (i), for a 6= b the operation obviously satisfies the following three conditions:
(a) a 6= a ∗ b 6= b;
(b) a ∗ b = b ∗ a;
(c) a ∗ (a ∗ b) = b.
What does the condition (ii) give? It claims that
(e’) x ∗ (a ∗ c) = x ∗ y = z = b ∗ c = (x ∗ a) ∗ c
for any three different x, a, c, i.e that the operation is associative if the arguments are different Now we can complete the definition of ∗ In order to save associativity for non-different arguments, i.e to make b = a ∗ (a ∗ b) = (a ∗ a) ∗ b hold, we will add to S an extra element, call it 0, and define
(d) a ∗ a = 0 and a ∗ 0 = 0 ∗ a = a.
Now it is easy to check that, for any a, b, c ∈ S ∪ {0}, (a),(b),(c) and (d), still hold, and (e) a ∗ b ∗ c := (a ∗ b) ∗ c = a ∗ (b ∗ c).
We have thus obtained that (S ∪ {0}, ∗) has the structure of a finite Abelian group, whose elements are all of order two Since the order of every such group is a power of 2,
we conclude that |S ∪ {0}| = n + 1 = 2m and n = 2m− 1 for some integer m ≥ 1.
Given n = 2m−1, according to what we have proven till now, we will construct a family
of three-element subsets of S satisfying (i) and (ii) Let us define the operation ∗ in the following manner:
if a = a0 + 2a1 + + 2m−1am−1 and b = b0+ 2b1 + + 2m−1bm−1, where ai, bi are either 0 or 1, we put a ∗ b = |a0− b0| + 2|a1− b1| + + 2m−1|am−1− bm−1|.
Trang 4It is simple to check that this ∗ satisfies (a),(b),(c) and (e’) Therefore, if we include in
F all possible triples a, b, a ∗ b, the condition (i) follows from (a),(b) and (c), whereas the condition (ii) follows from (e’)
The answer is: n = 2m− 1.
5 (a) Show that for each function f : Q × Q → R there exists a function g : Q → R such that f (x, y) ≤ g(x) + g(y) for all x, y ∈ Q.
(b) Find a function f : R × R → R for which there is no function g : R → R such that
f (x, y) ≤ g(x) + g(y) for all x, y ∈ R.
Solution a) Let ϕ : Q → N be a bijection Define g(x) = max{|f (s, t)| : s, t ∈ Q, ϕ(s) ≤ ϕ(x), ϕ(t) ≤ ϕ(x)} We have f (x, y) ≤ max{g(x), g(y)} ≤ g(x) + g(y).
b) We shall show that the function defined by f (x, y) = |x−y|1 for x 6= y and f (x, x) = 0 satisfies the problem If, by contradiction there exists a function g as above, it results, that g(y) ≥ |x−y|1 − f (x) for x, y ∈ R, x 6= y; one obtains that for each x ∈ R, lim
y→xg(y) = ∞.
We show, that there exists no function g having an infinite limit at each point of a bounded and closed interval [a, b].
For each k ∈ N+ denote Ak = {x ∈ [a, b] : |g(x)| ≤ k}.
We have obviously [a, b] = ∪∞k=1Ak The set [a, b] is uncountable, so at least one of the sets Ak is infinite (in fact uncountable) This set Ak being infinite, there exists a sequence
in Ak having distinct terms This sequence will contain a convergent subsequence (xn)n∈N convergent to a point x ∈ [a, b] But lim
y→xg(y) = ∞ implies that g(xn) → ∞, a contradiction because |g(xn)| ≤ k, ∀n ∈ N.
Second solution for part (b) Let S be the set of all sequences of real numbers The cardinality of S is |S| = |R|ℵ0 = 2ℵ2 = 2ℵ0 = |R| Thus, there exists a bijection h : R → S Now define the function f in the following way For any real x and positive integer n, let f (x, n) be the nth element of sequence h(x) If y is not a positive integer then let
f (x, y) = 0 We prove that this function has the required property.
Let g be an arbitrary R → R function We show that there exist real numbers x, y such that f (x, y) > g(x) + g(y) Consider the sequence (n + g(n))∞n=1 This sequence is an element of S, thus (n + g(n))∞n=1= h(x) for a certain real x Then for an arbitrary positive integer n, f (x, n) is the nth element, f (x, n) = n + g(n) Choosing n such that n > g(x),
we obtain f (x, n) = n + g(n) > g(x) + g(n).
6 Let (an)n∈N be the sequence defined by
a0 = 1, an+1 = 1
n + 1
n X
k=0
ak
n − k + 2 . Find the limit
lim n→∞
n X
k=0
ak
2k,
Trang 5if it exists.
Solution Consider the generating function f (x) = P∞
n=0anxn By induction 0 < an ≤ 1, thus this series is absolutely convergent for |x| < 1, f (0) = 1 and the function is positive
in the interval [0, 1) The goal is to compute f (12).
By the recurrence formula,
f0(x) =
∞ X
n=0 (n + 1)an+1xn=
∞ X
n=0
n X
k=0
ak
n − k + 2 x
n
=
=
∞ X
k=0
akxk
∞ X
n=k
xn−k
n − k + 2 = f (x)
∞ X
m=0
xm
m + 2 . Then
ln f (x) = ln f (x) − ln f (0) =
Z x
0
f0
f =
∞ X
m=0
xm+1 (m + 1)(m + 2) =
=
∞
X
m=0
xm+1
(m + 1) − x
m+1 (m + 2)
= 1 +
1 − 1 x
X
m=0
xm+1 (m + 1) = 1 +
1 − 1 x
ln 1
1 − x ,
ln f 1 2
= 1 − ln 2,
and thus f ( 1
2 ) =
e
2 .
Trang 6Let {a, b} be a solution and consider the sets A, B such that a · A = b · B Denoting d = (a, b) the greatest common divisor of a and b, we have a = d·a1, b = d·b1, (a1, b1) = 1 and a1·A = b1·B Thus {a1, b1} is a solution and it is enough to determine the solutions {a, b} with (a, b) = 1
If 1 ∈ A then a ∈ a · A = b · B, thus b must be a divisor of a Similarly, if 1 ∈ B, then a is a divisor of b Therefore, in all solutions, one of numbers a, b is a divisor of the other one
Now we prove that if n ≥ 2, then (1, n) is a solution For each positive integer k, let f (k)
be the largest non-negative integer for which nf (k)|k Then let A = {k : f (k) is odd} and
B = {k : f (k) is even} This is a decomposition of all positive integers such that A = n · B
5 Let g : [0, 1] → R be a continuous function and let fn : [0, 1] → R be a sequence of functions defined by f0(x) = g(x) and
fn+1(x) = 1
x
Z x 0
fn(t)dt (x ∈ (0, 1], n = 0, 1, 2, )
Determine lim
n→∞fn(x) for every x ∈ (0, 1] (20 points)
B We shall prove in two different ways that limn→∞fn(x) = g(0) for every x ∈ (0, 1] (The second one is more lengthy but it tells us how to calculate fn directly from g.)
Proof I First we prove our claim for non-decreasing g In this case, by induction, one can easily see that
1 each fn is non-decrasing as well, and
2 g(x) = f0(x) ≥ f1(x) ≥ f2(x) ≥ ≥ g(0) (x ∈ (0, 1])
Then (2) implies that there exists
h(x) = lim
n→∞fn(x) (x ∈ (0, 1])
Clearly h is non-decreasing and g(0) ≤ h(x) ≤ fn(x) for any x ∈ (0, 1], n = 0, 1, 2, Therefore
to show that h(x) = g(0) for any x ∈ (0, 1], it is enough to prove that h(1) cannot be greater than g(0)
Suppose that h(1) > g(0) Then there exists a 0 < δ < 1 such that h(1) > g(δ) Using the definition, (2) and (1) we get
fn+1(1) =
Z 1 0
fn(t)dt ≤
Z δ 0
g(t)dt +
Z 1 δ
fn(t)dt ≤ δg(δ) + (1 − δ)fn(1)
Hence
fn(1) − fn+1(1) ≥ δ(fn(1) − g(δ)) ≥ δ(h(1) − g(δ)) > 0,
so fn(1) → −∞, which is a contradiction
Similarly, we can prove our claim for non-increasing continuous functions as well
Now suppose that g is an arbitrary continuous function on [0, 1] Let
M (x) = sup
t∈[0,x]
g(t), m(x) = inf
t∈[0,x]g(t) (x ∈ [0, 1]) Then on [0, 1] m is non-increasing, M is non-decreasing, both are continuous, m(x) ≤ g(x) ≤ M (x) and M (0) = m(0) = g(0) Define the sequences of functions Mn(x) and mn(x) in the same way
as fn is defined but starting with M0= M and m0= m
Then one can easily see by induction that mn(x) ≤ fn(x) ≤ Mn(x) By the first part of the proof, limnmn(x) = m(0) = g(0) = M (0) = limnMn(x) for any x ∈ (0, 1] Therefore we must have limnfn(x) = g(0)
Trang 7Proof II To make the notation clearer we shall denote the variable of fj by xj By definition (and Fubini theorem) we get that
fn+1(xn+1) = 1
xn+1
Z x n+1
0
1
xn
Z x n
0
1
xn−1
Z x n−1
0
Z x 2
0
1
x1
Z x 1
0
g(x0)dx0dx1 dxn
xn+1
Z Z
0≤x0≤x1≤ ≤xn≤xn+1
g(x0)dx0dx1 dxn
x1 xn
xn+1
Z xn+1 0
g(x0)
Z Z
x 0 ≤x 1 ≤ ≤x n ≤x n+1
dx1 dxn
x1 xn
!
dx0 Therefore with the notation
hn(a, b) =
Z Z
a≤x1≤ ≤xn≤b
dx1 dxn
x1 xn and x = xn+1, t = x0 we have
fn+1(x) = 1
x
Z x 0
g(t)hn(t, x)dt
Using that hn(a, b) is the same for any permutation of x1, , xn and the fact that the integral
is 0 on any hyperplanes (xi= xj) we get that
n! hn(a, b) =
Z Z
a≤x1, ,xn≤b
dx1 dxn
x1 xn
=
Z b a
Z b a
dx1 dxn
x1 xn
=
Z b a
dx x
!n
= (log(b/a))n
Therefore
fn+1(x) = 1
x
Z x 0
g(t)(log(x/t))
n
n! dt.
Note that if g is constant then the definition gives fn = g This implies on one hand that we must have
1 x
Z x 0
(log(x/t))n
n! dt = 1 and on the other hand that, by replacing g by g − g(0), we can suppose that g(0) = 0
Let x ∈ (0, 1] and ε > 0 be fixed By continuity there exists a 0 < δ < x and an M such that
|g(t)| < ε on [0, δ] and |g(t)| ≤ M on [0, 1] Since
lim
n→∞
(log(x/δ))n
there exists an n0 sucht that log(x/δ))n/n! < ε whenever n ≥ n0 Then, for any n ≥ n0, we have
|fn+1(x)| ≤ 1
x
Z x 0
|g(t)|(log(x/t))
n
x
Z δ 0
ε(log(x/t))
n
1 x
Z x δ
|g(t)|(log(x/δ))
n
x
Z x 0
ε(log(x/t))
n
1 x
Z x δ
M εdt
≤ ε + M ε
Therefore limnf (x) = 0 = g(0)
Trang 86 Let f (z) = anz + an−1z + + a1z + a0be a polynomial with real coefficients Prove that
if all roots of f lie in the left half-plane {z ∈ C : Re z < 0} then
akak+3< ak+1ak+2
holds for every k = 0, 1, , n − 3 (20 points)
Solution The polynomial f is a product of linear and quadratic factors, f (z) = Q
i(kiz + li) · Q
j(pjz2+ qjz + rj), with ki, li, pj, qj, rj∈ R Since all roots are in the left half-plane, for each i, ki
and li are of the same sign, and for each j, pj, qj, rj are of the same sign, too Hence, multiplying
f by −1 if necessary, the roots of f don’t change and f becomes the polynomial with all positive coefficients
For the simplicity, we extend the sequence of coefficients by an+1 = an+2 = = 0 and
a−1= a−2= = 0 and prove the same statement for −1 ≤ k ≤ n − 2 by induction
For n ≤ 2 the statement is obvious: ak+1 and ak+2 are positive and at least one of ak−1 and
ak+3is 0; hence, ak+1ak+2> akak+3= 0
Now assume that n ≥ 3 and the statement is true for all smaller values of n Take a divisor of
f (z) which has the form z2+ pz + q where p and q are positive real numbers (Such a divisor can
be obtained from a conjugate pair of roots or two real roots.) Then we can write
f (z) = (z2+ pz + q)(bn−2zn−2+ + b1z + b0) = (z2+ pz + q)g(x) (1) The roots polynomial g(z) are in the left half-plane, so we have bk+1bk+2< bkbk+3 for all −1 ≤
k ≤ n − 4 Defining bn−1 = bn = = 0 and b−1 = b−2 = = 0 as well, we also have
bk+1bk+2≤ bkbk+3for all integer k
Now we prove ak+1ak+2> akak+3 If k = −1 or k = n − 2 then this is obvious since ak+1ak+2
is positive and akak+3= 0 Thus, assume 0 ≤ k ≤ n − 3 By an easy computation,
ak+1ak+2− akak+3=
= (qbk+1+ pbk+ bk−1)(qbk+2+ pbk+1+ bk) − (qbk+ pbk−1+ bk−2)(qbk+3+ pbk+2+ bk+1) =
= (bk−1bk− bk−2bk+1) + p(b2k− bk−2bk+2) + q(bk−1bk+2− bk−2bk+3)+
+p2(bkbk+1− bk−1bk+2) + q2(bk+1bk+2− bkbk+3) + pq(b2k+1− bk−1bk+3)
We prove that all the six terms are non-negative and at least one is positive Term p2(bkbk+1−
bk−1bk+2) is positive since 0 ≤ k ≤ n − 3 Also terms bk−1bk− bk−2bk+1and q2(bk+1bk+2− bkbk+3) are non-negative by the induction hypothesis
To check the sign of p(b2
k− bk−2bk+2) consider
bk−1(b2k− bk−2bk+2) = bk−2(bkbk+1− bk−1bk+2) + bk(bk−1bk− bk−2bk+1) ≥ 0
If bk−1> 0 we can divide by it to obtain b2
k−bk−2bk+2≥ 0 Otherwise, if bk−1= 0, either bk−2= 0
or bk+2= 0 and thus b2− bk−2bk+2= b2≥ 0 Therefore, p(b2− bk−2bk+2) ≥ 0 for all k Similarly, pq(b2
k+1− bk−1bk+3) ≥ 0
The sign of q(bk−1bk+2− bk−2bk+3) can be checked in a similar way Consider
bk+1(bk−1bk+2− bk−2bk+3) = bk−1(bk+1bk+2− bkbk+3) + bk+3(bk−1bk− bk−2bk+1) ≥ 0
If bk+1> 0, we can divide by it Otherwise either bk−2 = 0 or bk+3 = 0 In all cases, we obtain
bk−1bk+2− bk−2bk+3≥ 0
Now the signs of all terms are checked and the proof is complete