GAZETA MATEMATICĂ SERIA A. ANUL XXXIV (CXIII) Nr. 3 4/ 2016 ARTICOLE. The Euler-Maclaurin summation formula for functions of class C 3

GAZETA MATEMATIC ˘

A

SERIA A

ANUL XXXIV (CXIII) Nr. 3 – 4/ 2016

ARTICOLE

The Euler-Maclaurin summation formula for functions of class C3

Dumitru Popa1)

Abstract. We give a self-contained proof of the Euler-Maclaurin summa-tion formula for funcsumma-tions of class C3. We derive a general result from which, as applications, we prove: the famous Stirling formula, the as-ymptotic evaluations for the Stieltjes constants, the asas-ymptotic evaluation for the Glaisher-Kinkelin constant and an exercise in N. Bourbaki’s book regarding the asymptotic evaluation for Vandermonde determinant associ-ated to (1, 2, . . . , n).

Keywords: Euler-Maclaurin summation formula, asymptotic expansions, approximation to limiting values, Stirling formula, Stieltjes constants, Glaisher-Kinkelin constant.

MSC: Primary 65B15; Secondary 41A60, 40A25.

1. Introduction

The Euler-Maclaurin summation formula is one of the most powerful results to obtain various asymptotic evaluations for real sequences. From many books where this formula is proved we mention our favorites: N. Bour-baki’s book [2, pages 282–283] and K. Knopp’s book [3, Chapter XIV]. The aim of this paper is to give a self-contained proof of the Euler-Maclaurin summation formula only for functions of class C3, Theorem 1. We also prove a general result, Corollary 4, which, as the author knows, in concrete ap-plications gives, with minimum of calculations, strong enough asymptotic evaluations for a large class of real sequences. As applications, we prove: the famous Stirling formula, Theorem 5, the asymptotic evaluations for the Stieltjes constants, Theorem 6, the asymptotic evaluation for the Glaisher-Kinkelin constant, Theorem 7 and a proof for an exercise in N. Bourbaki’s

1)

Department of Mathematics, Ovidius University of Constant¸a, Bd. Mamaia 124, 900527 Constant¸a, Romania, dpopa@univ-ovidius.ro

book regarding the asymptotic evaluation for the Vandermonde determinant associated to (1, 2, . . . , n), Proposition 9. It is our hope that these results will be useful for students as a part of their mathematical culture and for other readers interested in this topic. The author used these notes for students at the master program Didactical mathematics at the Faculty of Mathe-matics and InforMathe-matics, Ovidius University of Constant¸a. Let us mention that the notation and notion used in this paper are standard, in particular N = {1, 2, . . . } is the set of all natural numbers and [x] is the integer part of

x∈ R.

2. The main results

Let B1, B2, B3 :R → R be the polynomials B1(x) = x−1₂, B2(x) =

= x2− x +1₆, B3(x) = x3−3x 2

2 +x2. These polynomials are called Bernoulli

polynomials. The Euler-Maclaurin summation formula for functions of class

C3 is given by the following theorem.

Theorem 1. Let k ∈ N ∪ {0} and let f : [k, ∞) → R be a function of class

C3_{. For every natural number n≥ k the following equality holds}

f (k) + f (k + 1) +· · · + f (n) =  _n k f (x) dx +f (n) + f (k) 2 + f(n)− f(k) 12 +1 6  _n k B3(x− [x]) f(x) dx.

Proof. For n = k the equality is trivial. Let n≥ k + 1 and ν ∈ N ∪ {0}. Since B₃ (x) = 3B2(x), B3(0) = B3(1) = 0, B2 (x) = 2B1(x), B2(0) = B2(1) =

= 1₆, B1(0) =−1₂, B1(1) = 1₂, integrating by parts, we have

 _ν+1 ν B3(x− ν) f(x) dx = B3(x− ν) f(x)  ν+1 ν − −  _ν+1 ν B₃ (x− ν) f(x) dx =−3  _ν+1 ν B2(x− ν) f(x) dx = =−3B2(x− ν) f(x)|ν+1_ν +6  _ν+1 ν B1(x− ν) f(x) dx = =−f _{(ν + 1)− f}_(ν) 2 + 6B1(x− ν) f (x) | ν+1 ν −6  _ν+1 ν f (x) dx = =−f _{(ν + 1)− f}_(ν) 2 + 3 [f (ν + 1) + f (ν)]− 6  _ν+1 ν f (x) dx, that is,  _ν+1 ν f (x) dx = f (ν + 1) + f (ν) 2 − f(ν + 1)− f(ν) 12 −

−1

6  _ν+1

ν

B3(x− ν) f(x) dx.

Since [x] = ν for x∈ [ν, ν+1) we have B3(x− ν) f(x) = B3(x− [x]) f(x)

for x∈ [ν, ν + 1) and since f is continuous  _ν+1 ν B3(x− ν) f(x) dx =  _ν+1 ν B3(x− [x]) f(x) dx

(see [1, Observat¸ie, pag. 61]). Thus  _ν+1 ν f (x) dx = f (ν + 1) + f (ν) 2 − f(ν + 1)− f(ν) 12 − −1 6  _ν+1 ν B3(x− [x]) f(x) dx. (1) Taking in (1), ν = k, ν = k + 1, . . . , ν = n− 1 we have  _k+1 k f (x) dx = f (k + 1) + f (k) 2 − f(k + 1)− f(k) 12 − −1 6  _k+1 k B3(x− [x]) f(x) dx  _k+2 k+1 f (x) dx = f (k + 2) + f (k + 1) 2 − f(k + 2)− f(k + 1) 12 − −1 6  _k+2 k+1 B3(x− [x]) f(x) dx . . .  _n n−1 f (x) dx = f (n) + f (n− 1) 2 − f(n)− f(n− 1) 12 − −1 6  _n n−1 B3(x− [x]) f(x) dx

and summing up it follows  _n k f (x) dx = f (k) + f (k + 1) +· · · + f (n) −f (n) + f (k) 2 − −f(n)− f(k) 12 − 1 6  _n k B3(x− [x]) f(x) dx

that is the Euler-Maclaurin summation formula for functions of class C3_{. }

We need in the sequel the following evaluation for the Bernoulli poly-nomial B3.

Proposition 2. For the Bernoulli polynomial B3 : R → R, B3(x) = x3 −

−3x2 2 +x2 we have|B3(x)| ≤ √ 3 36,∀ x ∈ [0, 1] and |B3(x− [x])| ≤ √ 3 36,∀ x ∈ R.

Proof. Indeed, B₃ (x) = 3x2_{− x +} 1 6



and B₃ (x) = 0 if and only if x =

3−√3 6 ∈ [0, 1] or x = 3+√3 6 ∈ [0, 1]. Also x 0 3− √ 3 6 3+√3 6 1 B₃(x) + 0 − 0 + B3(x) 0 √ 3 36 − √ 3 36 0 We deduce that|B3(x)| ≤ √ 3 36,∀ x ∈ [0, 1]. If x ∈ R, x − [x] ∈ [0, 1) and thus |B3(x− [x])| ≤ √ 3 36. 

Proposition 3. Let k∈ N∪{0} and let ϕ : [k, ∞) → R be a function of class

C3. Suppose that the improper integral _k∞|ϕ(x)| dx is convergent. Then

(i) the following limit exists L(ϕ) = lim n→∞  ϕ(k) + ϕ(k + 1) +· · · + ϕ(n) −  _n k ϕ(t)dt−ϕ(n) 2 − ϕ(n) 12  ∈R and L (ϕ) = ϕ (k) 2 − ϕ(k) 12 + 1 6  _∞ k B3(x− [x]) ϕ(x) dx;

(ii) for every natural number n≥ k the following asymptotic evaluation holds ϕ(k) + ϕ(k + 1) +· · · + ϕ(n) =  _n k ϕ(t)dt +ϕ(n) 2 + ϕ(n) 12 + L(ϕ) + Rn(ϕ) |Rn(ϕ)| ≤ √ 3 216  _∞ n |ϕ_(x)|dx. Proof. From Theorem 1 we have

ϕ (k) + ϕ (k + 1) +· · · + ϕ (n) −  _n k ϕ (t) dt−ϕ (n) 2 − ϕ(n) 12 = = ϕ (k) 2 − ϕ(k) 12 + 1 6  _n k B3(x− [x]) ϕ(x) dx, ∀ n ≥ k. (2) From Proposition 2, |B3(x− [x]) ϕ(x)| ≤ √ 3 36 |ϕ(x)|, ∀ x ∈ R. Since _∞

k |ϕ(x)| dx is convergent, by the comparison test for improper integrals

of positive functions it follows that_k∞|B3(x− [x]) ϕ(x)| dx is convergent

(see [5, page 60]), that is, the improper integral_k∞B3(x− [x]) ϕ(x) dx is

absolutely convergent, hence convergent (see [5, Teorema 2, pg. 59]). From (2) passing to the limit as n→ ∞ we deduce the existence of the limit

L(ϕ) = lim n→∞  ϕ(k) + ϕ(k + 1) +· · · + ϕ(n)−  _n k ϕ(t)dt−ϕ(n) 2 − ϕ(n) 12  ∈R and L (ϕ) = ϕ (k) 2 − ϕ(k) 12 + 1 6  _∞ k B3(x− [x]) ϕ(x) dx. (3)

From (2) and (3) we deduce ϕ (k) + ϕ (k + 1) +· · · + ϕ (n) −  _n k ϕ (t) dt−ϕ (n) 2 − ϕ(n) 12 = L (ϕ)−1 6  _∞ n B3(x− [x]) ϕ(x) dx,∀ n ≥ k.

Let us denote R_n(ϕ) =−1₆_n∞B3(x− [x]) ϕ(x) dx. Then, by Proposition

2, we get |Rn(ϕ)|= 1 6   ∞ n B3(x− [x]) ϕ(x) dx   ≤ 1₆ ∞ n B3(x− [x]) ϕ(x)dx≤ ≤ √ 3 216  _∞ n ϕ(x)dx.  From Proposition 3 and hence, from the Euler-Maclaurin summation formula Theorem 1, we derive the next Corollary. As the author knows, this is the result which, in concrete applications, gives, with minimum of calculations, strong asymptotic evaluations for a large class of sequences.

Corollary 4. Let k ∈ N ∪ {0}, ϕ : [k, ∞) → R be a function of the class

C3. Suppose that lim

x→∞ϕ

_{(x) = 0 and there exists a ≥ k such that ϕ} _has constant sign on the interval [a,∞). The following limit exists

L(ϕ) = lim n→∞  ϕ(k) + ϕ(k + 1) +· · · + ϕ(n) −  _n k ϕ(t)dt−ϕ(n) 2 − ϕ(n) 12  ∈ R and for every natural number n≥ a the following asymptotic evaluation holds ϕ (k) + ϕ (k + 1) +· · · + ϕ (n)=  _n k ϕ(t)dt +ϕ (n) 2 + ϕ(n) 12 + L (ϕ) + Rn(ϕ) |Rn(ϕ)|≤ √ 3 216ϕ _(n)_.

Proof. Since ϕ has constant sign on the interval [a,∞) we have two situ-ations. The case ϕ(x) ≤ 0, ∀ x ∈ [a, ∞). In this case for every t ≥ a we

have _ ∞ t ϕ(x)dx =−  _∞ t ϕ(x) dx =−ϕ(x)|∞_t = ϕ(t) ( lim x→∞ϕ

_{(x) = 0) and thus ϕ}_{(t) ≥ 0, ∀ t ≥ a. From Proposition 3 we get}

the evaluation from the statement and

|Rn(ϕ)| ≤ √ 3 216  _∞ n ϕ(x)dx = √ 3 216ϕ _{(n) =} √ 3 216ϕ _(n)_.

The case ϕ(x)≥ 0, ∀ x ∈ [a, ∞). In this case for every t ≥ a we have  _∞ t ϕ(x)dx =  _∞ t ϕ(x) dx = ϕ(x)|∞_t =−ϕ(t) ( lim x→∞ϕ

_{(x) = 0) and thus ϕ}_{(t) ≤ 0, ∀ t ≥ a. From Proposition 3 we get}

the evaluation from the statement and

|Rn(ϕ)| ≤ √ 3 216  _∞ n ϕ(x)dx =− √ 3 216ϕ _{(n) =} √ 3 216ϕ _(n)_.  3. Applications

The first application is the famous Stirling formula.

Theorem 5. The following asymptotic evaluation holds

ln n! =  n +1 2  ln n− n + ln√2π + 1 12n + Rn; |Rn| ≤ √ 3 216n2, ∀ n ∈ N. In particular, n!∼√2π·n_enn · √

n, for large values of n.

Proof. Let ϕ : (0,∞) → R, ϕ (x) = ln x. Then ϕ(x) = 1_x, ϕ(x) = −1

x2, ϕ(x) = _x23. From Corollary 4 there exists B ∈ R such that

ln 1 + ln 2 +· · · + ln n =  n +1 2  ln n− n + B + 1 12n + Rn, (4) |Rn| ≤ √ 3 216n2,∀ n ∈ N.

We prove that B = ln√2π which ends the proof. Let us denote β_n= _12n1 +R_n and note that β_n→ 0. Then from (4) we have

ln n! =  n + 1 2  ln n− n + B + β_n,∀ n ∈ N. (5) In the sequel we follow K. Knopp, see [3, page 528]. Let n∈ N. From (5) we deduce ln [(2n + 1)!] =  2n +3 2  ln (2n + 1)− 2n − 1 + B + β_2n+1. (6) Also by (5) 2 ln (2· 4 · · · (2n)) = 2 (n ln 2 + ln n!) = 2n ln 2 + (2n + 1) ln n− 2n + 2B + 2βn. (7)

By subtracting (6) from (7) we obtain

ln1· 3 · 5 · · · (2n − 1)

=  2n + 3 2  ln (2n + 1)− 2n ln 2 − (2n + 1) ln n − 1 − B + β_2n+1− 2β_n, that is, ln1· 3 · 5 · · · (2n − 1) 2· 4 · · · · (2n) · √ 2n + 1 = = ln1· 3 · 5 · · · (2n − 1) 2· 4 · · · · (2n) + 1 2ln (2n + 1) = = (2n + 1) ln (2n + 1)− 2n ln 2 − (2n + 1) ln n − 1 − B + β_2n+1− 2β_n= =  2n ln2n + 1 2n − 1  + ln2n + 1 n − B + β2n+1− 2βn. (8)

Now, from the Wallis formula

lim n→∞ 1 2 · 3 4 · · · 2n− 1 2n · √ 2n + 1 = 2 π,

see for example [6], lim

n→∞βn = 0, limn→∞2n ln 2n+1 2n = 1 and (8) we deduce ln 2 π = ln 2− B, B = ln √ 2π. 

The second application is related to the Stieltjes constants.

Theorem 6. (i) Let α∈ R \ {−1}. Then there exists

γ_α:= lim n→∞  lnα1 1 + lnα2 2 +· · · + lnαn n − lnα+1n α + 1  ∈ R

called the constant α of Stieltjes, and the following asymptotic evaluation holds lnα1 1 + lnα2 2 +· · · + lnαn n = lnα+1n α + 1 + γα+ lnαn 2n − (ln n)α 12n2 + + α (ln n) α−1 12n2 + Rn(α) , |Rn(α)| ≤ Cα(ln n) α n3 , ∀ n ≥ xα.

(ii) There exists γ₋₁ := lim

n→∞  ₁ 2 ln 2+· · · + 1 n ln n− ln (ln n)  ∈ R, called the constant −1 of Stieltjes, and the following asymptotic evaluation holds

1 2 ln 2+· · · + 1 n ln n = ln (ln n) + γ−1+ 1 2n ln n− 1 12n2_{ln n} − − 1 12n2_ln2_n + Rn; |Rn| ≤ 7√3 216n3_{ln n}, ∀ n ∈ N.

Proof. (i) Let ϕ_α : (0,∞) → R, ϕ_α(x) = lnα_xx. By calculations we have ϕ_α(x)=−(ln x) α−1_{(ln x− α)} x2 , ϕ_α(x)= (ln x)α−2  2 (ln x)2− 3α ln x + α (α − 1)  x3 , ϕ_α(x)=−(ln x) α−3 _{6 ln}3_{x− 11α ln}2_{x + 6α(α− 1) ln x − α(α − 1)(α − 2)} x4 . Since lim x→∞ 6 ln3x− 11α ln2x + 6α (α− 1) ln x − α (α − 1) (α − 2)=∞ there exists x_α ≥ 2 such that ϕ_α (x) < 0,∀ x ≥ x_α.

Also ₁nϕ_α(x) dx = ln_α+1α+1n. From Corollary 4 there exists

γ_α:= lim n→∞  lnα1 1 + lnα2 2 +· · · + lnαn n − lnα+1n α + 1  ∈ R

and moreover, the following asymptotic evaluation holds lnα1 1 + lnα2 2 +· · · + lnαn n = lnα+1n α + 1 + γα+ lnαn 2n − (ln n)α 12n2 + + α (ln n) α−1 12n2 + Rn, |Rn| ≤ √3 216|ϕα(n)| ≤ Cα(ln n) α n3 ,∀ n ≥ xα.

(ii) Let ϕ : (1,∞) → R, ϕ (x) = _{x ln x}1 . We have  _x 2 ϕ (t) dt = ln (ln x)− ln (ln 2) , ϕ(x) = −ln x + 1 (x ln x)2, ϕ _{(x) =} 2 ln2x + 3 ln x + 2 (x ln x)3 , ϕ(x) = −6 (ln x) 3 + 11 (ln x)2+ 12 ln x + 6 (x ln x)4 .

From Corollary 4 there exists

γ₋₁ := lim n→∞  1 2 ln 2+· · · + 1 n ln n− ln (ln n)  ∈ R

and moreover, the following asymptotic evaluation holds 1 2 ln 2+· · · + 1 n ln n = ln (ln n) + γ−1+ 1 2n ln n − 1 12· ln n + 1 (n ln n)2 + Rn |Rn| ≤ √ 3 216 ϕ _(n) ≤ 7 √ 3 216n3_{ln n},∀ n ∈ N. 

Let us note that γ0is the Euler constant. The next application is related

to the so called Glaisher-Kinkelin constant.

Theorem 7. (i) There exists a real number L ∈ R such that the following

asymptotic evaluation holds n  k=1 k ln k = 1 2  n2+ n +1 6  ln n−n 2 4 + L + Rn, |Rn| ≤ √ 3 216n, ∀ n ∈ N.

(ii) There exists

A := lim n→∞

1122· · · nn

nn22 +n2+121 _e−n24

∈ (0, ∞)

called the Glaisher-Kinkelin constant, and the following asymptotic evalua-tion holds n  k=1 k ln k = 1 2  n2+ n +1 6  ln n− n 2 4 + ln A + Rn, |Rn| ≤ √ 3 216n,∀ n ∈ N.

Proof. Let ϕ : (0,∞) → R, ϕ (x) = x ln x. We have ϕ(x) = ln x + 1,

ϕ(x) = 1_x, ϕ(x) =−_x12. From Corollary 4 there exists D∈ R such that

n  k=1 k ln k =  _n 1 t ln tdt + n ln n 2 + ln n 12 + 1 12+ D + Rn, |Rn| ≤ √ 3 216ϕ _(n)₌ √ 3 216n,∀ n ∈ N.

Since₁nt ln tdt = n2₂ln n−n₄2 +1₄ we deduce that there exists L = D +1₃ ∈ R such that n  k=1 k ln k = 1 2  n2+ n +1 6  ln n−n 2 4 + L + Rn,|Rn| ≤ √ 3 216n,∀ n ∈ N. This means that lim

n→∞ 1122···nn nn22 +n2 +121 _e− n 2 4 = eL lim n→∞e Rn _{= e}L_{= A}∈ (0, ∞), that

is, L = ln A which ends the proof of theorem. 

We need the next evaluation in which appear the Glaisher-Kinkelin constant. A less precise evaluation can be found in [4, Problem 2.14(b), page 21].

Proposition 8. The following asymptotic evaluation holds

n  k=1 ln k! =  n2 2 + n + 5 12  ln n−3n 2 4 + n  ln√2π− 1  + ln√2π + 1 12 − ln A + Rn, |Rn| ≤ √ 3 + 6 72n , ∀ n ∈ N,

where A is the Glaisher-Kinkelin constant. Proof. Let us note the following equality

n  k=1 (a1+· · · + ak) = a1+ (a1+ a2) + (a1+ a2+ a3) +· · · + (a1+· · · + an) = = na1+ (n− 1) a2+ (n− 2) a3+· · · + an= = n  k=1 (n + 1− k) a_k = (n + 1) n  k=1 a_k− n  k=1 ka_k which gives us n  k=1 ln k! = (n + 1) ln n!− n  k=1 k ln k. (9)

Now by Stirling’s formula (Theorem 5)

ln n! =  n +1 2  ln n−n+ln√2π + 1 12n+ R 1 n, R1n ≤ √ 3 216n2, ∀ n ∈ N (10)

and by Glaisher-Kinkelin evaluation (Theorem 7)

n  k=1 k ln k = 1 2  n2+ n +1 6  ln n−n 2 4 + ln A + R 2 n, (11) R2_n ≤ √ 3 216n,∀n ∈ N. From (9), (10) and (11) we get that

n  k=1 ln k! =  n2 2 + n + 5 12  ln n−3n 2 4 +n  ln√2π− 1  +ln√2π+ 1 12−ln A+Rn where R_n= (n + 1) R1_n− R2_n+_12n1 . Now we note that

|Rn| ≤ (n + 1)R1n+R2n+ 1 12n ≤ √ 3 (n + 1) 216n2 + √ 3 216n + 1 12n ≤ ≤ 2n √ 3 216n2 + √ 3 + 18 216n = √ 3 + 6 72n .  In our last application we give a proof for an exercise in N. Bourbaki’s book regarding the asymptotic evaluation for the Vandermonde determinant associated to (1, 2, . . . , n) (see [2, Exercise 6, page 328]).

Proposition 9. For every natural number n≥ 2 we denote V_n=      1 1 1 · · · 1 1 2 3 · · · n 12 22 32 · · · n2 · · · · · · · · · · · · · 1n−1 2n−1 3n−1 · · · nn−1     

the Vandermonde determinant corresponding to (1, 2, . . . , n). Then, the fol-lowing asymptotic evaluation holds

ln V_n = n 2 2 ln n− 3n2 4 + n ln √ 2π− 1 12ln n + 1 12 − ln A + Rn, |Rn| ≤ √ 3 108n, ∀ n ∈ N,

where A is the Glaisher-Kinkelin constant. In particular, V_n ∼ 12_A√e · nn

2 2 ·

e−3n24 · (2π)n2 · n−121 _{, for large values of n.}

Proof. As is it well-known we have

V_n =  1≤i<j≤n (j− i) = n  j=2 (j− 1) n  j=3 (j− 2) · · · n  j=n (j− (n − 1)) = = (n− 1)! (n − 2)! · · · 2! · 1!

Then, see the equality shown in the proof of Proposition 8, we have

ln V_n= n−1  k=1 ln k! = n  k=1 ln k!− ln n! = n ln n! − n  k=1 k ln k.

Now, by Stirling evaluation (Theorem 5) and Glaisher-Kinkelin evaluation (Theorem 7) we have ln n! =  n + 1 2  ln n− n + ln√2π + 1 12n+ R 1 n, R1n ≤ √ 3 216n2,∀ n ∈ N n  k=1 k ln k = 1 2  n2+ n + 1 6  ln n−n 2 4 + ln A + R 2 n, R2n ≤ √ 3 216n,∀n ∈ R. We deduce ln V_n = n  n +1 2  ln n− n2+ n ln√2π + 1 12 − 1 2  n2+ n +1 6  ln n + + 1 4n 2_{− ln A + Rn} = n 2 2 ln n− 3n2 4 + n ln √ 2π− 1 12ln n + 1 12 − ln A + Rn, where R_n= nR_n1− R2_n and |R_n| ≤ nR1_n+R2_n ≤ _108n√3 ,∀ n ∈ N. 

References

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[2] N. Bourbaki, Functions of a Real Variable: Elementary Theory, Springer, 2004. [3] K. Knopp, Theory and application of infinite series, Springer-Verlag, 1996.

[4] B.M. Makarov, M.G. Goluzina, A.A. Lodkin, A.N. Podkoryotov, Selected problems in real analysis, Transl. Math. Monographs, 107, A.M.S., 1992.

[5] M. Nicolescu, N. Dinculeanu, S. Marcus, Analiz˘a matematic˘a, vol. II, Edit¸ia a doua, Editura didactic˘a ¸si pedagogic˘a, Bucure¸sti, 1971.

[6] Bacalaureat 2007, Proba scris˘a la matematic˘a, varianta 20, subiectul IV.

On some perturbed Bullen inequalities Vlad Ciobotariu-Boer1)

Abstract. In this paper we establish some new general integral inequali-ties by perturbed Bullen type for twice differentiable functions. Then we apply these inequalities to obtain some inequalities for special means of real numbers and some new general quadrature rules of trapezoidal type. Keywords: convex function, integral inequalities, special means

MSC: Primary 26D10; Secondary 26D15, 41A55

Introduction

In [2], Minculete, Dicu and Rat¸iu gave the following perturbed Bullen inequality:  f (a) + f (b)₂ + f  a + b 2  − 2 b− a  _b a f (x)dx−(b− a)[f _{(b)− f}_(a)] 24   ≤ (M − m)(b − a)2 64 , (1)

for all twice differentiable functions f : [a, b] → R with the property there exist real constants m and M such that

m≤ f(x)≤ M for all x ∈ [a, b].

In this article, we first find an integral identity for twice differentiable functions. Then we use this identity to obtain new integral inequalities similar to and improving inequality (1). Finally, we give some applications for special means of real numbers and some numerical quadrature rules.

1)

1. Main results

In order to prove our main results, we need the following lemmas.

Lemma 1. Let f : [0, 1]→ R be a function defined by

f (t) =|1 − nt + nt2|, with n ∈ (0, +∞). Then we have _ 1 0 f (t)dt = v(n), where v(n) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 6− n 6 , if n∈ (0, 4], 6n− n2+ 2(n− 4)√n2_{− 4n} 6n , if n∈ (4, +∞). Proof. See [1]. 

Lemma 2. Let g : (0, +∞) → R be a mapping defined by

g(x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 6− x 6x , f or x∈ (0, 4], 6x− x2+ 2(x− 4)√x2_{− 4x} 6x2 , f or x∈ (4, +∞). Then we have min x∈(0,+∞)g(x) = g  16 3  = 1 16. Proof. See [1]. 

For the rest of the paper, I will denote a real interval and I◦ its interior. For p a positive real number and a < b, let L_p(a, b) denote the class of real functions defined on [a, b] for which the pth power of the absolute value is Lebesgue integrable. Moreover, L_∞(a, b) denotes the class of real functions defined on [a, b] which are essentially bounded.

Lemma 3. Let f : I ⊂ R → R be a twice differentiable function on I◦ with f∈ L1[a, b], where a, b∈ I◦, a < b. Then

1 b− a  _b a f (u)du− 1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 8n [f _{(b)− f}_{(a)] =} =(b− a) 2 16n  1 0 (1− nt + nt2)  f  1−t 2 a + 1+t 2 b  +f  1+t 2 a + 1−t 2 b  dt

Proof. By integration by parts, we get  1 0 (1− nt + nt2)f  1− t 2 a + 1 + t 2 b  dt = = 2 b− a  (1− nt + nt2)f  1− t 2 a + 1 + t 2 b   1 0+ +n  1 0 (1− 2t)f  1− t 2 a + 1 + t 2 b  dt  = = 2 b− a  f(b)− f  a + b 2  + 4n (b− a)2  (1− 2t)f  1− t 2 a + 1 + t 2 b   1 0+ +2  1 0 f  1− t 2 a + 1 + t 2 b  dt  and then, using the change of variables

u =1− t 2 a + 1 + t 2 b for t∈ [0, 1], we obtain  1 0 (1− nt + nt2)f  1− t 2 a + 1 + t 2 b  dt = 2 b− a  f(b)− f  a + b 2  − 4n (b− a)2  f (b) + f  a + b 2  + 16n (b− a)3  _b a+b 2 f (u)du. (2)

Then, multiplying both sides of (2) by (b−a)_16n2, we get 1 b− a  _b a+b 2 f (u)du−1 4  f (b) + f  a + b 2  +b− a 8n  f(b)− f  a + b 2  = = (b− a) 2 16n  1 0 (1− nt + nt2)f  1− t 2 a + 1 + t 2 b  dt. (3) Similarly, we find 1 b− a  a+b 2 a f (u)du−1 4  f (a) + f  a + b 2  +b− a 8n  f  a + b 2  − f_(a)₌ = (b− a) 2 16n  1 0 (1− nt + nt2)f  1 + t 2 a + 1− t 2 b  dt. (4)

Summing the above equalities (3) and (4), we get the desired equality.  Now, by using Lemma 3, we prove our main theorem. Recall that for

g∈ L∞[a, b] one putsg∞ = sup

u∈(a,b)

Theorem 4. Let f : I⊂ R → R be a twice differentiable function on I◦ such that f∈ L_∞[a, b], where a, b∈ I, a < b. Then the following inequality

  1 b− a  _b a f (u)du−1 2  f (a) + f (b) 2 + f  a + b 2  + b− a 8n [f _{(b)− f}_(a)]_ ≤ 1 8g(n)(b− a) 2_f_∞

holds for all n∈ (0, +∞).

Proof. From Lemma 3, using the properties of modulus, we have

 _{b− a}1  b a f (u)du−1 2  f (a) + f (b) 2 + f  a + b 2  + b− a 8n [f _{(b)− f}_(a)]_ ≤ f∞ 8n (b− a) 2  1 0 |1 − nt + nt2_|dt.

We obtain the desired inequality from Lemma 1 and the above inequality. 

Corollary 5. Under the assumptions of Theorem 4, for f(a) = f(b) we

have  _{b− a}1  b a f (u)du−1 2  f (a) + f (b) 2 + f  a + b 2   ≤ 1₈g(n)(b− a)2f_∞ for all n∈ (0, +∞).

Remark 6. By using Lemma 2, we deduce that the equation g(n) = ₁₂1 has the solutions n = 4 and n = 64₇ on the interval (0, +∞). We also obtain that

g(n)∈ ₁₆1,₁₂1  if n∈4,64₇ .

For n = 16₃ , we find the best inequality of this type:  _{b− a}1  b a f (x)dx−1 2  f (a) + f (b) 2 + f  a + b 2  +3(b− a) 128 [f _{(b)− f}_(a)]_{ ≤} ≤ 1 128(b− a) 2_f_ ∞. (5)

The inequality (5) improves inequality (1) if 0≤ 2m ≤ M. For n = 4 we obtain the inequality

 _{b− a}1  b a f (x)dx− 1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 32 [f _{(b)− f}_(a)]_{ ≤} ≤ (b− a)2 96 f _{∞, (6)}

and for n = 64₇ we obtain a new inequality as good as (6):   1 b− a  _b a f (x)dx−1 2  f (a) + f (b) 2 + f  a + b 2  +7(b− a) 512 [f _{(b)− f}_(a)]_{ ≤}

≤ (b− a)2

96 f

_

∞. (7)

For n ∈ 4,64₇ ∩ N, we find some important integral inequalities of perturbed Bullen type which are better than (6):

1) if n = 5, we have  _{b− a}1  b a f (u)du− 1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 40 [f _{(b)− f}_(a)]_{ ≤} ≤ 5 + 2 √ 5 1200 (b− a) 2_f_ ∞< (b− a) 2 96 f _ ∞; 2) if n = 6, we have  _{b− a}1  b a f (u)du− 1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 48 [f _{(b)− f}_(a)]_{ ≤} ≤ √ 3 216(b− a) 2_f_ ∞< (b− a) 2 96 f _ ∞; 3) if n = 7, we have   1 b− a  _b a f (u)du− 1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 56 [f _{(b)− f}_(a)]_{ ≤} ≤ 6 √ 21− 7 2352 (b− a) 2_f_ ∞< (b− a) 2 96 f _ ∞; 4) if n = 8, we have  _{b− a}1  b a f (u)du− 1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 64 [f _{(b)− f}_(a)]_{ ≤} ≤ 2 √ 2− 1 192 (b− a) 2_f_ ∞< (b− a) 2 96 f _ ∞; 5) if n = 9, we have  _{b− a}1  b a f (u)du− 1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 72 [f _{(b)− f}_(a)]_{ ≤} ≤ 10 √ 5− 9 1296 (b− a) 2_f_ ∞< (b− a) 2 96 f _ ∞.

Theorem 7. Let f : I⊂ R → R be a twice differentiable function on I◦ such that f∈ L1[a, b], where a, b∈ I, a < b.

If |f| is convex on [a, b], then the following inequality

  1 b− a  _b a f (u)du−1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 8n [f _{(b)− f}_(a)]_{ ≤} ≤ 1 16g(n)(b− a) 2_(|f_{(a)| + |f}_{(b)|) (8)}

Proof. From Lemma 3, using the properties of modulus, by the convexity of |f_{|, we obtain}  _{b− a}1  b a f (u)du−1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 8n [f _{(b)− f}_(a)]_{ ≤} ≤ (b− a)2 16n (|f _{(a)| + |f}_(b)|) 1 0 |1 − nt + nt2_{|dt. (9)}

Applying Lemma 2 in the inequality (9), we find (8). 

Corollary 8. Under the assumptions of Theorem 7, for f(a) = f(b) we

have _ _{b− a}1  b a f (u)du−1 2  f (a) + f (b) 2 + f  a + b 2   ≤ ≤ 1 16g(n)(b− a) 2_(|f_{(a)| + |f}_(b)|) for all n∈ (0, +∞).

Remark 9. By using similar considerations as in Remark 6, we see that the

inequality (8) provides interesting inequalities of perturbed Bullen type.

For n = 16₃ we get the best inequality of this type:  _{b− a}1  b a f (u)du−1 2  f (a) + f (b) 2 + f  a + b 2  +3(b− a) 128 [f _{(b)− f}_(a)]_{ ≤} ≤ 1 256(b− a) 2_(|f_{(a)| + |f}_{(b)|). (10)}

This inequality improves the inequality (1) if 0≤ 2m ≤ M. For n = 4 we obtain the inequality:

 _{b− a}1  b a f (u)du− 1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 32 [f _{(b)− f}_(a)]_{ ≤} ≤ (b− a)2 192 (|f _{(a)| + |f}_{(b)|) (11)}

and for n = 64₇ we obtain a new inequality as good as (11):   1 b− a  _b a f (u) du−1 2  f (a) + f (b) 2 + f  a + b 2  +7(b− a) 512 [f _{(b)− f}_(a)]_{ ≤} ≤ (b− a)2 192 (|f _{(a)| + |f}_(b)|).

For n∈4,64₇∩ N, we find some important integral inequalities of this type which are better than (11):

1) if n = 5, we have   1 b− a  _b a f (u)du− 1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 40 [f _{(b)− f}_(a)]_{ ≤}

≤ 5 + 2 √ 5 2400 (b− a) 2_(|f_{(a)| + |f}_{(b)|) <} (b− a)2 192 (|f _{(a)| + |f}_(b)|); 2) if n = 6, we have  _{b− a}1  b a f (u)du− 1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 48 [f _{(b)− f}_(a)]_{ ≤} ≤ √ 3 432(b− a) 2_(|f_{(a)| + |f}_{(b)|) <} (b− a)2 192 (|f _{(a)| + |f}_(b)|); 3) if n = 7, we have  _{b− a}1  b a f (u)du− 1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 56 [f _{(b)− f}_(a)]_{ ≤} ≤ 6 √ 21− 7 4704 (b− a) 2_(|f_{(a)| + |f}_{(b)|) <} (b− a)2 192 (|f _{(a)| + |f}_(b)|); 4) if n = 8, we have  _{b− a}1  b a f (u)du− 1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 64 [f _{(b)− f}_(a)]_{ ≤} ≤ 2 √ 2− 1 384 (b− a) 2_(|f_{(a)| + |f}_{(b)|) <} (b− a)2 192 (|f _{(a)| + |f}_(b)|); 5) if n = 9, we have  _{b− a}1  b a f (u)du− 1 2  f (a) + f (b) 2 + f  a + b 2  +b− a 72 [f _{(b)− f}_(a)]_{ ≤} ≤ 10 √ 5− 9 2592 (b− a) 2_(|f_{(a)| + |f}_{(b)|) <} (b− a)2 192 (|f _{(a)| + |f}_(b)|).

2. Applications for special means Recall the following means:

(a) The arithmetic mean

A = A(a, b) := a + b

2 , a, b≥ 0; (b) The geometric mean

G = G(a, b) :=√ab, a, b≥ 0;

(c) The harmonic mean

H = H(a, b) := 2ab

a + b, a, b > 0;

(d) The logarithmic mean

L = L(a, b) := ⎧ ⎨ ⎩ a if a = b > 0, b− a ln b− ln a if a= b, a, b > 0;

(e) The identric mean I = I(a, b) := ⎧ ⎪ ⎨ ⎪ ⎩ a if a = b > 0, 1 e  bb aa 1/(b−a) if a= b, a, b > 0;

(f) The p-logarithmic mean, defined for p∈ R \ {−1, 0} by

L_p= L_p(a, b) := ⎧ ⎪ ⎨ ⎪ ⎩ a if a = b > 0, bp+1− ap+1 (p + 1)(b− a) if a= b, a, b > 0, together with L₋₁ := L and L0 := I.

It is known that L_p is monotonically nondecreasing in p ∈ R. The following simple relationship are also known in the literature:

H ≤ G ≤ L ≤ I ≤ A.

Now, using the results of Section 2, some new inequalities are derived for the above means.

Proposition 10. Let p≥ 2 and 0 < a < b. Then we have the inequality



Lp(a, b)p− A(A(ap, bp), A(a, b)p) +

p(p− 1) 8n (b− a) 2_L p−2(a, b)p−2   ≤ ≤ p(p− 1) 8 g(n)(b− a) 2_A(ap−2_{, b}p−2_{) (12)} for all n∈ (0, +∞).

Proof. The assertion follows from (8) applied for f (x) = xp, x∈ [a, b]. 

Remark 11. Letting n = 16₃ in (12), we obtain the best inequality of this type:



Lp(a, b)p− A(A(ap, bp), A(a, b)p) +

3p(p− 1) 128 (b− a) 2_L p−2(a, b)p−2   ≤ ≤ p(p− 1) 128 (b− a) 2_A(ap−2_{, b}p−2_).

Proposition 12. Let 0 < a < b. Then we have the inequality



L(a,b)−1_{− A(H(a, b)}−1_{, A(a, b)}−1_{) +}(b− a)2A(a, b)

4nG(a, b)4   ≤ ≤ 1 4g(n)(b− a) 2_A(a−3_{, b}−3_{) (13)} for all n∈ (0, +∞).

Remark 13. Letting n = 16₃ in (13) we find the best inequality of this type:



L(a,b)−1_{− A(H(a, b)}−1_{, A(a, b)}−1_{) +}3(b− a)2A(a, b)

64G(a, b)4   ≤ ≤ (b− a)2 64 A(a −3_{, b}−3_).

Proposition 14. Let 0 < a < b. Then we have the inequality:



lnI(a,b) + lnG(G(a,b),A(a,b)) + _{8nG(a, b)}(b− a)22

  ≤ ≤ 1 8g(n)(b− a) 2_A(a−2_{, b}−2_{) (14)} for all n∈ (0, +∞).

Proof. The assertion follows from (8) applied for f (x) =− ln x, x ∈ [a, b]. 

Remark 15. Letting n = 16₃ in (14) we find the best inequality of this type:



lnI(a,b) + lnG(G(a,b),A(a,b)) + _{128G(a, b)}3(b− a)22

  ≤

≤ (b− a)2

128 A(a

−2_{, b}−2_).

3. Applications for composite quadrature formula

Let Δ be a division a = x0< x1<· · · < xm−1 < xm = b of the interval

[a, b]. Then the following results hold.

Theorem 16. Let f : I ⊂ R → R be a twice differentiable function on I◦

such that f∈ L_∞[a, b], where a, b∈ I, a < b. Then we have  _b a f (u)du = A(f, f, Δ) + R(f, f, Δ), where A(f, f, Δ) := 1 2 m  i=1 h_i 2  f (x_i−1) + f (x_i) 2 + f  x_i−1+ x_i 2  − − 1 8n m  i=1 h2_i[f(x_i)− f(x_i−1)] and h_i := x_i− x_i−1.

The remainder R(f, f, Δ) satisfies the estimation |R(f, f_{, Δ)| ≤} 1 8g(n)f _ ∞ m  i=1 h3_i, (15)

Proof. Apply Theorem 4 on the interval [x_i−1, x_i], i = 1, m, to get    _xi xi−1 f (u)du− 1 2· h_i 2  f (x_i−1) + f (x_i) 2 + f  x_i−1+ x_i 2  + + h 2 i 8n[f _(x i)− f(xi−1)] ≤ 1 8g(n)h 3 if∞.

Summing the above inequalities over i from 1 to m and using the generalized

triangle inequality, we get the desired estimation (15). 

Theorem 17. Let f : I ⊂ R → R be a twice differentiable function on I◦

such that f∈ L1[a, b], where a, b∈ I, a < b.

If |f| is convex on [a, b], then we have

 _b

a

f (u)du = A(f, f, Δ) + R(f, f, Δ). The remainder R(f, f, Δ) satisfies the estimation

|R(f, f, Δ)| ≤ 1 16g(n) m  i=1 h3_i[|f(x_i−1)| + |f(x_i)|] (16)

for all n∈ (0, +∞) and for all m ∈ N∗.

Proof. Applying Theorem 7 on the interval [x_i−1, x_i], i = 1, m, we find    _xi xi−1 f (u)du−hi 2  f (x_i−1) + f (x_i) 2 + f  x_i−1+ x_i 2  + + h 2 i 8n[f _(x i)− f(xi−1)] ≤ 1 16g(n)h 3 i(|f(xi−1)| + |f(xi)|).

Summing the above inequalities over i from 1 to m and using the generalized

triangle inequality, we get the desired estimation (16). 

References

[1] V. Ciobotariu-Boer, Integral inequalities of trapezoidal type, Gazeta Matematic˘a 121 (2016), no. 4, 171–178.

[2] N. Minculete, P. Dicu, A. Rat¸iu, Two reverse inequalities of Bullen’s inequality, Ge-neral Mathematics 22 (2014), no. 1, 69–73.

Traian Lalescu mathematics contest for university students Gabriel Mincu1), Radu Strugariu2)

Abstract. This paper deals with the problems proposed at the 2016 edi-tion of the Traian Lalescu naedi-tional mathematics competiedi-tion for university students, hosted in Sibiu, between May 19thand May 21st2016, by the ’Lu-cian Blaga’ University and the ’Nicolae B˘alcescu’ Land Forces Academy. Keywords: congruence, eigenvalue, eigenvector, idempotent matrix, in-tegral, limit, packing problems, quaternions, rank of a matrix, series, sub-group, vector

MSC: Primary 97U40; Secondary 97D50.

Introduction

The ninth edition of the ’Traian Lalescu’ national mathematics contest for university students was hosted between May 19th and May 21st 2016, in Sibiu, by the ’Lucian Blaga’ University and the ’Nicolae B˘alcescu’ Land Forces Academy.

The contest saw a participation of 83 contestants representing 13 uni-versities from Bucure¸sti, Cluj, Constant¸a, Craiova, Ia¸si, Sibiu and Timi¸soara, and was organised in four sections:

(1) Section A: theoretical, first and second year of study — for students from the faculties of mathematics and informatics (12 students); (2) Section B: electrical, first year of study — for students from the

fa-culties of automatics and computer science, physics, electronics, en-ergetics, including departments teaching in foreign languages having these profiles (30 students);

(3) Section C: nonelectrical — for students from the faculties of applied sciences, mechanics, chemistry, transports, metallurgy, constructions, economics (all sections), biology, including departments teaching in foreign languages having these profiles (22 students);

(4) Section D: engineering, second year of study — for students from all faculties teaching engineering (18 students).

We present in the sequel the problems proposed in Sections A and B of the contest, along with their solutions. More details concerning the competition can be found at the address

http://stiinte.ulbsibiu.ro/concurs-traian-lalescu/.

1)

Department of Mathematics, Faculty of Mathematics and Informatics, University of Bucharest, Romania, gamin@fmi.unibuc.ro

2)

Department of Mathematics and Informatics, Gheorghe Asachi Technical University of Ia¸si, Romania, rstrugariu@tuiasi.ro

Problems and solutions

Section A Problem 1. Find the integers x for which

2017| x2016+ x2015+· · · + x + 1.

Gabriel Mincu

The jury considered this problem to be easy. The contestants confirmed this opinion, more than half of them managing to fully solve the problem. The solutions they gave went along the lines of Solution 1 below.

Solution 1. Let us notice that 2017 is a prime number. Let x∈ Z be

such that 2017| x2016+ x2015+· · · + x + 1. Then 2017 | x2017− 1, so

x2017 ≡ 1 (mod 2017). (1)

This is obviously impossible if x divides 2017. If 2017  x, we derive form Fermat’s theorem that x2016 ≡ 1 (mod 2017), so x2017 ≡ x (mod 2017). From this and relation (1) we get x ≡ 1 (mod 2017), and since all these values actually satisfy the given condition, they form the solution of the

problem. 

Solution 2. Let p be a prime number. Then p | p_k for all k ∈

{1, 2, . . . , p − 1}, so, working in Zp[X], Xp − 1 = (X − 1)p, whence Xp−1+

Xp−2+· · · + X + 1 = (X − 1)p−1. Therefore, the only root of Xp−1+ Xp−2+

· · · + X + 1 in Zp is 1, so for x∈ Z we have p | xp−1+ xp−2+· · · + x + 1 if

and only if x≡ 1 (mod p).

Taking p = 2017, we get that 2017 | x2016+ x2015+· · · + x + 1 if and

only if x≡ 1 (mod 2017). 

Problem 2. Compute lim

h0 ∞  n=1 h 1 + n2_h2 . Ovidiu Furdui

Although the jury considered this problem to be easy-medium, only two contestants managed to give full solutions. We present the ideas of these solutions below.

Solution 1. For all n∈ N∗ and h > 0 we have

h 1 + (n + 1)2_h2 <  _n+1 n h 1 + x2_h2 dx < h 1 + n2_h2, so that ∞  n=1 h 1 + n2_h2 − h 1 + h2 <  _∞ 1 h 1 + x2_h2 dx < ∞  n=1 h 1 + n2_h2.

Since  _∞ 1 h 1 + x2_h2 dx = arctan(xh)  ∞ 1 = π 2 − arctan h, we get π 2 − arctan h ≤ ∞  n=1 h 1 + n2_h2 ≤ π 2 − arctan h + h 1 + h2.

Taking limits for h 0, we obtain lim

h0 ∞  n=1 h 1+n2h2 = π2. 

Solution 2. Let ε > 0. Since lim

x→∞arctan x = π 2, there is A > 1 such that π 2 − ε < arctan A < π 2, (2)

and we have lim

h0 ∞  n=1 h 1+n2h2 = lim_h0  1≤n≤A_h h 1+n2h2+ lim_h0  n>A_h h 1+n2h2. The first

term of the right hand side is a limit of Riemann sums of the function _1+x1 2

over the interval [0, A], while the second is nonnegative and, since h

1+n2h2 < < nh  (n−1)h dx

1+x2 for all n∈ N∗, is less than lim_h0

⎛ ⎝ ∞ h[A h] dx 1+x2 ⎞ ⎠. Therefore, arctan A≤ lim h0 ∞  n=1 h 1 + n2_h2 ≤ arctan A + lim_h0  π 2 − arctan h  A h  = π 2,

so, in view of relation (2), we get π₂ − ε ≤ lim

h0 ∞  n=1 h 1+n2h2 ≤ π2.

Since ε may be taken arbitrarily small, lim

h0 ∞  n=1 h 1+n2h2 = π2. 

Problem 3. We denote by K the subfield{a+bi+cj+dk : a, b, c, d ∈ Q}

of the fieldH of the quaternions. Prove that:

a) The order of any finite subgroup of (K \ {0}, ·) is of the form 2α _{· 3}β_, α, β ∈ N∗.

b) Any commutative finite subgroup of (K\ {0}, ·) has order 1, 2, 3, 4 or 6.

Victor Alexandru

The jury considered this problem medium-hard. This opinion has been confirmed by the contestants, only two of whom found suitable approaches, but none managed to solve the problem entirely.

Solution. For each x = a_x+ b_xi + c_xj + d_xk∈ K, we will call a_x the scalar part of x, and V_x = b_xi + c_xj + d_xk the vector part of x.

The following computational properties are immediate and will be used throughout the solution:

• If x, x_{∈ K, then xx} _{= a}

xax−bxbx−cxcx−dxdx+ (axbx+ bxax+ cxdx− d_xc_x)i + (a_xc_x− b_xd_x+ c_xa_x+ d_xb_x)j + (a_xd_x + b_xc_x − c_xb_x + d_xa_x)k.

• Denoting x = ax− bxi− cxj− dxk, we have xx = a2x+ b2x+ c2x+ d2x ∈ Q. • x2 _{= a}2

x− b2x− c2x− dx2 + 2axVx = 2axx− xx, • x, x_{∈ K commute if and only if V}

x and Vx are Q-linearly dependent.

a) Let H be a finite, nontrivial subgroup of the multiplicative group

G = K \ {0} and let p be a prime divisor of the order of H. According to

Cauchy’s theorem, there is x ∈ H of order p. Since xp = 1 = x, we derive that x, seen as an element of the commutative field L = Q(x), is a root of the irreducible polynomial f = Xp−1+ Xp−2+· · · + X + 1 ∈ Q[X].

On the other hand, x2− 2a_xx + xx = 0, so x is a root of the polynomial g = X2_{− 2a}

xX + xx∈ Q[X]. Since f is irreducible, we derive that f divides g, so p = 1 + deg(f )≤ 1 + deg(g) = 3.

Consequently, the order of H is of the form 2α· 3β, α, β ∈ N. b) Let x∈ G be an element of finite order n. We then have

xn= (a_x+V_x)n=  0≤2k≤n  n 2k  an−2k_x V_x2k+ ⎛ ⎝  0≤2k+1≤n  n 2k+1  an−2k−1_x V_x2k ⎞ ⎠V_x. Putting T_x=b2

x+ c2x+ d2x, we have Vx2 =−Tx2, so the relation above yields

xn=  0≤2k≤n  n 2k  an−2k_x (−T_x2)k+ ⎛ ⎝  0≤2k+1≤n  n 2k + 1  an−2k−1_x (−T_x2)k ⎞ ⎠ V_x= =  0≤2k≤n  n 2k  an−2k_x (iT_x)2k+ ⎛ ⎝  0≤2k+1≤n  n 2k + 1  an−2k−1_x (iT_x)2k ⎞ ⎠ V_x = = (ax+ iTx) n_{+ (a} x− iTx)n 2 + (a_x+ iT_x)n− (a_x− iT_x)n 2 Vx.

Thus, the condition xn= 1 is equivalent to the system 

(a_x+ iT_x)n= (a_x− iT_x)n (a_x+ iT_x)n+ (a_x− iT_x)n= 2,

and therefore to (a_x + iT_x)n = (a_x − iT_x)n = 1. Since the order of x is

n, there is k ∈ {1, 2, . . . , n − 1}, coprime to n, such that a_x = cos2kπ_n and

T_x= sin2kπ_n . But a_x∈ Q, so a_x∈−1, −1₂, 0,1₂, 1.

Let now H be a finite commutative subgroup of G. If H = {1}, then

|H| = 1. If H = {−1, 1}, then |H| = 2. If H ⊆ {−1, 1}, let x ∈ H \ {−1, 1}

and x ∈ H. Since x and x commute, there are μ, μ ∈ Q, μ2+ μ2= 0, such that μV_x+ μV_x = 0.

(I) If a_x = a_x = 0, then T_x2 = T_x2 = 1, so V_x2 = V_x2 = −1. We get

μ2 = μ2, so x =±x. But x2 = V_x2 =−1, so x3 =−x. Thus, −x ∈ x, so

(II) If a_x = 0 and a_x = ±1₂, then T_x2 = 1, T_x2 = 3₄, V_x2 = −1 and

V_x2 =−3₄. We obtain μ2=

4 3μ

2_{, contradiction.}

(III) If a_x = 0 and a_x =±1, then T_x2 = 1, T_x2 = 0, V_x2 =−1, V_x2 = 0,

so x =±1. Since x2 =−1 and x4 = 1, we get x ∈ x. As in Case (I), we obtain H =x and |H| = 4.

(IV) If a_x = ±1₂ and a_x = ±1₂, then T_x2 = T_x2 = 3₄, V_x2 = V_x2 = −3₄

and we get μ2 _{= μ}2 _{and x} _{∈ {x, −x, x, −x}. Since x = ±}1

2 + Vx, we obtain

x2 =−1₂ ± Vx = ∓x and x3 = ±1. If x3 =−1, then x4 = −x, x5 = −x2,

x6 = 1. We conclude that H =x or H = x, −x and |H| ∈ {3, 6}.

(V) If a_x =±1₂ and a_x =±1, then T_x2 = 3₄, T_x2 = 0, V_x2 =−3₄, V_x2 = 0.

We get x =±1 and, reasoning as in Case (IV), |H| ∈ {3, 6}.

(VI) If a_x=±1 and a_x =±1, then H ⊆ {−1, 1}, contradiction.

The other cases reduce to the ones above by swapping the roles of a_x

and a_x. 

Problem 4. Let (D_n)_n∈N∗ be a sequence of discs such that for each

n ∈ N∗ the radius of D_n is _n1. Find the minimum side of a square inside which one may arrange without superpositions all the discs D_n, n∈ N∗.

Gabriel Mincu

Figure 1

The jury considered this problem to be hard. Only one of the contestants ap-proached the problem, but his idea was not effective enough to yield a complete solu-tion.

Solution. Let us first notice that

if D1 and D2 lie inside a square of side

L, and the line of their centers makes an

angle α with one of the sides of the square, then, according to Figure 1,

L≥ RS ≥ 3

2(1 + cos α) and

L≥ T U ≥ 3

2(1 + sin α), so

L≥ 3

2(1 + max{sin α, cos α}) ≥ 3 2 1 + √ 2 2 ! = 3(2 + √ 2) 4 ,

equality holding only for α = 45◦. We will prove that 3(2+

√

2)

4 is the minimum we are looking for. To this

end, we arrange the discs inside a square of side l = 3(2+

√

2)

4 as shown in

Figure 2.

We will certify this is a valid configuration by proving the following claims:

1) D3 fits in the upper right corner

2) D6 fits in the lower right corner.

3) Between D2, D1, AD and CD we may fit a rectangle M N P D of sides 1

2 and l− 1.

4) We may fit D4 and D7 between D2, D1, D3 and AB.

5) We may fit all D_n, n≥ 8, in the interior of MNP D and outside D5,

M N P D being the rectangle from 3).

Figure 2

To prove 1), denote λ = l− 1. Using the notations from Figure 3, we have (1+x)2 = (1−x)2+(λ−x)2, relation that also writes as (λ−x)2−4x = 0. We

want to prove that x≥ 1₃; since the product of the roots of the last equation is λ2 _{> 1, we derive that x, which is smaller than unity, has to be the small}

root of that equation.

Therefore, in order for us to have x ≥ 1₃ it suffices to prove that 

λ−1₃2 − 4₃ ≥ 0. This relation is equivalent to 2+3 √ 2 4 − 1 3 ≥ 2 √ 3 or,

after computation, to 8√3 − 9√2 ≤ 2. But this relation is true, since 8√3− 9√2 < 8· 1.8 − 9 · 1.4 = 14.4 − 12.6 < 2, so claim 1) is proved.

We see in Figure 3 that 1 + z + z√2 =√2, so z = √√2−1

2+1 = 3− 2

√

2 > 3− 2 · 1.4143 > 0.17 > 1₆. This proves 2).

To prove 3), we notice that l− 2 = 3

√ 2−2 4 > 3·1.4−2 4 > 1 2. Since M N P D

is tangent to both D2 and CD, its length is l− 1.

We have 21₂+ 1₃+ 1₄+ 1₇ = 23₄ +10₂₁ < 5₂. On the other hand,

l = 3(2+ √ 2) 4 > 10.24 = 5.12 . Consequently, 2 ₁ 2 + 1 3 + 1 4 + 1 7  < l. We derive

that the length of the rectangle EF GH in Figure 2 is greater than 21₄+ 1₇ and, according to claim 3), its width is larger than 1₂, so D4and D7 fit inside

it. This proves claim 4).

To prove Claim 5), we divide the interior of M N P D into vertical strips that we denote, from right to left, by F3, F4, . . ., such as the width of Fk

is ₂k−11 (see Figure 2). Let f :N∗ → R, f(n) =

1 n + 1 n+1 +· · · + 1 2n−1, and

n∈ N∗. Then f (n + 1)−f(n) = _2n1 +_2n+11 −1_n < 0, so f is strictly decreasing.

Consequently, for all k ≥ 3 we have f(2k_{) ≤ f(8) =} 15 j=8

1

j < 0.728. Thus,

the sum of the diameters of D₂k, D₂k₊₁, . . . , D₂k+1₋₁ is less than 1.456, and

therefore less than the length of F_k. Consequently, we may arrange D₂k,

D₂k₊₁, . . . , D₂k+1₋₁ inside F_k vertically.

We will place them in this order from top to bottom and tangent to the left side of Fk, assigning to each D_na horizontal strip of height _n2 of the correspondingF_k, as suggested in Figure 2.

The disc D5 has a region in common with F3. Since ₁₃1 +₁₄1 +₁₅1 > 1₅,

the discs D8, D9, D10, D11 and D12 are above the upper horizontal tangent

to D5. Consequently, there is only left to prove that the distance d between

AD and the left vertical tangent to D5 is larger than 1₄ +₁₃2 .

But, as seen on Figure 4, y2=6₅2−4₅2= 4₅, so y = √2

5. Therefore,

Section B

Problem 1. a) Knowing that ∞

0 sin x

x dx = π2, compute the improper

integral I_n=∞

0 sin2 x_n

x2 dx.

b) Determine the sum of the series ∞

n=1 I_{f (n)}, where f (n) = 1 arctan 2 n2 . Florian Munteanu Solution. a) We notice that

 _∞ 0 sin2x x2 dx =− sin2x x  ∞₀ ₀∞2 sin x cos x_x dx =  _∞ 0 sin 2x x dx.

Via the change of variable 2x = t, we get

∞  0 sin 2x x dx = ∞  0 sin t t dt = π2. Now,

using the change of variable x_n = t, we get I_n=

∞



0 sin2t

nt2 dt = 2nπ .

b) Note firstly that I_{f (n)} = _{2f (n)}π = π₂ arctan_n22. Therefore, it is enough to

compute the sum of the series ∞

n=1

I_{f (n)} = π₂ ∞

n=1

arctan_n22 (the convergence

of which follows easily from a comparison test). To this end, we compute the sequence (S_n)_n of the partial sums of the series ∞

n=1 arctan 2 n2 : S_n= n  k=1 arctan 2 k2 = n  k=1 (arctan(k + 1)− arctan(k − 1)) = arctan(n + 1) + arctan n− arctan 1 − arctan 0 −−−→

n→∞ 3π 4 . It follows that ∞ n=1 I_{f (n)}= 3π₈2. 

Problem 2. Let A ∈ Mn(R) be a matrix having the distinct real eigenvalues λ1, λ2, . . . , λn, with−1 < λi ≤ 1, i = 1, n. Suppose that the sum

of the elements in each column is 1, and the elements in each line of the matrix sum to the same value.

a) Prove that 1 is an eigenvalue for A, and (1, 1, . . . , 1)T _{is a corresponding}

eigenvector.

b) For an arbitrary fixed (x1, x2, . . . , xn)T ∈ Rn compute

lim k→∞A k_(x 1, x2, . . . , xn)T  .