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ROMANIAN MATHEMATICAL SOCIETYROMANIAN MATHEMATICAL SOCIETY Mehedinți BranchMehedinți Branch SUMMER EDITION 2021 SUMMER EDITION 2021 R. M. M. - 29 R. M. M. - 29 ROMANIAN MATHEMATICAL ROMANIAN MATHEMATICAL MAGAZINEMAGAZINE ISSN 2501-0099ISSN 2501-0099 Romanian Mathematical Society-Mehedinți Branch 2021 1 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 ROMANIAN MATHEMATICAL SOCIETY Mehedinți Branch ROMANIAN MATHEMATICAL MAGAZINE R.M.M. Nr.29-SUMMER EDITION 2021 Romanian Mathematical Society-Mehedinți Branch 2021 2 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 ROMANIAN MATHEMATICAL SOCIETY Mehedinți Branch DANIEL SITARU-ROMANIA EDITOR IN CHIEF ROMANIAN MATHEMATICAL MAGAZINE-PAPER VARIANT ISSN 1584-4897 GHEORGHE CĂINICEANU-ROMANIA EDITORIAL BOARD D.M.BĂTINEȚU-GIURGIU-ROMANIA CLAUDIA NĂNUȚI-ROMANIA FLORICĂ ANASTASE-ROMANIA NECULAI STANCIU-ROMANIA MARIAN URSĂRESCU-ROMANIA MARIN CHIRCIU-ROMANIA DAN NĂNUȚI-ROMANIA IULIANA TRAȘCĂ-ROMANIA EMILIA RĂDUCAN-ROMANIA DRAGA TĂTUCU MARIANA-ROMANIA DANA PAPONIU-ROMANIA GIMOIU IULIANA-ROMANIA DAN NEDEIANU-ROMANIA OVIDIU TICUȘI-ROMANIA LAVINIU BEJENARU-ROMANIA ROMANIAN MATHEMATICAL MAGAZINE-INTERACTIVE JOURNAL ISSN 2501-0099 WWW.SSMRMH.RO DANIEL WISNIEWSKI-USA EDITORIAL BOARD VALMIR KRASNICI-KOSOVO Romanian Mathematical Society-Mehedinți Branch 2021 3 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 CONTENT INEQUALITIES WITH CONVEX FUNCTIONS-Daniel Sitaru,Claudia Nănuți……………….……………..4 ABOUT RMM INEQUALITY IN TRIANGLE 899 - Marin Chirciu …….………………………..……………..7 AN EXTENSION OF A TRIANGLE IDENTITY TO THE EUCLIDEAN N-SIMPLEX- Vasile Jiglău ......10 NEW INEQUALITIES WITH FIBONACCI AND LUCAS’ NUMBERS- Florică Anastase......................12 120 YEARS OF LALESCU SEQUENCES (I)- Bătineţu-Giurgiu, Neculai Stanciu................................. 16 SOME APPLICATIONS OF THE ROLLE SEQUENCES - Marian Ursărescu.......................................... 26 THREE FUNDAMENTAL THEOREMS FROM NUMBERS THEORY - Angela Niţoiu.....................29 ABOUT NAGEL AND GERGONNE CEVIANS (III)- Bogdan Fuştei …………………………..…………….32 ABOUT IONESCU-NESBITT INEQUALITY – D.M.Bătinețu-Giurgiu and Daniel Sitaru-……………40 GAKOPOULOS’ LEMMA (II) - Thanasis Gakopoulos …………………………………………………….......42 STRUCTURI ALGEBRICE(VI) – Vasile Buruiană…………………………………………..…………................45 ABOUT A TRIANGLE INEQUALITY - D.M.Bătineţu-Giurgiu, Claudia Nănuţi, Daniel Sitaru…………………………………………………………………………………………………………………….….......49 A BEAUTIFUL EQUIVALENCE-Daniel Sitaru………………………………………………………………………..50 ABOUT AN INEQUALITY BY GEORGE APOSTOLOPOULOS-Marin Chirciu…………………………….52 REVERSE HUYGENS’ INEQUALITY-Dorin Mărghidanu…………………………………………………..……53 HOLDER’S AND BERGSTROM’S INEQUALITIES-NEW APPLICATIONS- Florică Anastase....................................................................................................................................................................58 PROPOSED PROBLEMS…………………………………….………………………………………………………………...63 INDEX OF PROPOSERS AND SOLVERS RMM-29 PAPER MAGAZINE.……………………….……………119 Romanian Mathematical Society-Mehedinți Branch 2021 4 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 INEQUALITIES WITH CONVEX FUNCTIONS By Daniel Sitaru, Claudia Nănuți – Romania Abstract: In this paper it is proved a property of convex functions which can generate beautiful inequalities. Keywords: convex functions, inequalities Main result: Theorem 1(Hardy-Polya-Szego-Littlewood) If �, �, �, �, � ∈ ℝ; � ≤ � ≤ � ≤ � ≤ �; � + � ≤ � + �; �: [�, �] → ℝ; � convexe, then: �(�) + �(�) ≤ �(�) + �(� + � − �) (1) Proof: � ≤ � ⇒ 0 ≤ � − � ⇒ � ≤ � + � − � � ≤ � ⇒ 0 ≤ � − � ⇒ � ≤ � + � − � � ∈ [�; � + � − �] ⇒ (∃)� ∈ [0,1] � = �� + (1 − �)(� + � − �) (2) � ∈ �� + (1 − �)(� + � − �) (3) By adding (2); (3): � + � = (� + �)� + (� + � − �)(2 − � − �) (4) � + � = � + (� + � − �).Replace in (4): � + (� + � − �) = (� + �)� + (� + � − �)(2 − � − �) (� + � − 1)� + (� + � − �)(1 − � − �) = 0, (� + � − 1)(� − � − � + �) = 0 (� + � − 1)(2� − � − �) = 0 Case I: 2� − � − � = 0 ⇒ � + � = 2� (5) But � ≤ �; � ≤ � ⇒ 2� ≤ � + � (6) By (5); (6) ⇒ � = � = �. Inequality (1) becomes: �(�) + �(�) ≤ �(�) + �(� + � − �) (true) Case II: � + � − 1 = 0 ⇒ � + � = 1 ⇒ 2 − � − � = 1 � convexe; �, � ∈ [�, � + � − �] ⇒ (∃)�, � ∈ [0,1] such that: �(�) ≤ ��(�) + (1 − �)�(� + � − �) (7) �(�) ≤ ��(�) + (1 − �)�(� + � − �) (8) By adding (7); (8): Romanian Mathematical Society-Mehedinți Branch 2021 5 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 �(�) + �(�) ≤ (� + �)�(�) + (2 − � − �)�(� + � − �) �(�) + �(�) ≤ �(�) + �(� + � − �) Theorem 2 If �, �, �, �, � ∈ ℝ; � ≤ � ≤ � ≤ � ≤ �; � + � ≤ � + �; �: [�, �] → ℝ; � concave, then: �(�) + �(�) ≥ �(�) + �(� + � − �) (9) Application 1:If �, �, �, �, � ∈ ℝ; 0 < � ≤ � ≤ � ≤ � ≤ �; � + � ≤ � + �; � ∈ ℕ ∖ {0} then � �� + � �� ≤ � �� + � (�����)� (10) Solution: We take in (1): �(�) = � �� ; ��(�) = ������ ��� ��(�) = −� ���� ; ���(�) = �(� + 1)�� ����� = �(� + 1) ���� > 0 � convexe �(�) + �(�) ≤ �(�) + �(� + � − �) 1 �� + 1 �� ≤ 1 �� + 1 (� + � − �)� Equality holds for � = � = �. Application 2:If �, �, �, �, � ∈ ℝ; 0 < � ≤ � ≤ � ≤ � ≤ �; � + � ≤ � + �; � ∈ ℕ ∖ {0} then: �� � + √� � ≤ √� � + �� + � − � � Solution: We take in (9): �(�) = √� � ; ��(�) = � � � � � �� ���(�) = � � � � � − 1� � � � �� < 0; � concave �(�) + �(�) ≥ �(�) + �(� + � − �); �: (0, ∞) → ℝ �� � + √� � ≥ √� � + �� + � � − � (11) Equality holds for � = � = �. Application 3:If �, �, �, �, � ∈ ℝ; 0 < � ≤ � ≤ � ≤ � ≤ �; � + � ≤ � + �; � ∈ ℕ ∖ {0} then: arccot � + arccot � ≤ arccot � + arccot(� + � − �) (12) Solution:We take in (1): �(�) = arccot � ; ��(�) = �� ���� ; ���(�) = �� (����) > 0; � convexe �(�) + �(�) ≤ �(�) + �(� + � − �); �: (0, ∞) → ℝ arccot � + arccot � ≤ arccot � + arccot(� + � − �) Equality holds for: � = � = �. Application 4:If �, �, �, �, � ∈ ℝ; � ≤ � ≤ � ≤ � ≤ �; � + � ≤ � + �; � ∈ ℕ ∖ {0} then: Romanian Mathematical Society-Mehedinți Branch 2021 6 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 �� � + �� � ≤ �� � + �(�����) � (13) Solution: We take in (1): �(�) = �� � ; ��(�) = 2��� � ���(�) = 2�� � + 4���� � = 2�� � (1 + 2��) > 0; � convexe �(�) + �(�) ≤ �(�) + �(� + � − �) �� � + �� � ≤ �� � + �(�����) � Equality holds for: � = � = �. Application 5: If 0 < � ≤ �; � ∈ ℕ ∖ {0} then: 1 �√��� � + 1 � ��� � � � ≤ 1 � ��� ��� � � + 1 �√�� + ��� � − ��� ��� � � Solution: We take in (10): � = 2�� � + � ; � = √��; � = � + � 2 By AM-GM-HM: 0 < � ≤ 2�� � + � ≤ √�� ≤ � + � 2 ≤ � 1 �� + 1 �� ≤ 1 �� + 1 (� + � − �)� 1 �√��� � + 1 � ��� � � � ≤ 1 � ��� ��� � � + 1 �√�� + ��� � − ��� ��� � � Equality holds for � = �. Application 6: If 0 < � ≤ �; � ∈ ℕ ∖ {0} then: √�� �� + � � + � 2 � ≥ � 2�� � + � � + �√�� + � + � 2 − 2�� � + � � Solution: We take in (11): � = ��� ��� ; � = √��; � = ��� � By AM-GM-HM: 0 < � ≤ 2�� � + � ≤ √�� ≤ � + � 2 ≤ � �� � + √� � ≥ √� � + �� + � − � � �√�� � + � � + � 2 � ≥ � 2�� � + � � + �√�� + � + � 2 − 2�� � + � � Romanian Mathematical Society-Mehedinți Branch 2021 7 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 √�� � + � � + � 2 � ≥ � 2�� � + � � + �√�� + � + � 2 − 2�� � + � � Equality holds for � = �. Application 7: If 0 < � ≤ � then: arccot(√��) + arccot � � + � 2 � ≤ arccot � 2�� � + � � + arccot �√�� + � + � 2 − 2�� � + � � Solution: We take in (12): � = √��; � = � + � 2 ; � = 2�� � + � By AM-GM-HM: 0 < � ≤ ��� ��� ≤ √�� ≤ ��� � ≤ � arccot � + arccot � ≤ arccot � + arccot(� + � − �) arccot(√��) + arccot � � + � 2 � ≤ arccot � 2�� � + � � + arccot�√�� + � + � 2 − 2�� � + � � Equality holds for � = �. Application 8: If 0 < � ≤ � then: ��� + � � ��� � � � ≤ � � ��� ��� � � + � �√��� ��� � � ��� ��� � � Solution: We take in (13): � = 2�� � + � ; � = √��; � = � + � 2 �� � + �� � ≤ �� � + �(�����) � , ��√��� � + � � ��� � � � ≤ � � ��� ��� � � + � �√��� ��� � � ��� ��� � � ��� + � � ��� � � � ≤ � � ��� ��� � � + � �√��� ��� � � ��� ��� � � Equality holds for � = �. Reference: ROMANIAN MATHEMATICAL MAGAZINE – www.ssmrmh.ro ABOUT RMM-INEQUALITY IN TRIANGLE 899 By Marin Chirciu – Romania 1) In ���� the following relationship holds: � � � �� + � �� � �� ≥ ��� − �� Proposed by Mehmet Șahin – Ankara – Turkey Romanian Mathematical Society-Mehedinți Branch 2021 8 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 Solution: We prove the following lemma Lemma. 2) In ���� the following relationship holds: � � � �� + � �� � �� = �� + �� − ��� � Proof. Using �� = � ��� and ℎ� = �� � we obtain: � � 1 �� + 1 ℎ� � �� = � � � − � � + � 2� � �� = 1 2� �(� + �) �� = �� + �� − 2�� � which follows from ∑ ��(� + �) = 2�(�� + �� − 2��) Let’s get back to the main problem. Using the Lemma we write the inequality: �� + �� − 2�� � ≥ 28� − 2� ⇔ �� ≥ 27�� which follows from Mitrinovic’s inequality: � ≥ 3�√3. Equality holds if and only if the triangle is equilateral. Remark.The inequality can be strengthened: 3) In ���� the following relationship holds: � � � �� + � �� � �� ≥ ��� − �� Solution: Using the Lemma the inequality can be written: ��������� � ≥ 14� − 4� ⇔ �� ≥ 16�� − 5�� (Gerretsen’s inequality) Equality holds if and only if the triangle is equilateral. Remark. Inequality 3) is stronger than inequality 1): 4) In ���� the following relationship holds: � � � �� + � �� � �� ≥ ��� − �� ≥ ��� − �� Solution: See inequality 3) and Euler’s inequality � ≥ 2�.Equality holds if and only if the triangle is equilateral. 5) In ���� the following inequality holds: � � � �� + � �� � �� ≤ ��� � Romanian Mathematical Society-Mehedinți Branch 2021 9 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 Proposed by Marin Chirciu – Romania Solution �� + �� − 2�� � ≤ 6�� � ⇔ �� ≤ 6�� + 2�� − �� which follows from Gerretsen’s inequality: �� ≤ 4�� + 4�� + 3��. It remains to prove that: 4�� + 4�� + 3�� ≤ 4�� + 4�� + 3�� ⇔ �� − �� − 2�� ≥ 0 ⇔ (� − 2�)(� + �) ≥ 0 obviously from Euler’s inequality � ≥ 2�. Equality holds if and only if the triangle is equilateral. Remark.We can write the double inequality: 6) In ���� the following inequality holds: ��� − �� ≤ � � � �� + � �� � �� ≤ ��� � Proposed by Marin Chirciu – Romania Solution See inequalities 3) and 5).Equality holds if and only if the triangle is equilateral. Remark. If we split the above sum after �� and ℎ� we propose: 7) In ���� the following relationship holds: �(�� − �) ≤ � �� �� ≤ �� � (�� − �) Solution We prove the following lemma: Lemma. 8) In ���� the following relationship holds: � �� �� = �� + �� − ��� � Proof. Using �� = � ��� , we obtain: � �� �� = � � − � � ⋅ �� = 1 � � ��(� − �) = 1 �� ⋅ �(�� + �� − ��) = �� + �� − 8�� � which follows from ∑ �� (� − �) = �(�� + �� − 8��) Let’s get back to the main problem. Using Lemma and Gerretsen’s inequality: 16�� − 5�� ≤ �� ≤ 4�� + 4�� + 3�� Equality holds if and only if the triangle is equilateral. Romanian Mathematical Society-Mehedinți Branch 2021 10 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 9) In ���� the following relationship holds: ��� ≤ � �� �� ≤ �� Solution We use ∑ �� �� = 6� and Euler’s inequality � ≥ 2� Equality holds if and only if the triangle is equilateral. Remark. Between the sums ∑ �� �� and ∑ �� �� we write the following relationship: 10) In ���� the following relationship holds: � �� �� ≤ � �� �� Proposed by Marin Chirciu – Romania Solution: Using the sums ∑ �� �� = 6� and ∑ �� �� = ��������� � we write the inequality: 6� ≤ ��������� � ⇔ �� ≥ 14�� − ��, which follows from Gerretsen’s inequality �� ≥ 16�� − 5��. Equality holds if and only if the triangle is equilateral. Remark. We can write the sequence of inequalities: 1) In ���� the following relationship holds: ��� ≤ � �� � ≤ �� ≤ �(�� − �) ≤ � �� � ≤ �� � (�� − �) Solution: See inequalities 4) and 6) and Euler’s inequality � ≥ 2�.Equality holds if and only if the triangle is equilateral. Refference: ROMANIAN MATHEMATICAL MAGAZINE-www.ssmrmh.ro AN EXTENSION OF A TRIANGLE IDENTITY TO THE EUCLIDEAN N-SIMPLEX By Vasile Jiglău-Romania Abstract: In this note we’ll extended a geometric identity from triangle to an Euclidean n- simplex. Let ������ be an arbitrary triangle and denote by �� the radius of the circle which is tangent to the circumcircle of ������ and to the side ���� of the given triangle, simultaneously ({�, �, �} = {1,2,3}). In [1] the authors proved that: Romanian Mathematical Society-Mehedinți Branch 2021 11 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 1 �� + 1 �� + 1 �� = 2 � + 1 � where � and � are the circumradius and the inradius of ������, respectively. In the following we’ll prove an extension of this identity to the Euclidean n-simplex. Taking in the proposition enunciated below and in it’s proof � = 3 one obtains the corresponding identity for tetrahedron. Let ���� … �� be an Euclidean n-simplex.We’ll use the following notations: −�, �, �, � the centre if it’s circumscribed hypersphere, it’s volume, it’s circumradius and it’s inradius,respectively. −�′�, �′� the centre and the radius of the hypersphere tangent to the circumscribed sphere of ���� … �� in the vertex �� and to the hyperplane ���� … �������� … �� simultaneously. −�� the volume of the n-simplex ����� … �������� … �� −�′� the volume of the n-simplex �′����� … �������� … �� −�"� the volume of the (n-1)-simplex ���� … �������� … �� Proposition: With the above notations, the following identity holds: � 1 �� � ��� = � � + 1 � Proof: Since the hypersphere having −�′� as centre defined above is tangent to the circumscribed hypersphere of the given n-simplex, it results that ��, �′�, � are collinear. We recall that the volume of ���� … �� is given by the formula � = � � ℎ��"�, where ℎ� is the distance from the vertex �� to the hyperplane ���� … �������� … ��.Projecting �, �′� on the hyperplane ���� … �������� … ��, (� ≠ �) applying the Thales’ theorem, then the precedent formula, we remark that ��� �� = �� � . We have: � = � ��� � ��� ��� + ��� = � �� � �� � ��� ��� + ���"� � = �� ⎝ ⎜ ⎛1 � � �� � ��� ��� + �"� � ⎠ ⎟ ⎞ ⇒ � �� = 1 � � �� � ��� ��� + �"� � Since the sum ∑ ∑ �� � ��� ��� � ��� any �� appears n times, we have � � ∑ ∑ �� � ��� ��� � ��� = � � �� On the other hand �� = � ∑ �"� � ��� , therefore ∑ � �� � ��� = �� � + � � ⇒ ∑ � �� � ��� = � � + � � , q.e.d. Reference: [1] I.Isaev, Y.Maltsev, A.Monastyreva-On some geometric relations of a triangle, Journal of Classical Geometry, volume 4. Romanian Mathematical Society-Mehedinți Branch 2021 12 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 NEW INEQUALITIES WITH FIBONACCI AND LUCAS’ NUMBERS By Florică Anastase-Romania Let be (��)��� −Fibonacci sequences, �� = ���� + ����; �� = 0; �� = 1, ∀� ∈ ℕ, (��)��� −Lucas sequences, �� = ���� + ����; �� = 2; �� = 1, ∀� ∈ ℕ Application 1. �� (�� − �)� + �� (�� − �)� + ⋯ + �� (���� − �)� ≤ ���� + ���� − ��� ��� �(���� − ��) Solution: �� = � �� � ��� = �� + �� + ⋯ + �� = ���� − �� ⇒ ���� − 2�� = ���� + ���� − 2�� = �� + ����, ∀� = 1, � �� − 2�� = �� + �� = �� + �� + �� = 2�� + ��; �� − 2�� = �� + �� = 2�� + 2�� + �� ���� − 2�� = �� + ���� = 2�� + 2�� + ⋯ + 2���� + �� �� (�� − 2)� + �� (�� − 2)� + ⋯ + �� (���� − 2)� = = �� (2�� + ��) � + �� (2��+ 2�� + ��) � + ⋯ + �� (2�� + 2�� + ⋯ + 2���� + ��) � �� ��(�� + ��) + �� (�� + ��)(�� + �� + ��) = �� + �� ��(�� + �� + ��) Suppose: �� ��(�� + ��) + �� (�� + ��)(�� + �� + ��) + ⋯ + �� (�� + �� + ⋯ + ����)(�� + �� + ⋯ + ��) = �� + �� + ⋯ + �� ��(�� + �� + ⋯ + ��) ⇒ �� ��(�� + ��) + �� (�� + ��)(�� + �� + ��) + ⋯ + �� (�� + �� + ⋯ + ����)(�� + �� + ⋯ + ��) + Romanian Mathematical Society-Mehedinți Branch 2021 13 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 + ���� (�� + �� + ⋯ + ��)(�� + �� + ⋯ + ����) = = �� + �� + ⋯ + �� ��(�� + �� + ⋯ + ��) + ���� (�� + �� + ⋯ + ��)(�� + �� + ⋯ + ����) = = (�� + �� + ⋯ + ��) � + ��(�� + �� + ⋯ + ��) + ����(�� + �� + ⋯ + ��) + ������ (�� + �� + ⋯ + ��)��(�� + �� + ⋯ + ����) = = �� + �� + ⋯ + ���� ��(�� + �� + ⋯ + ����) From (� + �)� ≥ 4�� ⇒ � �� ≥ � (���)� we have: �� ��(�� + ��) ≥ 4�� (2�� + ��)� �� (�� + ��)(�� + �� + ��) ≥ 4�� (2�� + 2�� + ��)� �� (�� + �� + ⋯ + ����)(�� + �� + ⋯ + ��) ≥ 4�� (2�� + 2�� + ⋯ + 2���� + ��) � Adding up relationships, we have: 4�� (2�� + ��)� + 4�� (2�� + 2�� + ��)� + ⋯ + 4�� (2�� + 2�� + ⋯ + 2���� + ��)� ≤ ≤ �� ��(�� + ��) + �� (�� + ��)(�� + �� + ��) + ⋯ + �� (�� + �� + ⋯ + ����)(�� + �� + ⋯ + ��) = �� + �� + ⋯ + �� ��(�� + �� + ⋯ + ��) ⇒ �� (2�� + ��)� + �� (2�� + 2�� + ��)� + ⋯ + �� (2�� + 2�� + ⋯ + 2���� + ��)� ≤ �� + �� + ⋯ + �� 4��(�� + �� + ⋯ + ��) �� (2�� + ��)� + �� (2�� + 2�� + ��)� + ⋯ + �� (2�� + 2�� + ⋯ + 2���� + ��)� ≤ ���� − �� �� �(���� − ��) �� (�� − 2)� + �� (�� − 2)� + ⋯ + �� (���� − 2)� ≤ ���� − �� �� �(���� − ��) From: 5�� = ���� + ���� ⇒ ���� − �� = ������������� � ⇒ ������� �� �(�������) = ������������� ��� �(�������) �� (�� − 2)� + �� (�� − 2)� + ⋯ + �� (���� − 2)� ≤ ���� + ���� − 5�� 5�� �(���� − ��) Application 2. If (��)��� is in arithmetic progression with �� > 0, � > 0 then prove: Romanian Mathematical Society-Mehedinți Branch 2021 14 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 � ����� + � ����� + ⋯ + � ����� ≥ ��� � + ������ − �(���� − �� − ��) − ���� Solution: From A-G-M we have: √� ≤ ��� � ⇒ � √� ≥ � ��� , ∀� ∈ ℝ� ∗ = (0, ∞) then: 1 ����� + 1 ����� + ⋯ + 1 ����� ≥ 2 1 + ���� + 2 1 + ���� + ⋯ + 2 1 + ���� ≥ ��������̈� ≥ 2�� � + ���� + ���� + ⋯ + ���� From ���� = ���� + ��; �� = �� = 1 we have ���� = ���� + ��, ∀� = 1, ������ ⇒ ���� − ���� = ��, ∀� = 1, ������ ⇒ �� ∙ �� = (�� + (� − 1)�)�� = (�� − �)�� + ���� But: � ∙ �� = �(���� − ����) = (� + 2)���� − (� + 1)���� − 2(���� − ����) − ���� = = (� + 2)���� − (� + 1)���� − 2(���� − ����) − (���� − ����) ⇒ � � ∙ �� � ��� = �[(� + 2)���� − (� + 1)����] � ��� − 2 �(���� − ����) � ��� − �(���� − ����) � ��� = = (� + 2)���� − 2�� − 2(���� − ��) − (���� − ��) = ����� − ���� + �� � �� ∙ �� � ��� = (�� − �) � �� � ��� + � � � ∙ �� � ��� = (�� − �)(���� − ��) + �(����� − ���� + ��) = ������ − �(���� − �� − ��) − ����. So, � ����� + � ����� + ⋯ + � ����� ≥ ��� ����������(����������)����� Application 3. � ���� + � ���� + ⋯ + � �(� + �)�� ≥ ��� (� + ����)� + �� , ∀� ≥ � Solution: From A-G-M we have: √� ≤ ��� � ⇒ � √� ≥ � ��� , ∀� ∈ ℝ� ∗ = (0, ∞) then: 1 �3�� + 1 �4�� + ⋯ + 1 �(� + 2)�� ≥ 2 1 + 3�� + 2 1 + 4�� + ⋯ + 2 1 + (� + 2)�� ≥ ��������̈� ≥ 2�� � + 3�� + 4�� + ⋯ + (� + 2)�� For �� = 3, � = 1 we get: 3�� + 4�� + ⋯ + (� + 2)�� = (1 + ����)� + ��, then 1 �3�� + 1 �4�� + ⋯ + 1 �(� + 2)�� ≥ 2�� (1 + ����)� + �� , ∀� ≥ 2 Romanian Mathematical Society-Mehedinți Branch 2021 15 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 Application 4: � �2�� + � ���� + ⋯ + � ���� ≥ ��� (� + ����)� − �� + � , ∀� ≥ � Solution: From A-G-M we have: √� ≤ ��� � ⇒ � √� ≥ � ��� , ∀� ∈ ℝ� ∗ = (0, ∞) then: 1 �2�� + 1 �3�� + ⋯ + 1 ���� ≥ 2 1 + 2�� + 2 1 + 3�� + ⋯ + 2 ��� ≥ ��������̈� ≥ 2�� � + 2�� + 3�� + ⋯ + ��� For �� = 2, � = 1 we get: 2�� + 3�� + ⋯ + ��� = ∑ (� + 1) ∙ �� � ��� = = (� + 1)���� − (���� − �� − ��) − 2�� = ����� − ���� + 1 Then 1 �2�� + 1 �3�� + ⋯ + 1 ���� ≥ 2�� (1 + ����)� − �� + 1 , ∀� ≥ 2 Application 5: � �� � + �� � �� + � �� � + �� � + �� � �� + ⋯ + � �� � + �� � + ⋯ + ���� � �� ≥ ������� (� + �)���� − �� Solution: We have: ���� = ���� + ��, ∀� ≥ 0 ⇒ ���� − �� = ���� ∀� ≥ 0 ⇒ ���� ∙ ���� − ���� ∙ �� = ���� � ⇒ �� � + �� � + ⋯ + ���� � = ���� ∙ ���� − ���� ⇒ �� � + �� � + �� � + ⋯ + ���� � = ���� ∙ ���� ⇒ 1 �� � + �� � + �� � + ⋯ + ���� � = 1 ���� ∙ ���� ⇒ �� �� � + �� � + �� � + ⋯ + ���� � = ���� − ���� ���� ∙ ���� = 1 ���� − 1 ���� ⇒ � �� �� � + �� � + �� � + ⋯ + ���� � � ��� = 1 − 1 ���� From A-G-M we have: √� ≤ ��� � ⇒ � √� ≥ � ��� , ∀� ∈ ℝ� ∗ = (0, ∞) then: � �� � + �� � �� + � �� � + �� � + �� � �� + ⋯ + � �� � + �� � + ⋯ + ���� � �� ≥ ≥ 2 1 + �� �� ���� � + 2 1 + �� �� ���� ���� � + ⋯ + 2 1 + �� �� ���� ��⋯����� � ≥ ��������̈� Romanian Mathematical Society-Mehedinți Branch 2021 16 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 ≥ 2�� � + �� �� ���� � + �� �� ���� ���� � + ⋯ + �� �� ���� ��⋯����� � = = 2�� � + 1 − � ���� = 2������ (� + 1)���� − 1 Application 6: (� + �)��� � + (� + �)��� � + ⋯ + (�� + �)������ � ≥ �(������ − �) − ����� + � � Solution: (� + 1)��� � + (� + 2)��� � + ⋯ + (3� + 1)������ � = = [(� + 1)��] � � ��� + [(� + 2)��] � � ��� + ⋯ + [(3� + 1)�����] � � ���� ≥ ��������̈� ≥ [(� + 1)�� + (� + 2)�� + ⋯ + (3� + 1)�����] � � ��� + � ��� + ⋯ + � ���� But: � ��� + � ��� + ⋯ + � ���� < 2, ∀� ∈ ℕ from mathematical induction by � ∈ ℕ and (� + 1)�� + (� + 2)�� + ⋯ + (3� + 1)����� = = �(�� + �� + ⋯ + �����) + 1 ∙ �� + 2 ∙ �� + ⋯ + (2� + 1) ∙ ����� = = �(����� − ��) + 2������ − ����� + 2 = �(3����� − 1) − ����� + 2 (� + 1)��� � + (� + 2)��� � + ⋯ + (3� + 1)������ � ≥ �(3����� − 1) − ����� + 2 2 Refference: ROMANIAN MATHEMATICAL MAGAZINE-www.ssmrmh.ro 120 YEARS OF LALESCU SEQUENCES (I) By D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania Abstract. In this paper we present new methods to calculate certain limits from math journals from all over the world . MathEduc Subject Classification: D55, I35 MSC Subject Classification: 97D50, 97I30 Key words and phrases: Limits of Lalescu type sequences; Limits of Lalescu type functions; Problem Solving. 1. RESULTS – LALESCU TYPE LIMITS PROBLEMS FROM MATH JOURNALS FROM ALL OVER THE WORLD Romanian Mathematical Society-Mehedinți Branch 2021 17 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 The Stolz-Cesàro criterion (C-S). Let the sequences (��)��� and (��)��� such that(��)��� is strictly monotonous and boundless. If there exists the limit ��� �→� ������� ������� , then ��� �→� �� �� The Cauchy-D’Alembert criterion (C-D’A). Let the sequence (��)���with �� > 0 If there exists the limit ��� �→� ���� �� = � > 0, then ��� �→� ��� � = � We present new solutions of certain problems of Lalescu type limits from math journals from all over the world. In the next we use the following abbreviations: Crux Mathematicorum (CM); Gazeta Matematică Seria B (GMB); La Gaceta de la RSME (LG); Math Problems (MP); Pi Mu Epsilon Journal (PME); Mathematics Magazine from Timişoara (RMT); Mathematical Recreation from Iaşi (RM); Revista Escolar de la Olimpiada Iberoamericana de Matematica (REOIM); Romanian Mathematical Magazine (RMM); School Science and Mathematics (SSM); The Spark of the Mind (SM); The American Mathematical Monthly (AMM); The College Mathematics Journal (CMJ); The Pentagon (P); The Fibonacci Quarterly (FQ). Problem 1. Traian Lalescu’s limit, GMB, Vol. VI, 1900-1901, problem 579, p. 148. ��� �→� � �(� + �)! ��� − √�! � � =⏞ �!≅� � � � � ⇒ √�! � ≅ � � ��� �→� � � + � � − � � � = � � Problem 2. D. M. Bătineţu-Giurgiu’s limit, GMB, Vol. XCIV, 1989, problem C:890,p. 139. ��� �→� � (� + �)� �(� + �)! ��� − �� √�! � � =⏞ �!≅� � � � � ⇒ √�! � ≅ � � ��� �→� � (� + �)� ��� � − �� � � � = ��� �→� �(� + �)� − ��� = � Problem 3. D.M. Bătineţu-Giurgiu, Neculai Stanciu, GMB 5/2012. If (��(�))��� is defined by ��(�) = � ��� �(� + �)�� √� + � ��� � � − ��� √� � � � � , ∀� ∈ ℝ, then compute ��� �→� ��(�). Solution: ��(�) = � √�� � ∙ (�� − 1) = � √� � � � ∙ ���� ����� ∙ ����� �, ∀� ≥ 2, where �� = � � + 1 � � � ∙ � √� + 1 ��� √� � � � , ∀� ≥ 2; ��� �→� �� = 1 and then ��� �→� �� − 1 ����� = 1 ��� �→� �� � = ��� �→� �� � ∙ ��� �→� � √� + 1 ��� √� � � �� = �� ∙ ��� �→� � � + 1 � ∙ 1 √� + 1 ��� � � = �� ∙ 1 = ��. ��� �→� ��(�) = 1 ∙ 1 ∙ ��� � ��� �→� �� �� = ����� = �. Observation: For � = 1 we deduce lim �→� �� = ��� �→� ��(1) = 1, i.e. the limit of Romeo T. Ianculescu. Problem 4. D.M. Bătineţu-Giurgiu, Neculai Stanciu, CM 7/2013. Let (��)��� be a positive real sequence such that ��� �→� ������� � = � ∈ ℝ� ∗ .We define ��! by ���� = ��! ∙ ����, ∀� ∈ ℕ ∗. Compute ��� �→� � �����! ��� ��� − ���! � � � . Solution: ��� �→� �� �� = ��� ��� �→� ������� (���)���� = ��� �→� � ������� � ∙ � ���� � = � � . Romanian Mathematical Society-Mehedinți Branch 2021 18 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 ��� �→� ���! � �� = ��� �→� � ��! ��� � = ����� ��� �→� ����! (� + 1)�(���) ∙ ��� ��! = ��� �→� ���� (� + 1)� ∙ ��� �→� � � � + 1 � �� = � 2 ∙ 1 �� = � 2�� So, � = ��� �→� � �����! ��� ��� − ���! � � � = ��� �→� ���! � � ∙ (�� − 1) = ��� �→� � ���! � �� ∙ ���� ����� ∙ ����� �� ; ��� �→� �� = ��� �→� � �����! ��� � + 1 ∙ � ���! � � = ��� �→� � �����! ��� (� + 1)� ∙ �� ���! � ∙ � + 1 � � = � 2� ∙ 2� � ∙ 1 = 1; ��� �→� �� − 1 ����� = 1, ��� �→� �� � = ��� �→� � �����! ��� ���! � ∙ � � + 1 � � = ��� �→� � ����! ��! ∙ � � � + 1 � � ∙ 1 �����! ��� � = = ��� �→� � ���� (� + 1)� ∙ (� + 1)� �����! ��� � ∙ ��� �→� � � � + 1 � � = � 2 ∙ 2�� � ∙ 1 � = �. We obtain: � = � ��� ∙ 1 ∙ ���� = � ��� . Problem 5. D.M. Bătineţu-Giurgiu, Neculai Stanciu, RM 2/2012. If �� = (���)��� (���)� , ∀� ∈ ℕ∗ then compute ��� �→� � ����� … ������ ��� − ����� … �� � � Solution: �� = ����� … ������ ��� − ����� … �� � = ����� … �� � (�� − 1) = = ����� … �� � ∙ ���� ����� ∙ ����� �, ∀� ≥ 2, where �� = �����…������ ��� �����…�� � , ∀� ≥ 2; ��� �→� ����� … �� � � = ��� �→� � ���� … �� �� � = ��� �→� � ���� … ������ (� + 1)��� ∙ �� ���� … �� � = = ��� �→� � ���� � + 1 ∙ �� ��� = ��� ∙ ��� �→� (� + 3)��� (� + 2)���(� + 1) = ��� ∙ ��� �→� ���� ∙ � + 2 � + 1 = ��� ∙ � ∙ 1 = 1 Where we denote �� = �1 + � � � � ; so ��� �→� �� = 1, ��� �→� ���� ����� = 1. ��� �→� �� � = ��� �→� � ���� … ������ ���� … �� ∙ 1 ����� … ������ ��� � = ��� �→� � ���� � + 1 ∙ � + 1 ����� … ������ ��� � = � ∙ 1 = �. We obtain ��� �→� �� = 1 ∙ 1 ∙ ���� = 1. Problem 6. D.M. Bătineţu-Giurgiu, Neculai Stanciu, AMM 9/2012. Romanian Mathematical Society-Mehedinți Branch 2021 19 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 Compute ��� �→� ����� �� ��Г(� + �)� ����� ��� − �Г(� + �)� ����� � �� where � ∈ ℝ and Г is gamma function. Solution: ��� �→� �Г(���)� � � � = ��� �→� �∈ℕ∗ �Г(���)� � � � = ��� �→� √�! � � = ��� �→� � �! �� � = ����� ��� �→� � (���)! (���)��� ∙ �� �! � = = ��� �→� � � � + 1 � � = 1 � �(�) = ���� �� ��Г(� + 2)� ����� ��� − �Г(� + 1)� ����� � � = ���� ���Г(� + 1)� ����� � (�(�) − 1) �: ℝ� ∗ → ℝ, �(�) = � �Г(� + 2)� � ��� �Г(� + 1)� � � � ����� ; ��� �→� �(�) = ��� �→� � �Г(� + 2)� � ��� � + 1 ∙ � �Г(� + 1)� � � ∙ � + 1 � � ����� = � 1 � ∙ � ∙ 1� ����� = 1 ��� �→� �(�) − 1 ����(�) = 1. ��� �→� ��(�)� � = ��� �→� � �Г(� + 2)� � ��� �Г(� + 1)� � � � ������ = ��� �→� � Г(� + 2) Г(� + 1) ∙ 1 �Г(� + 2)� � ��� � ����� = = ��� �→� � � + 1 �Г(� + 2)� � ��� � ����� = ���� ��; ��� �→� �(�) = − ��� �→� ⎝ ⎜ ⎛ ���� �� � �Г(� + 1)� � � � ∙ �� ����� ∙ �(�) − 1 ����(�) ∙ ����(�) ⎠ ⎟ ⎞ = = − ��� �→� ⎝ ⎜ ⎛ � �Г(� + 1)� � � � � ����� ∙ ���� �������� ∙ �(�) − 1 ����(�) ∙ ����(�) ⎠ ⎟ ⎞ = = ���� �� ∙ lim �→� �(�) − 1 ����(�) ∙ log ���� �→� ��(�)� � � = ���� �� ∙ 1 ∙ ������� �� = ����� ∙ ���� ��; Problem 7. D.M. Bătineţu-Giurgiu, Neculai Stanciu, MP 2/2013. Romanian Mathematical Society-Mehedinți Branch 2021 20 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 If � ∈ ℝ and (��(�))��� is defined by ��(�) = � ����� �� �(� + �)! ��� � ����� − � √�! � � ����� � then compute ��� �→� ��(�). Solution: ��(�) = � ����� �� �(� + 1)! ��� � ����� − � √�! � � ����� � = = ���� ��� √�! � � ����� (�� − 1) = = � √�! � � � ����� ���� �������� ∙ �� − 1 ����� ∙ ����� � = � √�! � � � ����� � ∙ �� − 1 ����� ∙ ����� �, where �� = � �(���)! ��� √�! � � ����� and ��� �→� �� = ��� �→� � �(���)! � ��� ∙ � √�! � ∙ ��� � � ����� = � � � ∙ � ∙ 1� ����� = 1 ��� �→� �� − 1 ����� = 1 ��� �→� �� � = ��� �→� � �(� + 1)! ��� √�! � � ������ = ��� �→� � (� + 1)! �! ∙ 1 �(� + 1)! ��� � ����� = = ��� �→� � � + 1 �(� + 1)! ��� � ����� = ���� ��, We obtain ��� �→� ��(�) = � � � � ����� ∙ 1 ∙ ������� �� = ����� ���� �� . Observation: If � = � � , then ���� = 1, ���� = 0 so �� � � � � = �(� + 1)! ��� − √�! � ,i.e.we obtained the limit of Traian Lalescu: ��� �→� �� � � � � = ��� �→� �� = � � . Problem 8. D.M. Bătineţu-Giurgiu, Neculai Stanciu, RMM 2020. Let (��)���, �� ∈ ℝ� ∗ , ∀� ∈ ℕ∗ with ��� �→� (���� − ��) = � ∈ ℝ� ∗ . Compute ��� �→� ����� √� + � ��� − �� √� � �. Solution 1. �� = ���� √� + 1 ��� − �� √� � = (���� − ��) √� + 1 ��� + ��� √� + 1 ��� − √� � �; ∀� ∈ ℕ∗ − {1} ��� �→� �� = ��� �→� (���� − ��) ∙ ��� �→� √� + 1 ��� + ��� �→� �� � ∙ ��� �→� ��� √� + 1 ��� − √� � �� = = � ∙ 1 + ��� �→� ���� − �� (� + 1) − � ∙ ��� �→� � √� � (�� − 1) = � + � ��� �→� � �� − 1 ����� ∙ ����� �� Romanian Mathematical Society-Mehedinți Branch 2021 21 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 where �� = √��� ��� √� � , ∀� ∈ ℕ ∗ − {1}; ��� �→� �� = 1 then ��� �→� ���� ����� = 1; ��� �→� �� � = ��� �→� � � + 1 � ∙ 1 √� + 1 ��� � ; So, ��� �→� �� = � + � ∙ 1 ∙ 1 ∙ log � ��� �→� �� �� = � + � ∙ 1 ∙ 1 ∙ 0 = �. Solution 2. �� = ���� √� + 1 ��� − �� √� � = �� √� � (�� − 1), where �� = ���� �� ∙ √��� ��� √� � , ∀� ∈ ℕ∗ − {1} then ��� �→� �� = ��� �→� ���� �� ∙ ��� �→� √��� ��� √� � = ��� �→� ��������� ������� ∙ � � = � � ∙ 1 = 1 So, ��� �→� �� = 1, ��� �→� ���� ����� = 1 and ��� �→� �� � = ��� �→� � ���� �� � � ∙ ��� �→� � ��� � ∙ � √��� ��� � = = ��� �→� � ���� �� � � ∙ 1 ∙ 1 = ��� �→� ��1 + ���� − �� �� � �� ������� � (�������)∙ � �� = ��∙ � � = �. We obtain: ��� �→� �� = ��� �→� �� √� � � ∙ ��� �→� ���� ����� ∙ ��� � ��� �→� �� �� = ��� �→� �� � ∙ ��� �→� √� � ∙ 1 ∙ ���� = = � ∙ 1 ∙ ���� = �. Problem 9. D.M. Bătineţu-Giurgiu, Neculai Stanciu, RMM 1/2018. Let (��)��� be a positive real sequence such that ��� �→� (���� − ��) = � > � Compute ��� �→� � (���)���� �(����)!! ��� − ��� �(����)!! � �. Solution. Let (��)���, �� = (���)���� �(���)!! ��� − ��� �(����)!! � = ��� �(����)!! � (�� − 1) = = ��� �(����)!! � ∙ ���� ����� ∙ ����� �, ∀� ∈ ℕ∗ − {1} ,where �� = � + 1 � ∙ ���� �� ∙ �(2� − 1)‼ � �(2� + 1)‼ ��� = ���� � + 1 ∙ � �� ∙ (� + 1)� �� ∙ �(2� − 1)‼ � �(2� + 1)‼ ��� We have ��� �→� �� � = ��� �→� ������� (���)�� = � and ��� �→� �(����)‼ � � = ��� �→� � (����)‼ �� � = = ��� �→� � (2� + 1)‼ (� + 1)��� ∙ �� (2� − 1)‼ � = ��� �→� 2� + 1 � + 1 ∙ � � � + 1 � � = 2 � . So, ��� �→� �� = � ∙ � � ∙ 1 ∙ � � ∙ � � = 1, ��� �→� ���� ����� = 1; ��� �→��� � = ��� �→� � � + 1 � � � ∙ ��� �→� (2� − 1)‼ (2� + 1)‼ ∙ �(2� + 1)‼ ��� = = � ∙ ��� �→� ��1 + ���� − �� �� � �� ������� � (�������)∙ � �� ∙ ��� �→� �(2� + 1)‼ ��� 2� + 1 = Romanian Mathematical Society-Mehedinți Branch 2021 22 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 = � ∙ � ∙ ��� �→� � (2� − 1)‼ (2� − 1)� ��� = �� ∙ ��� �→� � (2� + 1)‼ (2� + 1)��� ∙ (2� − 1)� (2� − 1)‼ � = �� ∙ ��� �→� � 2� − 1 2� + 1 � � = = �� ∙ ��� = �. Then ��� �→� �� = ��� �→� �� �(����)‼ � ∙ ��� �→� ���� ����� ∙ ��� � ��� �→� �� �� = ��� �→� �� � ∙ ��� �→� � �(����)‼ � ∙ ���� = = � ∙ � 2 ∙ 1 = �� 2 . Problem 10. D.M. Bătineţu-Giurgiu, Neculai Stanciu, AMM 4/2014. Compute ��� �→� � �(�� − �)‼ � ���� � �(���)! ��� � √�! � − ���. Solution. ��� �→� √�! � � = �; ��� �→� �(2� − 1)‼ � � = ��� �→� � (2� − 1)‼ �� � = ����� ��� �→� (2� + 1)‼ (� + 1)��� ∙ �� (2� − 1)‼ = ��� �→� 2� + 1 (� + 1)�� = 2 � . where we denote �� = �1 + � � � � , ∀� ∈ ℕ∗ , �� = �(� + 1)! ��� √�! � , ∀� ≥ 2; ����→� �� = ��� �→� � �(� + 1)! ��� � + 1 ∙ � √�! � ∙ � + 1 � � = 1; ��� �→� �� − 1 ����� = 1; ��� �→� �� � = ��� �→� (� + 1)! �! ∙ 1 �(� + 1)! ��� = ��� �→� � + 1 �(� + 1)! ��� = �. We denote �� = � � ��, ∀� ≥ 1 and we have ��� �→� � �(2� − 1)‼ � ���� � �(���)! ��� � √�! � − 1�� = = ��� �→� �(2� − 1)‼ ��� � ∙ ��� �→� � ������ − ��� � 4 � = 2 � ∙ ��� �→� sin ��� − � � � �������� � � ∙ � = = 2 � ∙ ��� �→� � sin ��� − � � � �� − � � ∙ � ��� − � 4 �� ∙ 1 ���� � � = 4 � ∙ 1 ∙ ��� �→� � ��� − � 4 � = = 4 � ∙ ��� �→� � ∙ � 4 ∙ (�� − 1) = 4 � ∙ � 4 ∙ ��� �→� � � �(� + 1)! ��� √�! � − 1� = = � � ∙ ��� �→� � √�! � ∙ ����→� � �(� + 1)! ��� − √�! � � = � � ∙ � ∙ ��� �→� √�! � (�� − 1) = = � ∙ ��� �→� √�! � � ∙ ��� �→� �(�� − 1) = � � ∙ ��� �→� � �� − 1 ����� ∙ ����� �� = � � ∙ 1 ∙ ���� = � � . Problem 11. D.M. Bătineţu-Giurgiu, Neculai Stanciu, SSM 5/2014. Romanian Mathematical Society-Mehedinți Branch 2021 23 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 Let � ∈ ℝ� ∗ and (��)��� is defined by �� = ∑ � �! � ��� . Compute ��� �→� √�! � �� ��� � �� − ��. Solution. �� = √�! � �� ��� � �� − 1� = √�! � � ∙ � ∙ �� ��� � �� − 1� = = √�! � � ∙ � ∙ � ��� � − 1� ∙ � ��� � �� − 1 ��� � − 1 = √�! � � ∙ � ��� � �� ��� � − 1 ∙ ��� � − 1 � � ����� ∙ �����, ∀� ≥ 2. ��� �→� √�! � � = 1 � ; ��� �→� � ��� � �� ��� � − 1 = ���� ; ��� �→� ��� � − 1 � � ����� = 1 and ��� �→� �� = � We obtain ��� �→� �� = � � ∙ ���� ∙ 1 ∙ ���� = ���� � . Problem 12. D.M. Bătineţu-Giurgiu, Neculai Stanciu, RMM 2020. Let (��)���, �� = −���� + ∑ � � � ��� , with ��� �→� �� = � (�. �. � is Euler Mascheroni constant). Compute ��� �→� (����� − ����) √�! � . Solution. �� = (����� − ����) √�! � = 2 ∙ √�! � ∙ ��� ���� � ∙ ��� ���� � = = √�! � � ∙ ��� �� + � 2 ∙ ��� ���� � ���� � ∙ � ∙ �� − � 2 , ∀� ≥ 2; ��� �→� √�! � � = 1 � ; ��� �→� ��� �� + � 2 = ���� ��� �→� ��� ���� � ���� � = 1 and ��� �→� �(�� − �) = ��� �→� �� − � � � = ��� ��� �→� ���� − �� � ��� − � � = = ��� �→� �� − ���� � � − � ��� = ��� �→� − � ��� + ��� ��� � � �(���) = ��� �→� �� ���� � + 1 � − 1 � + 1 � = = ��� �→� �� �log(� + 1) − ���� − 1 � + 1 � = lim �→� ��� log �1 + � � � + ���� − � ��� �� = = lim �→� ��� log(1 + �) − � ��� �� = ��� lim �→� ��� � ��� − � (���)� 2� = lim �→� ��� � + 1 − 1 2�(� + 1)� = 1 2 ⇒ ��� �→� �� = 1 � ∙ 1 2 ∙ ���� = ���� 2� . Problem 13. D.M. Bătineţu-Giurgiu, Neculai Stanciu, REOIM 2013. Let �, � ∈ ℝ and (��(�, �))��� be a sequence defined by ��(�, �) = (� + �) � ∙ �((� + �)!)� ��� − �� ∙ �(�!)� � . Evaluate ��� �→� ��(�, �) = �(�, �) Solution. lim �→� √�! � � = �; ��(�, �) = � � ∙ � √�! � � � ∙ (�� − 1) = � � ∙ � √�! � � � ∙ ���� ����� ∙ ����� = Romanian Mathematical Society-Mehedinți Branch 2021 24 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 = ������ ∙ � √�! � � � � ∙ ���� ����� ∙ ����� �, ∀� ∈ ℕ∗ − {1} where �� = � ��� � � � ∙ � �(���)! ��� √�! � �, ∀� ∈ ℕ∗ − {1}; ��� �→� �� = 1; ��� �→� �� − 1 ����� = 1; ��� �→� �� � = ��� �→� � � + 1 � � �� ∙ ��� �→� � (� + 1)! �! ∙ 1 �(� + 1)! ��� � � = �� ∙ ��� �→� � � + 1 �(� + 1)! ��� � � = ����. So, �(�, �) = ��� �→� ��(�, �) = � ∙ log � ��� �→� �� �� ∙ ��� �→� ������ = ������� ∙ ��� ∙ ��� �→� ������ = = (� + �) ∙ 1 �� = � 0, if � + � < 1 ���, if � + � = 1 ∞, if � + � > 1 Problem 14. D.M. Bătineţu-Giurgiu, Neculai Stanciu, SM 1/2015. If �� = ∑ � �! � ��� compute ��� �→� (� − ��) ∙ (� + �)!. Solution. ��� �→� (� − ��) ∙ (� + 1)! = ��� ��� �→� (������)�(����) � (���)! � � (���)! = ��� �→� ������� ��� (���)! = ��� �→� � (���)! � (���)! = = ��� �→� � + 2 � + 1 = 1 Problem 15. D.M. Bătineţu-Giurgiu, Neculai Stanciu, P 2/2014. Compute ��� �→� √� ∙ � �(� + �)! �(���) − √�! �� � Solution 1. ��� �→� � √�! � = ��� �→� � �� �! � = ����� ��� �→� � (���)��� (���)! ∙ �! �� � = �. �� = √� ∙ � �(� + 1)! �(���) − √�! �� � = √� ∙ √�! �� ∙ (�� − 1) = √�! �� √� ∙ (�� − 1) ∙ � = = � √�! � � ∙ �� − 1 ����� ∙ ����� �, ∀� ≥ 2, �ℎ��� �� = �(� + 1)! �(���) √�! �� = �(� + 1)! �(���) √� + 1 ∙ √� √�! �� ∙ � � + 1 � So, ��� �→� �� = √� ∙ � � � ∙ 1 = 1; ��� �→� ���� ����� = 1. ��� �→� �� � = ��� �→� �� �(� + 1)! ��� √�! � � � = � ��� �→� (� + 1)! �! ∙ 1 �(� + 1)! ��� ∙ � ��� �→� � + 1 �(� + 1)! ��� = √�. Romanian Mathematical Society-Mehedinți Branch 2021 25 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 We obtain ��� �→� �� = � � � ∙ 1 ∙ log � ��� �→� �� �� = � √� ∙ ���√� = � �√� . Solution 2. �� = √� ∙ � �(� + 1)! �(���) − √�! �� � = √� ∙ �(���)! ��� � √�! � �(���)! �(��!) � √�! �� = �(���)! ��� � √�! � � �(���)! ��� ��� ∙ ��� � �� √�! � � , So, ��� �→� �� = ��� �→� �(���)! ��� � √�! � � �(���)! ��� ��� ∙ ��� � �� √�! � � = � � ∙ � � � � �� � � = � �√� . Solution 3. �� = √� ∙ � �(� + 1)! �(���) − √�! �� � = √� ∙ �(���)! �(���) � √�! �� ����� = √�! �� � ∙ ���� ���� ∙ √� = = � √�! � � ∙ �� − 1 �� − 1 = � √�! � � ∙ �� − 1 ����� ∙ ����� �� − 1 ∙ ����� ����� = � √�! � � ∙ �� − 1 ����� ∙ ����� �� − 1 ∙ ����� � ����� �, where �� = �(���)! �(���) √�! �� , �� = 1 + � � , so ��� �→� �� = ��� �→� �� = 1, then ��� �→� ���� ����� = ��� �→� ���� ����� = 1. We have ��� �→� �� � = √�, ��� �→� �� � = ��� �→� �1 + � � � � = �, �� ��� �→� �� = � √� ∙ 1 ∙ 1 ∙ ���√� ���� = � �√� . REFERENCES [1] D. M. Bătineţu-Giurgiu, Sequences, Albatros Publishing House, Bucharest, 1979. [2] D. M. Bătineţu-Giurgiu, Lalescu sequences, RMT, Vol. 20, No. 1-2, 1989, pp. 33-36. [3] D. M. Bătineţu-Giurgiu, Lalescu sequences and gamma function. Euler-Lalescu function, Romanian Mathematical Gazette - A, Vol. 11, No. 1, 1990, pp. 21-26. [4] D.M. Bătineţu-Giurgiu, The generalizations of Traian Lalescu sequence, Romanian Mathematical Gazette, No. 8-9/1990, pp. 219-224. [5] D.M. Bătineţu-Giurgiu, D. Sitaru, N. Stanciu, New classes of sequences-functions and famous limits, Octogon Mathematical Magazine, Vol. 27, No.2, October, 2019, 784-797. [6] D.M. Bătineţu-Giurgiu, D. Sitaru, N. Stanciu, Two classes of Lalescu’s sequences, Octogon Mathematical Magazine, Vol. 27, No. 2, October, 2019, 805-813. [7] D.M. Bătineţu-Giurgiu, M. Bencze, D. Sitaru, N. Stanciu, Some limits of Traian Lalescu type with Fibonacci and Lucas numbers (II), Octogon Mathematical Magazine, Vol. 27, No. 1, April, 2019, 101-111. [8] D.M. Bătineţu-Giurgiu, M. Bencze, D. Sitaru, N. Stanciu, Some limits of Traian Lalescu type (II), Octogon Mathematical Magazine, Vol. 27, No. 1, April, 2019, 184-209. [9] D.M. Bătineţu-Giurgiu, M. Bencze, D. Sitaru, N. Stanciu, Some limits of definite integrals of Lalescu’s type, Octogon MathematicalMagazine, Vol. 27, No. 1, April, 2019, 312-328. [10] D.M. Bătineţu-Giurgiu, M. Bencze, D. Sitaru, N. Stanciu, Some limits of Traian Lalescu type (II), Octogon Mathematical Magazine, Vol. 26, No. 2, October, 2018, 664-690. [11] D.M. Bătineţu-Giurgiu, M. Bencze, N. Stanciu, Some limits of Traian Lalescu type with Fibonacci and Lucas numbers, Octogon Mathematical Magazine, Vol. 26, No. 1, April, 2018, 54-87. [12] D.M. Bătineţu-Giurgiu, M. Bencze, N. Stanciu, Some limits of Traian Lalescu type (I), Octogon Mathematical Magazine, Vol. 26, No. 1, April, 2018, 116-136. Romanian Mathematical Society-Mehedinți Branch 2021 26 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 [13] D.M. Bătineţu-Giurgiu, M. Bencze, D. Sitaru, N. Stanciu, Certain classes of Lalescu sequence, Octogon Mathematical Magazine, Vol. 26, No. 1, April, 2018, 243-251. [14] D.M. Bătineţu-Giurgiu, N. Stanciu, Several results of some classes of sequences, The Pentagon , Volume 73, Number 2, Spring 2014, 10-24. [15] D.M. Bătineţu-Giurgiu, A. Kotronis, N. Stanciu, Calculating the limits of some real sequences, Math Problems, Issue 1, 2014, 252-257. [16] D.M. Bătineţu-Giurgiu, N. Stanciu, New methods for calculations of some limits, The Teaching of Mathematics, Vol. XVI, Nr. 2, 2013, 82-88. SOME APPLICATIONS OF THE ROLLE SEQUENCES By Marian Ursărescu-Romania In this article we want to highlight some types of problems that are elegantly solved with to the Rolle sequences. I.1.a) For any � ∈ ℕ∗, let the equation �� + ���� = 0. Prove that the equation admits a single root �� ∈ ℝ. b) Study the convergence of the sequences (��)���. Proof: a) Let be the function ��: (0, +∞) → ℝ, ��(�) = � � + ���� derivable with �� �(�) = ����� + 1 � > 0, ∀� > 0 � 0 + ∞ ��′(�) | + + + + + + + + + + + + + + + + ��(�) |−∞ ↗ ↗ ↗ lim�→� ��(�) = +∞; lim�→�� ��(�) = −∞, from the Rolle sequences ∃! �� > 0. b) ��(1) = 1 > 0 ⇒ ∃ �� ∈ (0,1), ∀� ∈ ℕ ∗ ⇒ (��)��� is limited. ��(��) = 0 ⇒ �� � + ����� = 0 (1) ����(����) = 0 ⇒ ���� ��� + ������� = 0 (2) ��(����) = ���� � − ������� = (�) ���� � − ���� ��� = ���� � (1 − ����) > 0 � ��(����) > 0 ��(��) = 0 ⇒ ��(����) > ��(��) ⇒ ���� > �� ⇒ (��)��� − increasing. So, (��)��� is convergent. Proposed problems: Romanian Mathematical Society-Mehedinți Branch 2021 27 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 1.a) For any � ∈ ℕ∗, let the equation �� + ������ = 1. Prove that the equation admits a single root �� ∈ ℝ. b) Study the convergence of the sequences (��)���. 2.a) For any � ∈ ℕ∗, let the equation �� − ���� − ⋯ − � − 1 = 0. Prove that the equation admits a single root �� ∈ ℝ. b) Study the convergence of the sequences (��)���. c) Find: lim�→� �� II.1.a) Show that it exists only one function �: ℝ → ℝ such that ��(�) + 3�(�) + � = 0, ∀� ∈ ℝ b) Prove to the function � is derivable on ℝ. Proof: a) Let �(�) = � ⇒ �� + 3� + � = 0 and let be the function �: ℝ → ℝ, �(�) = �� + 3� + � continuous and derivable with ��(�) = 3�� + 3 > 0, ∀� ∈ ℝ � −∞ + ∞ ��(�) + + + + + + + + + + + + + + + + �(�) −∞ ↗ ↗ ↗ +∞ lim �→� �(�) = ∞, lim �→�� �(�) = −∞ ⇒ ∃! � = �(�). b) To begin with, let’s show that the function � is continuous on ℝ. ⇔ ∀�� ∈ ℝ, lim �→�� �(�) = �(��) � ��(�) + 3�(�) + � = 0 ��(��) + 3�(��) + �� = 0 ⇒ ��(�) − ��(��) + 3��(�) − �(��)� + � − �� = 0 ��(�) − �(��)�(� �(�) + �(�)�(��) + � �(��) + 3) = −(� − ��) ⇒ �(�) − �(��) = −(� − ��) ��(�) + �(�)�(��) + ��(��) + 3 (1) ��(�) + �(�)�(��) + � �(��) + 3 = ��(�) + � � �(��)� � + � � ��(��) + 3 ≥ � � ��(��) + 3 (2) From (1),(2) we get: |�(�) − �(��)| ≤ |����| � � ��(��)�� → 0 by the criterion of the edge, lim�→�� �(�) = �(��) then � is continuous in any point �� ∈ ℝ. ��(��) = lim �→�� �(�) − �(��) � − �� (3) Romanian Mathematical Society-Mehedinți Branch 2021 28 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 From (1) we get: �(�)��(��) ���� = − � ��(�)��(�)�(��)���(��)�� then lim�→�� �(�)��(��) ���� = − lim�→�� � ��(�)��(�)�(��)���(��)�� , � −continuous in any point �� ��(��) = �� �(��(��)��) ∈ ℝ then � −is derivable in any point �� ∈ ℝ. Proposed problems: 1.a) Show that it exists only one function �: ℝ → ℝ such that �� − ��(�) + ��(�) = 0, ∀� ∈ ℝ b) Prove to the function � is derivable on ℝ∗. 2.a) Show that it exists only one function �: ℝ → ℝ such that ��(�) + ���(�) − �� = 0, ∀� ∈ ℝ b) Prove to the function � is continuous and derivable on ℝ. III. Let be the equation: �� − �� − � − 1 = 0. a) Prove to the equation admits only real root, noted ��. b) Find: lim�→�(�� � + �� �), where ��, �� −are the complex roots of the equation. Proof:a) Let be the function �: ℝ → ℝ, �(�) = �� − �� − � − 1 continuous and derivable with ��(�) = 3�� − 2� − 1 ��(�) = 0 ⇔ �� = − 1 3 , �� = 1 � −∞ − 1 3 1 + ∞ ��(�) 0 0 �(�) −∞ − −2 + ∞ lim�→� �(�) = ∞, lim�→�� �(�) = −∞ by the Rolle sequences ∃! �� > 1. b) If ��, �� ∈ ℂ ⇒ �� = �����. Let �� = �(���� + �����) ⇒ �� = ����� = �(���� − �����), where � = |��| = |��|. lim �→� (�� � + �� �) = lim �→� ����cos(��) + ����(��)� + ���cos(��) − ����(��)�� = lim �→� 2�� cos(��) (1) By Vieta’s relations ������ = 1 ⇒ |��||��||��| = 1 ⇒ � � = � �� < 1 ⇒ � < 1 (2) |�� cos(��)| = ��|cos(��)| ≤ �� → 0 (3) From (1),(2),(3) we have: lim�→�(�� � + �� �) = 0 Proposed problem: Let be the equation: �� − � − 1 = 0. Romanian Mathematical Society-Mehedinți Branch 2021 29 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 a) Prove to the equation admits only real root, noted ��. b) Find: lim�→�(�� � + �� �), where ��, �� −are the complex roots of the equation. IV. If �, �, � ∈ ℝ are such that � + � + � = 2 and �� + �� + �� = 1. Prove that ��� ∈ �0, � �� � Proof: The equation of degree (III) who admits the roots �, �, � is �� − ��� � + ��� − �� = 0, where � �� = � + � + � = 2 �� = �� + �� + �� = 1 �� = ��� ⇒ �� − 2�� + � − �� = 0. Because �, �, � ∈ ℝ we must show that all roots are real numbers. Let be the function �: ℝ → ℝ, �(�) = �� − 2�� + � − �� continuous and derivable on ℝ with ��(�) = 3�� − 4� + 1 = 0 ⇔ �� = 1 3 , �� = 1 � −∞ 1 3 1 + ∞ ��(�) 0 0 �(�) −∞ 4 27 − �� − �� + ∞ lim�→� �(�) = +∞; lim�→�� �(�) = −∞ by the Rolle sequences the equation admits 3 real roots, then � �� − �� ≥ 0 and −�� ≤ 0 ⇒ �� ∈ �0, � �� � ⇒ ��� ∈ �0, � �� � THREE FUNDAMENTAL THEOREMS FROM NUMBERS THEORY By Angela Niţoiu-Romania Euler’s theorem: If � ∈ �, � ∈ �, � ≥ 2 and (�, �) = 1 then ��(�) ≡ 1(mod �), where �: �∗ → �∗, �(�) = ����({� ∈ �∗|1 ≤ � ≤ � and (�, �) = 1 }) (that is,the number of natural numbers smaller or equal with � and prime with �) is the arithmetic function by Euler. Fermat Little theorem: If � ∈ �∗is an prime number and � ∈ � an integer number, then �� ≡ �(mod �). Wilson theorem: If � ≥ 2 it is an natural number,then the following affirmations are equivalent: a) p it is an prime number; b) (� − 1)! + 1 ≡ 0(mod �). Romanian Mathematical Society-Mehedinți Branch 2021 30 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 Aplications: Problem 1: Let p be an prime number. Let it look like the equation �� ≡ −1(mod �) it has a solution if and only if � = 2 or � ≡ 1(mod 4). Solution: From Wilson theorem, � prime number⇒ (� − 1)! ≡ −1(mod �). If � > 2 then p it is odd, ��� � it is integer and we have: (� − 1)! = �1 ∙ 2 ∙ … ∙ � − 1 2 � � � + 1 2 … (� − 1)� = � �(� − �) ��� ���� ≡ � �(−�) ��� � ��� (mod �). So, (� − 1)! ≡ (−1) ��� � �� and there it is the relationship (−1) ��� � �� ≡ −1(mod �). For � ≡ 1(mod 4), ��� � even number, so the relationship becomes: �� ≡ −1(mod �). If � ≡ 3(mod 4) and suppose it exists � with �� ≡ −1(mod �), we have: ���� ≡ (��) ��� � ≡ (−1) ��� � ≡ −1(mod �), But, how v���� ≡ 0(mod �) ⇒ � do not divide �, so,from Fermat little theorem, we get ���� ≡ 1(mod �) and we come to a contradiction: 1 ≡ (−1)(mod �), because p > 2. For � = 2 we have 1� ≡ 1 ≡ (−1)(mod 2), so the equation �� = −1 it has a solution. Problem 2: Let p be an prime number and k an natural number with the condition 1 ≤ � ≤ �. Prove that number (� − �)! (� − 1)! + (−1)��� it is divisible by p. ( A. Simionov) Solution: We have the congruences modulo p: 1 ≡ −(� − 1) 2 ≡ −(� − 2) …………………….. � − 1 ≡ −(� − � + 1) Which multiplies it gives: (� − 1)! ≡ (−1)���(� − 1)(� − 2) … (� − � + 1) . So, (� − �)! (� − 1)! ≡ (−1)���(� − 1)! ≡ (−1)� (from Wilson theorem). We get, obviously, to (� − �)! (� − 1)!−(−1)� ≡ 0(mod �), where, obviously, (� − �)! (� − 1)! + (−1)��� it is divisible by p. Problem 3: Prove that if p is an prime number and a is an integer numbers, then p it is divisible by (� − 1)! �� + �. Solution: From Fermat and Wilson theorems we can written: � ∣ �(� − 1)! + 1�, � ∣ (�� − �) , where � ∣ ����(� − 1)! + 1� − (�� − �)� or � ∣ (��(� − 1)! + �). Romanian Mathematical Society-Mehedinți Branch 2021 31 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 Problem 4: Let p be an prime number and a, b integer numbers. Prove that if �|(�� − ��) then ��|(�� − ��). Solution: From Fermat theorem ⇒ �� ≡ �(mod �), �� ≡ �(mod �), so � ≡ �(mod �). We can written � = � + �� and we get: �� − �� = (� + �� )� − �� = �� ������� + �� ��������� + ⋯ + �� � ���� = �������� + �� ������� + ⋯ + ������� where � ≥ 2 . So, �� − �� it is divisible with ��. Problem 5: Prove that for any natural numbers a,b relative prime, the number ���(�) + ��(�) − 1� it is divisible with the product ab. Solution: From Euler theorem,we get: ��(�) ≡ 1(mod �) and ��(�) ≡ 1(mod �). So, ��(�) = 1 + �� and ��(�) = 1 + ��, with �, � ∈ �. Then ���(�) − 1����(�) − 1� = (��)(��), so ���(�)��(�) − ��(�) − ��(�) + 1� it is divisible with ab and how ��(�)��(�) it is divisible with ab,result the conclusion. Problem 6: Prove the following characterization of the prime numbers: An natural number � ≥ 2 it is prime if and only if �(�)|(� − 1) and (� + 1)|�(�), where � is the Euler function and � is the function sum of divisors. Solution: " ⇒ " If p is an prime number, then �(�) = � − 1 and �(�) = � + 1. " ⇐ " Let be � ∈ �∗, with the property � ≥ 2, �(�)|(� − 1) and (� + 1)|�(�). We must to abserve that for any number � > 2, the number �(�) it is even number: from the expression by �(�), if n contains in the prime factorization ,prime numbers p different two by two 2 (so odd), then �(�) = (�� − ����)� = ����(� − 1)� , which is even number; if � = 2�, � > 2 then �(�) = 2� − 2��� = 2��� which it is even number. From the relationship �(�)|(� − 1) we get that n must be even number(contains in the prime factorization only powers of odd numbers). We show that all the factors from the factorization it is prime number (without exponent). If, through the absurd, it would exists �� �� therms by n, with �� ≥ 2 then ��� �� − �� �����|�(�)|(� − 1) so, �� ����|(� − 1) = �� ���� �� … �� �� − 1, which it is false. So, � = ���� … ��. Then, �(�) = (�� − 1)(�� − 1) … (�� − 1) and �(�) = (�� + 1)(�� + 1) … (�� + 1), with ��, ��, … , �� odd numbers and then result 2�|�(�) and 2�|�(�). Romanian Mathematical Society-Mehedinți Branch 2021 32 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 If � ≥ 2 then 4|�(�)|(� − 1), so 4�|(� + 1). From hypothesis, the numbers �(�) ��� it is integer, n+1 it is even numbers, not divisible with 4, and �(�) it is divisible with 2�. Then 2��� � �(�) ��� and result 2��� < �(�) � = (�� + 1)(�� + 1) … (�� + 1) ���� … �� = �1 + 1 �� � … �1 + 1 �� � < �1 + 1 3 � � = � 4 3 � � , And the relationship it is false; so � = 1, therefore � = �� it is prime numbers. References: 1. Romanian Mathematical Magazine-www.ssmrmh.ro 2. Vasile Pop, Viorel Lupşor- Matematică pentru grupele de performanţă, clasa a XII-a, Ed. Dacia Educaţional, Cluj- Napoca, 2004 3. P. Radovici- Probleme de teoria elementară a numerelor, Ed. Tehnică, Bucureşti, 1986 ABOUT NAGEL AND GERGONNE’S CEVIANS (III) By Bogdan Fuştei-Romania Let’s consider the triangle ABC with standard notations; From previous parts: ��� � = �� � + �� � + �����(��� �������); ��� � = ����� + (� − �) �(��� �������); ���� = �(� − �)(��� �������); �� + �� = �� � + �� � + ���� (��� �������); �� � = ���� + (� − �)� � � (��� �������); Romanian Mathematical Society-Mehedinți Branch 2021 33 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 �� = ���� + ��� (��� �������); ����� + (� − �) � = ���� + (� − �)� � � + �� � + ����(��� �������); It’s easy to prove that: �� � = �(� − �) + (� − �)� � (� − �) = (� − �) �� − (� − �)� � � (��� �������); We was show it that : �� � = � �� − � + (���)� � � (and analogs); �� � �� � = �(� − �) �� − (� − �)� � � �� − � + (� − �)� � � ≥ �(� − �)�(� − �) �(� − �) − (� − �) (� − �)� � + (� − �)� � � − (� − �)� � (� − �)� � ≥ �(� − �) (� − �)� � (� − � + �) ≥ (� − �)� � (� − �)� � (� − �)� � � ≥ (� − �)� � (� − �)� � (� − �)� ≥ (� − �)� � (� − �)� � If b=c then we get equality in up relationship; If b≠ � ⇒ 1 > (���)� �� ⇒ �� > (� − �)� ⇒ � > � − � or � > � − � ;which means the triangle inequality; Finally,we have the inequality: ���� ≥ ����(��� �������)(�);⇒����� + �� � + �� � ≥ ��� � So, we get: �� + �� ≥ ���(��� �������);(2) From up relationships, it’s easy to see that: �� + �� = �� � + �� � + ���� ≥ ����� + ���� ≥ ����� + ����; �� + �� ≥ �(���� + ���) ≥ ���(��� �������);(3) ����� = ��(�� + ��)(��� �������); ����� ≥ ��(�� + ��) ���� �� ≥ �� + �� � (��� �������); (�) Adding up relationships, we get: ∑ ���� �� ≥ ��+��+�� = �� + �;(5) and �� �� ≥ ����� ��� (��� �������);(6) Again, adding (5),(6) we get: ∑ �� �� ≥ � � ∑ ����� �� (7); ��� � � =� �(���) �� (and analogs); Romanian Mathematical Society-Mehedinți Branch 2021 34 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 �� = 2�� � + � ��� � 2 (and analogs); From: ���� ≥ �(� − �) = ���� = ��� � ��� � � �� ⇒ ���� �� ≥ ��� � ��� � � (��� �������);(8) We know that: �� ≥ ��� � ��� � � (and analogs) then, we get ∶ ���� ≥ �(� − �) = ����; So, we can write : ��������� ≥ ����(and analogs)(9); ∑ ���� = � � ; So, we have: ∑ ��������� ≥ � � = �� � + �����(��� �������);(10) From the abrove relationships, we have: ��������� �� ≥ ����� � (��� �������);(11) Adding up relationships, we have: ∑ ��������� �� ≥ ��+��+�� = �� + � ; (12) ��������� ����� ≥ �� � (and analogs); again, adding, we have: ∑ ��������� ����� ≥ � � (�� + �� + ��);(13) From: ������ ≥ ������ = � ��; (and analogs) ⇒ ������� ≥ �√�(��� �������);(14) ∑ ������� ≥ ��√�;(15) From: �� ≥ ��� � ��� � � and ���� �� ≥ ��� � ��� � � we get: � ������ �� ≥ � + � � ��� � � (��� �������)(��); Adding, we get: � � ������ �� ≥ � � + � � ��� � � (��) ������ � ≥ ��(��� �������) �� � = �� ����� (and analogs)so, we get: ������ ����� ≥ ��=> ���� ����� ≥ �� �� (and analogs) � �� + � �� + � �� = � � ; Romanian Mathematical Society-Mehedinți Branch 2021 35 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 So, we will get a newralationship: � ���� �� − �� ≥ �� � (��) �� = �� � + �����(��� �������) So, we will get a new relationship: ���� ����� ≥ �� � �� + ���(19) Adding up relationships, we get: � ���� �� − �� ≥ �(�� + �� + ��) + � �� � �� = �(�� + �) + � �� � �� (��) �� � �� �� − �� − �� �� � ≥ ��� => �� �� − �� − �� �� ≥ ��� �� (��� �������)(��) Adding, we get: � �� �� − �� ≥ � �� �� + � � �� �� (��) ������ (�� − ��)(�� − ��)(�� − ��) ≥ � � �� �� + ��� �� � (��) We know that: AI=�2�(ℎ� − 2�)= � ��� � � (and analogs); sin � � =� ��� �� (and analogs) ������� = � �; ��� = 4�� So, we have: (ℎ� − 2�)(ℎ� − 2�)(ℎ� − 2�) = 2�� � Then, the inequality (23) becomes: � �� ������ �� ≥ ∏ � �� �� + ��� �� � (��) From �� ����� − �� �� ≥ ��� �� (and analogs)we get the relationships: � � �� �� − �� − ��� �� � ≥ ������ ������ (��) � � �� �� − �� − �� �� � ≥ ������� ������ (��) From ���� ≥ ����(and analogs) => ���� �� � ≥ ���� �� � (and analogs); Romanian Mathematical Society-Mehedinți Branch 2021 36 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 �� � = 4�� (� + �)� ���� ; so, we will get the next relationships: ���� ��� = (� + �)� 4�� = 1 2 + 1 4 � � � + � � � ; Adding up relationships, we get: ∑ ���� �� � = � � + � � ∑ ��� � , so, we will write that: � ���� ��� ≥ � � + � � � � + � � (��) But: ��� � = ���� ���� (and analogs); so, we get: � ���� ��� ≥ � � + � � � �� + � �� − � (��) 2S=ℎ�� = ℎ�� = ℎ�� => � � = �� �� (and analogs) ⇒ � � + � � = �� �� + �� �� ⇒ ∑ � � � + � � � = ∑ � �� �� + �� �� � � � + � � = � ℎ� + ℎ� ℎ� , so, we get a new relationship: ∑ ���� �� � ≥ � � + � � ∑ ����� �� (29), but �� � = 1 + ��� � (and analogs), adding, we get: ℎ� + ℎ� + ℎ� � = 3 + � � + � � , following simple calculations, we have: � ���� ��� ≥ �� + �� + �� + �� �� (��) We know that : �� = 2�� �� + �� ��(and analogs) ⇒ �� �� = 1 2 � � � + � � � ⇒ ���� ��� = 1 2 (1 + �� �� )(and analogs) � ���� ��� = 1 2 (3 + � �� �� ) So, we get a new relationship: � ���� ��� ≥ � � (� + � �� �� ) (��) We know that: ��� � = ����� �� = 1 + �� �� (and analogs) ⇒ ∑ ��� � = 3 + ∑ �� �� From simple calculations, we have: Romanian Mathematical Society-Mehedinți Branch 2021 37 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 � ���� ��� = � � + � � � �� �� � ���� ��� ≥ � � + � � � �� �� From (� + � + �) � � � + � � + � � � = 3 + ��� � + ��� � + ��� � we have: � ���� ��� = 1 4 �3 + (� + � + �) � 1 � + 1 � + 1 � �� So, we get a new inequality: � ���� ��� ≥ � ���� ��� = � � �� + (� + � + �) � � � + � � + � � �� (��) (� + � + �) � 1 � + 1 � + 1 � � = 3 + � � + � � ; (ℎ� + ℎ� + ℎ�) � 1 ℎ� + 1 ℎ� + 1 ℎ� � = 3 + � ℎ� + ℎ� ℎ� But: ∑ ��� � = ∑ ����� �� then, we get a new inequality: � ���� ��� ≥ � � �� + (�� + �� + ��) � � �� + � �� + � �� �� ; (��) From: �� = 2�ℎ�(and analogs); �� + �� + �� = � � + �� + 4��, we have: ℎ� + ℎ� + ℎ� = �� + �� + 4�� 2� , then 3 + ℎ� + ℎ� + ℎ� � = �� + �� + 10�� 2�� So, we have a new inequality: � ���� ��� ≥ �� + �� + ���� ��� = � � + �� + �� ��� (��) If, we applying Gerretsen inequality: �� ≥ 16�� − 5��, we get: � ���� ��� ≥ �� + �� + 10�� 2�� = 5 4 + �� + �� 8�� ≥ 5 4 + 16�� − 5�� + �� 8�� = 5 4 + 2 − 4�� 8�� Finally, we get a new inequality: � ���� ��� ≥ �� � − � �� (��) We will improve this relationship. Let’s consider Yang inequality: �� ≥ 16�� − 5�� + ��(� − 2�) � − � (��� ������ ������������ �������� � − ���) Romanian Mathematical Society-Mehedinți Branch 2021 38 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 Following simple calculations, we get a new ralationship: � ���� ��� ≥ �� � − �(�� − ��) ��(� − �) ; (��) We know the following relationship: �� = �2�(ℎ� − 2�) ; (and analogs) �� = � sin � � (and analogs); ���� � 2 = � 2� �� ℎ� = �� − � 4� (and analogs) ��� ���� � � = 2�(ℎ� − 2�)4� �� − � = 8�� ℎ� − 2� �� − � ; (and analogs) 1 ���� � � = 8 � � � � � ℎ� − 2� �� − � (and analogs); 1 ���� � � = 4 � � � � � � ℎ� �� � � (and analogs); So, we have the following relationships: �� �� = √2� ����� ���� (and analogs); � ���� ��� ≥ 9 4 + 1 4 � ℎ� �� ; And from up relationships, we get: � ���� ��� ≥ � � + � �√� � � �� − �� �� − � (��) (� − �)� = �� � + �� � + 2��� − 2��� − 2���� (� − �)� = �� � + �� � − 2����(and analogs); (�� − ��) � = �� � + �� � − 2���� (and analogs); From up relationships, we get: |� − �| ≥ �� − ��(��� �������)(��) Adding up relationships, we get: |� − �| + |� − �|+|� − �| ≥ (�� + �� + ��) − (�� + �� + ��)(��) �� + �� ≥ 2�� ⇒ �� − �� ≥ 2(�� − ��), So, we have the following relationships: Romanian Mathematical Society-Mehedinți Branch 2021 39 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 � � |� − �| ≥ �� − ��(��)(��� �������) But: �� ≥ ��(and analogs) we have: � � |� − �| ≥ �� − ��(��)(��� �������); So, we have: | (� − �)(� − �)(� − �) | ≥ (�� − ��)(�� − ��)(�� − ��)(��) From the above relationships: � � | (� − �)(� − �)(� − �) | ≥ (�� − ��)(�� − ��)(�� − ��)(43) From the above relationships: � � | (� − �)(� − �)(� − �) | ≥ (�� − ��)(�� − ��)(�� − ��)(��) Let’s consider the triangle OIH by area : ���� = | (���)(���)(���) | �� . So, we have the following relationships: ���� ≥ (�� − ��)(�� − ��)(�� − ��) �� (��) ���� ≥ (�� − ��)(�� − ��)(�� − ��) � (��) ���� ≥ (�� − ��)(�� − ��)(�� − ��) � (��) �� + �� ≥ ��� => �� − �� ≥ �� − ��(��� �������)(��) (�� − ��)(�� − ��)(�� − ��) ≥ (�� − ��)(�� − ��)(�� − ��)(��) �� ≤ �� + � ≤ �� So, we get: (�� − ��)(�� − ��)(�� − ��) ≥ (�� − ��)(�� − ��)(�� − ��)(��) We know that: ∑ ���� = � � then, we have: � ���� ≥ � �(��) �� + �� ≥ 2�� then adding, we get: ∑(�� + ��) ≥ �(�� + �� + ��)(��) References: ROMANIAN MATHEMATICAL MAGAZINE-www.ssmrmh.ro Romanian Mathematical Society-Mehedinți Branch 2021 40 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 ABOUT IONESCU-NESBITT INEQUALITY By D.M.Bătinețu-Giurgiu and Daniel Sitaru-Romania Let be �, � ∈ ℝ� = [�, ∞); �, � ∈ ℕ, � ≥ � and �, �, �, � ∈ ℝ� ∗ = (�, ∞), � = �, ������, �� = ∑ �� � ��� such that � ∙ �� > � ∙ ��� ����� {��}. We want to proof the following inequalities: � ∙ � + � � ��� + ��� ��� − ��� + � � ���� ��� ≥ (� + �)�(�� + �)�� (�� − �)�� + �� ; (�) � + �� ∙ � � ��� + ��� ��� − ��� + � � ���� ��� ≥ (� + �)�(�� + �)�� (�� − �)�� + �� ; (�) Proof: � ∙ � + � � ��� + ��� ��� − ��� + � � ���� ��� ≥ ����� � ∙ � + 1 �� ∙ �� ��� + ��� ��� − ��� + � � ��� � ��� ≥ ����� ≥ (� + 1) �� ∙ � ∙ … ∙ �������� �� ������� ∙ 1 �� ∙ �� ��� + ��� ��� − ��� + � � ��� � ��� ��� = = (� + 1) ∙ � ��� + ��� ��� − ��� + � � ��� = (∗) = (� + 1) ∙ � ��� + ��� ��� − ��� + � � ��� + (� + 1)�� � − (� + 1)�� � = = (� + 1) ∙ � � ��� + ��� ��� − ��� + � + � � � � ��� − (� + 1)�� � = = (� + 1) ∙ �� ���� + ���� + ���� − ���� + �� �(��� − ��� + �) � ��� � − (� + 1)�� � = = � + 1 � �� (�� + ��)�� + �� ��� − ��� + � � ��� − ��� = = � + 1 � ��(�� + ��)�� + ��� ∙ � 1 ��� − ��� + � � ��� − ��� ≥ ��������� Romanian Mathematical Society-Mehedinți Branch 2021 41 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 ≥ � + 1 � � �(�� + ��)�� + ���� � ∑ (��� − ��� + �) � ��� − ��� = � + 1 � � �(�� + ��)�� + ���� � ���� − ��� + �� − ��� = = � + 1 � ∙ ������ + ��� ��� + ��� � − ������ + ����� − �� � (�� − �)�� + �� = = � + 1 � ∙ ������ + ���� (�� − �)�� + �� = � + 1 � ∙ (�� + �)�� (�� − �)�� + �� = (� + 1)(�� + �)��� (�� − �)�� + �� namely the inequality (1) is proved. � + �� ∙ � � ��� + ��� ��� − ��� + � � ���� ��� ≥ ����� � + �� ∙ 1 �� ∙�� ��� + ��� ��� − ��� + � � ��� � ��� = = � + �� ��� + ��� ��� − ��� + � � ��� � ��� ≥ ����� ≥ (� + 1) �1 ∙ 1 ∙ … ∙ 1������� �� ������� ∙ �� ��� + ��� ��� − ��� + � � ��� � ��� ��� = (� + 1) ∙ � ��� + ��� ��� − ��� + � � ��� (∗) �� � + �� �� ��� + ��� ��� − ��� + � � ��� � ��� ≥ (� + 1)�(�� + �)�� (�� − �)�� + �� namely the inequality (2) is proof. If in (1)or (2) we take � = 0, � = 0, � = � = � = 1, � = 0 we get: � �� �� − �� � ��� ≥ � � − 1 ; (� − �) For � = 3, �� = �, �� = �, �� = � we get: � � + � + � � + � + � � + � ≥ 3 2 ; (� − �) Namely Ionescu-Nesbitt inequality. If adding relations (1), (2) we get: �(� + 1) + (1 + ��) ∙ �� ��� + ��� ��� − ��� + � � ��� � ��� ≥ 2(� + 1)�(�� + �)�� (�� − �)�� + �� ; (3) References: ROMANIAN MATHEMATICAL MAGAZINE-www.ssmrmh.ro Romanian Mathematical Society-Mehedinți Branch 2021 42 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 GAKOPOULOS’ LEMMA (II) By Thanasis Gakopoulos – Larisa – Greece (A) Lemma (I): (I) �� �� �� �� = �� �� �� �� Proof. Let �� ∥ ��. Is: �� �� = �� �� (1) �� �� = �� �� (Thales’s theorem) (2) Is: (Menelaus’s theorem Δ���, ���): �� �� ⋅ �� �� ⋅ �� �� = 1 → (�) �� �� ⋅ �� �� ⋅ �� �� = 1 → (�) �� �� ⋅ �� �� ⋅ 1 �� ⋅ �� �� ⋅ �� = 1 → → �� �� �� �� = �� �� �� �� Note: the lemma (I) has been proved (with PLAGIOGONAL system) and so is known in my previous paper, which has been in RMM group. (B) Lemma (II) Let: �� �� = �, �� �� = �, �� �� = �, �� �� = �, �� �� = �, �� �� = �, �� �� = �. � � = �⋅� �⋅� (II) Romanian Mathematical Society-Mehedinți Branch 2021 43 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 Proof. Δ��� (����� �): �� �� �� �� = �� �� �� �� Δ��� (����� �): �� �� �� �� = �� �� �� �� ⎭ ⎪⎪ ⎬ ⎪⎪ ⎫ → (×) �� �� ⋅ �� �� �� �� ⋅ �� �� = �� �� ⋅ �� �� �� �� ⋅ �� �� → �� �� �� �� = �� �� ⋅ �� �� �� �� ⋅ �� �� → � � = � ⋅ � � ⋅ � (C) APPLICATIONS. Relationships that holds to any COMPLETE QUADRILATERAL. ��⋅�� ��⋅�� = ��⋅�� ��⋅�� (C1) Proof.From lemma (I), if � ≡ � then we have: �� �� �� �� = �� �� �� �� → ��⋅�� ��⋅�� = ��⋅�� ��⋅�� , [���] [���] = [���] [���] (C2) There is a harmonious relationship between the triangle’s areas. Proof. From (C1): ��⋅�� ��⋅�� = ��⋅�� ��⋅�� → ��⋅��⋅��� �� ��⋅��⋅��� � = ��⋅��⋅��� �� ��⋅��⋅��� � → [���] [���] = [���] [���] �� ⋅ �� = �� ⋅ �� (C3) Romanian Mathematical Society-Mehedinți Branch 2021 44 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 There is a harmonious relationship between the segment’s lengths. Proof. From (C2): [���] [���] = [���] [���] → ��⋅�� ��⋅�� = ��⋅�� ��⋅�� → ℎ� ⋅ ℎ� = ℎ� ⋅ ℎ� ��⋅�� ��⋅�� = ��⋅��⋅�� ��⋅��⋅�� (C4) Proof. From lemma (II), we have: � � � = �⋅� �⋅� � → �� �� �� �� = � �� �� �⋅� �� �� � � �� �� �⋅� �� �� � → ��⋅�� ��⋅�� = ��⋅��⋅�� ��⋅��⋅�� �� �� �� �� ⋅ �� �� = �� �� �� �� ⋅ �� �� = � (C5.1) ��⋅�� ��⋅�� = ��⋅��⋅��⋅�� ��⋅��⋅��⋅�� (C5.2) Proof. ���� ����� (�): Δ���: �� �� �� �� = �� �� �� �� → �� �� �� �� ⋅ �� �� = 1 ���� ����� (�): Δ���: �� �� �� �� = �� �� �� �� → �� �� �� �� ⋅ �� �� = 1 ⎭ ⎪⎪ ⎬ ⎪⎪ ⎫ → �� �� �� �� ⋅ �� �� = �� �� �� �� ⋅ �� �� = 1 Proof. Romanian Mathematical Society-Mehedinți Branch 2021 45 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 From Lemma (II): � � � = �⋅� �⋅� � → �� �� �� �� = � �� �� �⋅� �� �� � � �� �� �� �� �� � → ��⋅�� ��⋅�� = ��⋅��⋅��⋅�� ��⋅��⋅��⋅�� (D) Complete Quadrilateral ��� − ��� Cyclic Quadrilateral ���� ��⋅�� ��⋅�� = ��⋅�� ��⋅�� (D1) ��⋅�� ��⋅�� = ��⋅�� ��⋅�� (D2) Proof D1. Is �� = �� = �� = �� = � (1) From lemma (II) we have: � � � = �⋅� �⋅� � → �� �� �� �� = � �� �� �� �� �� � � �� �� �� �� �� � → (�) ��⋅�� ��⋅�� = ��⋅�� ��⋅�� Proof D2. From lemma (II) we have: � � � = �⋅� �⋅� � → �� �� �� �� = � �� �� �� �� �� � � �� �� �� �� �� � → (�) ��⋅�� ��⋅�� = ��⋅�� ��⋅�� References: ROMANIAN MATHEMATICAL MAGAZINE-www.ssmrmh.ro STRUCTURI ALGEBRICE(VI) By Vasile Buruiană-Romania 7. Să se rezolve în ℤ ecuația următoare: �� + ��� = ���� Romanian Mathematical Society-Mehedinți Branch 2021 46 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 Proprietatea 5|�, � ⇒ 5|� ± � și generalizarea ei s-au demonstrat întâi, apoi s-a menționat teorema: �|��, (�, �) = 1 ⇒ �|� ⇒ 5|7��, (5,7) = 1 ⇒ 5|�� ⇒ 5|� deci � = 5��. Înlocuind ⇒ � + 35��� = 320 și apoi ⇒ 5|� deci � = 5��. Înlocuind cercul �� + 7�� � = 64 și prin încercări se găsesc (57; 1), (36; 2), (1; 3) pentru (��, ��), de unde soluția {(285,5), (180,10), (5,15)} S-a atras atenția asupra noțiunii de rădăcină a ecuației cu două necunoscute. S-a schițat geometric soluția {(�, �)|�, � ∈ ℝ} și ce înseamnă că (�, �) este formată doar din întregi pentru verificarea ecuației date. 8. (clasa a X a, după construcția lui �, cerc de matematică) Cum găsim numere prime < 20 în (T) Cum putem găsi, cu un “ciur” al lui Erotostene, numerele prime din ℤ[�] cu norma < � dat? După ce s-a reamintit cum se procedează în primul caz, s-a introdus noțiunea de normă ca generalizare a modului, noțiunile de ireductibil, prim și s-a demonstrat că �(��) = �(�)�(�) cu �|� ⇒ �(�)|�(�). S-a definit unitatea și s-au găsit unitățile ±1, ±� arătând că � e unitate ⇔ �(�) = 1. S-a demonstrat că 2 este reductibil și factorii săi 1 ± � sunt ireductibili și asociați. Ca în cazul ciurului lui Eratostene (numai că acum se lucrează în plan) s-au șters unitățile și multiplii lui 1 + � dintr-un cerc dat (de ex. cercul cu centrul în origine și de rază 4). Cum �(�) = 4 ⇔ � = ±2 sau � = ±2� și cum deja au fost șterse ca multipli ai lui 1 + �, se caută cele de normă 5 (găsiți: �(1 + 2�), �(2 ± �) cu � ∈ {±1, ±�}. S-au scos cele ireductibile, s-au șters multiplii lor aflați în cerc ș.a.m.d. S-a insistat asupra aspectului geometric, s-a aplicat teorema împărțirii cu rest în acest caz, s- a arătat ce este aceea o generalizare. 9. (clasa a XI a, definiții, lecție). Fie �, � ∈ ℕ∗ astfel încât ����, �����, �����, �����, … . Să se arate că � = �. S-a insistat asupra scrierii unice � = �� �� … �� ��, � = �� �� … �� ��, a noțiunii de șir și faptului că �����|����� și �����|����� ⇒ (2� + 1)�� ≤ (2� + 2)�� și (2� + 2)�� ≤ (2� + 3)��; Romanian Mathematical Society-Mehedinți Branch 2021 47 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 � = 1, … , � ⇒ �� ≤ ���� ���� �� și �� ≤ ���� ���� ��. Cu definiția limitei avem că ���� ���� → 1, ���� ���� → 1 și deci trecând la inegalități obținem �� ≤ �� respectiv �� ≤ ��; ∀� = 1, … �. Deci �� = ��, � = 1 ⇒ � = �. 10. (clasa a XII a, după teorema Bezout, lecție). Să se descompună în factori în ℝ[�, �, �] polinomul (� + � + �)� − �� − �� − �� = �. Privind ca polinom în ℝ[�, �][�] avem întâi că: � ⋮ � + � fiindcă �(−�) = (−� + � + �)� − (−�)� − �� − �� = �� + �� − �� − �� = 0. Analog �: � + �, � ⋮ (� + �) ⇒ (factori comuni) � = �(� + �)(� + �)(� + �). Cum gradul în fiecare variabilă este 2 ⇒ � este o constantă și găsirea lui se face dând lui �, �, � valoarea 1 ⇒ 27 − 3 = � ⋅ 8 ⇔ � = 3. Deci � = 3(� + �)(� + �)(� + �). 11. (clasa a XII a, cap. Polinom, cerc). Criterii de ireductibilitate pentru polinoame din ℤ[�]. S-au reamintit noțiunile de prim, divizor, polinom ireductibil, primitiv, s-a enunțat și demonstrat criteriul Ersenstein și s-a aplicat pe cazuri: �� − 2 (pentru a dovedi că în ℚ[�] există polinoame ireductibile de orice grad) apoi �� − �� − �� − � ca polinom din ℚ[�][�]. S-a formulat teorema descompunerii unice pentru polinoame. 12. (clasa a XII a, cap. Polinoame, cerc). Să se descompună în fracții simple fracția � (���)������ . S-a definit noțiunea de polinom ireductibil în ℝ[�]și s-a arătat care sunt toate elementele ireductibile în acest inel, s-a format teorema de descompunere în factori primi în ℝ[�], s-a definit noțiunea de fracție simplă și s-a exemplificat pentru fracții ordinare. S-a formulat și demonstrat teorema în general.S-au făcut exemple și în particular problema de mai sus. S-au revăzut și fixat: noțiunile de polinoame identice, rădăcină, rădăcină multiplă. Teorema împărțirii cu rest în �[�] rădăcini conjugate, c.m.m.d.c, ideal, elemente compuse. Apoi cum din scrierea unei fracții �(�) �(�) și a �(�) = �� �� … �� ���� �� … �� �� în factori ireductibili peste ℝ (�� liniari, �� de gradul �� cu rădăcini complexe), deoarece polinoamele următoare sunt coprime: �� �� … �� ���� �� … �� �� = � �� ���� �� … �� ���� �� … �� �� = � Romanian Mathematical Society-Mehedinți Branch 2021 48 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 ⋮ �� �� … �� ���� �� … �� �� = � ⋮ �� �� … �� ���� �� … ���� ���� = � rezultă că există ��, … ��, �, … , �� astfel încât. � = ��� + ��� + ⋯ ������� + ����� deci � � = �� �� �� + ⋯ + �� �� �� + �� �� �� + ⋯ �� �� �� . Cum �� se dezvoltă după puterile lui �� și �� după cele ale lui ��, ajungem la scrierea cunoscută și ultilizată la integrarea funcțiilor raționale. Fiecare pas a fost exemplificat și pe fracții ordinare. 13. (clasa a X a, după definirea numerlor complexe, cerc). Să se găsească numerele pitagorice (care verifică ecuația �� + �� = ��) și sunt în ℤ. Mai întâi căutăm soluțiile cu (�, �) = 1, deci �, � nu sunt ambii pari și obligatoriu � este impar. În �[�], factorial, avem (� + ��)(� − ��) = ��. Fie � ireductibil astfel ca �|� + ��, �|� + �� ⇒ �|2� și �|2�; � fiind impar ⇒ (�) ≠ (1 + �) căci dacă sunt egale idealele ⇒ ��� = 2|�� ceea ce este imposibil căci ℤ este impar, deci � ⊺ 2 ⇒ �|� și �|� ⇒ �(�) = �|�, �|� ceea ce arată că nu avem că �, � sunt compuse, absurd. ⇒ � + ��, � − �� sunt coprime și dacă � = ��� ��, … , �� ��, � fiind unitate, datorită factorialității lui ℤ[�] prin scrierea unică ⇒ � + �� = ���. Dacă � = � + ��, luând � = 1 obținem � + �� = �� + 2��� + �� ⇒ � � = �� − �� � = 2�� � = − � � . Variantele pentru celelalte unități sau situații (� − �� = ��� sau (�, �) ≠ 1) se reduc la acestea, având pentru �, � semnele eventual schimbate. Deci orice soluție este de această forma. Orice soluție de forma respectivă, verifică ecuația �� + �� = ��, deci toate soluțiile sunt cu � � = �� − �� � = 2�� � = �� + �� ; �, � ∈ ℤ Mai întâi s-au precizat noțiunile de prim, ireductibil, inel ℤ[�], imagine geometrică, norma proprietății, unități, factorialitate, apoi s-a rezolvat. (Se poate compara eventual această soluție cu una care nu folosește numerele complexe) 14. (clasa XII, după cl de resturi mod �, cerc) Romanian Mathematical Society-Mehedinți Branch 2021 49 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 Să se construiască în ℤ[�] corpul claselor de resturi modulo 1 + � = � (sau mai general pentru � ireductibil). S-a arătat că în ℤ[�] este adevărată teorema împărțirii cu rest, că numărul elementelor în corpul respectiv este �(�), s-au figurat resturile posibile și s-a construit corpul finit corespunzător, apoi s-au făcut calcule în el. S-a remarcat numărul de elemente. S-a făcut paralela cu clasele de resturi mod �. Principial, relația de echivalență nu diferă. 15. (clasa X,lecție) Care este exponentul lui � în �!? S-a redefinit [�] și datorită factorialității, exponentul căutat este finit. Dacă ��(�) este exponentul căutat este clar că � > � ⇒ �� = 0, iar � = � ⇒ �� = 1. Deci e interesant pentru � < �. Atunci în �! multiplii lui � sunt 1 ⋅ �, 2�, 3�, … , � � � � �. Fie �� = � � � � ⇒ înmulțind acești multipli avem 1 ⋅ 2 … � � � � ⋅ ��� La fel în ��! ⇒ � întâi ca factor de � �� � � ori și cum � �� � � = � � � � � � � = � � �� � ⇒ ��(�) = �� + ��(��), apoi ��(��) = �� + ��(��) etc., unde �� = � � � � și în final α�(����) = �� ABOUT A TRIANGLE INEQUALITY By D.M.Bătineţu-Giurgiu, Claudia Nănuţi, Daniel Sitaru-Romania In ∆��� we’ll note with �, �, � −the lengths of the sides, � −semiperimeter and � −area.We will consider �, � ∈ ℝ� = [�, ∞); � + �, �, �, � ∈ ℝ� ∗ = (�, ∞), then: (� + �)� + �� + �� �� + �� ∙ �� + (� + �)� + �� + �� �� + �� ∙ �� + (� + �)� + �� + �� �� + �� ∙ �� ≥ �√��, (�) ��� �� + �� �� + �� + (� + �)� ∙ �� + �� + �� �� + �� + (� + �)� ∙ �� + �� + �� �� + �� + (� + �)� ∙ �� ≥ �√��; (�) Proof: We have: � �� + �� �� + �� + (� + �)� ∙ �� ��� = � �� + �� �� + �� + (� + �)� ∙ �� ��� + � �� ��� − � �� ��� =
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