Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
172 Articole şi Note Matematice Bibliografie [1] D. Andrica, Asupra unor şiruri care au mulţimile punctelor limită intervale, Gazeta Matematică nr. 11/1979, pp. 404-406. [2] V. Nicula, Asupra unor şiruri care au limita ∞, Gazeta Matematică nr. 3/1995, pp. 97-101. [3] G. Pólya, G. Szegö, Problems and theorems in analysis. Vol. l, SpringerVerlag, Berlin, 1998. [4] D. Tănasie, Asupra unor clase de şiruri, Gazeta Matematică nr. 6/1982, pp. 207-209. [5] A. Vernescu, Asupra unor subserii ale seriei armonice, Gazeta Matematică nr. 2/2006, pp. 59-62. A SIMPLE PROOF OF THE ERDÖS-MORDELL INEQUALITY FOR POLYGONS IN N-DIMENSIONAL SPACES Marian Dincă1) Abstract. The purpose of this article is a new proof of the generalization of the Erdös-Mordell inequality Keywords: Erdös-Mordell inequality MSC : 51M16 We start with the following: Theorem 1. For every xk ≥ 0, 1 ≤ k ≤ n, xn+1 = x1 and αk ∈ (0, π) n αk = π, the following inequality holds: with k=1 n n k=1 k=1 π 2 xk ≥ xk xk+1 cos αk . cos n (1) Proof. Let us consider, in the plane with the origin O(0; 0), the points Pk (ak ; bk ), 1 ≤ k ≤ n such that: (Pk OPk+1 ) = αk , and denote n−1 1 ≤ k ≤ n − 1, αk = π − αn . Denote: k=1 xk = OPk = 1) Profesor, Şcoala nr. 1, Pantelimon, Ilfov. a2k + b2k . 173 M. Dincă, Proof of the Erdös-Mordell inequality By using the cosine theorem in the triangle Pk OPk+1 , 1 ≤ k ≤ n − 1 and in triangle P1 OPn , we obtain: OPk 2 + OPk+1 2 − Pk Pk+1 2 = 2 1 2 ak + b2k + a2k+1 + b2k+1 − (ak − ak+1 )2 − (bk − bk+1 )2 = ak ak+1 +bk bk+1 , = 2 for k = 1, 2, . . . , n − 1 and n−1 n−1 αk = −x1 xn cos αk = x1 xn cos αn = x1 xn cos π − xk xk+1 cos αk = k=1 =− OP1 2 + OPn 2 Then: 2 = cos π n n n k=1 a2k + b2k − k=1 n−2 1 k=1 (k+1)π 2 sin kπ n sin n n−2 = −a1 an − b1 bn . π 2 xk − xk xk+1 cos αk = n k=1 = − P1 Pn n cos k=1 2 (ak ak+1 + bk bk+1 ) + a1 an + b1 bn = k=1 kπ π (k + 1)π ak − sin ak+1 + sin an sin n n n 1 n−1 2 (k + 1)π kπ π sin bk − sin bk+1 + sin bn n n n + 2 . (∗) (k + 1)π kπ sin 2 sin n n Let A1 , A2 , ..., An be the vertices of a convex n-gon and let P be an internal point. Let Rk = P Ak and let rk be the distance from P to√the side Ak Ak+1 . We consider An+1 = A1 . From the inequality (1) for xk = Rk , we obtain: n n n n π Rk ≥ Rk Rk+1 cos αk ≥ wk ≥ rk , cos n + k=1 k=1 k=1 k=1 k=1 where wk is the length of the segment of the angle bisector Ak P Ak+1 . In consequence we proved the well-known Erdös-Mordell inequality: n n k=1 k=1 π Rk ≥ rk . cos n Now we prove the identity: cos n n−1 k=1 k=1 π 2 ak − ak ak+1 + a1 an = n (2) 174 = Articole şi Note Matematice n−2 1 (k + 1)π kπ π sin ak − sin ak+1 + sin an n n n 2 (3) (k + 1)π kπ sin 2 sin n n by comparing the coefficients of a2k and ak ak+1 . For example, the coefficient of a2k is: k=1 (k − 1)π kπ π (k + 1)π sin 2 sin cos n n n n = cos π + = kπ kπ kπ n 2 sin 2 sin 2 sin n n n sin for k = 2, 3, . . . , n − 2 and the coefficient of a21 is: 2π n = cos π , π n 2 sin n sin the coefficient of a2n−1 is: 2π (n − 2)π sin n n = cos π = π (n − 1)π n 2 sin 2 sin n n sin and the coeficient of a2n is: π π n−2 sin kπ (k + 1)π n n = cot − cot = (k + 1)π kπ 2 n n k=1 2 sin k=1 sin n n n−2 sin2 π n cot π − cot (n − 1)π = cos π . 2 n n n sin = The coefficient of ak ak+1 is: kπ (k + 1)π sin n n = −1 (k + 1)π kπ sin 2 sin n n −2 sin for k = 1, 2, . . . , n − 2 and the coefficient of an−1 an is: π π (n − 2)π − sin sin n n n = π = −1. (n − 1)π (n − 2)π sin sin 2 sin n n n −2 sin 175 M. Dincă, Proof of the Erdös-Mordell inequality The coefficient of a1 an is: π 2π sin n n =1 π 2π 2 sin sin n n and the coeficient of ak an for k = 2, 3, . . . , n − 2 is: 2 sin π π (k + 1)π kπ n−3 sin sin 2 sin n n − n n = (k + 1)π (k + 1)π kπ kπ k=2 2 sin k=1 2 sin sin sin n n n n π π n−2 n−3 sin sin n n = 0. − = kπ (k + 1)π k=2 sin k=1 sin n n Similarly we can prove the identity for variabile bk : n−2 2 sin cos n−2 1 n n−1 k=1 k=1 π 2 bk − bk bk+1 + b1 bn = n (k + 1)π kπ π sin bk − sin bk+1 + sin bn n n n 2 . (4) (k + 1)π kπ k=1 2 sin sin n n Now the identity (∗) follows by summing the relations (3) and (4). We prove now the following generalization. Theorem 2. Suppose x = (x1, x2 , . . . , xN ) and y = (y1 , y2 , . . . , yN ) are vectors in RN , endowed with the inner product = x, y = x y cos θ = N xk yk , k=1 where θ is the angle between the vectors x and y, while x = x21 + x22 + ... + x2N is the corresponding norm. Then the following inequality of Erdös-Mordell type holds: n n n π Xk 2 ≥ Xk , Xk+1 = Xk Xk+1 cos θk , cos n k=1 k=1 k=1 where Xk = (x1k , x2k , . . . , xN k ), k = 1, 2, ..., n and n k=1 θk = π. When P is an exterior point, we have the following form of the ErdösMordell inequality for convex polygons: 176 Articole şi Note Matematice Theorem 3. Let A1 A2 ...An be a convex polygon, P be an exterior point and Ak P Ak+1 = αk , k = 1, ..., n. Let max(Ai P Aj ) = α, where n αk = 2α < 2π and i = j ∈ {1, 2, ...n, } and k=1 rk ≤ wk ≤ Then: Rk Rk+1 cos αk . 2 n n n n k=1 k=1 k=1 k=1 α π cos Rk − rk ≥ cos Rk − rk ≥ n n n n n n π αk π Rk − wk ≥ cos Rk − Rk Rk+1 cos . ≥ cos n n 2 k=1 k=1 k=1 k=1 Theorem 4. Let A1 , A2 , ..., Ap be p arbitrary points in the N-dimensional euclidian space RN , M an arbitrary point in the same space and Ak M Ak+1 = αk , k = 1, ..., p. Let max(Ai M Aj ) = α, i = j ∈ {1, 2, ...p} p αk = 2α < 2π. We consider Ap+1 = A1 . Then we obtain the following and k=1 inequality: cos p p p p π α Rk − rk ≥ cos Rk − rk ≥ p p k=1 ≥ cos π p p k=1 Rk − p k=1 k=1 wk ≥ cos k=1 π p p Rk − k=1 p k=1 k=1 Rk Rk+1 cos αk ≥ 0, 2 where Rk is the norm of Ak M , rk is the distance from M to the segment Ak Ak+1 , k ∈ {1, 2, ..., p} and wk is the length of the segment of the angle bisector Ak P Ak+1 . References [1] M. Dincă, Generalizarea inegalităţii Erdös-Mordell, G. M.-B nr.7-8 (1998) 269-273. [2] M. Dincă, Generalization of the inequality of Erdös-Mordell, Octogon Math. Magazine, 1 (1999), 143-147, (in romanian). [3] M. Dincă, A refinement of Erdös-Mordell’s inequality, Octogon Math. Magazine, 10 (2002), 77-86 (in romanian). [4] H. G. Eggleston, A triangle inequality, Math. Gaz., 42 (1958), 54-55. [5] P. Erdös, Problem 3740, Amer. Math. Monthly, 42 (1937), 396. [6] F. Faruk, Abi-Khuzam, A trigonometric inequality and its geometric applications, Mathematical Inequalities & Application, vol. 3 (2003), 437-442. [7] A. Florian, Zu einem Satz von P. Erdös, Elem. Math. 13 (1958), 55-58. L. Ţurea, Aproximări raţionale. Numere Farey. Fracţii continue (II) 177 [8] S. Gueron, I. Shafrir, A weighted Erdös-Mordell inequality for polygon, American Mathematical Monthly, 112 (2005), 257-263. [9] D. K. Kazarinoff, A simple proof of the Erdös-Mordell inequality for triangles, Michigan Math. J., 4 (1957), 97-98. [10] H. C. Lenhard, Verallgemeinerung und Verschärfung der Erdös-Mordellschen Ungleichung für Polygone, Arh. Math. vol. XII (1961), 311-314. [11] D. S. Mitrinović, J. E. Pečarić, On the Erdös-Mordell’s inequality for a Polygon, J. College Arts Sci, Chiba Univ., B-19 (1986), 3-6. [12] L. J. Mordell, D. F. Barrow, Solution of Problem 3740, American Mathematical Monthly, 44 (1937), 252-254. [13] N. Ozeki, On Paul Erdös’s inequality for the triangle, J. College Arts Sci., Chiba Univ., 2 (1957), 247-250. [14] G. R. Veldkamp, The Erdös-Mordell inequality, Nieuw Tijdshr. Wisk. 45 (1957/58), 193-196. [15] S. Wu, L. Debnath, Generalization of the Wolstenholme cyclic inequality and its application, Computers & Mathematics with Application, vol. 53, Issue 1, (2007), 104-114. APROXIMĂRI RAŢIONALE. NUMERE FAREY. FRACŢII CONTINUE (II)1) de Lucian Ţurea2) Abstract. This is the second part of an article about various approximations of real numbers using certain classes of rationals. Keywords: rational approximation, Liouville numbers, Farey numbers, continuous fractions, Kronecker density theorem. MSC : 11J68, 11J70. 4. Cele mai bune aproximări. Fracţii continue. La sfârşitul paragrafului anterior, după ce am studiat problema existenţei unor aproximări cât mai bune, am pus şi ı̂ntrebarea naturală despre cum arată aceste aproximări. Pentru cazul lui ϕ am ı̂ntâlnit şi nişte coincidenţe interesante. Un alt exemplu ı̂n care este interesant de studiat forma fracţiilor 22 este o aproximare foarte bună a acestuia (eroare este π. De exemplu, 7 de ordinul lui 10−3 ). Cel mai ,,simplu” număr raţional care ı̂l aproximează 333 179 . Alte aproximări foarte bune ale acestuia sunt pe π mai bine este 57 106 355 −5 −7 (eroare de ordinul lui 10 ). Un alt lucru (eroare de ordinul lui 10 ) şi 113 interesant este că, pentru 106 < q < 113, nicio fracţie nu ı̂l aproximează pe 1) Continuare din G. M. - B nr. 3/2009. 2) Student, Facultatea de Matematică şi Informatică, Universitatea din Bucureşti.