# GAZETA MATEMATICĂ SERIA A. ANUL XXXVI (CXV) Nr. 1 2/ 2018 ARTICOLE. Computing exponential and trigonometric functions of matrices in M 2 (C)

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1 GAZETA MATEMATICĂ SERIA A ANUL XXXVI CXV) Nr. 1 / 18 ARTICOLE Computing exponential and trigonometric functions of matrices in M C) Ovidiu Furdui 1) Abstract. In this paper we give a new technique for the calculation of the matrix exponential function e A as well as the matrix trigonometric functions sin A and cos A, wherea M C). We also determine the real logarithm of the matrix xi,whenx R, as well as the real logarithm of scalar multiple of rotation and reflection matrices. The real logarithm of a real circulant matrix and a symmetric matrix are also determined. Keywords: Square matrices of order two, matrix exponential function, matrix trigonometric functions, the real logarithm. MSC: 15A Introduction and the main results Let A M C). In this paper we give a new technique for the calculation of the matrix exponential function e A as well as the matrix trigonometric functions sin A and cos A. Our method, which we believe is new in the literature?!, is based on an application of the Cayley Hamilton Theorem and the power series expansion of the exponential and the trigonometric functions. The organization of the paper is as follows: in the first section we give formulas for e A,sinA and cos A in terms of the eigenvalues of A and in the second section we solve several exponential equations in M R) andwe compute the real logarithm of special real matrices. The next theorems are the main results of this section. Theorem 1. The exponential function e A. 1) Department of Mathematics, Technical University of Cluj-Napoca, Romania,

2 Articole Let A M C) and let α, β C be the eigenvalues of A. The following formula holds e α e β e A = α β A + αeβ βe α I, if α β, α β e α A +e α αe α ) I, if α = β. Theorem. The trigonometric function sin A. Let A M C) and let α, β C be the eigenvalues of A. The following formula holds sin α sin β α sin β β sin α A + I, if α β, sin A = α β α β cos α)a +sinα αcos α)i, if α = β. Theorem 3. The trigonometric function cos A. Let A M C) and let α, β C be the eigenvalues of A. The following formula holds cos α cos β α cos β β cos α A + I, if α β, cos A = α β α β sin α)a +cosα + α sin α)i, if α = β. Before we prove these theorems we collect a lemma we need in our analysis. Lemma 4. Let a C and let A M C). The following statements hold: a) e ai =e a I ; b) sinai )=sina)i ; c) cosai )=cosa)i ; d) if A, B M C) commute, then e A+B =e A e B ; e) if A, B M C) commute, then sina + B) =sina cos B +cosa sin B; f) if A, B M C) commute, then cosa + B) =cosa cos B sin A sin B. Proof. The proof of this lemma can be found in [4, pp ]. Now we are ready to prove these theorems. Proof. First we consider the case α β. We have, based on the Cayley Hamilton Theorem, that A αi )A βi )=O. Let X = A αi. It follows that XX +α β)i )=O which implies that X =β α)x. This

3 O. Furdui, Computing functions of matrices in M C) 3 in turn implies that X n =β α) n 1 X, for all n 1. We have, e A =e αi +X =e αi e X =e α X n n! n= ) =e I α β α) n 1 X + n! n=1 ) =e α I + eβ α 1 β α X ) =e α I + eβ α 1 β α A αi ) = eα e β α β A + αeβ βe α I. α β Now we consider the case α = β. The Cayley Hamilton Theorem implies that A αi ) = O.LetX = A αi. It follows that A = X + αi and X = O.Wehave, e A =e αi +X =e αi e X =e α I I + X ) 1! + X + X3 +! 3! =e α I + X) =e α I + A αi ) =e α A +e α αe α ) I, and Theorem 1 is proved. Now we give the proof of Theorem. Proof. One method of proving the theorem is based on the use of Euler matrix formula sin A = eia e ia i combined to Theorem 1. We leave these details to the interested reader. Instead we prove the theorem by using a power series technique. First we consider the case α β. As in the proof of Theorem 1 we have, based on the Cayley Hamilton Theorem, that A αi )A βi )=O.LetX = A αi. It follows that XX +α β)i )=O which implies that X =β α)x.

4 4 Articole This in turn implies that X n =β α) n 1 X, for all n 1. We have, sin A =sinx + αi )=sinxcosαi )+cosx sinαi ) = sinxcos α)+ cosxsin α). On the other hand, sin X = 1) n Xn+1 n +1)! n= n β α)n = 1) n +1)! X n= = 1 n β α)n+1 1) X β α n +1)! n= sinβ α) = β α X and cos X = 1) n Xn n)! n= n β α)n 1 = I + 1) X n)! n=1 = I + 1 n β α)n 1) X β α n)! n=1 cosβ α) 1 = I + X. β α Putting all these together we have, after simple calculations, that [ ] sinβ α) sin A = β α cos α)a αi cosβ α) 1 )+ I + A αi ) sin α β α sin α sin β α sin β β sin α = A + I. α β α β Now we consider the case α = β. The Cayley Hamilton Theorem implies that A αi ) = O.LetX = A αi. It follows that A = X + αi and X = O.Wehave, sin A =sinαi + X) =sinxcosαi )+cosx sinαi ) =sinxcos α +cosx sin α = X cos α + I sin α =cosα)a +sinα αcos α)i.

5 O. Furdui, Computing functions of matrices in M C) 5 We used that sin X = X and cos X = I. The proof of Theorem 3 which is similar to the proof of Theorem is left as an exercise to the interested reader. Corollary 5. Functions of reflection matrices. Let a, b R. The following equalities hold: ) a b 1) e b a = cosh ) a + b I + sinh a + b ) a b ; a + b b a ) a b ) sin = sin a + b ) a b ; b a a + b b a ) a b 3) cos = cos ) a b a + b I. Remark 6. Observe that ) ) a b cos θ sin θ = a b a + b, sin θ cos θ a where cos θ = and sin θ = a. a +b a Thus, any matrix of the form ) +b a b is a scalar multiple of a reflection matrix. The reason why the b a ) cos θ sin θ matrix is called a reflection matrix is given in [4, p. 14]. sin θ cos θ Corollary 7. Let a, b, c R. The following statements hold ) e a cosh b bc sinh bc bc a b c sinh bc cosh if bc >, bc e c a = bc cos b bc sin bc e a bc c sin bc cos if bc <. bc bc Corollary 8. Let a, b, c R. The following statements hold sin a cos b bc cos a sin bc bc c ) cos a sin bc sin a cos if bc >, bc a b sin = bc c a sin a cosh b bc cos a sinh bc bc c cos a sinh bc sin a cosh if bc <. bc bc

6 6 Articole Corollary 9. Let a, b, c R. The following statements hold cos a cos bc b sin a sin bc bc ) c sin a sin bc cos a cos if bc >, bc ab cos = bc ca cos a cosh bc b sin a sinh bc bc c sin a sinh bc cos a cosh if bc <. bc bc Corollary 1. Functions of rotation matrices. Let a, b R. The following equalities hold: ) a b ) 1) e b a cos b sin b =e a ; ) sin b cos b ) a b sin a cosh b cos a sinh b ) sin = ; b a cos a sinh b sin a cosh b ) ) a b cos a cosh b sin a sinh b 3) cos =. b a sin a sinh b cos a cosh b Remark 11. Observe that ) a b = ) cos α sin α a b a + b, sin α cos α a where cos α = and sin α = a a +b a. Thus, any matrix of the form +b ) a b is a scalar multiple of a rotation matrix. The reason why the b a ) cos α sin α matrix is called a rotation matrix is given in [4, Problem sin α cos α 1.61, p. 31]. Remark 1. We mention that if x R and A M R) the calculation of the matrix functions e Ax,sinAx)andcosAx) has been done, by a completely different method, in [4, Appendix A, pp ]. If f is a function which has the Maclaurin series fz) = f n) ) n! z n, n= z <R,whereR, ] anda M C) is such that its eigenvalues α and β verify the conditions α <Rand β <R,then fα) fβ) αfβ) βfα) A + I fa) =, if α β, α β α β 1) fα)a +fα) αf α))i, if α = β. For a proof of this remarkable formula, which is based on the calculation of the nth power of a square matrix of order two the reader is referred to [4, Theorem 4.7, p. 194].

7 O. Furdui, Computing functions of matrices in M C) 7 We mention that formula 1) holds in a more general case: when α β the formula holds for complex value functions f defined on speca) ={α, β}, while if α = β the formula is valid if f is differentiable on speca) ={α} see [, Notes p. 1] and [3, P.11, Sect. 6.1]).. The real logarithm of special matrices Let A M R). We say that B M R) is a real logarithm of A if e B = A see [1, p. 718]). In this section we solve in M R) various exponential matrix equations and hence we determine the real logarithm of special real matrices. Theorem 13. Two exponential equations. a) Let A M R). The solution of the equation e A = xi,where x R, is given by lnx 1) A = P k ) ) kπ kπ lnx 1) k P 1, ) where P GL R) and k is an even integer if x> and an odd integer if x<. ) x y b) The equation e A =,wherex, y R, does not have solutions y in M R). Proof. a) If α, β are the real eigenvalues of A, then there exists P GL R) such that ) ) α A = P P 1 α 1 or A = P P 1, β α according to whether the eigenvalues of A are distinct or not see [4, Theorem.1, p. 79]). It follows that ) α e A = P e β P 1 = xi and this implies that ) e α e β = xi. Thus, e α =e β = x which implies that x>andα = β =lnx. This shows that A =lnx)i. If the eigenvalues of A are α + iβ and α iβ, α R and β R,then there exists P GL R) such that see [4, Theorem.1, p. 79]) ) α β A = P P β α 1.

8 8 Articole This implies that α β ) e β α = xi and it follows, based on part 1) of Corollary 1, that ) e α cos β sin β = xi sin β cos β. Thus, e α cos β = x and e α sin β =. It follows that sin β = β = kπ, k Z. The first equation implies that e α = x 1) k α =lnx 1) k ), where k is an even integer when x>andk is an odd integer when x<. b) Passing to determinants we get that ) det e A x y =det e y TrA) = y, which is impossible over R. Remark 14. Part a) of Theorem 13 states that the real logarithm of the matrix xi is given by lnx 1) lnxi )=P k ) ) kπ kπ lnx 1) k P 1, ) where P GL R) andk is an even integer if x> and an odd ) integer if x y x<. Part b) of the theorem shows that the matrix, x, y R, y does not have a real logarithm. We mention that if A M C) the equation e A = zi, where z C, has been studied in [4, Problem 4.36, p. 19]. Also, the trigonometric equations sin A = I and cos A = I over the real square matrices have been studied in Appendix B of [4]. The equations sin A = xi and cos A = xi, with x R, can be solved similarly to the technique given in Appendix B of [4]. Theorem 15. The real logarithm of scalar multiple of rotation matrices. a) Let A M R). The solution of the equation ) e A y =, y R, y is given by A = ln 1)k y) k +1) π k +1) π, ln 1) k y) where k Z is an even integer when y> and an odd integer when y<.

9 O. Furdui, Computing functions of matrices in M C) 9 b) Let A M R) and let x, y R. The solution of the equation ) e A x y =, y x is given by ) ln x A = + y θ +kπ) θ +kπ ln x + y, where k Z and θ, π) is such that x y cos θ = and sin θ = x + y x + y. ) a b Proof. a) Let A = c d a b A = b a e a cos b sin b ) y we get that y.sinceacommutes with ). We have, based on part 1) of Corollary 1, that ) sin b = cos b ) y y and it follows that e a cos b =ande a sin b = y. The first equation implies that b =k +1) π, k Z, and the second equation shows that ea = 1) k y. Thus, a =ln 1) k y), where k is an even integer when y>andanodd integer when y<. ) ) a b x y b) Let A = a b A = b a. SinceA commutes with ) c d y x. We have, based on part 1) of Corollary 1, that e a cos b sin b ) ) sin b x y = cos b y x we get that and it follows that e a cos b = x and e a sin b = y. Thisimpliesthate a = x +y a =ln x + y x and cos b = and sin b = y.letθ, π) x +y x +y x be such that cos θ = and sin θ = y. It follows that x +y x +y { sin b =sinθ cos b =cosθ { b = θ +kπ or b = π θ +kπ, b = θ +nπ or b = θ +nπ, where k, n Z. Sincex, y these cases imply that b = θ +kπ, k Z. Remark 16. Part a) ) of Theorem 15 shows that the real logarithm of the y real matrix is given by y ) y ln 1) ln = k y) k +1) π ) y k +1) π ln 1) k, y)

10 1 Articole where k Z is an even integer when y> and an odd integer when y<. Part )b) of the theorem shows that the real logarithm of the matrix x y is y x ) ) x y ln x ln = + y θ +kπ) y x θ +kπ ln x + y, x where k Z and θ, π) is such that cos θ = and sin θ = y. x +y x +y Corollary 17. The real logarithm of a rotation matrix. Let α, π) \ { π,π, 3π }. The solution, in M R), oftheequation ) e A cos α sin α = sin α cos α is given by ) 1 A =α +kπ), k Z. 1 The previous corollary ) shows that the real logarithm of the rotation cos α sin α matrix is sin α cos α ) ) cos α sin α 1 ln =α +kπ), k Z. sin α cos α 1 Theorem 18. The real logarithm of a circulant matrix. Let A M R) and let x, y R, y. The equation ) e A x y = y x has solutions in M R) if and only if x> and x <y<xand in this case the solution is given by A = ln x y ln x+y x y ln x+y x y ln. x y ) ) x y a b Proof. Since A commutes with we get that A =.Wehave, y x b a based on Corollary 7, that ) cosh b sinh b e A =e a. sinhb cosh b It follows that e a eb +e b { = x> e a+b = x + y>, e a eb e b e = y a b = x y>.

11 O. Furdui, Computing functions of matrices in M C) 11 Thus, x <y<xand a calculation shows that a =ln x y and b =ln x+y x y. Corollary 19. Let A M R) and let y R, y> 1. The solution of the equation ) e A y +1 y = y y +1 is given by A =ln ) 1 1 y Proof. In Theorem 18 we let x = y +1>. Theorem. Let x, y R be such that xy <. The solution of the equation ) e A x = y is given by ln xy π x k +1) A = y ± π y k +1) x ln, xy where k is an integer. ) ) a b x Proof. Let A = M c d R). Since A commutes with we ) y a b have that A =. Observe that Corollary 7 implies that bc < and c a we get that e a cos b bc sin bc ) bc c sin bc cos x =, bc y bc which implies that cos bc =, e a b sin bc = x, bc c sin bc = y. bc e a ) The first equation implies that bc = π k +1), wherek isaninteger. It follows that e a b bc 1) k = x and e a c bc 1) k = y. These two equations imply that e a = xy a =ln xy. Dividing the last two equations in )

12 1 Articole we get that b c = x y b = x y c. On the other hand, bc = π k +1)) and it follows that c = y π x k +1)). This implies that c = ± π y x k +1) ) ) and b = x y c = x y ± π yx k +1) = π x y k +1). Theorem 1. The real logarithm of triangular matrices. Let x, y R. The equation ) e A x y = x has solutions in M R) if and only if x> and in this case the solution is given by ln x y ) A = x. lnx ) ) a b x y Proof. Let A =. SinceA commutes with we get that c d x ) a b A =. In this case both eigenvalues of A are equal to a. It follows, a based on Theorem 1, that e A e a be = a ) ) x y e a =. x Thus, e a = x>andbe a = y. This implies that a =lnx and b = y x. Remark. The preceding theorem shows that the matrix real logarithm if and only if x>. ) x y has a x Theorem 3. The real logarithm of symmetric matrices. Let X M R) be a symmetric matrix such that X ai, a R. The equation e A = X has solutions in M R) if and only if TrA) > and det A> and in this case the solution is given by A = ln λ 1 ln λ X + λ 1 ln λ λ ln λ 1 I, 3) λ 1 λ λ 1 λ where λ 1,λ are the eigenvalues of X. Proof. Since X ai and A commutes with X we get, based on Theorem 1.1 in [4, p. 15], that A = αx + βi,forsomeα, β R. The equation e A = X implies that e αx+βi = X e αx =e β X. Since X is symmetric and X ai, a R, we know that X has real distinct eigenvalues see [4, Theorem.5, p. 73]). Let λ 1,λ be the eigenvalues of X. Observe that, since A commutes with X, A is also symmetric and it has real eigenvalues. If μ 1 and μ are the eigenvalues of A, then {e μ 1, e μ } = {λ 1,λ }. Thus,

13 O. Furdui, Computing functions of matrices in M C) 13 both eigenvalues of X should be positive real numbers, i.e., TrA) > and det A>. The equation e αx =e β X combined to Theorem 1 show that e αλ 1 e αλ αx)+ αλ 1e αλ αλ e αλ1 I =e β X. αλ 1 αλ αλ 1 αλ This implies that e αλ 1 e αλ X + λ 1e αλ λ e αλ1 I =e β X λ 1 λ λ 1 λ and, since X ai, a,wehavethat e αλ 1 e αλ =e β λ 1 e αλ λ e αλ 1 and =. λ 1 λ λ 1 λ It follows that λ 1 e αλ λ e αλ 1 = e αλ 1 λ ) = λ 1 λ α = ln λ 1 ln λ λ 1 λ.on the other hand, β =ln λ 1 λ e αλ 1 e αλ λ 1 λ =ln λ1 λ λ 1 λ λ1 λ 1 λ =ln ) λ1 λ λ 1 ) λ1 λ1 λ 1 λ λ λ ) = λ 1 ln λ λ ln λ 1. λ 1 λ Thus, A = αx + βi = ln λ 1 ln λ X + λ 1 ln λ λ ln λ 1 I, λ 1 λ λ 1 λ and the theorem is proved. Remark 4. Theorem 3 shows that a symmetric matrix which is not a scalar multiple of the identity matrix has a real logarithm if and only if its eigenvalues are both positive real numbers and in this case the real logarithm is given by formula 3). References [1] D. N. Bernstein, Matrix Mathematics. Theory, Facts and Formulas, PrincetonUniversity Press, 9. [] S. R. Garcia and R. A. Horn, A Second Course in Linear Algebra, Cambridge University Press, 17. [3] R. A. Horn and C. H. Johnson, Topics in Matrix Analysis, Cambridge University Press, [4] V. Pop and O. Furdui, Square Matrices of Order Two. Theory, Applications, and Problems, Springer, 17.

14 14 Articole Stirling type formulas Ion-Denys Ciorogaru 1) Abstract. In this paper we prove two Stirling type formulas: and i+j n;i,j N n i + j) i,j=1 i + j) g k n n n 1 5 π e n 4 n g 5 k 1 eπ nn 1 4 n +n 6 e 3n where g k is the Glaisher-Kinkelin constant. Keywords: Stirling formula, Glaisher-Kinkelin constant. MSC: Primary 4A5; Secondary 41A6., 1. Introduction In this paper the notation and notion used are standard, in particular N = {1,,...} is the set of all natural numbers. Definition 1. Let b n ) n N be a sequence of real numbers with the property that n N such that b n, n n. We will say that the sequence of real numbers a n ) n N is equivalent with b n ) n N and we write a n b n if and only if lim n a n bn =1. From the well-known result that if a sequence of real numbers is convergent then every subsequence is convergent and has the same limit, we get: Proposition. If a n b n then a n+1 b n+1. We need the following two results. Proposition 3. Let f :, ) R be a function. The following formula holds n f i + j) = k 1) f k). i+j n;i,j N i+j n;i,j N Proof. We have: n 1 n i n 1 n f i + j) = f i + j) = f 1 + j)+ f + j) i=1 j=1 1) Mihai Viteazul Secondary School Nr. 9 Constanţa, Str. Cişmelei, Nr. 13, Constanţa and Department of Mathematics, Ovidius University of Constanţa, Bd. Mamaia 14, 957 Constanţa, Romania, k=1 j=1 j=1

15 I.-D. Ciorogaru, Stirling type formulas 15 n n 1) + + f n 1) + j) =f ) + f 3) + +n 1) f n) j=1 = n k 1) f k). k=1 Proposition 4. Let f :, ) R be a function. The following formula holds n n n+1 f i + j) = k 1) f k)+ n +1 k) f k). i,j=1 Proof. We have n f i + j) = i,j=1 + i=1 k=1 n n f i + j) = j=1 k=n+1 n f 1 + j)+ j=1 n f + j)+ n f n + j) =f ) + f 3) + +n 1) f n)+nf n +1) j=1 +n 1)fn+)+ +fn+n) = and so n k 1)fk)+ k=1 Changing n + k +1=i, weget n n k) f n + k +1)= k= n f i + j) = i,j=1 n k 1) f k)+ k=1 n+1 i=n+1 n+1 k=n+1 j=1 n n k) f n + k +1). k= n +1 i) f i) n +1 k) f k). We recall two well-known results. Their proofs can be found, for example, in [7]. The Glaisher-Kinkelin theorem. The sequence x n ) n N where x n = 11 n n n n + n e n 4 lim x n = g k is called the Glaisher-Kinkelin con- n is convergent and its limit stant.

16 16 Articole The Stirling formula. The following formula holds n i πn nn e n. i=1 We will use these results to obtain the two Stirling type formulas stated in the Abstract.. The main results Proposition 5. The following formula holds i+j n; i,j N i + j) g k Proof. Using Proposition 3, we get ln i+j n; i,j N i + j) = i+j n; i,j N n n n 5 1. π e n 4 n lni + j) = n k 1) ln k k=1 Thus, = i+j n; i,j N n k ln k k=1 i + j) = n ln k =ln k=1 n k=1 k k. n k k=1 n k k ln k=1 n k =ln k=1 n k=1 k k. n k From Glaisher-Kinkelin s theorem and Stirling s formula we obtain i+j n;i,j N i + j) g k n n + n e n 4 π n n e n n = g k k=1 n n n 5 1. π e n 4 n Proposition 6. The following formula holds n+1 k=n+1 k 4n n n+1 e n.

17 I.-D. Ciorogaru, Stirling type formulas 17 Proof. Using Proposition in Stirling s formula, we will have From the equality n+1 k=n+1 Since 1+ 1 n n+1 k=1 n+1 k=n+1 k π k = n+1 k=1 n +1)n+1 e n+1 n +1. k n k k=1 and Stirling s formula we obtain n +1) n+1 π k e n+1 n +1 n +1) n n n n π e n n = n n e n+1 n) n ) n n n = n n e n+1. ) n+1 e and + 1 n, we get n+1 k=n+1 k 4n n n+1 e n. Proposition 7. The following formula holds 1+ 1 ) n+1) n 4 e 3 e n. Proof. Indeed, we have lim n 1+ 1 ) n+1) n e n+1) 4n = lim n 1+ 1 n e ) n n +1) 4n = lim n n e ) n 1 n+1) 4n = e lim n 1+ 1 n) n 1 e ) n +1) 4n. lim n Also, 1+ 1 n e ) n ) n +1) 1 4n ) 1+ 1 n ) = lim n n n +1) 1 n e 4n

18 18 Articole ) 1+ 1 n ) = lim n n 1. n e From the L Hospital rule we have ) x) 1 x lim 1 x x e = 1 lim x e ln1+x) x 1 1 ln1+x) 1 x ln1+x) 1 x x = 1 lim x e ln1+x) x 1 1 x = 1 lim ln 1 + x) x x x 1 = 1 lim 1+x 1 = 1 x x 4. From the characterization of the limit of a function at a point with 1+ sequences, we deduce that lim 1 n 1+ 1 ) n+1) n n) n+1) e n+1) 4n = e 1 4,sothat e n+1) 4n 1 4 = e n n. Since e 1 4n 1, we obtain 1+ 1 ) n+1) n 4 e 3 e n. Proposition 8. The following formula holds n+1 k=n+1 k k e 1 n +3n n 3n e 3n 4 + 5n +1 Proof. Using Proposition in Glaisher-Kinkelin formula, we have n+1 k=1 Then from the equality n+1 n+1 k k g k n +1) n+1) + n e n+1) 4. k=n+1 k k = n+1 k k k=1 n k k k=1 we deduce that k k g k n+1) n+1) + n e n+1) 4 = g k n n + n e n 4 k=n+1 = n) n+1) + n ) n+1) n n n + n e n+1)3n+1) 4. + n +1)n+1) n n n + n e n+1)3n+1) ) n+ 7 1 n

19 = I.-D. Ciorogaru, Stirling type formulas 19 n +3n n 3n + 5n ) n+1) n From Proposition 7 and 1+ 1 n n+1 k=n+1 e n+1)3n+1) 4 ) n+ 7 1 e we obtain 1+ 1 ) n+ 7 1 n k k n +3n n 3n + 5n +1 4 e 3 e n e e n+1)3n+1) 4. =e 1 n +3n n 3n e 3n 4 + 5n +1. Proposition 9. The following formula holds 1+ n) 1 4n +5n+ 3 e e n. Proof. We have lim n 1+ 1 n ) 4n +5n+ 3 e n+ = lim n 1+ 1 n 4 e3 e n ) n+1) 1+ 1 n e ) n+1. By Proposition 7, 1+ 1 ) n+1) n 4 e 3 e n and so: lim n 1+ 1 n 4 e3 e n ) n+1) =1, and since lim n 1+ 1 n) n+1 e = 1 we will obtain that lim n Proposition 1. The following formula holds n+1 ) n+1 k e 1 e 4n +5n n n +3n+1. e n +n k=n n) 4n +5n+ 3 e n+ =1. Proof. From Stirling s Theorem, see for example [7, Theorem 5], we have n k=1 k = πn nn e n e 1 1n e R n,

20 Articole where R n n+1 k=1 where R n+1 n+1 k=n+1 k = 3 16n, n N. Changing n to n +1,weget k = π n +1) n +1)n+1 e n+1 1 e 4n+1 e R n+1, 3 16n+1), n N. Using these relations, we will obtain n+1 k=1 k n k k=1 n +1) n+1 1 π n +1) = e n+1 e 1n+1) e R n+1 n n πn e n e 1 1n e Rn 1+ 1 ) n+ 3 n n+ 1 ) n = n 3 +1)n+ n n+ 1 n) ) n+ 3 e n+1 e 1 1n 1 α n = 1n+1) e n+1 e 1 1n 1 1n+1) n ) n+ 3 n n+1 n = e n+1 n+1 α n, e 1nn+1) where α n = e R n+1 R n. Then we deduce that n+1 n+1 k) n+ 3 )n+1) 1+ 1 ) 4n +5n+ 3 n n+1)n+1) n = β n, k=n+1 e n+1)n+1) e n+1 1n where β n = α n+1 n = e n+1)r n+1 R n). Since R n R n n +1) Inthesameway,R n+1 n +1), so β n 1. Also, e n+1 1n using all that and Proposition 9, we get n+1 k=n+1 α n 3 16n, we will have 3n+1) 16n,andwhenn,wehavethatR n n +1). k ) n+1 n+ 3 )n+1) e n+ n n+1)n+1) e n+1)n+1) e 1 1 = e 1 e 4n +5n n n +3n+1. e n +n Proposition 11. The following formula holds n i + j) g 5 k 1 eπ nn 1 4 n +n. 6 i,j=1 e 3n e 1 1. By

21 I.-D. Ciorogaru, Stirling type formulas 1 Proof. UsingProposition4weobtainln = n k 1) ln k + k=1 and hence so n i,j=1 n+1 k=n+1 i + j) = n i,j=1 i + j) = n i,j=1 n ) k k k=1 n +1 k)lnk =ln n ) ) n n+1 n+1 k ) k k k=1 k=n+1 n ) ). n+1 k k k k=1 k=n+1 k k=1 lni + j) = n+1 k=n+1 n+1 k k=n+1 ) n+1 k k ) From Glaisher-Kinkelin, Stirling, Propositions 8 and 1, we obtain ) g k n n n + n e n e ) 4 1 e 4n +5n n n +3n+1 e n +n i + j), πn n n ) e 1 n +3n n 3n + 5n +1 i,j=1 e n e 3n 4 n i + j) g 5 k 1 eπ nn 1 4 n +n. 6 i,j=1 Acknowledgments. The author thanks Prof. Univ. Dr. Dumitru Popa for his helpful suggestions that make possible to obtain these formulas. References [1] Gh. Gussi, O. Stănăşilă, T. Stoica, Matematică, Manual pentru clasa a XI-a, Editura Didactică şi Pedagogică, Bucureşti, [] D. Popa, Exerciţii de analiză matematică, Biblioteca Societăţii de Ştiinţe Matematice din România, Editura Mira, Bucureşti 7. [3] D. Popa, The Euler-Maclaurin summation formula for functions of class C 3, Gazeta Matematică SeriaA, Nr. 3 4, 16, 1 1. e 3n

22 Articole Some generalizations and refinements of the RHS of Gerretsen inequality Marius Drăgan 1), Neculai Stanciu ) Abstract. This paper presents some generalizations of the RHS of the Gerretsen inequality. Keywords: Blundon inequality, Gerretsen inequality, geometric inequalities, the best constant. MSC: Primary 51M16; Secondary 6D5. In [5] appears the inequality a + b + c 8R +4r due to J.C. Gerretsen. In [6] L. Panaitopol proves that the inequality of Gerretsen is the best if we suppose that a + b + c αr + βrr + γr, where α, β, γ R and β =. In [7] the same result as in [6] is shown for α, β, γ R and β. In [8] R.A. Satnoianu gave the following generalization of Gerretsen inequality a n + b n + c n n+1 R n + n 3 1+ n n+1) r n, for any n. The purpose of this article is to give a proof of Satnoianu s inequality in the case n = 6, i.e., a 6 + b 6 + c 6 18 R 6 38 r 6, then we prove that this inequality is the best if we suppose that a 6 + b 6 + c 6 α 1 R 6 + α R 5 r + + a 6 Rr 5 + α 7 r 6, where α 1,α,...,α 7 R and α,α 3,...,α 7. Also we give some refinements for this inequality and we shall prove these refinements are the best of their type. Theorem 1 fundamental inequality of Blundon s inequality). For any triangle ABC the inequalities s 1 s s hold, where s 1,s represent the semiperimeter of two isosceles triangles A 1 B 1 C 1 and A B C, which have the same circumradius R and inradius r as the triangle ABC and the sides a 1 = R + r d)r r + d), b 1 = c 1 = RR + r d), a = R + r + d)r r d), b = c = RR + r + d), where d = R Rr. A proof of this theorem is given in [3]. 1) Mircea cel Bătrân Highschool, Bucureşti, Romania, ) G.E. Palade Secondary School, Buzău, Romania,

23 M. Drăgan, N. Stanciu, The RHS of Gerretsen inequality 3 Theorem some minimal and maximal bounds for the sum a 6 + b 6 + c 6 ). In any triangle ABC is true the double inequality 4 R + r d) 3 [ 4R r + d) 3 + R 3] a 6 + b 6 + c 6 4 R + r + d) 3 [ 4R r d) 3 + R 3]. 1) Proof. If we replace x = a, y = b, z = c in the identity x 3 =3xyz + x x ) yz, cyc cyc cyc cyc then we obtain a 6 =3a b c + a a 4 ) b c. cyc cyc cyc cyc We consider the functions f,g,h,f :[s 1,s ] R, fs) =3a b c =34Rrs),gs) =s r 4Rr), hs) = a 4 bc) cyc cyc ) 4 = a 4 ) ab a +6abc a + ab cyc cyc cyc = s 4 r4r+7r)s +4R+r) r, F s) =fs)+gs)hs). Since h s) =4ss 4Rr 7r )ands s 1 16Rr 5r > 4Rr +7r, it results that h is increasing on [s 1,s ]. Also f and g are increasing on [s 1,s ]. Hence, F is increasing on [s 1,s ], and then F s 1 ) F s) F s )or a b c 6 1 a 6 + b 6 + c 6 a 6 + b 6 + c 6. ) If we replace a 1,b 1,c 1,a,b,c from Theorem 1 in ) we obtain 1). Theorem 3 some maximal bound for the sum a 6 + b 6 + c 6 ). In any triangle ABC holds the inequality a 6 + b 6 + c 6 18R 6 38r 6. 3) Proof. From ) it follows that to prove 3) it will be sufficient to prove that 16R + r + d) 3 [ 4R r d) 3 + R 3] 18R 6 38r 6. 4) If we put x = R, then the inequality 4) can be written as r ) [ 3 ) 3 16 x+1+ x x 4 x 1 x x + x 3] 18x 6 38, x, cyc cyc )

24 4 Articole which is successively equivalent to 64x 5 +64x 4 +48x 3 48x + 56x 384) xx ) 64x 6 +48x 5 64x x 19x 944, 16x )4x 4 +1x 3 +7x 74x + 1) xx ) 16x )4x 5 +8x 4 +19x 3 91x + 16x + 9), 4x 4 +1x 3 +7x 74x + 1) xx ) 4x 5 +8x 4 +19x 3 91x + 16x +9, 48x 8 +4x x 6 +78x 5 +41x 4 177x 3 94x +197x+8464, x, which is true since 816x 6 94x = 816x x 4 94 ) 816x x 4 16) 816 and 78x 5 177x 3 = 78x 3 x 177 ) 78x 3 x 4), x. 78 Theorem 4 the best maximal bound for the sum a 6 +b 6 +c 6 is 18R 6 38r 6 ). If α 1,α,...,α 7 R and α,α 3,...,α 7 with the property that the inequality 5) is true in every triangle ABC a 6 +b 6 +c 6 α 1 R 6 +α R 5 r+α 3 R 4 r +α 4 R 3 r 3 +α 5 R r 4 +α 6 Rr 5 +α 7 r 6, 5) then we have the inequality α 1 R 6 +α R 5 r +α 3 R 4 r +α 4 R 3 r 3 +α 5 R r 4 +α 6 Rr 5 +α 7 r 6 18R 6 38r 6 in any triangle ABC. Proof. In the case of equilateral triangle, from 5) we have that 6 α α + +α 6 + α ) If we consider the case of isosceles triangle ABC, b = c =1,a =,R = 1, r =, then by 5) we deduce α ) Taking into account R r, 6) and 7), we successively obtain that α 1 18)R 6 + α R 5 r+ α 3 R 4 r + α 4 R 3 r 3 + α 5 R r 4 + α 6 Rr 5 +α 7 +38)r 6 [α 1 18) 6 + α 5 + α α α 5 + α 6 + α 7 +38]r 6,

25 M. Drăgan, N. Stanciu, The RHS of Gerretsen inequality 5 or α 1 R 6 + α R 5 r+ α 3 R 4 r + α 4 R 3 r 3 + α 5 R r 4 + α 6 Rr 5 + α 7 r 6 18R 6 38r 6. Theorem 5 the best constant for certain Gerretsen type inequality). The best real constant k such that the inequality a 6 + b 6 + c 6 18R 6 + krr 5 k + 38)r 6 8) is true in every triangle ABC is k 1 = Proof. Inequality 1) yields that if 8) is true in any triangle ABC, then 16R + r + d) 3 [4R r d) 3 + R 3 ] 18R 6 + krr 5 k + 38)r 6 9) is true in any triangle ABC. The inequality 9) is successively equivalent to 16R+ r+ d) 3 [4R r d) 3 + R 3 ] 18R 6 +38r 6 kr 5 R r), 1 [ 16x+1+ x x x) 3 [4x 1 ] x x) 3 + x 3 ] 18x k, x, 16 4x 5 8x 4 19x 3 +91x 16x 9) +16 x x 4x 4 +1x 3 +7x 74x + 1) k, so the best constant is the maximum of the function f 1 :[, ) R, f 1 x) = 16 4x 5 8x 4 19x 3 +19x 16x 9) +16 x x 4x 4 +1x 3 +7x 74x + 1), i.e., k 1 =max f 1x) = x Remark 1. The best integer constant for which the inequality 8) is true in every triangle ABC is k 1 = 569. Hence, in any triangle ABC holds the following inequality a 6 + b 6 + c 6 18R 6 569Rr r 6. Theorem 6. In any triangle ABC holds the following inequality a 6 + b 6 + c 6 18R 6 + krr 5 k + 38)r 6, for each k [k 1, ), where k 1 represents the best constant for the inequality 8). Proof. If we consider the increasing function F :[k 1, ) R, F k) = 18R 6 38r 6 + kr 5 R r), then by 8) we infer that a 6 + b 6 + c 6 F k 1 ) F k) 1) for any k k 1.

26 6 Articole Remark. If we take k =, then by 1) we obtain a 6 + b 6 + c 6 F k 1 ), i.e., a refinement of inequality 3). A generalization of Theorems 5 and 6. For n {1,, 3, 4, 5}, find the best real constant k such that the inequality a 6 + b 6 + c 6 18R 6 + kr n r 6 n n k + 38)r 6 11) is true in any triangle ABC. Proof. By 1) we have that if the inequality 11) is true in any triangle ABC, then 16R + r + d) 3 [4R r d) 3 +R 3 ] 18R 6 +kr n r 6 n n k + 38)r 6 1) is true in any triangle ABC. If we denote x = R, then the inequality 1) r becomes successively 16x+1+ x x) 3 [4x 1 x x) 3 + x 3 ] 18x x n n k, x, or 16 4x 5 8x 4 19x 3 +91x 1 16x 9)+4x 4 +1x 3 +7x 74x+1)16 x x k. x n 1 + x n + + x n + n 1 So, the best constant is the maximum of the functions f n :[, ) R, f nx)= 16 4x5 8x 4 19x 3 +91x 16x 9)+164x 4 +1x 3 +7x 74x+1) x x x n 1 + x n + + x n + n 1. If n = 1, then we find the best constant from Theorem 5. If n =, then using Wolfram Alpha) we find k = If n = 3, then using Wolfram Alpha) we find k 3 = If n = 4, then using Wolfram Alpha) we find k 4 = 96. If n = 5, then using Wolfram Alpha) we find k 5 =. Remarks. 1) The best integer constant for 11) in the case n =isk = 197, so the inequality a 6 + b 6 + c 6 18R 6 197R r r 6 13) is true in any triangle ABC; ) The best integer constant for 11) in the case n =3isk 3 = 419, therefore the inequality a 6 + b 6 + c 6 18R 6 419R 3 r r 6 14) is true in any triangle ABC; 3) The best integer constant for 11) in the case n =4isk 4 = 96, thus the inequality a 6 + b 6 + c 6 18R 6 96R 4 r 3 147r 6 15)

27 M. Drăgan, N. Stanciu, The RHS of Gerretsen inequality 7 is true in any triangle ABC; 4) Also we proved that the inequality a 6 + b 6 + c 6 18R 6 569Rr r 6 16) is true in any triangle ABC; 5) If we compare the inequalities 13), 14), 15) and 16) we observe that there are not two of them with right-hand side ur, r) andvr, r), respectively, such that ur, r) vr, r) foreach R r. Theorem 7. If n {1,, 3, 4, 5}, theninanytriangleabc holds the following inequality a 6 + b 6 + c 6 18R 6 + kr n r 6 n n k + 38)r 6 for any k [k n, ), where k n represents the best constant for 1). Proof. We consider the function G :[k n, ) R, Gk) = 18R 6 38r 6 + kr 6 n R n n r n ). Since G is increasing function, from 8) we have that a 6 + b 6 + c 6 Gk n ) Gk) for any k k n. Theorem 8. If n {1,, 3, 4, 5}, α 1,α,...,α 7 R, {i 1,i,i 3,i 4 } = {, 3, 4, 5, 6} \{6 n +1} such that α i1,α i,α i3,α i4, α 6 n+1 k n, with the property a 6 +b 6 +c 6 α 1 R 6 +α R 5 r+α 3 R 4 r +α 4 R 3 r 3 +α 5 R r 4 +α 6 Rr 5 +α 7 r 6, 17) then we have that the inequality 18R 6 +k n R n r 6 n n k n +38)r 6 α 1 R 6 +α R 5 r+ +α 6 Rr 5 +α 7 r 6 18) is true in any triangle ABC, wherek n represents the best constant for 11). Proof. In the case of equilateral triangle from 17) we get α 1 R 6 + α R 5 r + α 3 R 4 r + α 4 R 3 r 3 + α 5 R r 4 + α 6 Rr 5 + α 7 r ) In the case of isosceles triangle with the sides b = c =1,a =,R = 1, r =, from 17) we have α )

28 8 Articole Taking into account 19) and ), we obtain that α 1 18)R 6 + α i1 R 6 i1+1 r i1 1 + α i R 6 i+1 r i 1 + α i3 R 6 i3+1 r i3 1 +α i4 R 6 i 4+1 r i 4 1 +α 6 n+1 k n )R n r 6 n +α n k n )r 6 [ α 1 18) 6 + α i1 6 i α i 6 i +1 +α i3 6 i α i4 6 i 4+1 +α 6 n+1 k n ) n + α n k n ] r 6 = [ α α 5 + α α α 5 + α 6 +α 7 7 ] r 6, which yields that α 1 R 6 + α R 5 r+ +α 6 Rr 5 + α 7 r 6 18R 6 + k n R n r 6 n n k n + 38)r 6. References [1] W.J. Blundon, Problem 1935, The Amer. Math. Monthly ), 11. [] W.J. Blundon, Inequalities associated with triangle, Canad. Math. Bull ), [3] M. Drăgan and N. Stanciu, A new proof of the Blundon inequality, Rec. Mat. 1917), [4] M. Drăgan, I.V. Maftei, and S. Rădulescu, Inegalităţi Matematice Extinderi şi generalizări), E.D.P., 1. [5] J.C. Gerretsen, Ongelijkheden in de Driehoek, Nieuw Tijdschr. Wisk ), 1 7. [6] L. Panaitopol, O inegalitate geometrică, G.M.-B 87198), [7] S. Rădulescu, M. Drăgan, and I.V. Maftei, Some consequences of W.J. Blundon s inequality, G.M.-B 11611), 3 9. [8] R.A. Satnoianu, General power inequalities between the sides and the circumscribed and inscribed radii related to the fundamental triangle inequality, Math. Ineq. & Appl. 5),

29 SEEMOUS 18 9 Olimpiada de matematică a studenţilor din sud-estul Europei, SEEMOUS 18 1) Gabriel Mincu ), Ioana Luca 3), Cornel Băeţica 4), Tiberiu Trif 5) Abstract. The1 th South Eastern European Mathematical Olympiad for University Students, SEEMOUS 18, was hosted by the Gheorghe Asachi Technical University, Iaşi, România, between February 7 and March 4. We present the competition problems and their solutions as given by the corresponding authors. Solutions provided by some of the competing students are also included here. Keywords: Diagonalizable matrix, rank, change of variable, integrals, series MSC: Primary 15A3; Secondary 15A1, 6D15. Introduction SEEMOUS South Eastern European Mathematical Olympiad for University Students) este o competiţie anuală de matematică, adresată studenţilor din anii I şi II ai universităţilor din sud-estul Europei. A 1-a ediţie a acestei competiţii a avut loc între 7 februarie şi 4 martie 18 şi a fost găzduită de către Universitatea Tehnică,,Gheorghe Asachi din Iaşi, România. Au participat 84 de studenţi de la 18 universităţi din Argentina, Bulgaria, FYR Macedonia, Grecia, România, Turkmenistan. A existat o singură probă de concurs, cu 5 ore ca timp de lucru pentru rezolvarea a patru probleme problemele 1 4 de mai jos). Acestea au fost selectate de juriu dintre cele 35 de probleme propuse şi au fost considerate ca având diverse grade de dificultate: Problema 1 grad redus de dificultate, Problemele, 3 dificultate medie, Problema 4 grad ridicat de dificultate. Pentru studenţi, însă, Problema 3 s-a dovedit a fi cea cu grad ridicat de dificultate. Au fost acordate 9 medalii de aur, 18 medalii de argint, 9 de medalii de bronz şi o menţiune. Un singur student medaliat cu aur a obţinut punctajul maxim: Ovidiu Neculai Avădanei de la Universitatea,,Alexandru Ioan Cuza din Iaşi. Prezentăm, în continuare, problemele de concurs şi soluţiile acestora, aşa cum au fost indicate de autorii lor. De asemenea, prezentăm şi soluţiile date de către unii studenţi, diferite de soluţiile autorilor. 1) ) Universitatea din Bucureşti, Bucureşti, România, 3) Universitatea Politehnica Bucureşti, Bucureşti, România, 4) Universitatea din Bucureşti, Bucureşti, România, 5) Universitatea Babeş-Bolyai, Cluj-Napoca, Romania,

30 3 Articole Problema 1. Fie f :[, 1], 1) o funcţie integrabilă Riemann. Arătaţi că 1 1 xf x)dx f x)dx 1 < 1 fx) +1)dx fx)dx Vasileios Papadopoulos, Democritus University of Thrace, Grecia Juriul a considerat că această problemă estesimplă. Concurenţii au confirmat în bună măsură această opinie, mulţi dintre ei reuşind să găsească calea spre rezolvare. Juriul a considerat că partea mai delicată asoluţiei este demonstrarea faptului că inegalitatea cerută este strictă, motiv pentru care punctajul maxim pentru această problemă a fost acordat doar acelor studenţi care au argumentat în mod satisfăcător această chestiune. Soluţiile pe care dorim să le propunem diferă în esenţădoarîn acest punct. Din acest motiv, în loc să prezentăm mai multe soluţii care coincid la nivelul detaliilor simple, am optat pentru varianta unei singure soluţii, în cadrul căreia chestiunea inegalităţii strictevafitranşată într-o lemă pentru care vom prezenta patru demonstraţii, pentru a căror ordonare am ţinutcontdecât de elaborate sunt elementele teoretice utilizate. Soluţie. Faptul că orice funcţie integrabilă Riemann pe un interval compact ale cărei valori sunt pozitive are integrala pozitivă este o consecinţă imediată a definiţiei integralei Riemann şi se va folosi în mod repetat în cele ce urmează fără vreomenţiune explicită. Începem prin a proba o lemă caresevadovediutilăatât pentru a legitima scrierea fracţiilor din enunţ, cât şipentruaarăta că inegalitatea finalăeste strictă. Lemma 1. Fie a, b R, a < b, şi h : [a, b] R ofuncţie integrabilă b Riemann care are toate valorile strict pozitive. Atunci hx)dx >. Demonstraţia 1. Presupunem că d b a a hx)dx =. Remarcăm că în această situaţie hx)dx =pentruoricec, d [a, b] pentrucarec < d, căci c altminteri am ajunge la contradicţia b a hx)dx = c a hx)dx + În continuare avem nevoie de d c hx)dx + b d hx)dx >. Lemma. Fie α, β R, α < β, n N şi g : [α, β] R o funcţie integrabilă Riemann care are toate valorile pozitive şi integrala nulă. Atunci

31 SEEMOUS există intervale închise nedegenerate I [α, β] astfel încât gx) 1 n pentru orice x I. Demonstraţia lemei. Presupunem contrariul. Atunci orice interval închis nedegenerat inclus în [α, β] vaconţine elemente x pentru care gx) > 1 n. Considerăm şirul de diviziuni Δ q ) q N ale lui [α, β] cu Δ q = α, α + β α,α+ β α ),...,β q q şi sistemele de puncte intermediare ξ q [ =ξ q1,ξ q,...,ξ qq ), unde pentru ] fiecare q N şi k {1,,...,q} avem ξ qk α +k 1) β α q,α+ k β α q şi gξ qk ) > 1 n. Pentru fiecare q N, suma Riemann corespunzătoare lui g, Δ q şi ξ q va fi q β α q gξ qk ), care este mai mare decât β α n, de unde β β α α gx)dx n >, k=1 contradicţie. Revenim la demonstraţia lemei 1: Conform lemei, există uninterval nedegenerat I 1 =[a 1,b 1 ] [a, b]astfelîncât hx) 1 pentru orice x [a 1,b 1 ]. Constatăm că restricţia lui h la I 1 se încadrează larându-i în ipotezele lemei. Continuând inductiv, construim un şir de intervale nedegenerate I n ) n N astfel încât I n+1 =[a n+1,b n+1 ] [a n,b n ]şi hx) 1 n+1 pentru orice x [a n+1,b n+1 ]. Atunci c not = sup{a k : k N } I n, deci hc) 1 n N n pentru orice n N, de unde hc) =, contradicţie. Demonstraţia. Întrucât [, 1] = h 1 [ 1 n N n, + )), există m N not pentru care F m = h 1 [ 1 m, + )) nu este neglijabilă Lebesgue. Există prin urmare un număr A>cu proprietatea că F m nu este conţinută în nicio reuniune de intervale pentrucare sumalungimilor nu-l întrece pe A. Din acest motiv, oricum am lua o diviziune Δ = a = x,x 1,...,x n = b) alui[a, b], printre sumele Riemann n hξ i )x i x i 1 ) asociate acesteia se vor număra i=1 şi unele care conţin subexpresii i L hξ i )x i x i 1 )satisfăcând condiţiile L {1,,...,n}, i Lx i+1 x i ) >Aşi hξ i ) 1 m pentru orice i L. Aceste sume Riemann vor fi mai mari decât A m, deci b a hx)dx A m >, contradicţie. Demonstraţia 3. Funcţia h fiind integrabilă Riemann, mulţimea punctelor sale de discontinuitate este neglijabilă Lebesgue. Cum intervalul [a, b] nueste neglijabil, h admite puncte de continuitate pe intervalul a, b). Fie c un astfel de punct. Atunci există δ>astfelîncât c δ, c + δ) a, b) şi hx) > hc) pentru fiece x c δ, c+δ). Se obţine b a hx)dx c+δ c δ hx)dx δhc) >, contradicţie.

32 3 Articole Demonstraţia 4. Fiind integrabilă Riemann pe interval compact), funcţia h este şi integrabilă Lebesgue; cum valorile lui h sunt pozitive şi b a hx)dx =, deducem că hx) = aproape peste tot pe [a, b], contradicţie. Revenind acum la soluţia problemei 1, constatăm, aplicând lema 1, că numitorii fracţiilor din enunţ sunt nenuli, deci aceste fracţii au sens. Avem xfx) <fx) pentru orice x [, 1); aplicând lema 1 punând în loc de 1 f1) pentru a ne încadra în condiţiile acesteia), obţinem Evident, <) 1 1 xfx) dx< fx) +1)dx 1 1 fx) dx. 1) fx)dx, ) ultima integrală fiind strict pozitivă conform lemei 1. Din relaţia ) obţinem 1 < 1 fx) +1)dx 1 1 3) fx)dx, iar din 1) şi 3) obţinem inegalitatea dorită. Observaţii. 1) În cele precedente s-a probat de fapt inegalitatea din enunţ pentru orice funcţie integrabilă Riemannf :[, 1], + ). ) La această problemă 5 studenţi au obţinut punctaj maxim. Problema. Considerăm numerele naturale m, n, p, q 1şi matricele A M m,n R), B M n,p R), C M p,q R), D M q,m R), aşa încât A t = BCD, B t = CDA, C t = DAB, D t = ABC. Demonstraţi că ABCD) = ABCD. Această problemă nu este originală, fiind dată în anul 4 la concursul universităţii Taras Shevchenko din Ucraina. Din păcate acest lucru a fost remarcat foarte târziu, iar juriului i-a fost adus la cunoştinţă în dimineaţa concursului, când era dificil ca problema să fiescoasădinlistăşi înlocuită. Soluţia 1. CuP ABCD = AA t M m R) avem P 3 =ABCD)ABCD)ABCD) =ABC)DAB)CDA)BCD)= = D t C t B t BCD)=BCD) t BCD)=A t ) t BCD)=ABCD = P. Apoi, P 3 = P P P )P + I m )=O m. 4) Vom demonstra cădetp +I m ). Presupunem cădetp +I m ) =. Aceasta implică existenţa lui X M m,1 R) \{O m,1 },aşa încât P + I m )X = O m,1. Atunci,

33 SEEMOUS P + I m )X = O m,1 AA t + I m )X = O m,1 = X t AA t + I m )X = A t X) t A t X)= X t X. 5) Dar, pentru orice Y =y 1,y,...,y m ) t M m,1 R) avemy t Y = în timp ce 5), în care X O m,1,aratăcă m yi, i=1 A t X) t A t X)= X t X<. Această contradicţie ne asigurăcădetp +I m ),şi astfel relaţia 4) implică P = P. Soluţia Dragoş Manea, student, Facultatea de Matematică, Universitatea Bucureşti). Matricea P ABCD = AA t M m R) estesimetrică AA t ) t =A t ) t A t = AA t ), şi astfel diagonalizabilă cu valorile proprii λ 1,...,λ m reale). Deci, există o matrice inversabilă S M m R), aşa încât Pe de altă parteavem P = S diag λ 1,...,λ m ) S 1. 6) P = P t ABCD =ABCD) t, ABCD) t = D t C t B t A t =ABC)DAB)CDA)BCD)=ABCD) 3, deci P 3 = P. Astfel, din 6) obţinem P 3 = P diag λ 3 1,...,λ3 m )=diagλ 1,...,λ m ) λ 3 k = λ k, k =1,...,m, ceea ce implică λ k { 1,, 1}, k =1,...,m. Dacă arătăm că λ k 1, vom avea λ k = λ k,şi cu aceasta justificarea egalităţii P = P este încheiată, căci S diag λ 1,...,λ m ) S 1 = S diag λ 1,...,λ m ) S 1 P = P. Să demonstrăm, deci, că λ k 1; de fapt, vom arăta că λ k. Valorile proprii λ 1,...,λ m ale matricei P sunt rădăcinile polinomului caracteristic asociat lui P, detxi m P )=X m σ 1 X m 1 + σ X m + 1) m σ m. Aici, pentru fiecare k =1,...,m, coeficientul σ k este egal cu suma minorilor principali de ordin k ai lui P. Considerăm un astfel de minor, care provine din liniile j 1,...,j k. Remarcăm că, deoarece P = AA t, acest minor se scrie

34 34 Articole ca L j1 L j1,l j... L j1,l jk L j,l j1 L j... L j,l jk det... L jk,l j1 L jk,l j... L jk L j1 ) =det. L t...l t, j1 jk = 7) L jk unde L j1,...,l jk sunt linii ale matricei A, iar, şi reprezintă produsul scalar euclidian, respectiv norma euclidiană în R m. Folosind faptul că detmm t ) pentru orice M M m,n R), din 7) rezultă că minorii principali ai lui P sunt nenegativi, şi cu aceasta σ k, k = 1,m. Acum, presupunând că polinomul caracteristic are o rădăcină λ<, ajungem la contradicţia =detλi m P ) = λ m σ 1 λ m 1 + σ λ m + 1) m σ m = = 1) m { λ) m + σ 1 λ) m σ m ) }. Cu aceasta demonstraţia egalităţii P = P este încheiată. Observaţii. 1) Justificarea faptului că deti m + P ) este nenul, echivalent, matricea P AA t nu are ca valoare proprie pe λ = 1, poate fi făcută remarcând că P este matrice pozitiv semidefinită, AA t v, v = A t v, A t v = A t v, v M m,1 R), şi apelând la rezultatul cunoscut conform căruia o astfel de matrice are toate valorile proprii. Această observaţie a fost folosită decătre unii studenţi la soluţionarea Problemei. ) Într-un spaţiu vectorial V cu produs scalar matricea având componentele v i,v j, unde v 1,...,v n V,senumeşte matricea Gram corespunzătoare sistemului de vectori v 1,...,v n, iar determinantul acesteia determinantul Gram. Astfel, dacă A M m,n R), matricea AA t este matricea Gram corespunzătoare liniilor lui A, iar determinantul din formula 7) este un determinant Gram. O matrice Gram este pozitiv semidefinită şi orice matrice pozitiv semidefinită este o matrice Gram), în particular, un determinant Gram ca produs al valorilor proprii ale matricei corespunzătoare) este nenegativ. Această proprietate este apelată în soluţia a problemei, prin remarca det MM t. Inegalitatea det MM t se mai poate demonstra folosind formula Binet- Cauchy pentru calculul determinantului produsului a două matrice. 3) La această problemă de studenţi au obţinut punctaj maxim.

35 SEEMOUS Problema 3. Fie A, B M 18 R) cu proprietatea că AB = BA şi A 18 = B 18 = I, unde I este matricea unitate. Arătaţi că dacătra) = 18, atunci tra) =trb). Vasileios Papadopoulos, Democritus University of Thrace, Grecia Soluţia 1 a autorului). Deoarece A 18 = B 18 = I şi AB = BA avem că AB) 18 = I. De aici obţinem că valorile proprii ale matricelor A, B şi AB sunt rădăcini de ordinul 18 ale unităţii. Din faptul că trab) = 18 deducem că valorile proprii ale lui AB sunt toate egale cu 1. Mai general, dacă z 1,...,z 18 sunt numere complexe de modul 1 şi 18 z k = 18, atunci z 1 = = z 18 = 1. Aceasta rezultă imediat: scriemz k =cosa k + i sin a k şi z k = 18 deducem că 18 cos a k = 18, deci cos a k = 1 pentru orice din 18 k=1 k=1 k =1,...,18 şi, în consecinţă, sin a k =pentruoricek =1,...,18.) Aşadar polinomul caracteristic al matricei AB este P AB =X 1) 18. Pe de altă parte, polinomul minimal al matricei AB, μ AB, divide pe X 18 1, respectiv pe X 1) 18, deci μ AB = X 1 iar aceasta înseamnă că AB = I. Astfel am obţinut că B = A 1. Se ştie însă că valorile proprii ale inversei unei matrice sunt inversele valorilor proprii ale acelei matrice. În cazul nostru inversele valorilor proprii ale matricei A sunt egale cu conjugatele lor deoarece au modulul 1), iar de aici deducem că trb) = tra) = tra), ultima egalitate având loc pentru că matricea A este reală, deci şi urma sa este tot un număr real. Soluţia. Fie λ k, k =1,...,18, valorile proprii ale matricei A, respectiv μ k, k =1,...,18, valorile proprii ale matricei B. Deoarece AB = BA, matricele A şi B sunt simultan triangularizabile peste C), adică existăomatrice inversabilă U M 18 C) cu proprietatea că matricele UAU 1 şi UBU 1 sunt superior triunghiulare. Acest lucru se demonstrează prin inducţie după dimensiunea matricelor observând că AB = BA implică faptulcă cele două matrice au un vector propriu comun.) Aşadar valorile proprii ale matricei AB sunt eventual după o renumerotare) λ k μ k, k =1,...,18. Se putea, de asemenea, argumenta că matricele A şi B sunt diagonalizabile, deoarece valorile lor proprii sunt printre rădăcinile de ordinul 18 ale unităţii, deci distincte, şi atunci sunt simultan diagonalizabile.) Din trab) = 18 se obţine,calasoluţia 1, că λ k μ k =1pentruorice k = 1,...,18, adică valorile proprii ale lui B sunt inversele valorilor proprii ale lui A şi ne găsim astfel în situaţiadelasoluţia 1. Observaţii. 1) Soluţiile studenţilor care au rezolvat complet această problemă aufostîn spiritul celor douăsoluţii prezentate mai sus, diferenţele apărând doar la nivel de detalii. ) La această problemă 15 studenţi au obţinut punctaj maxim. i=k

36 36 Articole Problema 4. a) Fie f : R R o funcţie polinomială. Să se demonstreze că e x fx)dx = f) + f ) + f ) +. b) Fie f : R R o funcţiecareadmitedezvoltareîn serie Maclaurin cu raza de convergenţă R =. Să se demonstreze că dacăseria f n) ) este absolut convergentă, atunci integrala improprie şi are loc egalitatea e x fx)dx = f n) ). n= n= e x fx)dx este convergentă Ovidiu Furdui, Universitatea Tehnică, Cluj-Napoca, România Soluţii şi comentarii. Toţi concurenţii care au rezolvat punctul a) au dat una dintre următoarele două soluţii. Surprinzător, au fost unii concurenţi şi chiar dintre studenţii români) care au primit puncte la acest,,exerciţiu de seminar. a) Soluţia 1. Fied gradul lui f. Integrând prin părţi, obţinem e x fx)dx = e x ) fx)dx = e x fx) + e x f x)dx = f) + e x f x)dx. Repetând integrarea prin părţi şi ţinând seama că derivatele lui f sunt tot funcţii polinomiale, se obţine e x fx)dx = f) + f ) + e x f x)dx = = f) + f ) + + f d) ) + = f) + f ) + + f d) ), e x f d+1) x)dx deoarece f d+1) =. Soluţia. Deoarece f este funcţie polinomială, conform formulei lui Taylor avem fx) = d n= f n) ) n! x n oricare ar fi x R,

37 SEEMOUS unde d este gradul lui f. Drept urmare, avem e x fx)dx = = d n= f n) ) n! d f n) ). n= e x x n dx = d n= f n) ) n! Γn +1) Punctul b) s-a dovedit a fi cea mai dificilă problemă din concurs. El a fost rezolvat complet doar de patru studenţi Ovidiu Neculai Avădanei, George Kotsovolis, Georgios Kampanis şi György Tötös). Soluţiile celor patru studenţi s-au bazat, în esenţă, pe teoremele clasice de convergenţă teorema convergenţei uniforme, teorema convergenţei monotone sau teorema convergenţei dominate). Prezentăm mai jos soluţia autorului, precum şi două dintre soluţiile date în concurs de studenţi din Grecia. b) Soluţia 1 a autorului). Pentru orice număr natural n notăm n n S n := f k) f k) ) ) precum şi T n x) := x k, k! k= cel de-al n-lea polinom Maclaurin al lui f. CumT n k) ) = f k) ) oricare ar fi k {, 1,...,n},în baza lui a) avem k= e x T n x)dx = S n pentru orice n N. Pentruadovedicăintegrala improprie e x fx)dx este convergentă, v vom arăta că limita lim v e x fx)dx există şi este finită. Folosim teorema lui Bolzano. Fie ε>arbitrar. Deoarece seria f n) ) este convergentă, există unn N astfel ca n=n +1 n= f n) ) < ε. Cum integrala improprie e x T n x)dx este convergentă conform punctului a)), limita lim v v e x T n x)dx există şi este finită. În baza părţii de necesitate a teoremei lui Bolzano, există unδ>în aşa fel încât pentru orice v, v δ, ) săavem v e x T n x)dx < ε. v

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