GAZETA MATEMATICA Seria A N0 1-2/2018 by Leon Petrakovsky

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GAZF,TA MATEMATICA SERIA A Nr. 1 2/ 2or8

ANUL XXXVI (CXV)

ARTICOLE Computing exponential and trigotrotnetric functions of matrices in .\ 1- -. ()rrr,ir F.

.' - :' -:,' r:iji rtlation o{' ... '. ':.:''1. '111o-1ie ' .' :'-'l:r:ilti lhc'real

Abstract. [rL this papcr \\1:' givr ,. :-' ' tir, rrralr'ir ,,r1,,,1' ttlial frtn, ti,,r flrnctions sin;1 tucl cos jl . \\'lrcr(i 1a.'. 1- l,,girtitlrtrr,rlllr,'ttralrix.r'l.r,1,-:-:.'lr'r :.alal tililllil)le ,,l tut al jull ;rlr,l , . |eal cilt'trlaltt tttatrix aii(l a i\'llllir,r': ll -: Keywords: S, 111;114 ltr r' .' ttiatLir lIig()l11)lIlatli( tlltri ti,,lr- 'l-. :-'. NISC: lr \ lti.

.-1

':.: .

:-_r'-.-:

.--"

1-l-i(111('

fol the calt'ttla-

....--','Hatlriltc,tr Theoretn anci .' ratrrrell. i. 1,r.., . :tlre yrou,er s('ri(': ri-:- -'.1-:- l- . : - I ---- ,. nllll tllc tf igonometric fl-ulctiotis. The orgarliZtttiL-rl- - -- ' :l :l ' - -' 'ii'ru':: in the first section we gir-e fornulas lirr e1. sttt -f ':, . - -- -t -'ttr-: ll the cigcrlr.tilues of ,4 and in ..' r-,riiial ecllt.Itiotls h ,Mlz (R.) arrd lve tlrt' seconcl s('ttiorl \\'c :r ,,\r - ' ( ()lt)lln[e t]Lc lt,trl logzit'it1rtr, - -: - -,. 1..r,- irliitfi('eS. Tlro riCxt thc'orelnS are 1rr, rrLain r('slllt,ii of tltis s('( l i( )li. 1

Tlreorenr 1. The efrporlerltial functicttt, ,'\. 1)Departrrrcrrt of NlatherrLatics. -ftr'lurical l-rrirersit.v of Clrr,i-Napoca. Ronttrttia. Ovj.diu. Furdui@math. utcluj . ro , ofurdui@yahoo . com

ARrrcoln

Let A e Mz (C.) and let a. B e C be the e-igenL'alues ctf -\. The followi,ng forrnula holds

(eo_ eB

oeJ-Jc'

jf a+l' i.f a.: 0.

I:' 11-.1A-o- J "o:l [e^A * (e" - oeo)12. I

Theorem 2. The trigonometric function sin-{. Let Ae Mz@) andleta,li e C be the tigtnialues of A. Thefollou'ir,-c formula holds

- sin 6.0 o sin J ..irr o ,^. il a t' i. a- ) cr-] I il o:J. [(coso)A+(sin0-ocoso,ll.

sinA:1I

sino

Theorem 3. The trigonornetric function cos--1. Let A e Mz (C) and let a, P e C be the tigent'alues of A.

The followi'ng

formula holds COS

cosA: Before analr.sis.

COS

u- l) (sina)A *

p--A +f- clc .r-

eolz

sin(a12)

,'f a:

a ierlnra ure need in

our'

The followi,ng statements hold:

(sina)12; (coso)12;

(d) if A,B e Mz(C) commute, then

(") if A,B e Mz(C) (f ) i.f

eol2;

: (r) cos(a12) : (b)

12.

p rove these tI bheo rems we collect

"ve

(cCOS (a-v*o a sino)12,

Lemma 4, Let a e C. and let A e Mz (C).

(o)

-cos

commute, then

eA+B

e"{eB;

sin(A*B) : sinAcosB*cosAsinB;

A,B e Mz(C) commute, then cos(A+B): cos,4cosB - sinAsinB.

Proof. The proof of this lemma can be found in [4. pp. 189 192]. tr Now we are readl-to prove these theorems. Proof. First we consider the case e + P. \A/e have. based on the Cayley Hamilton Theorem, that (A - aIz)(A - 0Iz) : Oz. Let X - A - aI2. It follows that X(X + (a 8)12) : Oz ivhich implies that X2 : (0 - a)X. This

FUNCTIoNS oF N'IATRICES tN

O. FuRour, Colr.lpurrxc

in turn implies that

X" :

- o)"-r X, for all n )

Mz (C)

7. We have,

eA :e aIc*X : eol2ex Do vn

" /-/ -oo\-^ n,:o

n).

/,x"\ eo r1" 't--

- n)"-'x

\'=t/ / eB-o-1

:e" (12+#X) J-a \/

: e' I 1., +

ea-o-t

o-a

\ - olr) -'/)

_eo-"dr,oej-0"nr: -r -i-4 t2' a-3 o-0 \orv we consider the case a : P. The Cayley-Harnilton Theorem irn:--:'-. tliat (.A- aI)2:Oz.Let X - A-a12. It follows that,4: X +alz '.:, - -Il : Oz.\Ve have,

atldTiir,-,:::-,-..-I-''\ot- r','t :-'.': --: - - . l---- t:It- L

Proof. Onenreti--r ,::'.:--,-.-tl--:-:.-:t-tsba-.edolttheuseof Eulerrnatri,r formula {_ t, 1 _ . "-r.1

cornbined to Theorem l. \\-e ieave these details to the interested reader. Irrstead we prove the theorenr b]-using a power series technique. First we consider tirc case a + .3.-{s in the proof of Theorem l we have, basecl on the Cayley Hamilton Theoreur. that (-'l -aIz)(A- 0Iz) - 02. Let X : A-alz. It follows that X(X + (a - ,3)12): Oz which implies t'hat X2: (1j - a)X.

ARrrcolB

This in turn implies that Xn sin,4 :

sin(X -f aI2) :

- a\n-lx, sin

for all n

) l.

We have,

X cos(al2) * cos X sin(al2) sinX(cos a) + cos X(sin a).

On the other hand,

sinx: i,-, L^. n:u

YJ":(2n + t)l

: i' -1)'(9-:t:x (2n + 1)! '. ?' n:0 !

t ) ^r2rr-l I \-r-1,n,t \J - u/ o ,L B_a4' (2n_r-1)l ' n,:o

- o) ll-a

sin(6 and

cosX

tz2rt

^ : \-r-1)' .L^, (2n)l n:(l

: rz *i,-

,,"

n:I

i,- t2t^ -r-L-1 i-

,-l .),'r' l2n )l

,,n(o '

ll' - "

- a)2' (zn)l. n*

cos(P : r.2 , ___;__-\. ' J-a Putti,g ali these together we have, after simple calculations, that l_ cos(J \vU\u-- q/ a)- 1(A1 I - o)'(.coso)(A sin-{: sin(J aI2)l sinn /\ - alz) -/,+ lb + .l_n \

- sin0, * a-A

sina

a sin0

0_a

L-"

- 0,srnsfr. o-b Norv n.e consider the case a : 13.The Cayley Hamilton Theorem implies that lA-aI2)2 -Oz. Let X - A-a12. lt follows that,4 : X+alz and X2 - Oz. \Ve have, sin,4:sin(ol*X) : sin X cos(o12) * cos X sin(a/2) _

:sinXcosa*cosXsina :Xcoscr*12sina

(cos a)

A 1- (sin a

cos

12.

O. Funorr. Cortptrt\G FL'\crIo\s oF \I.{TRICES IN M2 (C)

\Ve nsed that sin X

: X and cos X : Iz.

The proof of Theorem 3 'ivhich is similar to the proof of Theorem 2 is Ieft as an exercise to the interested reader.

Corollary 5. Functions of refl,ecti,on rnatrices. Let a,b e R.. The followinq erlualzti,es hold:

(1)

:") : (,osn,/pr.) ,, * "(;

b\ (' (1 b\ -'i"gly \u -u) lo2 y fiz 16 -o )' (3) ."= (l l,): (,,"JF +E) u

(2) .i,

Rernark 6.

Obserr-e

that

la t \ r- ^ lcos? sin0 \ \/ o' + b' ' (.in o (; -") "oro) : Thus- any matrix of the form r'-r,ete cos H : ;4,zL and sin 0 i#. (': ' ) t. a scalar multiple of a reflection matrix. The reason why the \b a) tll^) is called a reflection matrix is given in p. 1a]. rnarrix ('?"1, [a. \srnU -cosU) Corollary 7. Let a,b,c e iR.

The followi,ng statements ltold

( cosh v& { "i,h /ta \ _ t/bc I ,t bc>0, ",f ? sinh v6r cosh v47 ) \ m / l, .-\ t1n u--6" \ / "ot J --il' -+= t/-bc c sin t/-bc _ l rt bc<0. ", 1 cos t/-b, \ /I /_br t \ -+

::):l

Corollar5r E.

Iit

,:.:,,.c

1?.

.*(::):l //

.i,,

The fol,louti,ng statements hold n

"os

\;.o.

sirr a cosir

lbc a"o.ori, Jbr""-- v47\

a"in

r - 6r' -

r -il t \*,=cosa:inh

sinacos,,/bc

cos a

) .-r sinh ,/

- sinacosh l-fr

-bc

ifbc>o. \

)

ARrrcor,s

Corollary

9. Let a,

.".(: 2): Corollary

c â&#x201A;Ź IR.. The followi,ng statements hold

( ;H.::; f"-:'::7,*)

i,f bc

lH;:t,:-J; r;.:;#=)

ifbc<0.

Functi,ons of rotation rnatrices. Let a,b e IR. The following erlualities hold: 1-O.

la -b\

(1)

"(,

o):""("?"? "

/o, -f\ _ (2) t'' (o o ): / o -a\ (3) tot(, o): Remark 11.

\sin b

_',1b), cosb

(sinacoshb -cososinhb\.

\"r.osinhb .;;;;.h; /' / cos acosh b sin a sinh b\ (-.ino.inna ."."."rr.a/

Observe that

lo -b) _ (coso -sina) (a u/ G +b, \srno coso/ where cos o : and sin * : Thus. any matrix of the form ffi ffi. la " I ; -a\I is a scalar multiple of a rotation matrix. The reason why the \D a) -^::":) is catlecl a roration matrix is given in [4. probtem matrix (:::: \slno cosa ) .

1.61,

31].

Remark 12. We mention that if r â&#x201A;Ź IR and A e Mz (R) the calculation of the matrix functions eA", sinlAr) and cos(Ar) has been done, by a completely different method, in [4, Appendix A, pp. 354-358].

If / is a function which has the N{aclaurin lrl < R, where

series

/(z) : i frPr",

A e Mz (C) is such that its B verify the conditions lal < R and l0l < ,8, then -R

(0, oo] and

n-0

eigenvalues

( f(a)-f_@).0*"f(il-Qf@)r,. ir 1+o' a_ tl f@):1'a-a (f (a) - of'(a))12, if a: 0. t /(")a + (

a and

(1)

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. 1941.

O. Funour. Cor,lpurrNc

FUNCTToNS

oF MATRICES IN -Mz

(C)

We mention that formula (1) holds in a more general case: when (t + 0 the formula holds for complex value functions / defined on spec(,A) : {a.3}. while if n : .J the formula is valid if / is differentiabie on spec(A) : {a} (see [2. Notes p. 221] and [3, P.11, Sect. 6.1]).

2. Let A e

if eB : A

TTTB REAL LOGARITHM OF SPECIAL I\{ATRICES

,A4z

(R). We say that B e Mz (R) is a real logarithm of A p. 718]). In this section we solve h Mz (R.) various

(see [1,

exponential matrix equations and hence we determine the real logarithm of special real matrices.

Theorem 13. Tuo etponential equat'ions. (a) Let I e Mz (lR). The soluti.on of the equat'ion eA rt

tr12, where

lK 1LSgtllen0a

-4:p(tn1r1-t)fr1ln(r(-kr,t)k) \p-, \ -r" )' where P e GLz (R) and k is an euen'integer i,f r > 0 and an odd i.nteger if r<o' / \ (I) Tlreequat'ion"': (1 X) u/ ,wherer,aelR, does nothauesolutions i,n

M2 (R).

Proof . (a) such that

If a,6 are the real

eigenvalues of A, then there exists

P e GL2 (R)

1)p-' 9)p-' o, A-P(1 -\:P(7 '\0 o)B)\0 ..-, : --:-.

r.'hether the eigenvalues of A are distinct or not (see [4. Theorem

I:-.-,:.'>:l-at

-. and this implies

that

,,11

lg 0 \ (.; .o):'r''

eo:&: o which impliesthat z ) 0 and a: g: lnr. This shows tb.at A: (lnr)12. If the eigenvalues of A are a*i,fl and a -i0, a e IR. and B â&#x201A;Ź lR*, then

Thus,

there exists P e GLz (lR) such that (see [4, Theorem 2.10, p. 79])

o^ \-p

o) p-t. a)

ARucor-p

This implies that

ll\ o)

: *b "(-B and it follows. based on part (1) of Corollary 10, that sinJ) :.rtc. srn l, cos ) ""('o?8. \Thus,eocos0:r ande*sinp:0. It followsthat sinB:0+ 0- kn, k e Z. The first equation implies that ea : r(-1)k + a - in(r(-1)k), where k is an even integer when r > 0 and k is an odd integer when r < 0. U

(b) Passing to determinants we get that

(' 1\ " :- a"t ""'\u o)

det eA

which is impossible over

er'(A)

-y2,

IR..

Remark 14. Part (a) of Theorem nratrix rI2 is given by

13 states

that the real logarithur of the

kr \ -(t"('(-,-])*' ln(r(-l\k))' -f" where P e GL2 (lR) and k is an even integer if r ) 0 and an r < 0. Part (b) of the theorem shows that the -at.i* (l : P ln(r12) 2)-1\

does not have a real logarithm.

odd integer if

'o)'

''9

€ R'

We mention that if A € Mz@) the equation eA : zI2, where z € C*, has been studied in [4, Problem 4.36, p. 219]. Also, the trigonometric equations sin A : 12 and cos A : 12 over the real square matrices have been studied in Appendix B of [4]. 'Ihe equations sinA : rI2 and cosA: rI2, with r e IR, can be solved similarly to the technique given in Appendix B of t4l.

Theorem 15. The real logari.thn'r of scalar rnultiple of rotation rnatrices. (a) Let a e Mz (R). The soluti,on of the equat'ion "A

:(; #) ,

g€R*,

i,s gi,uen by

f q1-r1ky.. A: | " 7r + t;

\(zr

;

(2k +

t);

In((-1)fre)

where k e Z 'is an euen i,nteger when'y>0 and an odd i,nteger when y

O, FuRnuI, Cotlpuumc

FUNCTIoNS oF i\.{ATRICES rN ,AZz (C)

(b) Let A e Mz (R) and let r,y e R*. The soluti'on of the equati'on

-Y),

"o:('\v i,s gr,uen by

where k e

tnzT? -lo + 2kr)\ o:\ ^ (rn o+2ktr 11r/Pafr)'

Z andO e (0, 2r) ct.,sd:

i,s such

-L t/",

that

nnfl -sinl:

-L.

t/*'

+ a'

- "\ A: (: d) s,,r." A commutes witn (f f). \e i )*" \r part (1) of Corollary 10, that A: (? -b) a/ rv" have, based on Proof. (a) Let

get that

: (l

(:nl

;,) ;:,:') ", and it follows that eocosb:0 and eosinb: g. The first equation implies that b : (2k+l)5, k e Z, and the second equation shows that e" : (-7)kA. Thus, a : ln((-1)kg), where k is an even integer when y > 0 and an odd integerwheng{0.

: (: f). s,r.. A commutes *itr, (j \v \c d/

(b) Let A -+

/^ :

( \0

r.\ _" Q )

-') *" x/

get that

\\'e have, based on part (1) of Corollary 10. that

(.?= | .. '\-inb

-'ilb\ _ f, -v) r)

co=h)-\'J 6r-'i -- -,:----.-. -: a: at --=: : : ai,i e'srni : ;,. This impLie: that e2o : r2*A2 tr: -#. Let d e (OrZr1 - c: -,- 1 .- - - ,--,i . .:'-,- ---=+- al,l si: r, rlil'# ,nr, be such thar cos , :------- u-,-'...1: :-.

\^ -

fsinb : sind /.=1 [cosb:coS0 where

k,ne Z.

Stn.cer,A

\l', :, - ]i:: or b: r - 0 +2kr, ltr-n-2n- or b:-012rur, T 0 these ca-=es imply that b:0*2kr,ke Z. -'

Rernark 16. Part (a) of Theorem

/o real mat,rix [ "

that the real logarithm of the

-,,\] i. gir-en b1-

^' \Y t)/ r,.

15 shows

(o -y\ :\s o )

y) - (2k * l) ro) t1; in((-1)kil )'

(tn-1.t-tf \rz,t +

ARrrcoln

where k e Z is an even integer when y > 0 and an odd integer u''hen g ( 0. Part (b) of the theorem shows that the real logarithm cif the rnatrix

(x -v) r \v r/

( * -y\ : (tt, 1/r2 + s2 -\0 n "'\y ,/:\

-r 2A'n)

o+2kr tnF+i)'

where k e Z and d

e (0,2tr) issuchthat cos0: -"{rr*0,

ancl sin0

Corollary L7. The real logarithrn of a rotation rnatrir. Let a â&#x201A;Ź (0.2n) \ {i,", f}. fne solut'ion,, i,n Mz(R), ,/

----L6+rr.

th,e r:quation

("g..i -sin,r\ \srn 0 "or., /

"r: ts g'iuen by

(a + 2k7Tt (0.

-^l) . k e z. u

The previous corollary shows that the real logarithm of the rotation

matrix

(cosn - sino) '. \srno cosG /

(:il; ;l:') -

(? l,')

t' e z 'r"^) Theorern 18. The real logarithm. of a circulant rnatrir. Let A e Mz (R) ond let r,y â&#x201A;Ź R, U 10. The equat'ion ',,

: (' Y\ '-\v ,) no

has solut'ions'in Mz(R) i/ and only case the soluti,on ts gi,uen by

if r > 0 and -r < y < r

and

this

(t,,r/ir] t,,\P, \ '- h\F, ntP:u,) ^, \ Proof. Sin.e A commutes,irt, based on corollarY 7'

(' 1) *" set rhat .4: (i '). \v r) " \b ,)

that^ /coshb e' : e'^ (slnh b

sinhb\ coshb)

follows that

( f

, :r)o "'' 1 eb+e-b

lpa "b --"-' :Y ("' 2 -

1""*o:r-l-y>0. \e'[z-a:r-y>0.

,,0 nur".

O. Funour, CoMpurtt{c

Thus,

-r < A ( r

b:rn,H

FUNCTIoNS oF MATRICES IN ,AZz (C)

and a calculation shows that o

:Int/*2 -V

und

Corollary L9. Let A e Mz(lR) and let g e IR, y ) -1. fne solution of the equat'ion

'o l.*

q1r.r

: (';'

,t. r)

bLl

A:tnJfr-(+ i) Proof. In Theorem 18 we let

Theorem 20. Letr,lJ e R

y -F 1

bc such

> 0.

thatry <0.

"': (: r,s

The solut'ion of th,e equati,on

gluen by

"- Tx.-. -\ hJ-rY *;l-:(zk*')l

^-l[ aL,f '^l,J-'a \ 2u-_.12a,-1) 'r' where k

Proof. have

)

> 0 i,s an integer.

Let,4: (: 3) . rr,u,.

that

we get that

(o O). \r' a/

Sirrce

-{ col,mutes *r,r,

(! 6) *"

Onr.r.-e that Corollalr-7 irnplies that br < 0 and

//-h-\cos J-bc

sin

y'-bc'l

-r\ "'[#sinr/-bc ';';: J-* _: f,(; ;) i which implies that

f.,F-0

r,/-bc : r. \" @sin sin ,/ - bc : s. 1"" i-tu

The first equation inrplies that J-6r': tQt * 1), where k > 0 is an integer. It foliows tnat e"fr(-1)k : r and eal+fre\o : a.These two equations imply that e2" - -:xA =) a : ln 1/-ry. Dividing the last two equations in (2)

ARrroor-p

| : t +b:1". Ontheother hand,bc: -(tQn 1r))'and it follows that c2 : -v (;Qn 1 r))'. This implies that c : t|1[-yen * t1 ! and b: - Gtr)' G;r[+ek + 1)) : +; yffoer *

weget that

i,:

Theorem 2L, The real logarithm of tri,angular matrices. Let

r,g e R*. The equat'ion

(* v\ ":\o *) ^o

has solut'ions i,n gi,uen by

Let A--1.---Proof.

Mz (R) i/ and only if

#\ o-lnro lnr)' ^-\

q\ since A commures *itn (l ') *" eet that \0 r/ \c d) ," this case both eigenvalues of A are equal to o,. It follows,

: ('

A: (2 a) 1)

based on Theorem 1,

that

,o Thus, ea

r > 0 and i.n thi's ca,se th,e solut'ion 'is

:t)

0 and beo

be'\ 1r u\ - ("o r)' 0 e'/-\0

A. This impiies that

: lnr and U : *.

Remark 22. Thepreceding theorem shows that the -utri* real logarithm if and only if r > 0.

(f f) \u r)

no.

Theorem 23. The real logarithm of syrnrnetri,c rnatrices. Let X â&#x201A;Ź Mz(R) be a symmetric matrir such that X f aI2, a e IR.*. The equati,on eA : X has solut'ions i,n Mz (R) i/ and only i,f Tt(A) )> 0 and det A > 0 and i,n thi,s case the solution i,s gi,uen by

- l,)2, , )'1'|z - ]zl')rrr. .4: l,]r)r-,\z lr-)z where

)r, )z

are the e'igenualues of

(3)

Proof. Since X * alz and ,4 commutes witli X we get, based on Theorem 1.1 in [4, p. 15], that A: aX * 0Iz, for some a,p e 1R.. The equation eA : X implies that eox+glz - X + etX : e-1X. Since X is symmetric and X * alz, a â&#x201A;Ź IR*, we know that X has real distinct eigenvalues (see [4, Theorem 2.5, p. 73]). Let Ar,,\z be the eigenvalues of X. Observe that, since A commutes with X, A is also symmetric and it has real eigenvalues. If p1 and p2 are the eigenvalues of A, then {e+l,euz} : {,\1,)z}, Thus,

O. FuRour, CorrlpuuNc

FUNCTIoNS oF MATRICES rN

Mz (C)

both eigenvalues of X should be positive real numbers, i.e., TY(A) ) 0 and det,4 > 0. The equation eoX : e-A X combined to Theorem 1 show that a)1ed^z - a,\2eolr 12

a),1-

a),2

s-0 Y.

This implies tliat

e^lr - eo), _- );g^^z - )2g^)r lz:'- _R., x-r

)r-),

and, since

-4

aI2, a + 0, we have that -

"a)r

)1eo)z - )2eotrr

"a.\z

)r-)z

follows that )1ea)z the other hand,

- l2eolt :0 + )r-)z u_

tLt-

"a()1-)2)

eaAt _ eo

-0.

- \

), -ln )c )r

-)z

)r-lz

:ln

.\,

.\"

(+)^'-^' - (f )^' "

: .)r In / )r \ ^r-lz ---;\E/ )rln)z-.\zln)r _

)r-)z

Thus,

ax + sI2-

)r -

ln )z

A1-AZ

x + I!]?-*Elrr, At-A2

"rd the theorem is proved. Remark 24. Theorem 23 shows that a symmetric matrix which is not a sr:a--:-: :r,,rltiple of the identity matrix has a real logarithm if and only if its eisr:-'..r--.re= are both positive real numbers and iS g-',-t:- :,-.' -,-,:nlffla (3).

in this case the real logarithm

R,pppRpNcns f1]

D \.

Berr..:=::-. ,l-lrrirt.z' Jlathernatics. Theory, Facts and Forrnulas, Princeton Uni-

versitl Pr, -:.

f2] S. R. Garcia anrr R.

Horn, .4 Second Cou,rse i.n Linear Algebra, Cambridge Uni-

versit;'Press. li'11.

[3] R. A. Horn and C', H. J,:hnson, Topics in Matri,u Analgsi,s, Cambridge University Press, 1994.

[ ] V. Pop and O, Furdui. -i4uare -\Ialrtces of Order Two. The.ory, Appli,catzons, Problents, Springer. 2017,

and,

ARrrcole

Stirling type formulas IoN-DpNys CroRocaRu') Abstract. In this

paper we prove two Stirling type formulas: r,2 n 9k 'nz z u+il- J2"

i+j<n;i,j€N

and

' ^ nn'-i.4"2'" -------,.,-. ti r jt - aH f[ v'(tt e i..i=1 2

*.here gr is the Glaisher-Kinkelin constant.

I(eywords: Stirling formula, Glaisher-Kinkelin constant. MSC: Primary 40A25; Secondary 41A60.

1. INrRooucrroN N

In this paper the notation and notion used are standard. in particular {1,2, ...} is the set of all natural numbers.

Definition l. Let (b")rerv be a sequence of real numbers wi,th the property that lrls € N szch that b, I O,Yn ) no. We ui,ll say that the sequence of real numbers (a,,)rex is equivalent wi,th (b,),,e x and we write an - bn i,f and : .lonlu if lim 9o on " " ,-+"o

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 2. If

bn then a2na1

b2n+r.

We need the following two results.

Proposition 3. Let "f ,

(0,

oo) -+

IR. be

a functi,on. The following formula

holds

itj

<nri.j

f (i.+ il:i(k -

€N

f (k)

&:1

Proof. \\'e have: n-1

n-L

n-2

f (i,+i) : I (-r(c+i)) :I,ro+i) +I re+i) j= t 2fj<n;j,j€N i:1. 1|

1) N{ihai Viteazul Secondary School

Nr. 29 Constanla. Str. Cigmelei. Nr. 13. Constanta and Department of Nlathematics, Ovidius University of Constanla. Bd. \,Iamaia I24. 900527 Constanla. Romania. cidenys@gnail . com

I.-D.

n-(n-r)

+ + tj:r

CIOROGARU, STIRLING TYPE FORMULAS

f(("- 1) +:) :

f (2) +2f (3) +...+ (n-t) f (n)

:I(,k-1)

f (k).

k-1

Proposition 4. Let "f ,

(0,

co) -+ R be a functi,on. The following formula

holds

iJ:t

k-7

Irtn+7) :Itr

1)/(k)

2nl-L

(2n-t7 k)f (k).

k:n] l

Proof. We have

ro+r) :I (t L, j-t ,:1 rL

f (i+r) ) :Irrr+i) j:l

\J:1

-1 ln+ j): (2) +2f (3) +...+ (, +>_f f j-r

+(n-1)/(n+ 2)+...+f(n+n): Changing n + k + 1

+\,tt j:r

+i) +

f (n) +nf (n+t)

t,u-1)/(k)*Ir, A-1 &:0

f @+k+

1).

: i, we get

2ntl

(n+k+1)

It"-k)f ,k:0

: D Q"+1-i,)f (i,) i:n11.

and so n

L t tn+ i) :Ir* A'-

- 1)/(k) +

2n*7

(2n+r - k) f (k). t k:nI l

tr We recall two well-known results. Their proofs can be found, for example, in [7].

The Glaisher-I(inkelin theorern. The sequence 11 .22...Tln n2,n,7

is conuergent and i,ts stant.

limit

(r,,,)r.er,l wltere

rt-Tt2-r rz . g- a lim r, : gr is called the Glai,sher-Ki.nkeli,n

ARrrcor-B

The Stirling formula.

follouing formula holds

The

;1, -

i:1.

Jr;; X

\['e will use these results to obtain the tu,o Stirling type formu]as stated in the Abstract.

2. TUB ruArN RESULTS

Proposition 5.

The followi,ng formula holds

n2-n 2n2

fI t,*r)- +

'Iifu' '/zi "*-' Proof. Using Proposition 3, we get

i*j<n: i,j€N

i) :

t, +

rn(i +

i+j<n;

:ifn-

-!r,4, :

k:1

ft:1

l)rn

k:l

t,i€N

ktnk

,b:1

- h. II k : k:7

r,H f[r k:1

l[r

iljln:

t,i€N

From Glaist

".-Xi*#fin's

theorem and Stirling's fbrmula we obtain

il (i+i)-'ffi:hffi ,r2-,r-l

v att

I f.i <n:i..le N

,r2

n2-n-5

Proposition 6.

The followi,ng formula holds

2n-ll

1'o -2'/, k:n*1

!#:

I.-D. CIoRoGARU, STIRLINc typE

FoRI\4Lrt,AS

Proof. Using Proposition 2 in Stirling's formula, we will have 2nll .,2n+l

r- l-'2" Y!++ f[ k:1

An+t

2n*7

2n*1 ll k From tlre equality II k: +k:n*r ll k k:t 2n11.

Ik-

,k:n*

J-2-

(2n

and Stirling's formula we obtain

112n+1

(2n+ 1,2n'1.,12+,VN

T.\2rJ

nn .

en+L

,'2+\n ,n Since (1 +*)'"11

"n-1

-eurd f + - r'f.

"11'u-2va

2r"+I

k:n*l

Proposition 7.

\i'eger

+ 2t-1

The following

formula holds

(l - +)----- - {3 .e".

Proof. Indeed, we have (2n

(,*l)@'

+ 1)2 n"

_ i_ [(,.;r)"1 -:'lTI[-] J*\-##e--z; en+'z

:,li--**

"*-

l, -l'.. i)^,)

AIso,

:,'B(('*#)"-r )@#

))

(2"+r)2: (2"+r)2 (O+*)'" _r\ :,'5'I"\ ,.^ Hm n((r+*)'" _,\ -') -')--W;5'J\ , * ,

ARrrcor,p

lim

_1\. ,(o+*)r" e

n--+@

\ Flom the L'Hospital rule we have

)

,i* t (l+*li _r\:1u,,, -L):ril,i

i'If2.\ "

lnfl .J)

Inrl+/J_l

- |'

rr ,. t'/ f, - 1 :-llm rn(r+r)_1 i1O 2

1*c) i

In{

--1

I- ,. :-[lm

-I

ln(1

*r)-r

r_+O

l+r-1:-1.

:1rr,,,

4 2 r-+0 21 From the characterization of the limit of a function at a point with

that

sequences, we deduce

1. we

Proposition 8.

so that

7 \ -|

e-4.

\2n-l

/ It+Since ea"

r,* , tGSd

lim

rue

2n)

i2n+r)2 1

| 4:e,, -_E_ 4 4n.

t2n-112

2 - i/e3.e'.

obtail (t + #)

The following formula holds

2T!t

e2n2-3n ,**'+*, _ 2e,:/i .___________

k:n*7

Proof. Using Proposition 2 in Glaisher-Kinkelin formula, we have 2nl1 ^ _, "n'

k:7

- 9t, . (2n + 1)

2n*t Then from the equality ll lxk : x:*t

2n*L

**fl k:n*7

Qn+Lt2

gx. (2n-1)-------

'fi'no J=J-,!,

2n {

r+ I

2 L

2n+r r

(2n)----z-* 2

r-e'+t

we deduce that r2nt

t\2

(2n

l2n*l t2 , 2n+1 , r + l)----a-t--T--r n

n2 , n

n 2T2-t2.e-

(n-t)(3n*t]

l2rt1-1)2

1\--r-

-n (, * X) n2 ,,

t2 . e-

lp.nzrzrn.e-q en-rf

*'**i

u#f

, I

n,-TazTu.c

(n*1)(3n-1) 4

(,*

*)*r

I.-D. Ctonoca.Ru, SIIRI-IItG TypE

22ri2-3n. ]3

FoRMULAS

., *-'; -, (, * ,l

\'++

2")

#)"+* - ,/i *" obtain 2n+r r - 22n2+s,*# . n**!+r . {@ .en . rt

From Proposition 7 and (r +

k:n*l

7 e'-t-..l-llr-]L

- ,Proposition 9.

22n: - 3n .

'.'/.)

,''r'

formula hold- (l

Proof. We have

By Proposition

The fotlowi,ng

/ I 1 an2+sn+t t1+-l \ 2") : Iim --en;+n-+@

, ";

* *)n"'-Snr'3 -

e2.e2n.

--)'(,-i)'.'

Iim

z-+oo

l(,-i) t'e'" t,

''G

)

7, \Zn+tl2 12

('- *)....-

---or-) Proposition 10.

The followi,ng

-1,

fonnula holds

Proof. From Stirling's Theorem. see for example [7, Theorem 5], we have D-n

f{ r : J2r; .'j .",-- . "*".

Anrrcor,r,;

where lErl

< u{ub,Vn â&#x201A;Ź N. Changing nto2nf 1, we get 2nll (2n -t t12"+t . sR2n+7, . t' : \/,;e;TT "rr*E k:t

"2n*7

wlrere lRz,,-tl S

â&#x201A;Ź N. Using these rerlations) we u,ill obtain

*#;,f.Vn

2ni.1

2rtt7

ilk:

[:-n*l

f{ t' j#-:

2r (2n

(2n

1)2"+1

. e 12\2n+1) ett2n+t

":D l2n ' gttn

l{r

k:1

C",r!

Qn)'"-i (' . *)'"'

(2n t- t)2n-s . n-l'-i) :fr.(ln:

en+| . eTr;- t2t2bt'

ari-1 . e12n- t2t2*1

'-@+i) .an '

,2n*j

'.rntl

22'-i./r*a\2"/ \ ::'on,

en+l.et2nt2n.t.

where

{Yn, :

sRzn+r

-8 . Then

: LT.,u) / 2n*r 1 2n*r

gn:

we deduce that

2lz,t'))tzn vrl ( , *:

"' )

-i .

,,(n- t)(2rt*t)

we will have r(2n-t1)(Rzn+t-R"). Since 16r] < #, .(2n and when n -) oo, we have that R, + 1) -+ 0. gl' Irr the salrle way, Rzn+r ' (2n -t1) -+ 0, so 13, - 1. Also, e# "#. get that Proposition 9, we using all and

a2*+r

lR^1.(2n+ 1) < W,

where

/ 2:!t 1zn+t \o -r*,

2Qr. i)rzn ri1

zer/2

w Proposition 11.

24'n2

"2rt

n- l tl2n

t2. n(n I l)(2rr 1

' ' "S

+5rr . rr2n2 +:ln+l

+tt "2n,2

The folloui,ng formu,la holds

' ""'-# 'n'*n ;1 ,n * i) - *F"*o vtctt e i'j:l 2

-1)

I.-D. CIoRocARU, STIRLING

l" fl (i+i): i,i:t TL

Proof. Using Proposition 4 rvc obtain

T.trpE FORMULAS

i.j:r

ln(z + 7)

: i(k - 1)lnk r T rr*+ 1 - k)lnk : ln k--n*l k:1 and hence

ij:r

(n kk) (-+i'n)'"*' \ A-:1

-!,-) (-Tj,--)

+j) :

From Glaisher-Kinkelin, Stirling, Propositions 8 and 10, we obtain

t ,,, \ 2't2.â&#x201A;Ź'i

\0r.,22

|,-w ----er;rt,

(oa T) (*'* ;1 t ,o * j)

L'J:L

21n2+5n . n'2n2 -3rtt1

/ zrr/2

22n2+3n

.n**+*, e

Jn' 4

- ,!=i. n"'- l'#" vLttt

Acknowledgments. The author tharrks Prof. Univ. Dr. Dumitru Popa for his helpful suggestions that make possible to obtain these formulas. RopBRBNcps t1l 121

t3l

Gh. Gussi, O. Stdnaqild, T. Stoica, Matematicii, Iv[anual pentru clasa a XI-a, Editura

Didactici qi Pedagogicd, Bucureqti, 1996. D. Popa, Erercilii de analizd matematicd". Biblioteca Societalii de $tiinle l\'Iatematice din Romdnia, Editura 1\4ira, Bucureqti 2007. D. Popa, The Euler-Maclaurin summation forrnula for functions of class C3, Gazeta Matemati,cd. Seria A, Nr. 3 4, 2016. 7-12.

Asrrcor,o

Some generalizations and refinements of the RHS of

Gerretsen inequality N{lnrus DRXcaNt), NBcuLl.r Sramcru2) Abstract. This

paper presents some generalizations of the RHS of the

Gerretsen inequaiity.

Keywords: Blundon inequality, Gerretsen inequality, geometric inequalities, the best constant. MSC: Primary 51M16; Secondary 26D05.

In [5] appears the inequality a2 + b2 *

+ 4r2 dte to

J.C.

In [6] L. Panaitopoi proves that the inequality of Gerretsen is the if we suppose that a2 + b2 * c2 < aR2 + pRr. + 1r2,

best

8R2

Gerretsen.

lR. and 0 :0. the same result as in [6] is shov,.n for a, 13,1 e IR and p > 0. [7] In [8] R.A. Satnoianu gave the following gerrera,lization of Gerretsen inequality

where

a,0,1â&#x201A;Ź

+bn

*c' < 2n-tpn +z'(!+; \/

_ 2n- l).". for anv n

)

The purpose of this article is to give a proof of Satnoianu's inequality in the case ?z : 6, i.e., o6

128 . ,R6 - 3oo8 .16,

then we prove that ttris inequality is the best if we suppose that

+ b6 * c6 < a1R6 + a2Bbr+ ... + a6Rr5 * a7r6, where (tttc\2,...,(17 â&#x201A;Ź ]R. and (\2,a3,...,oZ ) 0. Also we give some refinements for this inequality and we shall prove these refinements are the best of their type. a6

Theorem 1 (fundamental inequality of Blundon's inequality). For any triangle ABC the i,nerlualitzes s1 ( s ( s2 ltold,, where s7)s2 represent the semi,peri,meter of two ,isosceles tri,angles ArBtCt a,nd A2B2C2, wh,ich haue the same c'ircumrad,,iu,s R, and ,inrad,ius r as the triangle ABC and the std,es at : 2l(Rt r - d,)(R - r * d), bt : ct : + r j, '/ZA1A a2:2J(R* r * d)(R - r - d), b2: c2: t/zA1A +, + a7, where d

: V@-Rr.

A proof of this theorem is given in 1) Mircea cel

[3].

B5trin Highschool, Bucuregti, Romania, 2) G.p. Palade Secondary School, Buzdu, Romania, marius.dragar2OO5@yahoo.con

tN,{.

DnXceN, N. SraNcru. THs RHS

GeRRorsEN INEeUALITY

Theorem 2 (some minimal and maximal bounds for the sum a6 + In any tri,angle ABC is true the double tnequali,ty

+ c6).

24(R+r-d)3 [+(n- r+d)3 +E'] <16+b6+c6 <211R-r-Fd)3 [4(E -r-a)3+n3] . Proof. If we replace r: a2. lJ:b2. :: c2 in the identity

r'- !rr) fc]'c r'' : jJr,,. - I,.i : (f cyc \ cvc / then

r.",e

(1)

obtain

\-,':

-.=.*i

.,).-,.---\-

-itn,-f

a,",)

c]'c

: (t,)'- +\ab(X,) '*u,o,Lo- (I,o)' \ .r'.

'.r.

,y, \ft )

\ ...

: s4 - 2r(4R+Tr)s2 + (4R+r)2r2, F(s) :/(")+e(s)h(s). Since h'(s) :4s(s2-4Rr-7r2) and s2 > > l6Rr -5r2 > 4Rr *7r2. "? results that h is increasing on [s1, s2]. Also / and g are increasing on lsr, sz]. Hence, F is increasing on ["r, "z], and then F(rr) < f.(") S F(s2) or

a!+u6r+"f < o6+bG *c6< If

,*'e replace a1,b1,ct,a2,b2,c2

"$+u$+"$. from Theorem 1 in (2) we obtain (1).

Theorem 3 (some maximal bound for the sum ABC holds the i.nequali,ty a6 +bG

*c6 <

c6).

(2)

In any triangle (3)

t28R6 -300816.

Proof. From (2) it follows that to prove (3) it will be sufficient to prove that 16(.R+ r+d,)s

we put tr

: i,R

(r+ s

1/az

lt@-, - d)3 +E']

then the inequality

- n)'

128-R6

( ) can be written

- 300816.

(4)

("- ; 1/*'-2,)' * "'] < L2Br6 -2008, vr )

AR'rrcor,p

which is successively equivalent to (6415

{

+ 64ra +

4Br3

64rG

4Br5

206413

-t

- JBq\[r@ -t 406g12 - 1920r - 2944,

204812

2b60r

* 2T* - T4r + t2)\/r@ - 2) < 16(r - 2)(415 I Bra * r9r3 - g7r2 f 106r + g2),

16(r

2)@ra

(4ra

I r

+ t2r3 + 2Tr2 - T4r-t t2)\/,4r -

4r5

Bra

tgr3

+24017 +8163.6 +TB0r5 +24tra

4818 Y

t2r3

)> 2,

gtr2

-t|T2rs

1062

-920412

92,

+lgT2r*8464 >

which is true since 81616 _ 92o4rz:81612

(. _ #) > 81612(."4 _ 16) > 0 816 ) \

and 78025

- r712r3:780.rr (r, \

g) > 780r3(r2 4) > o, vt.z2. 780)f

Theorem 4 (the best maximal bound for thc sum a6+b6+co is 12g,R6-300g16). I/ (1r,a2,...,a7 â&#x201A;Ź R. ond a2,a3,... ,(t7 > 0 with the property that the i,nequali,ty (5) is true i,n euery tri,angle ABC a6 +bo

+c6

atB6

+a2R5rla3R4r'+onr3r3 +o5R2ra +a6Rr5 +a7r6,

(b)

then we haue the i,nequali,ty a1 R6

a2 R5

a3 Ra 12

a a R3

r3 + a5 R2 ra + a6 Rrs

az rG

128R6

- 300816

i,n any tri.angle ABC.

Proof. In the case of equilateral triangle. from (5) we 2uol + 2'o, + ...

2a6

we consider the case of isosceles triangle

r:0,

+ a7 > ABC,

ha'".e

that (6)

222.

b: c:

a:0,

then by (5) we deduce a1

Taking into account

) 128.

R> 2r. (6) and (7), we successively obtain that

(a1-128)rB6+ azRsr+ asRar2+ a4R3r3+ a.sR2r4* a6Rrt'+ (crz+soog)r6

> [(o,

-728)26

o2.

a3. 2a

aa. 23

o.s. 22

aa.

a7+3008]r6 > 0.

\I.

Dn.c.cA,ri.

N. Sr.qNcIrr. Tuo RHS or GsRRsrsEN

INEQUALITY

a1R6l .,2Rlrl a3R4r2+ aaRsr3+ a5R2ra* a6Rr5+ a7r6

)

728R6

-300816. tr

Theorem 5 (the best constant for certain Gerretsen type inequality)- The best real constant k such that the inequali'ty (8) o6 + b6 * c6 < 128R6 * kRrs - (2k + 3008)16 'is trure 'in euery trtangle ABC i,s h = -5269-275' Proof.Inequalitl- (1) yields that if (8) is true in any triangle ABC, then 16(E+r+d)3[4(R -r-d,)3 +]?'l < 128R6 +kRrt'-(2k+3008)16 (9) is true in an1. triangle ABC. The ineclualit\' (9) is successively equivalent to 16(n+ r+ d)3[4(.R - r- d)3+ /?3]- 128R6+300816 < krs(R-2r),

-i ftu,;r- t* /*z-n1314(r-1-v,r'2,

i't- r3l-

12816+30081

< k'

Yr)r2,

+16"F - Z* l4ra -t l2r3 + 2712 16(-415

Bira

1913

9112

106r

Jr) : i6(-4r5 -

Bra

7913

+rc.,81m(ua i.e.. kr

Tgf "fr(r) =

l9r2

92)

71r -t 12) < k.

so the best constant is the maximunr of the function :f

ft,1.2.

106r

+7213 +2712

R..

92)

-7)t

+12). tr

-5269.275,

Rernark 1. The best integer constant for u,hich the inequality (8) is true in

everytriangleABCisA'l:-5269'Hence'inarrvtriangleABCholdsthe following inequalitl'

a6+b6 *c6 <128R.6 -5269Rr5 *753015.

Theorem 6. In any triangle ABC holds th,e followi'ng i'nerluali'ty o6 + b6 + cG l-128R6 * ARr5 - (2k + 3008)16. for eaclr,,k e [k1, oo), where k1 represents the best constant for the i'nequali'ty (8) ' Proof. If we consider the increasing function F:lk1,oo) +iR, F(,k) :128R6-300816 infer that

+kr5(R-2'), thenby

o6+b6*c6<r(kr) <r'(k)

for any

k> h.

(8) we (10) D

ARnlcor.p

we take k : 0, then by (10) we obtain a6 + i.e., a refinement of inequality (3).

Remark 2. If

< F(k),

A generalization of Theorems 5 and 6. For n e {1,2,3,4,5}, find the best real constant a6

is true

'i,n

such that the inequality

L28R6

1 7rpnr6-n

(2*k

* 3008)16

(11)

any triangle ABC.

Proof . By (t) we have that if the inequality (11) is true in any triangle ABC, then

t6(R + r +

d13

1+1a

- r - a;3 + R3l < 128,86 a ABC.If

is true in any triangle

we denote

becomes successively

rc(r+t+t/M)eV@-t-t/7 or r6(4rs -8r4-1913+

rn -2n

pn r6-n

fr: {,r

- (2" k + 9008)16

th"r, the inequality

12)

(12)

-u)3 + r3l-para +zooa <.k,Yr)2,

t06r-9\+(aaa a12as a27a2 -74r-ll2)l6t/@ -2" rn 2 . 2+... + r. 2n-2 + 2n-7 So, the best constant is the maximum of the functions fn i l2,oo) -+ R, r, _ \ _ 16(-4;15 - 8r4 - 19u3 + gL* - to6r -92) + i6(4ra + 1273 +22.12 - ar + D) tff 9r:r2

(

-l-

xtn-t +

lf n:1, then we find the best constant from Theorem 5. lf n:2. then (using Wolfram Alpha) we find k2- -1297.57.

If n : 3, then (using \\bifram Alpha) If n :4, then (using Wolfram Alpha) If n:5, then (using Wolfram Alpha)

we find ks > -479.402. we find k+ -96. = we find ks 0.

Remarks. 1) The best integer constant for (11) in the case

n:2

kz:

-7297.

so the inequality a6

+b6

r2BR6

r29TR2ra

* 218016

is true in any triangle ABC; 2) The best integer constant for (11) in the case therefore the inequality

(13)

n:3 is kz:

+ b6 * c6 < 128.86 - 4tgRsr} * J44rG is true in any triangle ABC; 3) The best integer constant for (11) in the case n : 4 is kt thus the inequality

(14)

r28R6

96RarB

- t4z2r6

-419.

-96. (15)

\1. Dn,lcax. N. St'.tNcn.r, Tuo R,HS op GpRRersEN

INtrQUALITY

is true in any triangle ABC: 4) AIso we proved that the inequality a6

+b6 + c6 ( 128R6

5269Rr5

* 753016

(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 u,ith right-hand side u(B,r) and u(R,r), respectively, such

that u(-R,r) < u(fi. r) for each !ar.

Theorem 7. If n

{7,2,3,4,5}, then in any trtangle ABC holds the fol-

lowing i,nequaltty

+kRnr6-" - (2'A- 3008)16 for anyke lkr,n), where kn represents the best constant for tl)). o6

+bo

*c6 <

128R6

Proof. \Ve consider the function G: Lk". -)

-:-. G(k) : 128R6 - 3oo816 - l'''6 R' -'2'r'"

Since G is increasing function, froni (8) ri-e har-e that a6

< G(k") < G(,k) for

any k

)

k,,.

Theorem 8. If n â&#x201A;Ź {7,2,3,4.5}, at,(72,...,a7 â&#x201A;Ź R, {ir. ir. zr. l+} : {2,3,4,5,6} \ {6 - n * 1} (tit, c\it. oi:. crz+ ) 0, a6-21 | ) kn, with the property such, that s6 as6 a s6

(

o 1 -R6

n z R5 r + az Ra 12 + o n R3 13

o5 R2 ra + a6 Rr1

ru,

(

then we haue that the inequahty l,z8 R6 'is

+k,

Rn 16-

n- ( 2',k, +3008)16 < ay R6 I

R5r*

...

ae&rs +

a7r6 1ra)

true r,n any trr,angleABC. where kn represents the best constant for (71).

Proof. In the case of eqr-rilaterai triangle from (17) we get

arR6+ a2?br +ci3/?lr2+alE3r3 +asR2ra+o,6Rr5 laTrG >722. In the case of isosceles triangle r.vith the sides from (17) we have cr1

)

128.

7, a

:0,

'r,

(19)

(20)

ARucoln

Taking into account (19) and (20), we obtain that

aitR6-ittT ri't-r + airR6-iz*7riz-r * airftl-iz+t rnr-, *ainR6-it+trit-r * (ao-r,+r - kn)R"r6-" * (o, + 3008 *2,'kn)r6

(a1

128)ft6 +

) [(", - 12q261rl,ir 26-i'+t * oi, ' 26-i'+t +r,i:j 26-it+t + oi4 26-i'+t * (ao-,,+r - k,r)2" + CI7 + 3008 + 2"kn) 16 : fo,' 26 + a2 . 25 + as . 2a + at . 23 + a5 . 22 + aa . 2 * at - T2')16 ) 0, which yields that o1,R6+ azRt'r+... 1ai;-Rr.r'* a7r6

128R6

krR,16-,,

(2,rk,-* 300g)16.

RupBRpNcBs f1] W.J. Blundon. Problcm 1935, Tlte Amer'. Math,. Monthly Z8(1966). t122. [2] w.,I. Blunrlon, Inequalities associated with triarrgle. canad,. Math. Bull. g(1g6b),

618

626,

[3] M. Drdgan and N. starrciu, A new proof of the Blundon inequality, Rec. Mat. 19(2017). 100 104. Drdgan, I.V. Maftci, and S. Ridulescu. Ineqalitd"Ii Matematice (Ert,indrcr.t, q,i generalizdri), E.D.P., 2012. l5l J.c. Gerretsen, ongelijkheden in rle Drichoek. Nieuu Tijdsr:hr. wisk.41(1953), 1-7. [6] L. Panaitopol, O inegalitate geometric.i. G.Xt.-B 8Z(1982), i13 115. t7l S. Rddulescu, N,l. DrAgan. and I.\,'. \Iaftei. Some corrsequences of W.J. Blundon,s irrequality, G.ltf.-B 116(2011), 3 9. l8l R..\. Satrroianu. Gettcral pos-er itrr:qualities bctri'een the sitles arrd the circunrscribefl arrd inscribed radii rciated to the fundanrental triangle ineqrralitv, Math. Ineq. â&#x201A;Źt Appl. 5(2002). 715 751,

[4]

N4.

SEEIIOUS 2018

Olimpiada de matematicl a studentilor din sud-estul EuroPei, SEEMOUS 2018') B'[olrcaa) . TtsnRtu TnIr'5) G-.reRrBr N'ItNcu2), IoaNa Llica3), CoRNotOlympiad for 12tr' South Eastern European Nlathematical Gheorghe- Asachi the bv hosbed *'as 2018' SOEITOUS Students. Li;;;y 27 and March 4' Teclrnical University. taqi Ro*unia' bets'een February as giverr by the sol,tions their a,d probl"r-,* iv;^|i.-,;a tt " .orrip"tliio. competing stuthe of some by providecl authors. Solutions

Abstract. 'fhe

"orrirpo.ai,rg dents are also inclutled herc'

I{eywords: Diagorralizablc matrix' rank

change of variable' integrals' se-

ries

MSC: Primary 15A03;

Secondar-v-

15A21' 26D15'

INrnonucrioN SEENIoUS(SouthEasternEuropearr\IathematicalolympiadforUniadresat5 studen{iversity stuclents) este o competilie anuala de matematicx. A 12-a editie a aceslor din anii I gi Il ai universit[1ilor c,in sud-estul Europei' gi a fost gSzduit5 201-8 qi 4 martie tei competi(ii a ar-ut loc intre 27 februarie Romania' Au Iaqi' din de cxtre tiniversitatea Tehnic5 ..Gheorghe Asachi"

participat8ldestudenlidelal8uttiversitSlidinArgentina.Bulgaria,FYR N,Iacedonia. Grecia. RornAnia' Turknlenistan'

AexistatosirrguriprobSdecorrctrrs.Cu5orecatimpdeiucrupentru mai jos)' Acestea au fost rezolvarea a patl-tl p'Jf"t"" (problemele 1 4 de selectatedejuriuaintrecele35deprobleniepropusegiaufostconsiderateca avAnddiversegladecledificultate:Problerrralgradredusdedificultate, problemele 2, 3 dificultate medie. Problema 4 - grad ridicat de dificultate' cu grad ridicat de Pentru stuclenli. insa. Problema 3 s-a dovedit a fi cea dificultate. Au fost acorciate 9 lredalii

cle

aur, 18 medalii de argint, 29 de medalii de

bronzgiomenliune.Llrsingurstudentme<laliatCuaulaob{inutpunctajul ..Alexandru Ioan cuza" maxim: ou,id"iu l,lecrLlai -lt,dianet de Ia universitatea din Iagi. prezentim. in continuarc. problemele de concurs gi solutiile acestora, asemenea. prezentS,m qi soluliile aqa cum a* fbst indicate de autorii 1or. De aate ae c5tre unii studengi' diferite de soiu{iile autorilor' 1)

http: //rnath.etti.tuiasi ro/seemous201B/

gamin@fni' unibuc' ro Universitatea din Bucuregti' Bucuregti' RomAnia' 3)l]niversitatea politehnica Bucureqti, Bucuregti. Romirria, ioana.luca@mathem'pub'ro 4)Universitatea din Bucuregti. Bucuregti, Romania. cornel .baeti.ca@fni.unibuc'ro 5)universitatea Babeg-Bolyai. cluj-Napoca. Rornania. ttrif @math'ubbcluj 'ro

ARrrcor,B

Problema 1. Fie ,f [0,1] + (0,1) o func{ie integrabilS Riemann. Ardtali '

z *f'l*1a*

lo'

* l,' oar

1)dr

y'g1a,

nl-

t tco'

Vasileios Papadopoulos, Democritus University of Thrace, Grecia

Jurtul a cons'iderat cd, aceastd, problemd, este s,impld. Concurenlii au confir'rnat tn bund, mdsurd, aceastd, op,inie, mul{t, d,intre et, r'eugtnd sd, gd,seasca calea spre rezoluare. Juri,ul a cons'iderat cd, partea rnaz delicatd, a solu{te,t este demon,strarea faptului, cd, i,negali,tatea cer"utd. este strtctd, moti,u pent,ru care prr,nctajul marim pentru aceastd, problemd, a fost acordat doar acelor studenli co,re au argumentat tn mod sati,sfd,cd,tor aceastd, chest,iune. Solu[ii,le pe care dori,m sd le propunem di,ferd tn esenld. doar tn acest punct. Din acest moti,u, tn loc sd, prezentd,m mai, multe soluli,i, care coi,nc,id la ntuelul detali,i,lor si,mple, am optat pentru uarianta une'i si,ngure solulr,z, tn cadrul cd,re,ia, clrcstiunea i,negali,td,lii, stri,cte ua fi tranqatd, tntr-o lemd, pentru care r)orn prezenta patru demonstral'i'i, pentru a cd,ror ordonare am lrnut cortt de c6,t de elaborate sunt elementele teoret'i ce uti,lizat e.

Solutie. Faptul c5 orice functie integrabilS Riemann pe un interval compact ale c5rei valori sunt pozitive are integrala pozitivd este o consecin!5 imediati a definiliei integralei Riemann qi se va folosi in mod repetat in cele ce urmeazd fdrd vreo mentiune explicitd. Incepem prin a proba o lemd care se va dovedi util5 atAt pentru a legitima scrierea fractiilor din enunt, cAt gi pentru a ardta cd inegalitatea finald este strictS.

Lemma L. Fie a,b e IR, a

( b, gi, h:

[a,b] -+ R o func{ie

R'iemann care are toate aalorile stri.ct poziti,ue.

Demonstra{ia 1. Presupunem d

situatie

"d J"fo

Atunci

lr(r)dr:

h(r)dr : 0 pentru orice c, d e

i,ntegrabi,ld

ttlrlar > O. \

I J,,

0. RemarcXm c5 in

[o,,b] pentru care

aceastX

c ( d, cdci

altminteri am ajunge la contradictia

l"o

n1r1o,

in continuare

f,'

h(r)dr *

l.o

h(r)dr +

loo

n{dar

) 0.

avem nevoie de

Lemma 2. Fi,ea,0 e

IR.,

o ( 0, n e N* fz g:lc-,P) -+R

i,ntegrabi,ld, R'iemann care are toate ualorile pozi,tzue

qi,

ofunc{i,e

i,ntegrala nuld. Atunct

SF]EN"IOUS 20i8

eri,std, interuare tnchi,se ned"egenerate

ori,ce

la. pl astfet tncat

g(r) a I

nentr"u

r € I.

Demonstralialemei2.Presupunerncontrariul.Atuncioriceinterval r pentru care 9(r) > ] ' inchis neriegenerat inclus in [a, 6] va conline elemente 6] cu Consider[m qirul de diviziuni (An)qex. ale iui [ri'

l.,:(o'"r+ 1-\"'-

o+zi a^o"

(, (€ql. (qz' ' ' ' , {rr), unde pentru fiecare a € + (k- l)+'o +k?] aie((sk) > s €N-ri A'e { 1.2..-..q} avem€nr Lo

qi sisternele de puncte intermediar"

corespunz[toare lui 9,

|.q^pentru fiecare Q € N*, suma Riemann 2 e:f s6ol). care este mai mare decAt #, k:1

a" u"de

qi (n va

ff g@)d'r' *

'o'

contracliclie.

Rer'eninriademonstralialemeil:Conformlemei2,existSuninterval € ["'b']' < nedegenerat.Il: lor,bt] c [a,b] astfelinc6't h(t) lpentru.oricer la r6,ndu-i i, ipotezele lemei Constat6m ca restriclia lui h'la 11 se incadreaz5 nedegenerate (1/l)'ex. 2. continuand incluctiv. construim un qir de intervale pentru orice r € astfel incAt In'r : [a,+r.b,+r] C lo,,,bn) qi h(r') < # Pt .rp{, k i k €N.} e,!n-''' deci h(c) S'1 pentru lanat,b,+r]. Atunci . tr orice n € N*' de unde h(c) :0' conttadiclie'

Demonstratia

intrucAt [0,t]

!].-n-'(-* +-))'

existd rn

€

g, n-1 ([#,+*)) ,,, neglijabild Lebesgue. ExistS, prin pentru care F".;" nu este con{inut5, in nicio urmare un numar a >'d"1., proprietatea ca FreuniunedeintervaleperrtruCaresumaiunginrilornu-lintrecepeA.Dinacest : (a : J0' jtl: "''rn: b) a lui [a'b]' motiv. oricum a- Iru o diviziune A se vor numdra printre sumele Riemann i n((r)(', - ri-r) asociate acesteia i:l nta,)(ri - ri-r) satisfXcand condiliile 'L c qi unele care conlin suUe*presil .E > Aqi h(€i) > p"'ttt'oricei € L' Acestesume 11,2,...,n),

D(r,*t-,,)

i€L

Riemann vor fi mai mari rlec6t

' deci [! n61a" >

*'0'

contradiclie' D

Demonstralia3.FuncliahfiindiritegrabildRiemann,mullimeapunctelor Cum intervalul [4, b] nu este sale de discontinuitate esie neglijabil[ Lebesg,.e. (o', b)' Fie c un astfel h admite puncte de. continuitate pL intervalul

negiijabil, 6,c*d) : (.,,b) qi h(r) de punct. Atunci exist5 d > 0 astfel incat (c > dh(c)'+> 0' pentrufiecer e (c-d,c*d). Seobline [!n61a"> t:::h(r)dz

contradiclie.

ARucol,n

Demonstralia 4. Fiind integrabilX Riemartn (pe interval compact), functia h este gi integrabilS Lebesgue; cum valorile lui h sunt pozitive qi # h@)dr : tr 0, deducem cE" h(r):0 aproape peste tot pe [a,b], contradictie. Revenind acurn Ia solu(ia problemei 1, constatdrn, aplicand lema 1, ci nurnitorii fracliilor din enun! sunt nenuli, cleci aceste fraclii au sens. Avem r.f (r)2 < f (")'pentru orice r e [0,1); aplicAnd lenia 1 (pun6nd 0 in loc de l. f Q)2 pentru a ne incadra in condiliile acesteia), oblinem (o

.)l

@)2d,r

lo'

f {il'a,

(1)

/(r)d.r.

(2)

Eviclent.

lo'ora'*

1)dr

lo'

ultima integralS fiind strict pozitiv6 conform lemei 1. Din relaqia (2) obtinem

(3)

<-l-l

Jirtal'

+ 1)dr

iar din (t) qi (e) oblinem inegalitatea

- 2 I; /(r)dr

doritS.

Observalii. (1) in cele precedente s-a probat de fapt inegalitatea din enun! pentru orice fr-rnclie integrabil5 Riemann "f , [0,1] -+ (0. +co). (2) La aceastd problemi 5 studenti au obtinut punctaj maxim. Problema 2. Consider5rn numerele naturale m.t).p.q > | qi matriceie A e M,.,.,,(R), B â&#x201A;Ź Mr,,e(R), C e IZr,.,,(lR), D e Mq,,,r(IR), ar;a inc6,t At Demonstra\i

: BCD, cd,

(ABCD)'

: CDA,

: DAB.

ABC,

ABCD.

Aceastd problenr,d, nu este orr,ginald, fii.nd datd, irt, anul 2004 la concursul uniuersitalti, Taras Sheuc.h,enko di,n Llcrai.na. Dzn pd"cate acest l'ucru" a fost remarcat foarte tarziu, iar juriulu'i i,-a fost adus la cunoEttnld l,rt, dimi,nea{a concursulu'i, cd,nd era difi,ci,l ca problema sd, fie suasa din listd" qi, tnlocur'td,.

Solulia 1. Cu P: ABCD: AAt e .M-(R.) avem p3 : (ABC D)(ABC D)(ABC D) : (ABC)(DAB)(C DA)(BC D)

: : DtCtBI(BCD): (BCD)I(BCD): (At)t(BCD): AaCD:

Apoi,

: p <==+ (p' - P)(P + 1,,,) : O*. (4) Vom clemonstra cd det(P+I-) f 0. Presupunem c5, det(P+I,") : 0. Aceasta implicd existenla lui X e M,,,r(lR) \ {O-,r}, aqa incAt (P + 1",)X : O,m1t. p3

Atunci,

SEENIOUS 2018

: O*,1 â&#x201A;Ź+ llAt + I*)X : O*; + xt(AAt + I*)x : o +=+ (AtX)t(Atx) : -xtX. (P +

Dar, pentru orice Y

I*)X

(At,A2,... ,Arr)t

eM-,r(R)

avem

YtY: i

i':t

* O,n,r,arat5, cd (Atx)t(Atx): -xtx <0.

in timp ce (5). in care X

Aceastd contradiclie ne asigurS, cX det(P+1-)

P2:P.

O, qi

(5)

A? >-

astfel relalia (4) implicd

Solutia 2 (Dragoq Manea, student, Facultatea de Matemati,cd,, Uni,uersitatea Bucuregti). Matricea P: ABCD: AAt e M-(lR) este simetricd ((AAt)t : (At)tAt: AAt), gi astfel diagonalizabil5 (cu valorile proprii

)r,... . ).

reale). Deci, existd o matrice inversabilS

Sdiag

(,\1,...,

)-)

M-(R),

aqa incAt

S-1.

(6)

Pe de alta parte avem

p:pt (ABC D1t deci P3

DtCtBtAt

<+ :

ABCD:(ABCD)t,

(ABC)(DAB)(C DA)(BC D)

(ABCD)3,

: P. Astfel, din (6) obtrinem P3 : P <==+ diag(,\f,...,):") : diag()r,...,)-) <+ )f, : )r, Y k:1-,...,n'1,

ceea ce implicd )e e {-1,0,1}, V k:7,...,ffi. Dac[ aritlm cb )r # -1, vom avea A? : )r, gi cu aceasta justificarea, egalitS.lii P2 : P este incheiatS, cXci

- Sdiag ()r.... , )-)S-1 <+ P2 - P. S5. demonstr5m, deci, cE" ),61-1; de fapt, vom ardta cX ,\p ) 0. Valorile proprii )r, . . . , )- ale matricei P sunt rS,ddcinile polinornului ,Sdiag ()?,. . .,^?*) 5'-1

caracteristic asociat lui P,

: X* - o1f,m-1 + o2X^-2 - ...+ (-l)'"o,n. Aici, pentm fiecare k:1,... )m) coeficientul ok este egal cu suma minorilor d,et(Xln

principali de ordin k ai lui P. Consider[m un astfel de minor, care provine din liniile .jt,...,.7k. RemarcS,m c5, deoarece P: AAt, acest minor se scrie

ARrrcor-e

34,

/r,,

a"t

| , I (il,

,1-)

\t,^ /

reprezinta prounde -Lr, , . . . .L jo sunt linii ale matricei A. iar ;i Folosind faptul :-'-'. in dusul sialar euclidian, respectiv norma euclidiana c5 det(I/,l/r) > 0 pentru orice 1J.f e M,,.n(-(.). din l7i rezulta ca minorii principali ai lui P sunt nenegativi, qi c1 aceasta o;, ) !. A': 1.m' Acum, presupunind c5 polinomul caracteristic are o radacina ,\ < 0. ajungem la contradic!1a

0:detl^t,r- P) : )* _ ot\- | +o2^"' I - " - r-1\'no^ : : (-1)," {(-l)"' * or(-.\ r"r-L - ... 1o,n)} + O. Cu aceasta demonstralia egalit[1ii P2 : P este incheiatS.. observalii. (1) Justificarea faptului c5 det(1,,, -

matricea P: AAt nu are ca valoare proprie pe remarcAnd c[ P este matrice pozitiv semideflnita.

(AAtu.u)

\At'u. Atu\1

llAtt'll2 2

P I este nenul. echivalent, ) : -1. poate fi fScutd

0. t = .!4,,.1(lR.)

rezultatul cunoscut conform cdruia o a-.tfel de matrice are toate r-alorile proprii ) 0. Aceasti observalie a fost folosita de catre unii studenti la solu(ionarea Problemei 2. (2) intr-un spa(iu vectorial V cu prodlts scalar matricea ar.And componentele (r';. i_,r). unde utt...,r1r, â&#x201A;ŹV, se numegte matricea Gram corespunzfltoare sistemultii de vectori ul :....ur.,. iar determinantul acesteia - determinantul Gram. Astfel. dacd A â&#x201A;Ź M-,r(R.), matricea --llt este matricea Gram corespunzdtoare liniilor lui ,4. iar determinantul din formula (7) este un determinant Gratn. O matrice Gram este pozitiv semidefinita (gi orice matrice pozitiv semidefinit5 este o matrice Gram). in particular. un determinant Gram (ca produs al valorilor proprii ale rnatricei corespunzS,toare) este nenegativ. Aceasta proprietate este apelati in solulia 2 a problernei, prin remarca Ei apelAnd la

det

MMt > 0.

Inegalitatea det 11,41t ) 0 se mai poate demonstra folosind formula Binetcauchy pentru calculul determinantului produsului a douS matrice. (3) La aceastd problem[ 22 de stridenli au oblinut punctaj maxim.

SEEN{OUS 2018

Problema 3. Fie A,B e y\lzors(iR) cu proprietatea ci AB: BA\i 32018 - 1. unde 1 este matricea unitate. Ar[ttrli cd dac6 tr(A) atunci tr(A) : tr(B).

4.201'8

2913,

Vasileios Papadopoulos, Democritus University of Thrace, Grecia Deoarece l2oLB - 32018 - I ai AB: BA averrr De aici oblineur cX valorile proprii ale matricelor A, B qi ,48 sunt radacini de ordinul 2018 ale unitdtii. Din faptul ci tr(AB) : 2018 deducem cA valorile proprii ale lui /8 sunt toate egale.:l , (l\'Iai general,

Solu{ia 1 (a autorului). cd"

(AB)2018

: L

dac[:1.....:2018 sunt numere complexe de modul 1qi

I scriem ,ll

zr,:201.8, atunci

zt:... : :lurr : 1. Aceasta rezulti imediat: qi ".o"aplisinal 201f 2018 din f :r. :2018 deducem c5 D cosal.:2018, deci cos ak: I pentru orice A:1 &=1 k:1.....2018 qi, in consecint5, sinotr :0 pentru orice k - 1,...,2018.) Aqadar polinonrul caracteristic al matricei AB este P.q6 : (X - 1)zots. Pe de alta parte. polinomui minimal al matricei AB, pta, divide pe X2018-1, respectiv pe if - 1)z0ts. deci p.ne - X - l iar aceasta insearnni cd" AB: I.

Astfel am oblinut cd. B : A 1. Se qtie ittsA ca valorile proprii ale inversei unei matrice sunt inversele valorilor proprii ale acelei matrice. in cazul nostru inversele valoriior proprii aie matricei -1 sunt egale cu conjugatele lor (deoarece au modulul 1), iar de aici dedr-icenr ca tr(B) : t(,4) : tr(A), ultirna egalita,te avAnd loc pentru c[ matricea -{ este real5, deci gi urma sa este tot un num5.r real.

Solulia 2. Fie )1.. A': 1,...,2018, valorile proprii ale rnatricei A. respectiv H*. k : 1. . . . . 2018. r'alorile proprii ale matricei B. Deoarcce AB : BA, :ntatricele -1;i B sunt simultan triarrgularizabile (peste C), adicX existd o matrice inversabilA L- e -\12613(C) cu proprietatea cd, matricele LtALi-| Ai UBU-l sunt snperior triunghir"rlare. (Acest lucru se demonstreazl prin induc(ie dup5 dinren,.iunea uratricelor observAnd ca lB: BA implicA faptul cd cele dou[ matrice au un vector propriu conlurl.) Aqadar ralorile proprii ale matricei AB sunt (eventual dupa o renumerotare) l6pr6. A : 1.....2018. (Se putea. de asemeneal argumenta cd matricele -{ qi B sunt diagonalizabile. deoarece valorile lor proprii sunt printre rad5cinile de ordinul 2018 ale unit5tii, deci distincte, gi atunci sunt sinrultan diagonalizabile.) Din tr(,48) :2018 se obline. ca la solulia 1. ca \tltt :1 pentru orice k : 7,. . . , 2018, adicd valorile proprii ale lui B sunt inversele valorilor proprii ale lui A rsi rre g5,sim astfel in situalia de Ia solutia 1.

Observatii. 1) Solu{iile studcnliior care an rezolvat complet aceast6 problemd, au fost in spiritul celor doua solulii prezentate rnai sus, diferentele ap5rAnd doar la nivel de detalii. 2) La aceastd problemd 15 studenli au ob{inut punctaj maxinr.

ARucol-R

Problema

(u) Fie

]R.

lo* "-'f (b) Fie

IR.

(,r)clr

o funclie polinomialS,. S5 se demonstreze

: /(o) + /'(o) + /"(o) + ' "

-+ R o func{ie care aclmite dezvoltare in serie N{aclaurin cu Iaza

de convergen!5 1? : oo. s5 se demonstreze

lut

convergentd,. atunci

gi are

l<,rc

c5, clac5,

seria

; /t"l {o) este abson:0

i,tegraia improprie

[- ,-'f

(r) dr

este converge.ti

egalitatea Y,x

*f(r\a..:I-f""iu). n:0

Ovidiu Furdui,

Universitzrtea Tehnic[' Cluj-Napoca' RomAnia

concurenli,'i ca're au rezoluat Ttrr'nctul (a) a'u dat (gi una d,intre urmd,toar.ele d,oui solu{,i,i,. Surpri,nziitor, aIL Jost uni,i' conatren[t' acest la puncte 0 primit ,,ereTci'l'iu d'e chr,ar rli,ntre stutlenli,i rorn6"ni,) care au

Solulii qi coment arii. To[i sem'inar'".

(a) Solulia 1. Fie d gradui lui

IntegrAnd prin p6rti, oblinem

[- "-'l(r)dr : Jo[- ,-" ')'./(r)dr : Jn

(

-' Ilr')l; .

rJ'

-'l'1r;d;'

: f(o)+ | e-'f'(r)dr. Jo RepetAnd integrarea prin pdrli gi {inand seama c6 derivatele 1ui furrclii polirromiale' se obline

f .lu* "-"

a' :

/(o) + /'(o)

lo*

: /(0) + /'(0) + "'+ j

deoarece .1(d+t)

Solulia 2.

/(o)+ /'(o)

"-' ," f

('t)

('L')

(0)

: ''

sunt tot

(r)dr "-r /d'+r)

+ ..+,f('r)1s;. n

Deoarece

este

funclie polinomial5, conform forrnulei lui

Taylor avem --------- I" tlrl:.' J_ ) lrnr(0) /-/ n! n,-O

oriCare ar

fi r â&#x201A;Ź R,

SEEI\{OUS 2018

unde d este gradul lui

lo* "-' f

(')

d''

Drept urmare, avem

*ry

fo*

"-*

"a"

*+9 r(n +

1/t"){o).

n:o

tr Punctul (b) s-a douedi,t a fi cea mai, di,fici,ld, problemd' d'in concurs. El a fost rezoluat complet doar de patru studen{i. (Oui.di,u Neculai, Aud,danei', George Kotsouol'is, Georgi,os Kampan'is 9i, Gydrgy Tdtds). Soluliile celor patru, studenli s-au bazat, tn esen{d,, pe teoremele clasice de conuergenld' (teorema conuergenlei, un'iforme, teorema conuergenlei monotone sau teorema conuergen[ei dominate). Prezentdm mai, jos solu[i,a autorulu'i, precurn Ei doud di,ntre solulii,le date tn concurs de studen[i. di,n Greci,a. (b) Solutia 1 (a autorului). Pentru orice numdr natural n not5m

s, ::

l,k:0ftkrtol precum gi r,(r),: ik:0

cel de-al n-lea polinom Maclaurin al lui fi k e {0,1.... . n}, in baza lui (a) avem

/(1(o) ,r,

Cum Zlo)(O) :1@)(0) oricare ar

I e-'Tr(r) dr : Sn pentru orice n € N. JO Pentru a dovedi cX integrala improprie If f (") dz este convergentd, "-* qi (r) finit5. Folosim teorema existd este e-'f dz vom ardta c5 limita #

J11

lui Bolzano. Fie e ) 0 arbitrar. Deoarece."riu existd un ri.o € N astfel

n:o

n:no*l

l/(,)(o) |

l/(")(0)l

este convergent5,

dr este convergenti (conform puncCum integrala improprie ff "-rTno@) exist5 qi este finit5,. in baza pdrlii de tului (a)), limita J516 "-'Tno@)dr necesitate a teoremei lui Bolzano, existS un d > 0 in a4a fel incat pentru orice 't),'u' € (6, oo) sX avem

[ ."' '-.',,(')d'l < ; l"u

ARrrcor,r.;

Atunci pentru orice u, u' e (6,co) cu u < u', avem

2' '24

n:rr.o*1 T rt: t Lol x

: 1+ 2' t./'

qg l,'' ,-.,,d, r+qf(nr1)

r1t,)10)l< r

n-no*1

Cum e a fost arbitrar, in baza pir{ii de suficientra a teoremei lui Bolzano rezultd c[ limita,[1# e 'f (r)dr existS Ei este finitd. adic5 integrala improprie

If "-" f(") Pentruoricenâ&#x201A;ŹNavem

d;r este convergent5.

lr- e-'.f(r)dr-S,l| : lr* e-''(/(.rt-Tn(r)) lllJ,' I lll/o j

drl < I

rn '

L-

A.:n I

kl 1

k:nll CC

Convergenla seriei

t l/(k)(0)l a:0 av6nd limita [f, e-'f (r)dr.

impiicX faptul ca girul (S") este convergent, D

(b) Solu{ia 2 (datd, tn concurs de un student funclii 9, : [0, oc) -+ IR (, > 1), definite prin

t r(k)

grlr)::.-'t +' (o)l _r A-0 t-

di,n Greci,a).

Fie qirul de

SEEMOUS 2018

Evident, qirul (g"(r)) este cresc5tor pentru orice z € )o

S:: I

notand

&:0

[0,

oo) fixat. IVIai mult,

lyrkttO)1. avem

n_k

O<g,(r) (,5e-'Ih Prin urmare. existi

fr:0

pentruoricen> 1Eioricere

)yyS"@) €

IR oricare

ar fi u

€

[0,

[0,oo).

oo). Fie funclia

9: [0,co) -+ ]R. definitS,prin 9(r):: j!1O"(";.

Deoarecegiruldefunc{ii (g") converge uniform pe compacte, rezultd cd g este local integrabilS Riemann. In baza teoremei convergentei monotone, avem f'xfoofca lim g,(r)dr : lim I S,(r) dr. IJ g,*)d, : Jo I n-+x n-oc Jo ' Dar

I Ja

s,@)ar:

Ik=o l/(fr)(o)l

oricare ar fi n

conform celor demonstrate la (a). Deducem de aici

)

ut{or; :5. IJo glrldr: n-+x -linr f tf ,t,:O /,

deci /o- 9(r) dr converge. Fie acurn qirul de functii .f" : [0, co) -+ R f ,@) Deoarece

::"-"

ik:0

(, > 1), definite

lior

prin

in serie Ntlaclaurin pe IR., rezult5 ci e-" f (r) oricare ar fi z € [0, m)'

este dezvoltabil5

,,19

/"{"; :

Avem gi

pentru orice n > 1 gi orice z € [0,oo). Aplicdnd teorema convergenlei dominate. deducem cX fx f@ *f

f"(*) Sg"(r) ag(r)

l,* u

(r) dr

: / J0

lirn (r) dr n-)- /,

: n+ooJo lim I f"(r) dr.

Dar rd)

f,(r)d,r: t/(fr)(o)

oricare ar fi n

)

ft:0

conform celor demonstrate 1a (a). Drept urmare, avem ,oo

I e-" f(r)ar : Jo

Jg5!/tk){o) nk:o

: t/(k)(o). k:o

ARrrcor,B

tr (b) Solutia 3 (datd, tn concurs de un student di,n Greci,a). fn: l},oo) -+ IR (ri, > 1), definite prin

Se aplic5

girului de func{ii

fn,),:.-" /(;lo) ,,. urm5toarea variantd a teoremei lui Tonelli (a se r-edea R. Gelca gi dreescu, Putnam and Beyond. Springer, 2007. p. 178): dacX )c

roo @

/ f V*@)ldz ( ro sau t J\) ,

n:0

f).

'fn\.rt d.r

(

T. An-

>c.

n:gJU

atunci are Ioc egalitatea

/-; r*@)dr:tr-|,- l"@)dr. REmAne doar s5, notim x

c[. in ipotezele problemei. roo

I1Jo/

\'' ' ' -''' :Il/'"'o' lf,@)ldr rr "

'r:u

x)orx

./, (.r)

dr : /

f,/r.r r dr'.

J l)

n=O

precum

ar-em

,crxx

t / f,(r)dr:f /" urL'

n o'u

Observa$ie. Pentru Problema de 9 dintre studenti.

n:0

ror.

punctajul marirn (10 puncte) a fost obgimrt

A. B6ruvr, I. Cegu, A snppv cASE oF MATHEMATICaI (lr,lts)tNouc'rtou

NOTE MATEMATICE

A happy Abstract. a journer-

case of mathematical (mis)inductionl) AnpLo B6Nvl2), IoAN Cagulr)

From a rnissed attempt to mathenratical induction and through

in calculus, we arrive to a stronger inequalitv conccrning partial

sunts of the harmonic series.

Keywords: hrequalities. mathematical induction, sequences. MSC: Primarv 26406, 26D06, 26D15; Secorrdary 40A05.

series

The topic of an exercise iri [1] is the following inequality: for all n e N, 2. \\'e have

t*;"('- hl

(1)

k--2

This folloii's straightforwardly from the AN,I-GM inequality if we re-write it as

,*ik.' uA. L.'

,IT , V;

However. a-. n-ith many of the statements that contain the phrase "for all n â&#x201A;Ź N...". iots of our students will be tempted into using the Principle of Mathematic'al Licluction to prove it. It does not quite work for this statement for reasons that n'ill become clear soon, but our story has a happy ending triggered bv a series of fbrtunate calculus events. If ri'e clenote by P(n) the statement in (1), one easily checks first that P(2) is trLre being equivalerrt to ,/, < 1.5). Showing that P(k) + P(k + l) is at the c'ore of the method atrd, in this case, it comes down to proving that for all i: : I i: k

k+7 k ,ffi-m'iu

(2)

The statement expressed in (2) is quite strong in the serise that the difference bet*'een the left and right-hand sides is getting srnaller as k gets larger' This is alreaclr' apparent for small values of k: when k : 1, (Z) is Jl- 1 = 0.'11 < 0.5 : 1/2. ri-hile rvhen k : 2, (2) is ffi - 1/Z x, 0.665 < 0.666 < 213. A computer algebra s1'stem sucli as Mathematica easily verifies that (2) holds 1)This work is partially supported by a grant from the Simons Foundation (No. 246024

to Arprid B6nyi).

2)Department of Mathematics, 516 High St, Western Washington University, Bellingharn, WA 98225, USA, arpad.benyi0wwu.edu 3)Departrnent of Mathematics, West I-lniversity of Tirnigoara, Bd. Va^sile Pdrvan, no. 4, 300223 Timigoara, Romania, ioan. casu@e-uvt . ro

Norr:

N{arRruATrcE

for other randomly chosen natural values of k, so...(2) must be true?! Yet, it is not an obvious inequality, and the considerations above do rrot amount to a rigorous proof of it. We extrapolate that a typical student would conclude at this point that showing (1) bV induction is either not possible or, if possible. not an easily accomplishable task. The moral of our tale is that doing matheniatics is a fun roller coaster and perseverance pays off in the end. We u'ill indicate below how basic ideas from calculus combine to prove (2). In s-hat foilou,s, we will assume that k e N and k > 20. This is iust a technicaiitl' that r,vi1l simplify some of the calculations; as mentioned already, the fact that (2) holds for k e {1,2,.. . , 19} carr be checked directly by a computing device. The inequalin'u'e q'ant to prove can be Let f : [1,oo), f(r): re-written

f lk +1)

/(tu)

. #.

2(.r

(3)

First of all, this inequality is ureaningful in tlie sense that the expres-siott otr the left is always strictly positive due to the fact that .f ::::::ctlv increasing. Incleed, since In(/(r)) : (1- *)t"r, by irnplicit diffe:er.:.aii,-,Ili'e find tltat. for all r ) 7, f'(r) : ,-t-tlr(r - 1+ Inr) > 0. \lore,,,r'er. br- the \Iean Value Theorern we see that proving (3) is equivaient to prc\-ilrg that for some 0 e (0, 1) we have f'& + 0) < #.Now. a tedious calculation further shows that, for all r ) 1,

f" (r1 :r-s-tlr((htr)2 - 2lnr * r - 1 - *-3-r1rS.((lnr)2 - Inr + 1i - ,r'- In,,')] > 0. Thus, / is concave up (that is, that for all k â&#x201A;Ź N and k > 20,

is increa.sittg r. atid so it suffices to show

f'(k+1) This last inequality can be rewritten

(r+

t)-1-1 l(k+1)(k*ln(k+ t)) <

< A1'

(4)

A' -+-

o ln(k+ 1) < k((k* -f K+ -I

t)1/(t+rl

- t;.

Denote

ttl.: ffi- I > 0 <+,k: (1-

n1.)l <+

P: -ln(1EGI+ o1)

With this notation, (4) is equivalent to (k + 1)ln(l

a*+r)

Intuitively, we see that (5) that, assurning we have

/l[

(

,kar.-r

I.(1-+ al+r)

k+ a[.+1 . -!-

(5) 1

just a quantitative version of the '*'ell-known fact 1. In the remainder of ak:0,

!+#P :

_____________.

A. B6wvr, I. Cagu, A uappv

cASE oF MATuEMATrcal (urs)rNoucuoN

this note we show that (5) holds for k e N with k be based on two sirrrple observations. The first observation is that. in<leed,

)

20. Our argument will

0, and, more impor-

quantita,tive way. Wb can show tantly, we can express this convergence in a ^llrl this by using the Binomial Theorcm, which, in particular, plives that for all

z)0andkâ&#x201A;ŹNwehave (l * r)k > l + If

we plug

tr.,

1k(k

r:-.

in.r: 2lvq iu the last displayed inequality, A-(A"-1) 4_ /" 2\r >1+-r-k>A'.

we see that

(r*m/

which gives

o<W-t<

+*n^.4'l'

'/l' lbr sequences) we obtain '/from herc Our second observation is that. for all :l ) S, -,ve have ^.lj1nu ;2 13

br- the Squeeze Principle

ln(l*.rtr-.r -1+ o.) ZJ

(o)

(7)

The student that has encountered seriers in his or her studies will recognize that the right hand-side of (7) is just a partial sunr of tlic series expansion of ln(1 + r). Regardless of this, wo car] prove (7) easily as follows. Let h : [0, oc) -+ R, h(0) : 0, and. for r > 0.

h(r) :lrr(1 *,r')

- z. + - +

Since

h'(r)

:.fI

1+r-

r.r' -

: -' ; S0. .r" r 1 -

h is decreasing ancl hencc l,(r) < 0 for r > 0. \&'e are readv to irrrplenrent our two observations in obtaining (5). Applying (7) to r : .ra-1 ri'e get ln(1 + a611) ak+t Thus,

suffices to show that

. that

, o'k+, <1- ak+t 2- J

ak+t

T-\r

^2 uk+t

k+t

(rt. r\ tl (k+l),,r"+r(.:-;) r,

Norp iVIa.'rpuetrcl:

Notice now that, bV (6) and for k

47.-,-1

20. we have

- 1

2-i>r-

We are left to show that for Rewriting, we have

zr/E+f i'

27 we have nan

> 3 (, :

1 above).

na,)Z.'anr*orr(r*:)' But since

(l+yltlu < e for all 9 > 0, letting ?J:31n,

we get

(r+1)'.",., \ n,/

for

21. The proof of (5) is complete.

An immediate corollary of (2) is that, as observed at the beginning of this note, we have

-. (\/ k+7

k-{/F+ 1 -

ktrm or, equivalently,

k m-

,. / k+l

A._ 1 )

ii1(rffi-Tk) :r

We note first that for ali k e N,

(k+t)t/t't+1) < l'1 ' ,.

''j;! a Y,which '3.

in turn follows easily from the fact that the function 9 : x),5@): be it (0forr)3. \ori.br i21 andsince(VE; it decreasing; sinceg/(r) :ry a decreasing sequence, we have 1

111. -*-y[+1

k,-:i a : +. L, W'{T - iF-V?

and the conclusion follows from the fact obserr-ed before tfrat and the Squeeze Principle for sequences.

*l*"

{/E

RpppRpNcss [1]

Zvonaru and N. Stanciu, Problema 2'/269. Gazeta \,{atematicd (Seria B), no.

(2016).

L.-I.

C,q.r,q.N,q,

45 N,IoNoToNY oF N,IIIUfIDIMENSIONAL DISTRIBUTIONS I,.AN'III,IES

The monotony in hazard rate sense of some families of mult idimensional distributions LutcI-IoNuT

C.q.reNat)

we will prove that the fronotony in hazard rate of multidimensional distributions. families some for holds sense risk theory, hazard rate, distribrrtion orclers, Stoc}rastic I{eywords:

Abstract. Irr this note MSC:

60E15'

INrRooucuon

: For a random vector X : Q -+ IR.d we consider its distribution p(B) P(X e B). its distribution function F(r):P(x < r)' and F.('):P(X2t' x I t). In this article, the distribution function for another random vector Y u'ill be denoted bY G. In this note tve are using the notation ancl some well known results from l1l and l2l. Definition l. Let X and,Y be two rand,om uectors. we say that x 'is smaller thanY tn hazarcl rate sense (and' we denote X <h,Y ) if F-(r)G.(y) < F-(, Av)G.(rv v), Vr,y â&#x201A;Ź IRd' Definition 2. Let X and, Y be two random uectors. we say that x :: ,f srnaller than Y in uteak hazard, rate sense (and we denote X {*h. Y ) fr is i,ncreas'ing on SuPP(G.).

X,Y be two rap,d,om uectors. If X <nrY then X l*h. Y' Theorem 4. Let X : (X)i_-,rr1 and'U : (Yi)*17 be two ro'ndom uectors' If X <n Y then Xt <n,Yi. i': l,d'' Theorem 5. Let (Xt)*Ti and (Y1)r-* be random uariables' If Xt <,n, Y, i' : 13, the, 8!:rX, <r,. ef:rYl'

Theorem 3, Let

For the sake of completeness, let us recall the definition for some mult iclinreusional distributions. The multtuaTiate unzform distribution Unif (1) on some product of in: ffi. Its tervals I : It x ... x 16 C IRd has the clensity function 1@) marginals are Unif (/i). 1)Departa-errt of Mathematics, Faculty of Mathematics and Informatics, University of Bucharest, Romania, luig j- -catana@ya-hoo' com

NorB Ntllrplrerrcn

The multiuariate Poi,sson distribution of ) : ()a, )a_r,... , )r, )o), Poisson(,\), where each ,\1 is nonnegative, has the density function _$,

/(r)

o?'"*

min(

II ^;^ *,

k:1

r4'

i=0 lit'',' i:1

The marginals are Poisson(); + )o). The multi,uariate normal distribution lf (p.

f (,) : (a.t (2, t))-L where p : (EXt, EXz,. . . , EXi and ! The marginals are l/(p,;, D).

fi;)' \ r!

has the density function

".lt'-.ri'L-'"-P), deuotes the covariance matrix.

The bi,uari,ate Bernoulle distribution B(1.p) has densitv

P(X:at):Pi. where a; e {(0,0) , (1,0) , (0,1) bers with sum one.

(1,1)} and pt are four nonnegative real num-

1. MarN RBsur-rs

Proposition 6. Let a, b e IR{ . i,fa<b.

Then Unif

(10.

a]) <n.

L nif ([0. b))

if and onlg

Unif(I) <5. Unif(J) + Unif([0.o,]) <6. Unif([0,bt]),i :

Proof. \Ve have

T,d+a;{bi+a1b.

a /-b. Then Unif([0.or]) <5. Unif(l0,bi]).1 :7,d.. It follows thar Ef:1Unif([0,ar]) <r,. Af:, Unifl[O,b;]), in other words. we Conversely, assume

have Unif(fO,rl) <r,.

Unif([O,b]).

Proposition 7. Let a.P â&#x201A;Ź re1*t. f/Poisson(o) {6. Poisson(p). then ai+ ao 1 ,& * go, i : 7.1. Th,e conuerse h,olds if the marginals are inclependent. Proof. Suppose that Poissor(a) 3r.. Poisson(B). Then for i -- ln one has Poisson(altao) ln. Poisson(p;1B6), anci therefore oi*n6 < Ai+ 39. i :7,d". For the converse. if fc,r each i : 7.d it holds .r, + o0 a Ji - B6 then Poisson(a; f ao) ln, Poisson(3; +,30). Assuming moreover that both random vectors have independent marginals, it follows that dd

Ii:l

Poi.tot (ar + )o) tu.

that is, Poisson(a) <5. Poisson(B).

Poisson(/ii 1 )o),

L.-I. Carl.Ne, N,'IoNorouv oF

IvIULTIDIIUENSIoNAL DIS'lRIBI.rrloNS IAI\'IILIES

l-t

l{(fr,I)

<1,. l{(p',1) then m,arg'inals. zndependent haue l_t,. The conuerse holds i,f both dtstrzbutions

Proposition 8. Let b" p, t' € IRd and D > 0. If

Proof. From I'(p.

<n, l/(lr', D) it follows

N(pn,I) <n' I{(p'1!),

which

z : tJ, so lhat p'11"tt. I Conversell,. l-t'means Ltr I lt'i for each i :7,c1- Therefore, one has l/(p;, D) -n. l{jr't,f), and hence Ai!:rl'{(pu,I) <n, \f:rN(tt'r,I), i, ii ii

in burn implies that 7-r; l pt,for ail

it other words .\'(4.

Propositio n

ii <n,

l/(lr',I).

9. We haue B(7,p) <n. B(1, s) e ff

) max (, X,#,ffi)

Proof. Let X and Y be the random vector for B(1.p) and B(l,q), respectivel1,. If X <6. Y then X l*,h. Y, so that fr i. ir."t"asing on Supp(G-), o, #(r) < F (g), for all r,y e Supp(G.) with r { A. \Ve specialize the vectors J. g to conr,'enient values. (L, i) *" get that ;1 > 1 For r ,U (+,i) *" get that (1, For r ,y (+,j) and y > *;+.t:2.3. But it is easv to verify the follou'ing implication:

: (--, --) : : (-i,i) :

P4 -

Pt-rPl

For

q4 \ q2+q3 pn' pr+pt' I ) So -

-i),a:

* % .i :2.1 + 9! > 9_.i : 2.3. Pt Pi P; * P,t' (0.0) ,u : (L, i) *" obtain # > m. Qt, P1, -

ancl this implies

q' *u* (1. . :qt .-9'o?'' ) \ P2 P3 P2-P;t/ Conversely, it is easy to r,'erifv that F*(.")G.(A) ( F*(r Ay)G*(rv P4 -

Vz, g €

y),

1R.d.

RnrsRPNcPs [1] NI. Shaked. J. G. Shanthikumar. Stochastic Orders, Springer, New York' 2007' Bucuregti, [2] Gh. ZbAganu. lfetode llatematice in Teorza Riscu,hn q'i Actu,ariat, Ed. Univ. 2004.

PRoeLprrs

PROBLEMS Arrthors shoirlcl submit prol>osed problenrs to gmaproblems(Qrms.unibuc.ro.

fortiat.

Once a problem i-s ircceptcd and considered for publication, thc author will be asked to submit the TeX filc also. Tlie referee process will usually take betweerr several weeks ancl tt'o rnonths. Soiutions ntay also be submitted to the same e-mail address. For this is-sue. solirtions should arrive

Files should be in PDF or DVI

before 15th

of May 2019.

PROPOSED PROBLEI\,IS 472. Let a,b, c e

[0,

$] such that

a*b*

ri. Prove the foljowing inequality:

/a*b\ /b-c\ /c-a\l sina*sinb*sinc22--1lsin( 'L )sittI o ,)sinl-tt2 I \ " / \ \ )l Proposed by Leonard Giugiuc, National College Traj.an, Drobeta Turnu Severin, Romania and Jiahao He, South Chrna Un:.versity of Technology, People's Republic of Chi.na.

473. Let â&#x201A;Źtt...,enbe the elcurentary symnletric poll-n',nrials in the variables

Xr,...,Xn, ex(Xt,...,Xrr) : 1(i1(...(i1(n

.-\r,. and let ^9 be the ideal generated by â&#x201A;Ź1,....e, iu l.-fThen every monomial Xf''..Xn'" with the degree n7: n71l...*mn strictiy greater than ([) belongs to S. On the other hand. there exists a monomial of degree ([) which does not belong to S. Proposed by George

474. Calcutate

f{2n n=r

Stoica,

New

Brunswick,

Ca:rada.

r +... \r} lrt+*r----.--

1) I (

L\n' (n-1\'

Proposed by Ovi.diu Furdui and

1l " "-)

Alina Slntdm5.ri-a::, Tech:ri-caI Uni-

versity of Cluj-Napoca, Cluj-Napoca, Roroaaia. 475. We say that afunction /: R -+ IR. has the propertv (P) if it is continous and

a) Prove that

zf (f (")) :3f (r) - r for a1l ;r e R' / has property (P) then ll,l[ - {;r e R. : f

(r): r} is a

nonempty interval. b) Find all functions with property (P). Proposed by Dan Moldovan and Bogdan Moldovan, Cluj-Napoca, Roma-

nia.

PH<

lt,ost.tD I)ttoBLnt\ts

476. Calutlate the integral

/m arctgr , Jn {7 + Lo"c ' Proposed by

Vasile Mircea Popa, Lucian Blaga University, Sibiu,

Romania.

we denote by 1* its adjoint, i.e., the AT . If A is square and A: A* we say that transposed A is self-adjoint (or Hermitian). In this case for every complex vector r we have r*Ar â&#x201A;Ź R. If A, B are self-adjoint we say that A > B if r*Ar) n*Br for every complex vector r. For a complex matrix A we denore lAl2 : AA*. Note that 1,412 is self-adjoint and ) 0. (If A*r : A : (h,.. . ,An)r then r"lAl2r : U*A : (at,... ,an)(at,... ,un)r lal2 + ''' + la"l2 >- o.) Let A be a square matrix with complex coefficients and 1 the identity matrix of the same order. Then the following statements are equivalent: (i) 11+ zAl2 : lI - zAl2 for all z e C;

477. For every complex matrix,4 of its conjugate, A* :

(ii) 11 *zAl2 (iii) ,4: o.

)Ifor allze C;

Are these statements still equivalent if we replace "complex" by "real" throughout? Proposed by George Stoica, New Brunswick, Canada.

478. Determine the largest positive constant k such that for every a' b, c ) with a2 + b2 + c2 :3 we have

(a+b +

c)2

+ kl@ -

b)(b

- c)(c- r)l S g.

Proposed by Leonard Giugiuc, National College Turnu Severin, Ronania.

Traian, Drobeta

and A e Mp(Q) a matrix such that and det(A + I) * 0. Prove that: a) t(,a) is an eigenvalue of A + Ir. b) det(.A -det(,A - Io) fu t) a\(A) + 2. Proposed by Vlad Mihaly, Technical University of Cluj-Napoca, Cluj-Napoca, Ronania.

479. Let p be an odd prime number

det(An

* Ir):0

t Ii

480. Let k,nbe natural numbers, rL,fr2,... ,fik be distinct complex numbers and the matrix Ae M*(C) such that (A -r11")(A-*zln)"'(A-r6ln): O,.,. Prove that rank(A -rJn) *rank(A -rzln) +"'+rank(,4, -npl,-): n(k - 1). Proposed by Dan Moldovan and Vaslle Pop, Technical University of Cluj-Napoca, Cluj-Napoca, Romania.

Pn.oei,sNis

481. Let 1r be a field and let n

) 1. Let A,B e LI,,(I{) such that IA,B)

commutes u,-ith A or B.

If char lf : i.e.. iA,Bl" :0.

0 or char

K > n then it is knou,n that lA. B] is nilpotent,

Prove that this result no longer hoids if 0 < char I{ I n. (Here b1' 1.,'] u,e mean the comrnutator, [A.Bl : AB - BA.) Constantin-Nicolae Be1i, IMAR, Bucharest, Romania.

SOLUTIONS 455. Let n ) 2 be an integer. Determine the largest number of real solutions the equatio, ar/i16+...+ a,r1/7lbu:0 can have. HeIe &tt... jan ate real nurnbers, not all zero. and b1 , ...,bn are mutnallr-distinct numbers. Proposed by Marius Cavachi-, Ovldius

University, Constanla,

Romania,

Ed'itor's note. Unfortunately, the solution plor-ided by the author is rrong. r A reformulation of the problem is the follor-ing: Given ri ) 1 'uve rn'ant to find the largest m for which there are mutually distirrct bt,...,br, mutually distinct ix7t....r. arrd not all zero &71 ...,an l?_

such

tlrat Lnrl/.,'t+bi -:0 for 1< z < m. Equivalentll,, if A e ,4.f-.r(1R.), j:1

A: (\E+4)..3, therr lhe equation AX:0 has a non-trivial real soiution, viz., X: (ar, ...,ar)T. l'or this to happen one neecls that rank A I n - l.

Hence rn is the largest rrurrrber such that there are mntualll- distinct bl, . . . ,bn and mutualh. distinct 2:1,...)rr77u sltch that the rank of the rn, x n matrix A: (\/Gi+, hj)i; is < n - 1. Obviousl1,.. when nr ( rr - 1. regardless of the values of r; and t.,;, x.e harre rank A { m.( n - 1. so it is po-ssible to have rn solutions. The author claims that r,r,hen 'm,: n. i.e.. I is a square matrix, we always have det A+0 so rank A: n, so it is not possible to irave n solutions. Therefore the rerquired maxirnum is n - 1. N{ore precisell.. if we rearrar}ge the ri's and br's such that 1r ( . .. 1r,

on tlre fact that the function./ : l0.oo) -+ [0,-), f (.r): v6.ir strictiy concavâ&#x201A;Ź. Unfortunatell,, rvhile the proof is verl, ingenious. it contains an error. It seems that the result is true urhen n { 3. \\'e don't know what Iiappensi,vhenn)4.

456. Let .f ,g,h be non-negative continuous frinctions or1 i0,1] satisfying the inequalitv f (tr + (1- t)g) > gt(l)h,l-r(y) for all r,;q â&#x201A;Ź [0, 1] and some (fixed)

PRoposBo PRoBLEMS

â&#x201A;Ź (0.1). If. in a<lditorr.

;1r;a" >

we havc

New

I 11lt')d.r:JtI h@)dr: .lo

1. prove rhar,

Proposed by George Sto!.ca,

John,

University of

New

Brunswick, Saint

Brunswick, Canada.

Solution by the author. \\'e assume first that g,h are positive. not merely non-negzrtir-e on [0,1]. Then y,+ [i g(s)ds and y ,+ I& h(s)ds are strictly increasing ancl differentiable lirnctions sending [0.1] to [0,1]. We denote bv u. r' : 10.1] -+ [0,1] their inverses. Then u,tr are strictly increasing arrd clifferentiable rvith u(0) : u(0) :0 and z(1) : u(1) :1.

\\e

har-e:

l,^o e(s)ds Bv differentiating

lo"o'

h(s)ds: r.

rve get

'u,'(t:)g(u(r))

t,'(:r)h(t,(r))

(1)

Define u.'(r) : tu(r) * (1 - t)u(r,). Along rvith u and u, ur : [0.1] -+ [0,1] is strictly increasing and differentiable rvith u(0) :0, u,(1) : 1. Using the inequality bettveen the arithmetic and geometric means, we have

(r) : tu'(r) +

t)ri (,r) > (u' (r))t (u'(t ))'-t . Fionr (1) and (2), the hvpotheses and tlie change of variables ,s : follows that w'

I Jo

(2) ?r,(",-),

/ /(r'(r)) 'u,'1.r;dr

,ttt)cls>

a [' Jo

: ['

' (u'\.r)g(u(x)))t (u'(r\g(u(rrr) \u\ \w\d,v\

@(r))h'-'(,'(r)) (u'(r))t(u'(r))1-tdz

ta.r: l. ln[' If g.h are only non-negatir-e then for any 0 < d < 1we define ga(r): d).q(r) * d arrd h5(r) : (t - ri)h(r) + d. Same as for g,h, we have Ju

(1 17 I O;t.r|rt.,

'dr:

: t h5(.r)dr: Jo

1. Howevâ&#x201A;Źr, 9d,h5 are positive, not merely nonnegative. Then. bv coutirruitr-and the compactriess of 10,1] x [0,1], from l(tr + (1 - i)y) > gt|)h'-'(y) we get that for any r ) 0 there is some 0 < d ( 1small enough such that f(tr+(t-t)y) *e > gi@)hl-'(y) ( < V0 r:, y 7. Then, by applying the case we have just proved to 95, h5 ancl we s"t J;(/(") +e)da; > 1, i.e., .fi t@)a, *e r6, 1. since rhis f, f happens for cvery e ) 0, we get our result. tr

)

PRoeLsNrs

LtZ

457. Lt:t A, B e M,,(C) bc so that A2 + 82 + thr. follt-ru-irrg eqrralit ies: a) Tr((A - B)(A- B + 1,,)) : 2r.

b) det((A

- B)(t-

Proposed by

B+

A* B:2(AB

+ 1,,). Prove

1,,)):2".

Vasile Pop, Tehni-cal University of Cluj-Napoca,

Cluj -Napoca, Roman:.a. Solutzon bu tltr: author. We use the

[x, ]'] :

notatio"

i' ] for the commutator

XY -YX. A,'I: (A- B)(A- B + 1,,): (-{ - q2 + A- B u,,rites as LI : A2 + 82 - AB - BA+ A - B : (AB - B-\) +21,, : lA,B)t21,,. Since Tr[A.B] :0 we get TrIVI: Tr(21,,):2n. i.e.. we have a). Let C : A B, so that lI : C (C */,). \ote that [C, B) : lA- B, B) lA, Bl. Then '*,e have The matrix

1r'+

1,,)

* )t

lC. B)

(3)

and equivalently

lC.B] -C'2+C'-21.. If we note D : IC,B) : C2 + () +1,, then D commutes u,ith C. \\'e that 'I\'(Dfr) :0. k: Ln.Inrleed. s'e have Dk _ Dk_t.D: Dk tlCn _ nC)

- Dk tcB -

(4) prove

- c(Dk tBi - tDk-rB'1c, so that Dk : iC,Du 'B), which implies fr (Dft) :0. In conclusion Tr(D) : f. (D') : -' : Tr(D"t : 0. so

Dk-rBC

that, if ,\1.)2,...,),, are the eigenvalues of the nratrir D. frorn the svstem

)r*)z+"'+ln:0. tl + x] + ... + \?*: o. :

.\i+)|+...+)fi:0, we obtain

)r : lz : " r : tr, :

0 (frorn the Newton forrriulas the poly'nornial

/p with the roots )r, )2,..., ),. is unique, more precisell-/p(A) : )".) From tlre equality (3) it results that the eigenvalues of tlie ntatrix C (C +.f,) : D +21, are p1 : lt2: "': F,n:2 and their prodpct is det (C (C

i.e., the eclualitv b).

In11

:2",

PRoposBp

PROBLEMS

to prove that )r - ' r ' : )r, : 0, i.e', that prove a rrrore general rcsult: If X,Y are square l) is nilpoterit, calr l.tc usecl to rnatriccs of thc same climertsion n and [X, Y] commutes witir X then [X, Y] is niJ.potetrt. Inotrr casc X : C.Y: B antl we have that [C, B): C2+C-2In comrmrt,es with C. The resrtlt holr1s, with the same proof, over every field K of charactelistic 0. If the c:haracteristic is p ) 0 then it holds only if n < p. See problem .181 florn the current issue of GNIA.

Note. Tlie method

458. (Corrected) Fol

ttsetd liere

a continuous and non-negative function

on [0,1] we

deflne the Hausdorff moments LL,

i: /t o

r"y1..1.1.,', z

1,2,. . .

Provc that t-tr+zt lt|

-f

p,,,,p,1

)

Proposed by Cezar Lupu,

PA,

2ptylarl.Ltltot fr

0,1,2,.

University of Pittsburgh, Pittsburgh,

USA

Solut'ton by th,e auth,or. Let us recall thc celebra,ted Schur's inequality

rn(,r-y)(:r- z)+a"(y- r)(y -r) for all fr,l1, z ) 0 and n, ) 1. Then we have

z"(z-")(r-

,: I I1t I1l (X -r Y t Ju Ju Jo

Zlf

tJ'ty)l(:)drdydz

g) >

)

(5)

X ::f," (" - y)(.r - z'),Y -- y"('!l - z)(y - z) arrd Z : z"(z-r)(z-'a). we lrave ,I : ,Ix -t Jv -r J2, wttet'e.tu : #,C'# xf @)f (y)f (z)d,rd,yd,z and similarly for ./v. .I 7. But, by s"-vrrrmetrY reasons, Jx : Jv: Jz. Thus 3Jx : J > 0, so ..fu ) 0. We have wlrere

11 17 rl

: Ittl I I Jo Jo .lo

7r"+2 + :r"yz

- r"*ry - *"+r r)f (r)f (y)f(z)drdydz

I r'-2f111d..,.Jo I fttidy JoI fGl: - Jo

1t rt ft .r" !1.r1c).r I ttf fy)rls / zl'1:)d: I .lo Jo Jo tt 11 fl - I ,,'"'rf\.rt,lr I g,l'\tt)tlg I ftrla, Jt ;1

I - lo l.Ln

''-1l1.r

+z ll} I Lo

so 1t",r12p,f,1 p,,.lt'i

1t,"

.lLt 11

Jrt

ldr J,," I f ls),ta " ,Jo/ :/(z)d: 1t.y 1-11

- lln+t

2lr,,,a1ltsl.1,t

- LLn+t l1,O llt t 0. Hence the conclusion.

l-Ll l-to

: Jx )

PRosr,oN4s

Solution by Leonard Gi.ugi,uc, Nati,onal College Tra'ian, Drobeta Tut-nu Seuerin, Roman'ia. If f :0 then we have nothing to prove. So we assume that / I 0, which implies that all lli are positive. The relation to prove writes as

Pn+z.lto- n .!!-,2, Pn+l l-tt Pn+t lto 11

By the Arithmetic X.fean Geometric Mean inequality

rne have

F:)2 . ro t ttn . ltr pnil pn*7 2

ltt

Z, @. l_to \l l1i+t

Hence it suffices to prove that p,n12pr, > F?,+t. Cauchy-Bunyakovsky-Schwarz inequality: Fn*2t1n

But this follows from the

\ z rl / ft \ - (/ r"'2f ir)dt) (/. f t(r)dt')

(1,' "n+zf(r)rnf(r)dr )': The proof is complete.

(/;' ,n-, rt )dr) :

F2n,,.

459. The faces of arr icosahedron ale colored ri ith blue or white such that a blue face cannot be adjacent to more than trr-o other blue faces. What is the Iargest rnmber of blue faces that can be obtained folloiving this rule? (Two faces are considered acl.jacent if ther- share an edge.) Proposed by Eugen J. Ionaqcu, Coli:mbus State University, Columbus, GA,

USA.

Solution by the au,thor. \\ie prove that the ans\\-er is 1.1 and the coloring that gives this maxirrrum is the one shornn in Figure 1(a). t'hich is a Schlegel diagram of the icosahedron in u,hich the projection is done from a point close to a blue face. (That is. there is an extra blue face tirat cannot be seen.)

Figure 1(a)

Figure 1(b)

Figure 1. Icosahedral graph

PRoposBo PRoBLEMS

Indeed, for the each of the 20 faces fi, Tz,. .. ,T2s we assign a Boolean variable tri,i:1,...,20, which is equal to 1if the face I is blue or 0 if the f.ace Ti is white, given a certain coloring. Let us denote by I/, the indices j for which f.aceTi is adjacent to 4. The condition we require is then written AS

rt,* D "r( j€N(i)

3, i

1,...,20.

(6)

S : f ,0. After summing all these I i : 1,.. .,20, we obtain S + 35 ( 60. Hence, it follows

Clearly, we want

maximiz

inequalities up over that S ( 15. Since we have arrangement that accomplishes 14 blue faces we only need to prove that 15 blue faces are not possible to be arranged without violating one of our requirements. So, by way of contradiction, let us assume that it is possible to have 15 blue faces satisfying the requirement of a 2-dependence set (as in the text of our problem). Then all the inequalities in (6) have to be equalities. So, for every face that is white we need to have exactly three blue around it and for every face that is blue we need to have exactly two around it. Then we are forced to have a coloring as in Figure 1(b) if we start with a white face (the one in the middle) and then we end up with too many white faces (at least 7), which is in contradiction to the number of faces left possible (i.e., 20

- 15:5).

460. Let X be a set with at least two elements, and fix a,b e X, a I b. We define the function f : X3 -+ X by

r,y,z*a, ", : ( alf flr,a.r) t b ff a i {r',a,r}. Is there a binary operation ',: X2 -+ X such that f (r,A,z): (**y)* z for all r,y,z € X? Proposed by George Stoica, Uuiversity of New Brunswick, Saint John, New Brunswick, Canada. Solutzon by the author. The answer is no. Indeed, let us assume that, forer-ery- r,ye X,r,af a,we havez*A*a. Then,forevery r,y,z€X,

r,U, z

a. '*,e have a

: f(r,A,z) :

(* * A) * z f

a contradiction. Hence, there exists u,u e FYom here it follows that

u,tL

o, such that z

a)Fa: (u*u)+a: f (u,u,a):g

*'t):

PRosr,pN,rs

and then, for every

r e X we have b * r : (a * a) * r : f(a,a,r) :

Finally, a

.f(b,u.u)

a contradiction with a +

(b* u) * u : b* u :

461. Let A,B e M"(R) so that A2: A, 82: B, and det (2A+B) Prove that det (A + ZA1 :0. Proposed by Vasile Pop, Tehnical Universi-ty of CIuj-Napoca,

:0.

Cluj-Napoca, Romaaia. solutton by Francisco Perd,orruo and Angel Plctza. Departamento d,e lr,tatemdti,cas, Uniuersidad de Las Palmas de Gran Canaria, Espafi,a. If det(2A+B) : 0 then there exists u I 0 in lR." such that (2A*B)u : 0. From (1 - A)(2A+ B)u:0 and (I - A)A:0 ne get (1 A)Bu:0 and

so Bu : ABu. Similarly. from (1 - B)B : 0 and (I B)(2A* B)u : 0 we get Au : BAu. Then (A -l2B)2u : A2r: -f2ABu -f2BAu * 4B2u : Au *2Bu * 2Au * 4Bu: (3A*6f')u and so (A+28)(A+28 31)u:0, where 1 is the identity matrix. It (A+28 -31)u : 0, then (A+2f)u: 3t,. Together with (2A+B)u : 0, this implies that Bu :2u. But B is idenpotent so its only possible eigenvalues are 0 and 1. Hence (A + 28 - 31)t, I 0. Thus det(,4 t 28) : g, becarrse (A+28}u : 0 has the non trivial solutio, (A+28 - 31)u in 1R.,.. x

The atfihor's solution is essentially the same up to the relation (-4 f 2B'Sz, : 3(/ -l2B)u : 0. Here he assumes that clet(.A + 28) I 0 and concludes that (A *28)u:3u. Together rvith (2-,1 -r B\t:0, this implies that Au : -'u. Hence A2u: -Au:,u. But l2L': Ar,- -u. so u:0. Contradiction.

Soluti,on by Moubi,nool O,marjee, Lyc€e Henri IV. Paris, France. We prove that if A+28 € GI"(R) then ker(2,4 -F B) : g. Take z e ker(2A + B). We have 2Ar * Br : 0. i.e.. Br : -2Ar. When we multiply to the ieft by A we get ABr : -2A2r : -2Ar : Br. When we multiply it to the left by B u'e get that B2r: -2BAr. But B2r: Br: -2AX, so Br: -2BAr arrd Ar : BAr. Then (A+2 B)2r : (A2+ZAB+2BA+48\r : Art2Br*2Ar*4Br : 3(/ + 2B)r. Since ,4 -t 28 e GI,, (R.) we have (A + 2B)r : 3r. We mrrltipllto tlre left by,4 and we get (A+ 2AB)r, : 3Ar, rnhence Ar : ABr : Br. Since also 2Ar -t Br :0, we have Ar : Br: 0 and so 3e : 2Ar * Br : 0. So we proved that if A + 28 € Gr"(lR) then ker(2,4 + B) : {0}, i"e.,

2A+B€Gr"(R).

By contraposition, if det(2A + B)

tr.

0 then det(4

2F)

: g.

PRoposen

PRoBLEN4S

462. If .f , [0,1] -+ IR is a convex function with /(0)

then prove that

/(')o') lu''' rr(r)dr' '

'r(1,' ,r(r)d'r - lo''' Proposed by

Florin stdnescu, $erban cioculescu school, GSegti,

D5mbovila, Roroania. scthtt,ton by the o,uthor. we need two well known properties of convex functious. Lernma f. if 1isanintervaland /:1-+IR.is aconvex functionthen for every a e l the function ro :.I\{r} -+ IR.. r,,(r) : tkFIL, is itrcreasing' Lemma 2. If I is an interval arrd / : 1 -+ IR. is a convex firnction then for cvery r,?J,z,f â&#x201A;Ź l with r <A 1z <f rzu'e itavc

fty)-fl,) _ /(1)-,/(:) -

A-t B1' Lemma 1 vr,itir a

t-;

0 the frrnction

,r increasing. It follows thrit

e l},ll2)we have + < $f. .., rf (r) +'yf (r) < rf (r * v) Vt,.11 e 10,112).

for every' z e (0. tl2l, a

0 we have equalitY, 0 : 0.) this inequality first u,ith respect to g. then rvith respect integrate we to r. We get

(When

),.r,*,+

jtt,r ,-

./(.r+

y)da:*

f ,'-"

fr,,n,

tr. [,;]

; 1,"'rl(r)cr-r -l

lr"

l(,,',u,

/,',,/(

= lo''(;',)

)dr

(1,'

'' '/,'iot) n'

-; 1,"' ,' (, (, - ;) - rtr)) 'r

Here in the right hand side of the inequtrlity we used iritegration bv parts ancl the fact that the derivative of r+ ir+tlz f (t)dt is /(r +112) - f (r). \\'e now use Lernma 2. If r'y e [0,1/2] with r < y then 0 I :r I y I

.r-12(ur1l2<l.so

*;)

-t

(".;)

(,. ;) - (,. ;)

which implies that /(r +112)- f (r) < f @+112)- f (y).Hence the function r ,+ f (r + 112) - f (r), r e [0,1,12), is increasing' Then, bv applying the

PRoBLIiNIS

Chebyshev inequality

U! t(t)g@a t >

the same monotony) we get

f,,,'''

(' (".

.t t(tat Il g(ilat if /

and s have

(t (,.;) - y1,,;) o,

n,> , 1) - r1,;) fr't',ra, lr'''

i(1,: ,/(r)dr - lo''' rt,).,) In conclusion, we get '= [''' rf(r)d.r 2lo

t"' -:8Jo

s1r1o,

a:8Jrp"' l' ./(r)d., '

n1/n

:(l 2a\Jiz" fe)d.r-l .lo

/(r)dr) /

and equivalently

r1/2

rtl2

I r/(.r)dr1zl /(r)dr'-2/ Jo .lo Jt1z"'

When we divide by 12 we get the required

/(r)dr.

result.

Solutzon by the edttor. There is an alternative approach using a general property of convex functions. If a < b we denote br-C([a,b]) the set of all continuous functions on fa, b] with values in R. Then C([4, b]) is a vector space over IR and it has a metric topology, u'ith the distance given by the

norm ll . ll, where

ll/ll

max,,â&#x201A;Źla,bl

l/(")1. Lemma. If T : C([a, b]) -+ IR. is a linear continuous operator then T(f) > 0 for every convex function f ,lo, b] -+ R if and on1;, if the following hold:

(1) 7(1) : r@) : 0. Equivalently, T(f) i.e., when ./ has the form /(r) : mtr + c. (2) T(f ,) ) 0 Vc e (a, b), where

,, \

[0. J'"(r):tr_"

r' e

:0 --a.

for any' affine futtction

c).

fr.bl.

This result is already known but we couldn't provide a reference. For the sake of self-containment we sketch a proof here. For the necessity note that if / is affine so is -/. Hence f , -f are both convex and we have T(f) > 0 and -T(f) : T(-f) ) 0, i.e., f (f) : O. Since 1 and z form a basis for all affine functions, in order that 7(/) : 0 for every affitte function on [a,b] it sufliccs to have 7(1) : T(r):0. Since /" are convex we also have 7'(/") ) 0 Vc e (a,b).

PRoposBo PRoBLEMS

Before proving the reverse implication, note that fbr every c e (a,b), is continuous and it is differcntiable evcrywhere erxcept at r: and wc liave

r< candft(r):1if iL>c. Alsonotethatl"(4,) :Q. Assume first that / is a convex function a,rrci its graph is a lrroketi line with the vertices at (c6,do),...,(c,',dr), whele a: (:0 < q < "'4 c:.r: I is a division of the interval [o,b]. Let ttr; : 'FH bc the slope of the graph on the interval lct-t,cil. Since / is cclnvex wc ltavr: rlz,1 ( " ' { rn". The function / is continuous ori fa, b] and diffcrerttiable evcrywlterc except dt cs,....c-.. Namely, if 1< i < s then .f'(r): Ini.Yr € (c; 1,c1). We ciairn f'(r):0if

that

(tn2- mt)J",(") + " * (nz" rn,-1)f,,",,(r).

J@):-r(,-a)+ds1

If we denote by g(r) the right side of the above equalitv therr the easiest way to verify that / - g is to prove that g has the same properties with f : g is continuous, g(a) : /(a) and g'(r) : f'(r) Yr e [a.b] , t f c11,....c.. T]re contirruity is obvious sirrce each /" is contintrous. Since J,,(a):0 Vc e (a,b) we have g(a):m1(a- a)+d,s- d,o:,/(c0) :./(a). For the third corrditiort let r € la,b), r I co,...,c". Then ci-1 < r < c,; for sorrle 1 ( I ( s, so f'(r) : rni. we have

'(*) : mt l (mz -,nr)f|.,(r) + " * (rrr, - nr,,-),f '"" ,(:r). Ifj < i - 1 then c, I c;-r ( -r'. so J!,1.?') : i. Ifj > I tlren t'i )> t'; > r'. so f '.,(") : o. It follows that f

g'(r): mtI (rnz-

rrr) +

"'+

(rr,

- rr"i-,):rfti: f'(")

By the linearity of 7 we get

arj):T(ms(r -o) + do)*(m2-rrt1)T(f1,) +

(rn,

-rru, 1)T(f"",,).

But rne(r-o,)+d6 is affine. so by property (1) T(mo(g-a)+do) :0, and by property (2) T(f.r) ) 0 for 1 < i < s - 1. BuL u''e a,Is<-r have m,1 t "' { Trlst so each m,i - mi*L is non-negative. We conclude that 7("f ) 2 0. Suppose now that J , lo,b] -+ R is convex arbitrary. For each n ) 7 we define the function f,, : [n, b] -+ lR. whose glaph is the broken lirre rvith vertices al (o,+;(b-a).f la+;(b-a)))u'itlr U < i < n. Notc that o -;lb- o) n'ith 0 < i, < n make a partition of [4, b] into n er1ua,l intervals. Also note that (a + *(A - o,),f \a+ *(b - u))) are points on tlie gra,ph of f , which is convex. so /,, will bc convex. as well. Since the graph of f, is a broken liue, by the particular case we have .just proverl, we havc T(f ,,) > 0. Since / is corrtinuor:s on a compact interval, it is rrrrifbrrnll, contintrotts. Then for an). t > 0 there is n., suclt thtrt for alty n ) rr. we have 1f@)

f@)1 <

Yr.y €la,b)

rvith

lr'

1!1t,-

PRoeLsl.rs

r e la,bl. Then r e lo * *fU - a),a * ;\b - a)lfor some 7 { i, { n. of [a* *tb- a),at *(U- o)) is ](b-a) we have lr - (a+ *tu-o))1,1"- @+fi(b-a))l < *(b a), so l/(r) f (a,++(b-o))1, lf @)-

Let

Since the Iength

f("+*(u-

o))l < e. It foilows that

f("+*tu-a)),f(a+ fi(b-o)) â&#x201A;Ź

(f (r)-e , f (r)+e). But r belongs to the intervai la++(b-a), at i(b-a)), where /, is affine. It follorvs that f"(r) takes an intermediate value betr,r,een I,@+ *O -a)) : f (a+ f ta-a,)) and I,b +;(b-a)1 : [(o + )(b-a)). since /(a +*(u -o)),f (a+ ft(b- a)) e (f (r) e ,f (r) +e), this implies f"(r) e (f (")-e,f (r)*e), i.e., 1f"@)-f (r)l < a. Inconclusion llf"-fllma-x. l/"(r) - /(.r)l < E.

râ&#x201A;Źla,bl

lim f" : 1. Since 7 ]lf" - fll < e. for n ) n- u'e har.e rr-+oo : is continuous, this implies JIr1f11,, T(f\. But 7(/") ) 0, Vn, so 7(/) > 0. tr Since

We now return to our problem. \Ve must pror-e that T(.f ) convex function [0,1] -+ R with /(0) : 0. u-here

> 0 for every

r(r)::6 ( t'

y1,1o,

\Jrtz"'

- Jo [''' /r, ra,)/ - ro| '

rr(r),b.

we try to prove that T(f) ) 0 for all convex functions / we see that this is not true. We have 7(1) : -* I 0, so the condition (1) of the Lemma is not satisfied. Therefore w-e need the information that /(0) : 0. The idea is to find a ) e IR such that TU) > 0 for every convex function,f ,[0,1] -+ IR., where T(f) : rU) + )/(0). If we prove that 7 has this property then for a convex function / that has the additional propertr.that /(0) : 0 we have T(f) : rff) +^/(0) : 7(/), so TU) > 0 implies T(f) >0 and we are done. Whenwetake f =7weget7(t) :7(1) +)-1: *+f,soinorder that ?/(1) : 0 ne need to take ) : *, so T(f ) - T(f ) + *f tOl \\,e have:

r(r)

: ; (1,',1d2 -

l,''' ro,) - lo'''r:crr * i, :

r(,)

: i (|,',,rd.r -

l,''',o,) - lo''',2d* +i

so the condition (1) of the Lemma is satisfied.

We now prove the condition (2), i.e.. that 7(/.) ) 0 Vc e (0.1). u,here f.(*):0 ifr < c and f.(r): r- cif r) c. First note that /"(0):0, so T(f T(f + *r"tol T(f .) vc e (0, 1). we have tu,'o cases: ") ") Case 1. c < ll2. Then on the interval lll2,1l we have f,(r) : r - c, while on the interval l0,7l2l we have f ,(r) :0 for r e [0, c] and /"(r) : r - c

PRoposop PRoBLEI\{s

for

r € lc,ll2). It follows that

T(f ,,)

/ 11 rtl2 \ r1l2 : r(fi :u1 (/,,r,, - c)dr - J. (.r - c)a.r) - J" ,r(r - c)dr : - cI3 +2" - 31. 6 - n+ E: -hrnr, c2

But we have 0 I c .-7f 2, so -& < 0 and 4c2t2c-3 -1 < 0. Thus ?/(/") : --uU"2 + 2, - 3) > 0. Case welra,ve

712. Thenon 10,1,12) we have

{

4(1142 +2Q,12)-3

f.(*):0,

f"(r):0if z€ lll2,c) and/"(r) :r,*cif re

while onl7l2,7) [c,1]. Weget

\ t/c2 1\ (r-c)dr-o)-o:; T(f,,):r(f"):;t/rt (/" (; -"*;): ?

(c-1\2

Hence

thatT(.f)

rn.

satisfies ihe conditions (1) and (2) of the Lemma, which irnplies ) 0 for every convex / : [0,1] -+ R., ancl we are donc. n

463. Prove that there exists n6 € N such that Vn )

-l-*.L*...+ l*r 2lr has a unique solution in the interval

v1o

the equation

:rnrz

n*r (0,-), cienoted by 2,,, and that

,\,!'y'n,

- o,

where a e (0, 1) is the unique solution in the interval (0, 1) of the equtrtion

, i"

'\.+t1 "*=

1, where 1 is the Euler constant.

Proposed by Dumitru Popa, Ovidius

University, Constanla,

Romania.

Soluti,on by the author. For every natural number n let us define 7, : 1+ + +...+ L - lnn. As it is well-linown,0 ( 1r, .--7, Vn € N, and :1€ (0,1). For every nai,ural number nlet hn: [0,oo) J R, ,,fg1l, .+*+" lnn. Let us note that all h,, are continuous and It,,(.r) : #+#* strictly decreasirrgon [0,m). We have h,.(0) : 1+]+ +*-lnn: ?, ) 0,

: i+i+...+#-Inn:1,,.-l

+#.

Since,,l5gir,,1t; 1 - i < 0. it follows that there exists n6 € N such that Yn, ) ns we have h,,(1) < 0, Tlms there exists ns € N sr.rch that Vn, ) n6 the equatiort h,,,(n) : 0 has a unique solution in the interval (0, *) and this soit-ttion, denoted by ,n, is in the interval (0,1), that is 0 ( r,n l l,Yn ) no. For evely n ) n11 wt ltave hn(;:,,) - 0. T+ + r+ +...-t- -]- : lttn. or h,,11;

PRosr-sNrs

(r+i+ +*l -(++7i+ *"

.I-l *

11,i

+"h) *S: 1n

For every natural number n define

-i,.andrhus

gn: {0,1] +

)

ns-

(/,1

iR b1'

p,(r)-,f+, -ln :ir1---r =i\i-;*r)-1'' ) -t\2+x)

Define also p : [0,1] -+

IR.

)o-x

p(r)

:.L, *+;-1 i-l

Let us observe that Vr

€

[0,

1]. Vz

i:l

(i\ - *) / -,

e N 0 S,h-+,, < *

and the

series

fc-)c

D i Ir convergent, by the Weierstrass

i:l

criterion. so the series

! ,z{; it

i:t uniformly convergent on [0, 1] ancl its sum is a continuous function, that is, p is a continuous function. o , is a strictll- increasing function. Since

i:l

equation

p(r) : 0 has a unique

sequelbya€ (0,1).

ocr'rll-l 0(r\--=-1--r\--j-:.. *Li(i "-*Li(i+rl i:l

follows

solution in (0. 1). ri'hich is denoted in the foreverl-n € Nandr € [0,1] it holds

N{oreover, since

i=t

*r) '

* \/ ___( l(i+r)-' i:n.t

that ro-+cn lim q"(r) : p(r) uniformly rvith '

respect to

\,L ii2'

i_nll

z € [0,1]. From

the uniform com€rgence. by a general result. rnhich rl,e rvill prove in the end of the proof, it follows that 1im [er,(r,)- p(r")):0. Since by (7) Yn) no gn (rn) : 0, we deduce that g@") -+ 0 : g(a). Since ,p is strictly increasing it follows that rn-+ o. Indeed, let e > 0. Take 0 < ry < min(e. a.l-a):such a numberexistssince0 <a < 1. Then0 < a-q <a <a+71 < l and.sincerpis strictly increasing, q(a-q) < p(a): 0 < p(a+q).From p(r,) -+ 0 it follows that there exists n, ) ns such that Yn ) ne , p(a-n) < p(r") < p(a+4), that is, since rp is strictly increasing, Yn ) nu, a-n < rn I alT,|xr- al < n < €.

i.".,

,l$o

rn:

It remains to show that if ,{X f"(r) : p(r) uniformly rn'ith respect to r € [0,1], then for every sequence (rr,)r.ex c [0,1] it follows that j5g[r"("") - p(*")) : o. Indeed, we have: VE > 0, 1n.. € N such that Yn ) nu and Vr e [0,1] we have lp"@) - p(r)l < e . In particular. for

PRoposnp

fr : fin we deduce that Vn ) ri.

J$[r,

(r,

)

- p(r, )] :

PRoBLIJI\4s

we have lpr(r".)

- p(r")l ( 6, i.e., !

INsrRuc'rIUxI pEt{'t'RL AIrTonI

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Trparul executat la tipografia Editurii Paralela 45

CARAGEA

SUMAR

CONTENTS - Vol. XXXVI (CXV), nr.

1-2, 2018

AHTICOLE 1. Computing exponential and trigonometric functions of matrices in Mz(C), by Ovidiu F\rrdui 2. Stirling type formulas, by fon-Denys Ciorogaru

NOTE MATEMATICE 5. A happy case of mathematical (mis)induction, by Arpdd B6nyi and Ioan CaSu. The monotony in hazard rate sense of some farnilies of multidimensional distributions, by

Luigi-Ionu!

........

Catana. ....

PROBLEMS o

PROPOSED

PROBLEMS

48 50

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