2014 Decembirie Revista Electronica MAteInfo.ro

8/10/2019 2014 Decembirie Revista Electronica MAteInfo.ro

http://slidepdf.com/reader/full/2014-decembirie-revista-electronica-mateinforo 1/22

REVISTA ELECTRONICĂ MATEINFO.RO

ISSN 2065-6432

DECEMBRIE 2014

REVISTĂ LUNARĂ DIN FEBRUARIE 2009

DE PESTE 4 ANI Î N FIECARE LUNĂ WWW.MATEINFO.RO

[email protected]

COORDONATOR: ANDREI OCTAVIAN DOBRE

REDACTORI PRINCIPALI ŞI SUSŢINĂTOR PERMANENŢI AI REVISTEI

NECULAI STANCIU, ROXANA MIHAELA STANCIU ŞI NELA CICEU

Articole:1. Solutins and hints of some problems from the Octogon Mathematical Magazine (IV) – pag. 2

D.M. Bătineţu-Giurgiu, Neculai Stanciu, Titu Zvonaru2. Other solutions from some problems from School Science and Mathematics journal – pag. 8

Nela Ciceu, Roxana Mihaela Stanciu3. Exerciţii cu progresii aritmetice. Generalizări - pag. 11

Ciobîcă C. Constantin, Ciobîcă Elena 4. Metode de integrare numeric:

polinomul de interpolare Lagrange, formula lui Simpson, aplicaţii - pag 19 Boer Elena Milena

http://www.mateinfo.ro/

mailto:[email protected]

mailto:[email protected]

http://www.mateinfo.ro/



REVISTA ELECTRONICĂ MATEINFO.RO ISSN 2065-6432 – DECEMBRIE 2014 www.mateinfo.

1

1. Solutions and hints of some problems from theOctogon Mathematical Magazine (IV)

by D.M. Bătineţu-Giurgiu, Bucharest, Romania

and

Neculai Stanciu, Buzău, Romania

PP. 20660. Solve the following system:...)2(341)2(343 3

43232

332

221

x x x x x x x x

0)3(343 321

21

x x x xn .

Solution. Adding the equations of the system we obtain:

n

k k k k k x x x x

1

32 0)2(343 .

We prove that for any 3 x we have the inequality:32 )3(343 x x x x , (1)

Denoting 2 x y , the inequality (1) becomes:32 11 y y y , which after some algebra becomes:

01)1)(1(

22 y y , true, with equality if and only if01)1)(1( 232 y y y y y

,01 y

251

3,2 y . But only 12

51 y , i.e.

255

x .

So, the system has the unique solution

2

55,...,

255

,2

55 , and we are done.

PP.20661. In all acute triangle ABC holds A ActgAtgA cossin4 .

Solution. The inequality from the statement is not true, for e.g. if triangle ABC isequilateral we should have 399 .We will prove that

A ActgAtgA cossin4393 .Indeed by Bottema we have




2

233

sin A (the item 2.1),23

cos A (the item 2.16),

33tgA (true in all acute triangle) and 3ctgA (the item 2.38).Therefore, we get

A ActgAtgA cossin4393 , and we are done.

PP. 20668. If 0, y x , then:

nxy xy y xn y xy y x x n nnnnn 11221 )1(... .

Solution. The right inequality yields by AM-GM inequality. Indeed

nxy xy y xn xy y x x x xy y xn n nn nn n

n

n n

111

1

1 ...)1(

.

The left inequality is not true. For e.g. if we take 4n ,31

x , 1 y we should have

44 33223

31

2131

91

271

)3( xy y x y xy y x x

72934002732031

22740 44 , but 700

47

4003400 .

PP.20669. If 0k x ),...,2,1( nk , then n x x

x x x x

x xcyclic

3

21

2221

21

221 2

1 .

Solution. By AM-GM inequality3

21221 3

2

x x

x x and the inequality

0)(31

)2)(2(2

22

baabba

baba ,

yields that:

cyclic cycliccyclicn x x x x

x x x x x x

x x x x x x 1)2)(2(

)(32

13

2121

2

221

2

1

3

21

2

221

2

1221

,and we are done.

PP.20670. If 0k x ),...,2,1( nk , then cycliccyclic

x x x x x x

x x x x21

2121

22221

21

)2)(2()( .




3

Solution. By

0)(31

)2)(2(2

22

baabba

baba ,

and the fact that abbaba 322 we obtain

ababbaba

babababa

babababa

31

3)2)(2(

)()2)(2(

)( 2222

222

,

and then by adding cyclic yields the conclusion, and we are done.

PP.20671. Solve in R the following system:

3 222

3 222

3 222

)2(

)2(

)2(

zy x z z zx z

yx z y z yz y

xz y x y xy x

.

Solution. We solve the system in R . By AM-GM inequality we have:

323 2 z x

xz

, and other two similar.

Adding the equations of the system yields: 3 23 23 22 )2()2()2(2 zy x z yx z y xz y x xy x

3)2)(2(

3)2)(2(

3)2)(2( y z x z x y z y z x y x

3

72 2 xy x, i.e.

xy x xy x xy x 222 7236 , but xy x2 .Therefore, z y x , and we obtain the solutions ),,( aaa , with Ra .

PP.20672. If 0, y x then

162

12

1113 23 2

22

xy

x y

y x

y x x y

y x .

Solution. We prove the right inequality with AM-GM inequality, i.e. we have3 232 y x y x , 3 232 xy x y .

So,




4

16)31)(31(2

12

13 23 2

xy

x y

y x

y x .

After some algebraic manipulations the left inequality is equivalent with:

02 3 22

3 23 2

3 43 4

3 55

3 83 8

3 43 4

y x

y x y x

y x

y x y x , which is true

because the expressions 3 43 4 y x and 3 83 8 y x , respectively 3 43 4 y x and3 23 2 y x has the same sign.

We have equality if and only if y x .

PP.20677. In all nonisosceles triangle holds 6

)4()4(2

22

r s

sr Rr R

hh

h

h

hh

ba

c

c

ba .

Solution. The RHS is not correct. A solution and the correction for the RHS is given byM. Bencze and is the following

ba

c

c

ba

hhh

hhh

222

2223222223

1648)4()4(1648

r R sr R s Rr r s Rr r s Rr s R

.

(a proof by M. Bencze, can be found in math journal Sclipirea Minţii – Vol. 6, No. 11,2013, p. 9).In fact, also in math journal Sclipirea Minţii – Vol. 6, No. 11, 2013, p. 9, was given byBencze a proof for this identity

6)4()4(

2

22

r s

sr Rr Rr r

r r

r r

ba

c

c

ba ,

which holds in all nonisosceles triangle

PP.20678. In all nonisosceles triangle holds

C B A

C C

C B A2

2

sin)sin(cos

cossin)sin(

))2((448))4(4)(4(

6 22

2222222222

r R s sr r R s Rr r sr sr Rr s .

Solution. A proof for the identity from the statement was givenby M. Bencze in math journal Sclipirea Minţii – Vol. 7, No. 12, 2013.




5


2sin2sin

2cos

2cos

2sin

2sin

2

2

2

2

C B A

C

C

C B A2

165

r s

r R .

Solution. See the math journal Sclipirea Minţii – Vol. 7, No. 12, 2013.


Rr

C B

A

A

C B2

1

2

sin

2cos

2

cos

2sin

.

Solution 1. We have2

cos2

sin A

acbC B

, so

abcaccbba

acb

A

C B))()((

2cos

2sin

, and

))()((

)(

2sin

2cos 322

accbba

abcacaba

cba

C B

A

.

Since abcaa saabcacaba 3)2()( 32322

abcabcaa s 9322 32 Rrs Rr r s Rr r s s Rr r s s 36)4822)(2(2)4(4 222222

)2(4)91234(4 22222 Rr r s Rr Rr r s Rr r s s .Therefore,

))()(()2(4))()((

2sin

2cos

2cos

2sin

accbbar Rrs

abcaccbba

C B

A

A

C B

Rr

Rr R 2

12 , and the proof is complete.

Solution 2. See the math journal Sclipirea Minţii – Vol. 6, No. 11, 2013, p. 9.




6


216

5

22sin

2sin

2cos

2sin

2sin

22sin

r s

r R

Atg

BC

A BC

A BC

Atg

BC

.

Solution. See the math journal Sclipirea Minţii – Vol. 6, No. 11, p. 9.

PP.20684. If 0,, cba , then2

22

22

4

ab

ba

bab

bba

baa

aba .

Solution. Because 22

2 baa

aba and 2

22 ba

bb

ba the left inequality yields

immediately. For the right inequality we have:

0)2()(0322

22323

bababbaaab

ba

baa

aba

, true.Similar, we have

ab

ba

bab

bba 2

2, from where by multiplying yields the desired

result.

PP.20690. Solve in R the equation 0773 x x .

Solution. The equation 03 q px x , has three real roots if 032

32

pq .

In our case 7,7 q p and108409

27343

449

32

32

pq .

We have37

37

27343

27

3 pr and

73

23

27

2cos3 p

q

.

The three roots are:

3cos2 3

1

r x ,

03

2 1203

cos2

r x and

03

3 2403

cos2

r x ,

and we are done.

PP.20705. If 0k a ),,...2,1( nk then

n

k k

cyclic a

na

aa

a

aa

a

aa

1

2

23

2122

3121

32 6 .

Solution 1. We have the inequality:




7

cbabac

acb

cba 18

222 (1)

which is Problem L222 from Rec.Mat. 1/2012, proposed by Florin Stanescu.Solutions, refinements and generalizations you can find in:[1] D.M. Bătineţu-Giurgiu, Neculai Stanciu, I.V. Codreanu, Problema L222 din nr.1/2012 revizitată, Rec.Mat. nr. 1/2013; [2] TituZvonaru, Câteva soluţii la problema L222 din Rec.Mat. nr. 1/2012, Rec.Mat, nr.2/2012.By (1) and the inequality of Harald Bergström we obtain that:

cycliccycliccyclic aaa

naaaa

aaa

aa

a

aa

321

2

32123

2122

3121

32 1818

n

k k

n

k k a

n

a

n

1

2

1

2 6

318 ,

and first solution is complete.Solution 2. We prove that:

cbabac

acb

cba 222

222 (2)

For (2) we give also two solutions.

(i)

2222

212

ccb

cca

ccba

acba

0))(( 22

2222

cacac

aac

cca

ccb

cca .

(ii)cac

a 212

;bab

a 212

;cbc

b 212

;aba

b 212

;aca

c 212

;bcb

c 212

which by

adding yields (2).By (2) and the inequality of Harald Bergström we obtain that:

cycliccyclic aaaa

aa

a

aa

a

aa

32123

2122

3121

32 1112

n

k k cycliccycliccyclic

a

na

na

na

n

1

2

3

2

2

2

1

2 6222 ,

and we are done.




8

2. Other solutions from some problems fromSchool Science and Mathematics journal

By Nela Ciceu, Roşiori, Bacăuand Rox ana Mihaela Stanciu, Buzău

Solution:

We denote y x 5 , and after squaring we obtain)1)(1(22 223 y y y y y , and squaring again we obtain

0)1(0)122( 2222342 y y y y y y y y , which yields that

0 y ,2

51 y ,

251

y .

Therefore we have to solve in complex number the equations

05 x ,2

515 x and

2515

x .

We obtain the solutions

0 x ,

52

sin5

2cos

2515 k

ik

xk , 4,3,2,1,0k and

52

sin5

2cos

2515 m

im

xm , 4,3,2,1,0m .




9

Solution:




10

Solution:

The values modulo 13 of2n for 13 consecutive values ofn are:0,1, 4, 9, 3, 12, 10, 10, 12, 3, 9, 4,1.

Since )13(mod133 yields that )13(mod3 n it can take only the values 1, 3, 9.The expressions 23 nn and 21 )1(3 nn are simultaneously divisible by 13 onlyif )13(mod122n and )13(mod10)1( 2n , but then )13(mod10)2( 2n which addedwith )13(mod93 2n does not give )13(mod0 .




11

3. EXERCI ŢII CU PROGRESIIARITMETICE.GENERALIZĂRI.

Prof. Ciobîcă C. Constantin - COLEGIUL „ VASILE LOVINESCU”,FĂLTICENI Prof. Ciobîcă Elena- COLEGIU TEHNIC MIHAI BĂCESCU; FĂLTICENI

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4. Metode de integrare numerică:polinomul de interpolare Lagrange, formula lui Simpson,

aplicație.

Prof. Boer Elena Milena, Școala Gimnazială Vulcan, Jud.Brașov

1. Polinomul de interpolare Lagrange.

Considerăm o funcție y=f(x) pe care dorim să o interpolăm. Pentru aceasta, presupunem cunoscute valorile corespunzătoare argumentelor. Construim polinomul care are în punctele aceleași valori ca și funcția

f(x).(1)

Fie familia de polinoame (2)reprezintă simbolul lui Kronecker. se anulează în toate punctele mai

puțin . Îl putem scrie pe astfel:(3)

– este un polinom de ordin n-1.Deoarece , coeficientul este:

, ceea ce înseamnă că(4)

Polinomul unde m = n-1, îl putem scrie ca o combinație liniară a polinoamelor

(5)Observăm că .Înlocuindrelația (4) a polinoamelor , se obține polinomul de interpolare al lui

Lagrange:(6)

2. Formula lui Simpson.

Fie integrală definită pe intervalul [a,b],împărțitîntr-un număr den-1




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subintervale de lungime , prin punctele Se presupuncunoscute valorile funcției f(x) în punctele . Aproximăm funcția f (x) prin polinomul de interpolare Lagrange.

Introducem mărimea adimensională și atunci vom avea:

Astfel pentru polinomul de interpolare Lagrange se obține relația: (7)

Rezultă astfel, următoarea aproximație pentru integrala dată: (8)

unde

Notăm Din relația (8) obținem formula de cuadratură Newton – Cotes:(9)

Astfel pentru n=3 se obțin relațiile:

;

Rezultă formula lui Simpson: (10)

Pentru o mai bună acuratețe, generalizăm relația (10) și împărțim intervalul [a,b] printr-un număr impar, n=2m+1 de puncte echidistante . Aplicăm formulalui Simpson pentru fiecare din cele m subintervaleduble de lungime 2h: [ ], [ ],…, [ ], șiastfelintegraladefinităpeintervalul [] se poate scrie:

Regrupândtermenii, se obține formula lui Simpson generalizată. (11)

Unde: (12)(13)

3. Aplicație.




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În încheierea acestui articol, prezint un program realizat în C, care calculeazăfuncția de eroare prin integrare numerică, bazată pe formula lui Simpson.

#include <stdio.h>#include <math.h>

float Func(float x){return exp(-pow(x,2));

}float Simpson(float Func(float), float x){float h, h2, s1, s2, y;int i;

const float pi = 4.0 * atan(1.0);int n = 51;

h = x/(n-1); h2 = 2*h;s1 = 0.0; s2 = Func(h);

for (i=1; i<=(n-3)/2; i++) {y = i*h2;s1 += Func(y); s2 += Func(y+h);

}return 2*(h/3)*(1 + 4*s2 + 2*s1 + Func(x))/sqrt(pi);

}int main(){float x;int n; printf("x=");scanf("%f", &x);getchar; printf("\nerf(%.2f)=%2.6f\n", x, Simpson(Func, x));getchar();return 0;

}Bibliografie:

1. Titus Beu, “Calcul numeric în C”, EdituraAlbastră, 2004. 2. S. Dumitru, A. Armășelu, A. Boer, “Fizicăprobabilistă – îndrumar delaborator”,EdituraUniversității Transilvania, Brașov, 2004.

2014 Decembirie Revista Electronica MAteInfo.ro

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