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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
revistaelectronica@mateinfo.ro
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
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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
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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()( .
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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,
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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.
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PP.20679. In all nonisosceles triangle holds
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.
PP.20680. In all nonisosceles triangle holds
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.
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PP.20681. In all nonisosceles triangle holds
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:
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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.
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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 .
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Solution:
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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 .
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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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533311381
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n
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n
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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.