INTERNATIONAL JOURNAL OF MATHEMATICAL MODELS AND METHODS IN APPLIED SCIENCES
Spectrum of Fibonacci and Lucas numbers
Asker Ali Abiyev
Abstract It has been achieved polynomial function, depending on 3 3 2 n n a− b = a − b a + b− ab arguments a + b and ab arguments of expression a +b for the ()() biggest and smallest numbers which are in the centre of natural geometrical figures (line, square, cube,…,hypercube). The coefficients of this polynomial are defined from triangle tables, But there are no formulae for big values of n in literature. written by special algorithm by us. The sums of the numbers in The purpose of the article is to write new identities for the each row of the triangles make Lucas and Fibonacci sequences. binomial expression an± b n and study the fields of their New formulae for terms of these sequences have been suggested by application. us (Abiyev’s theorem). As the coefficients of the suggested polynomial are spectrum of Fibonacci and Lucas numbers they will opportunity these number’s application field to be enlarged. 2. F( a+b), ab EXPRESSION 2 Keywords: Lucas, Fibonacci, sequences, identity, Binet formula, If set of natural numbers {1, 2, 3, … , n}, {1, 2, 3, … , n }, 3 4 p polynomial. {1, 2, 3, … , n }, {1, 2, 3, … , n }, … , {1, 2, 3, … , n } is written accordingly in line, square, cube,…, hypercube of p 1.INTRODUCTION order one after another , the amount of numbers in their centres is defined with q=2p formula. Let’s mark the n n maximal number with A and minimal number with B . The binomial expression a± b is one of the most used p p Where, n even number and order of geometrical figure, p expressions in algebra. It has evidently been seen up to now dimension of space. If p=3 and n=6, it is cube of 6 order in that one of the powers of a and b coefficients increases space of 3 dimension. and another one decreases in the identity of this expression The following formulae Ap(n) and Bp(n) achieved after [1]. The following formula can serve as an example. doing some algebraic operations by using numbers of figures’ centres for p=1, p=2, p=3, n=6 and n=8: abn− n =( aba −) n−1 + abab n − 2 + n − 3 2 +... + abn−2 + b n − 1 (1) ( ) n p−1 p − 2 p − 3 Ap () n= n +( n + n +... + 1) + 1 2 (2) When a and b letters are conjugate expressions, algebraic n calculus become more difficult by using (1) identity. As the p−1 p − 2 p − 3 Bp () n= n −( n + n +... + 1) (3) sum and multiplication of the conjugate expressions are real 2 numbers, writing the right side of (1) identity as a function depending on a+b and ab arguments simplifies the Let’s calculate A2(6), B2(6), A3(6), B3(6) by these formulae: calculus. This function can be easily written for small values 6 6 A ()6=[] 6 + 1 + 1 = 22 ; B ()6= 6 − 1 = 15 ; of n : 2 2 2 2 6 A 6= 62 + 6 + 1 + 1 = 3.43 + 1 = 130 3 3 2 3() () , a+ b =()() a + b a + b − 3 ab 2 6 B ()6= 62 −() 6 + 1 = 3.29 = 87 . 3 2
And now let’s find out sum and product of Ap(n) and Bp(n) :
n p−1 p − 2 p − 3 n p−1 p − 2 p − 3 p Ap ()() n+ Bp n = n +( n +++++ n ... 1) 1 n −( n +++=+ n ... 1) n 1 2 2 (4)
n n p n A n. B n= n p−1 +1 −n p−2 + n p − 3 +... + 1 n p−2 + n p − 3 +... + 1 + 1 p()() p ( ) ( ) 2 2 2 (5)
By these formula let’s calculate A2(6) + B2(6), A2(6). B2(6) and A3(6)+ B3(6) , A3(6).B3(6):
3 A3(6) + B3(6)=6 +1=217=130+87; A3(6) B3(6)=3{36.109 (4) and (5) formulae have been proved by mathematical 7.[3.7+1]}=3.3770=11 310=130.87 induction method.
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After using Binomial theorem and doing some algebraic operations, we get the following equalities for
Ap+ 1 + B p + 1 = F A + B, A B expression: if p odd p p ( p p) p p p+1 p + 1 1 p+1 Ap +B p= E p+1() A p + B p − A p B p × p+1 p+3 2 p−1 3 p−3 2 2 2 (6) ×EABABEABABEABEABp+1() p + p − p p p+1( p+ p ) − ... − p p p+1 ( p+ p ) − p+1 p p ... if p even p+1 p + 1 Ap +B p 1 p =TABABp+1() p + p − p p × ABp+ p
p p (7) +1 2p− 2 3 p − 42 2 2 TABABTABABTABTABp+1() p+ p − p p p + 1( p+ p ) − ... − p p p+1 ( p+ p ) − p + 1 p p ...
p+3 p Table1, drawn with special algorithm by us [2]. Where +1 EEE1, 2 ,..., 2 TTT1, 2 ,..., 2 1 1 p+1 p + 1 p +1 and p+1 p + 1 p + 1 1 TE pp ++ 1 == 1 coefficients in (6) and (7) expressions are found from the Table 1.
If to use (2) and (4) expressions, a new equality is found for ApBp product:
p p s−1 s + 1 n n n p−2 p − 3 n p−2 p − 3 ( p) ( p ) ABp p = +−1 (n +++ n .... 1) (n ++++= n .... 1) 1 −()AADp−1 −1 p−1 = p 2 2 2 2 4 (8)
If to use (4) and (8) equalities, (6) and (7) will be as following:
p+1 p+3 p+1 p+1 p+11 −=+ p−12 − p−33 −− (... 2 2 − 2 DEsEDsEDsEDsEBA )... p p +1 pp +1 ppp +1 ppp pp +1 pp +1 p (9)
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p p +1 p+1 p+1 =+ 1 p − p−22 − ( p−43 −− (... 2 2 − 2 DTsTDsTDsTDsTsBA )..) p p +1 ppp +1 ppp +1 ppp pp +1 pp +1 p (10)
2 2 2 2 2 1 2 2 2 2 (n+ 2) n (n + 1) + 1 n n A1+=−=−=+− B 1 E 2 s 1 E 2 D 1 s 12 D 1 ( n 1) 2 −−= (AA0 1) 0 =++1 (11) 4 2 2 2 Let’s calculate (10) and (11) equalities for p=1, p=2 , p=3.
n n Where A =1, p=1, A = +1 , B1 = . 0 1 2 2
For p=2
3 3 1 2 2 2 2 ()()s2 +1 s2 − 1 (12) ABsTsTD2+ 223232 =( −)() = ss 22 −3 D 2 = ss 22 − 3 −()AA1 − 1 1 = 4 n2+ 2 n 2 2 2 2 () n n =()()n +1 n + 1 − 3 − +1 4 2 2
For p=3
n3 + 2 n 3 4 4 3 4 ( ) n n A3+ B 3 =() n +1 − −()n + 1 () n + 1 + 1 × 4 2 2 n3 + 2 n 3 3 2 () n n ×4 ()n + 1 − 2 −()n +1 () n + 1 + 1 4 2 2 (13)
3. ABIYEV TRIANGLES
As it seen from the formulae, as dimension of space It should be mentioned that, formulae (6) and (7), which increases, (12) and (13) equalities become more complicated suggested for natural numbers in centres of geometric depending on order of geometrical figure. Only p=1, figures, are true as well as for any numbers. Pythagorean theorem is satisfied. Because, the orders of If p+1=n, Ap=a and Bp=b in (6) and (7) equalities and expressions are the same in both right and left of equality open the brackets, they will be written as follows: [3].
n−1 n−3 n 12 n 3 n+ 1 n 1 abn+ n = abTab +1 + − Tab2 + ab +... + Taby2 + 2 − Tab 2 2 ()( n ()()()n n () n () ) (7’) n n The identity a± b = F( a+b), ab was side and figure 1 to the left side of this column; write figure 2 before even numbers at the right and figure 0 at the left. In first revealed and a special table was compiled to determine consequence, 2 columns formed at the right and at the left of the coefficients in this identity [4]. Let’s explain the column 0. algorithm first to write the table: insert columns to the right nd rd Afterwards, write figure 1 in the 2 and 3 rows of the and left of 0 at the top of the table 1 and write numbers nd nd 2 columns; by summing the figure 1 in the 2 row of the beginning from 1 in sequence in column 0. Write odd nd rd st 2 column with the figure 3 in the 3 row of the 1 column numbers to the right
Issue 1, Volume 6, 2012 167 INTERNATIONAL JOURNAL OF MATHEMATICAL MODELS AND METHODS IN APPLIED SCIENCES write figure 4 in the 4th row of the 2nd column. So by If the method used to form the columns at the right side continuing this method figures 4, 9, 16, 25 etc. squares of of the table1is applied symmetrically with respect to column sequent numbers are written at the right in the even rows of 0, columns can be written at the left side too. the 2nd column. The compiled table 1 is composed of right and left Summing the figure 1 in the 3rd row of the 2nd column triangles. The numbers in the rows of the right and left with the figure 4 underneath figure 5 is written in the 5th triangles specify the coefficients of the polynomial row; later summing this last figure 5 with the figure 9 n n n n th expressions to which a+ b and a− b expressions underneath figure 14 is written in the 7 row. At the result, equal respectively. They were put in sequence from right to the 2nd column is formed at the right. rd left in the right triangle and from left to right in the left In order to write the 3 column figure 1 is written in the triangle. One of the interesting properties of this table is that 4th and 5th rows of the 3rd column again; afterwards the nd the sum of the numbers at the right and at the left rows above mentioned method is applied to form the 2 column. forms Lucas and Fibonacci sequences correspondingly. So a number of columns and rows of figures can be formed.
4. IDENTITIES Let’s accept the following replacements in order to simplify 1n− 1 the method of writing polynomial expressions: a+b≡x and is E =n ≡1 an even power, n 0 ( −1 ) is accepted. ab≡y. Let’s show the identities of even and odd powers. If If n is an odd power, nis even power n n 1n− 1 2 n − 3 3n− 5 2 n n1 n 2 n− 2 3n− 4 2 a+ b = xTx( n − TxyTxy n +n −... a+ b = En x − E n x y + En x y −... n 1 n 3 n+1 n 1 n n 2 n+2 n 2 2 2 2 2 2 2 2 2 2 ... +Tn x y − Tn y (16) ... − En x y+ En y (14) ) here here k+1n n − k − 1 n−1 T =;k = 0,1,2,3,..., (17) k+1 n n − k −1 n nk ( k−1 ) 2 En= k ( k−1 );k = 0,1,2,3,...,2 ; (15) 1n n− 1 and Tn =0 ( −1 ) ≡1 is accepted.
TABLE 2. ABIYEV’S TRIANGLES (LEFT AND RIGHT)
7 6 5 4 3 2 1 0 1 2 3 4 5 6 7
F n L
1 1 1 1 1
1 1 0 2 2 1 3
2 1 1 3 3 1 4
3 1 2 0 4 2 4 1 7
5 1 3 1 5 5 5 1 11
8 1 4 3 0 6 2 9 6 1 18
13 1 5 6 1 7 7 14 7 1 29
21 1 6 10 4 0 8 2 16 20 8 1 47
34 1 7 15 10 1 9 9 30 27 9 1 76
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55 1 8 21 20 5 0 10 2 25 50 35 10 1 123
The written E k+1 and T k+1 coefficients can be described sum of the numbers in the rows forms Lucas (at the right) n n and Fibonacci (at the left) sequences. Though in the in another way too: literature [5] it’s shown that the sum of the numbers in a k+1 n − k n −1 − k n diagonal line in Pascal triangle forms Fibonacci sequence, En=( k) +( k−1 );k = 0,1,2,..., 2 ( n even) what values these numbers have is unknown.
k+1 n − k n −1 − k n−1 Tn=( k) +( k−1 );k = 0,1,2,..., 2 ( n odd) Let’s display that in Abiyev’s triangles these singularities are not random.
Let’s write the polynomial expressions for an− b n expression. If n is an even power, For this purpose the coefficients of (14), (16), (18) and (20) polynomial expressions and Binet
n n αn− β n 1+ 5 1 − 5 formula LFn=α + β ;;; n =α− β α =2 β = 2 n n a− b 1n− 2 2n− 4 3n− 6 2 of Lucas and Fibonacci sequences are used [3]. Here =x( Mn x − Mn x y + Mn x y −... a− b x =α + β =1;y = αβ = − 1 . If these values are put into n−2 n − 4 n n−2 (14), (16), (18) and (20) polynomial expressions, the 22 2 2 2 (18) ...−Mn x y + Mn y ) following equations can be obtained: n n+2 n n n n 1 2 3 2 2 a+ b =α + β =EEEEELn + n + n +... +n + n = n k+1 n − k − 1 n here M n=( k );k = 0,1,2,3,...,2 − 1 (19) n –even; (Lucas term). n−1 n+1 n n n n 1 2 3 2 2 If n is an odd power, a+ b =α + β =TTTTTLn + n + n +... +n + n = n ; n –odd; (Lucas term). n n n n n n a− b 1n− 1 2n− 3 3n− 5 2 a− b α − β n−2 n =N x − N x y + N x y −... = =MMMMMF1 + 2 + 3 +... +2 + 2 = ; a− b n n n a− b α − β n n n n n n n−1 n − 3 n+1 n − 1 ...+N2 x2 y 2 − N2 y 2 (20) n –even; (Fibonacci term). n n n n n n a− b α − β 1 2 3 n−1 n+1 k+1 n − k − 1 n−1 2 2 here N =;k = 0,1,2,3,..., (21) = = NNNNNF n + n + n +... +n + n = n ; n( k ) 2 a− b α − β k+1 k+1 n –odd; (Fibonacci term). The written M n and Nn coefficients can be described in another way too: So taking into account (15), (17), (19) and (21) polynomial k+1 n − k n −1 − k n M n=( k) −( k−1 );k = 0,1,2,...,2 − 1 (n even) expressions new formulae can be written for Lucas and Fibonacci triangles: k+1 n − k n −1 − k n−1 n−1 Nn=( k) −( k−1 );k = 0,1,2,...,2 ;( −1 ) ≡ 0 n n 2 2 (n odd). LE=k +1 = n n−1 − k ; Note: As the signs in (14), (16), (18) and (20) polynomial n ∑n ∑ k ()k (22a ) n k =0 k =0 expressions always change in sequence, ()−1 symbol n−1 n−1 2 2 hasn’t been used in them. k +1 n n−1 − k (22b) LTn=∑ n = ∑ k ()k ; Let’s show 2 examples for the identities: k=0 k=0 6 6 6 4 2 2 3 n −1 n −1 If n = 6 , a+ b = x −6 x y + 9 x y − 2 y ; 2 2 FM=k+1 = n−1 − k ; (23a) a7− b 7 n ∑n ∑()k If n = 7 , =x6 −5 x 4 y + 6 x 2 y 2 − y 3 . k =0 k=0 n−1 n−1 a− b 2 2 See the 6th row in the right triangle and the 7th row in the left k +1 n−1 − k (23b) FNn =∑n = ∑ ()k ; triangle of the table 2. k = 0 k = 0
Let’s give examples for L6 and F7 formulas: 3 3
LE=k +1 =6 5−k ⇒EEEE1 + 2 + 3 + 4 = 6 ∑6 ∑ k ()k −1 6 6 6 6 k =0 k =0 5. PROPERTIES OF COEFFICIENTS 64 6 3 6 2 =1 +1( 0) + 2( 1) + 3 ( 2 ) =1 + 6 + 9 + 2 = 18 .
3 3 It’s known that the sum of the numbers in each row of k +1 6−k 1 2 3 4 FN7 =∑7 = ∑()k ⇒FNNNN7 = 7 + 7 + 7 + 7 = Pascal triangle equals to 2n−1 . But in Abiyev’s triangles the k =0 k=0
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6 5 4 3 n+ 2 n+ 2 n+ 4 = + + + =1 + 5 + 6 + 1 = 13. n 2 n +1 2 n + 2 2 ( 0) ( 1) ( 2) ( 3 ) ⇒ + = ⇒ n− 4 ( 4 ) n− 2 ( 3 ) n− 2 ( 4 ) The above formed formulae enable us to express the 2 2 2 following theorem. n+ 2 nn − 2 n− 4 n+ 2 n n − 2 n 2 2 2 2 n +1 2 2 2 Theorem. There are Lucas and Fibonacci numbers ⇒ n− 4 1.2.3.4 +n− 2 1.2.3 = 2 2 n n 2 2 n n−1 − k n−1 − k n+4n+2 n n − 2 n n+2 2 2 2 2 2 2 expressed by Ln= k−1 and F = ∑ k () n∑ k () k = n−2 1.2.3.4 ⇒n +6 n + 8 = n + 6 n + 8. k=0 k=0 2 formulas for any positive whole number. ΙΙ −l + l = l Proof. Taking into consideration the peculiarities of the n−1,2 n ,2 n+ 1,2 algorithm used for formulating Abiyev’s triangles, let’s | prove the theorem applying a mathematical induction method. According to (15) and (17) formulae let’s form right Abiyev’s triangle for (n−1) , n ,( n + 1) ,( n + 2) n n n+ 2 n −1 2+n 2 =n +1 2 ⇒ rows: n −4( n −6) n − 2( n − 4) n − 2 ( n − 4 ) 22 2 2 2 2 In accordance with the singularities of the right and left n n n+ 2 triangles the following expressions can be written: ⇒n −1 2 +n 2 =n +1 2 ⇒ n −4( 3 ) n − 2( 2 ) n − 2 ( 3 ) 2 2 2
Ι −lij + l i+1, j − 1 = l i + 2, j ; ΙΙ −li−1, j − 1 + l i , j − 1 = l i + 1, j − 1 ; 2≤i ≤ 2 p ; 2 ≤ j ≤ p + 1; p ≥ 1. (right) nn− 2 n − 4 n n − 2 n −1 2 2 2 n 2 2 ⇒n − 4 1.2.3 +n − 2 1.2 = 2 2 n+2 n n − 2 Ι −fij + f i+1, j − 1 = f i + 2, j ; ΙΙ − fi−1, j − 1 + f i , j − 1 = f i + 1, j − 1 ; n+1 2 2 2 2 2 =n−2 1.2.3 ⇒n +3 n + 2 = n + 3 n + 2. 2≤i ≤ 2 p ; 2 ≤ j ≤ p + 1; p ≥ 1. (left) 2 here l and f are elements of the right and left triangles respectively. Let’s conduct the following algebraic operations using this triangle: These operations specify that the algorithm of formation in the right triangle is defined for any n. The same assumption Ι −ln,3 + l n+ 1,2 = l n + 2,3 concerns the left triangle too. By this way, the theorem is
n+ 2 n+ 2 n+ 4 proved. n 2 +n +1 2 =n + 2 2 ⇒ n− 4 ( n−6 ) n− 2 ( n− 4 ) n− 2 ( n− 4 ) 2 2 2 2 2 2
TABLE 3. PART OF THE ABIYEV’S RIGHT TRIANGLE col. → 1 2 3 ... n−4 n−2 n n+2 n + 4 ↓ row 2 2 2 2 2 n−2 n n+2 n −1:n−1 2n−1 2n−1 2 ... n−1 n−4 n−1 n−3 1 n−2()n−4n − 4()n − 6n − 6 ()n − 8 2 ()1 1 ()0 22 2 2 2 2
n−2 n n+2 n: n 2n 2n 2 ... n n−4 n n−3 n n−2 1 n ()n−2n−2()n − 4n − 4 ()n − 6 3 ()2 2 ()1 1 ()0 22 2 2 2 2
n n+2 n+4 n +1: n+1 2n+1 2 n+1 2 ... n+1 n−3 n+1 n−2 n+1 n−1 1 n ()n−2n−2()n− 4n− 4 ()n− 6 3 ()2 2 ()1 1 ()0 22 2 2 2 2
n n+2 n + 4 n + 2: n+2 2n+2 2n+2 2 ... n+2 n−3 n+2 n−2 n+2 n−1 n+2 n 1 n+2()n n() n−2 n − 2 () n − 4 4 ()3 3 ()2 2 ()1 1 ()0 22 2 2 2 2
Xn± Y n = a 6. APPLICATIONS X+ Y = b The solution of equation system is simplified, i.e. the power of the system
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n is easily calculated. is reduced to a half and brought to the solution of 2 C =1 − 7 + 14 − 7 = 1 equivalent equation system. 7. CONCLUSION n n n n The calculation of ()()X+ iY ± X − iY = Z ± Z expression mostly used in linear algebra is rather facilitated The content of the article is new. The coefficients of the 7 7 polynomial expressions can especially be accepted as a 1 3 1 3 For instance, let’s show C=( 2 + i 2) +( 2 − i 2 ) . spectrum of Lucas and Fibonacci sequences. As these sequences have a significant importance in mathematics, Here taking into account their spectrum can also be applied in different fields. The
identities proposed in the article will allow simplifying
mathematical calculations in various fields of science and
technology.
x= a + b =1 ; y = ab = 1 and a7+ b 7 = x( x 6 −7 x 4 y + 14 x2 y 2 − 7 y 3 ) ,
REFERENCES [1] I.N.Bronstein, K.A.Samandiyev. Handbook on Mathematics, Name: Asker Ali Abiyev Moscow, Science, Vol.XIII, Place and date of birth: Baku, Azerbaijan, 28.06.1934. 1986, p.141 (in Russian); Education: primary school No 79, and 208 (1944 1954), Baku; [2] A. A. Abiyev, M. Sahin, A. A. Abiyev, A study on Derivation Azerbaijan State University (1954 1957); Moscow State University p+1 p + 1 named M. V. Lomonosov, Physic faculty (1957 1961); post and Explanation of ABFABAB+ = + , p p ( p p) p p graduate student of Institute Atomic Energy named I. V. Kurchatov Formula For Some Geometric Figures, Journal of Marmara for (1963 1966); Ph.D in physics mathematics (1970); Doctor in Pure and Applied Sciences, 18 (2002), 73 80. physics mathematics (1989); Professor (1990). [3] Zhang Zaiming, Problem #502, Pell’s Equation Once Again, Scientific activity: neutron spectroscopy, radiation physics of Two year College Math. Inl, 25 (1994), 241 243. semiconductors, magic squares and cubes. [4] S.Goktepe, A.A.Abiyev. A New Algorithm to Form Lucas and Career/Employment: Fibonacci Sequences, XIX International Symposium on 1.Head of the laboratory radiation physics semiconductors in the Mathematics, August 22 25, 2006, Kutahya, pp.600 604 (in Institute of Radiation Problems of Azerbaijan National Academy Turkish); Sciences 1975 1993; Consultant in the School Boys, Ankara [5] Kennets H. Rozen, Handbook of Discrete and Combinatorial Turkey 1993 2000; Professor of Mathematics in the Gaziantep of Mathematics, CRC Press LLC (New York, 2000), pp. 98, 104, University in Turkey, 2000 2007; Head of Experimental 140 143. Department of the Electron Accelerator, 2009 up today. nd Participated Conferences: 1. The 2 International Conference on Applied Informatics and Computing Theory (AICT’11), Prague, Czech Republic, September 26 28, 20011. 2.IMS'2008 6th International Symposium on Intelligent and Manufacturing Systems «Feature, Strategies and Innovation”, October 14 17, 2008, Sakarya, Turkey. 3. Fourth International Conference on Soft Computing, Computing with Words and Perceptions in Systems. Analysis, Decision and Control, Turkey, August 27 28, 2007. 4. School of Information Technology and Mathematical Sciences, University Ballarat (Australia). 14 28 June 2006. 5. Joint International Scientific Conference ‘’New Geometry of Nature’’, August September 5,2003, Kazan State University, Kazan, Russia. 6. The Third International Conference On Mathematical And Computational Applications, September 4 6, 2002, Konya,Turkey. 7. 2nd International Conference On Responsive Manufacturing, Gaziantep, Turkey, 26 28, June, 2002. 8. The Second International Symposium on Mathematical and Computational Applications, September 1 3, 1999, Baku, Azerbaijan.
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9. Research Conference on Number Theory and Arithmetical A.K. Abiyev, Sayılı Sihirli Karelerin Doğal ifresi The Natural Geometry, San Feliu de Guixols, Spain 24 29 October, 1997. Code of Numbered Magic Squares, Enderun Publications, Ankara, International Conference on Radiation Physics of Semiconductors (ISBN 975 95318 3 6), p.77, 1996, (in Turkish and in English ). and Related Materials, TBĐLĐSĐ, USSR, 1978. http:/www1.gantep.edu.tr/~bingul/php/magic/ Publications: Number of paper in refereed journals:120 Language Skills: Azeri, Russian, English, Turkish. Number of communications to scientific meetings: 25 Contact: [email protected] The Book:
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