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International Journal of Computational Engineering Research(IJCER) is an intentional online Journal in English monthly publishing journal. This Journal publish original research work that contributes significantly to further the scientific knowledge in engineering and Technology.

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### C04502009018

1. 1. ISSN (e): 2250 – 3005 || Vol, 04 || Issue, 5 || May – 2014 || International Journal of Computational Engineering Research (IJCER) www.ijceronline.com Open Access Journal Page 9 Extensions of Enestrom-Kakeya Theorem M. H. Gulzar Department of Mathematics University of Kashmir, Srinagar 190006 I. INTRODUCTION AND STATEMENT OF RESULTS A famous result giving a bound for all the zeros of a polynomial with real positive monotonically decreasing coefficients is the following result known as Enestrom-Kakeya theorem [8]: Theorem A: Let    n j j j zazP 0 )( be a polynomial of degree n such that 0...... 011   aaaa nn . Then all the zeros of P(z) lie in the closed disk 1z . If the coefficients are monotonic but not positive, Joyal, Labelle and Rahman [6] gave the following generalization of Theorem A: Theorem B: Let    n j j j zazP 0 )( be a polynomial of degree n such that 011 ...... aaaa nn   . Then all the zeros of P(z) lie in the closed disk n n a aaa z 00   . Aziz and Zargar [1] generalized Theorem B by proving the following result: Theorem C: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some ,1k 011 ...... aaaka nn   . Then all the zeros of P(z) lie in the closed disk n n a aaka kz 00 1   . Gulz ar [4,5] generalized Theorem C to polynomials with complex coefficients and proved the following results: Theorem D: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some 10,1  k , 011 ......   nn k . Then all the zeros of P(z) lie in the closed disk ABSTRACT: In this paper we give an extension of the famous Enestrom-Kakeya Theorem, which generalizes many generalizations of the said theorem as well. Mathematics Subject Classification: 30 C 10, 30 C 15 Keywords and Phrases: Coefficient, Polynomial, Zero
2. 2. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 10 n n j jn n n a k a kz     0 000 2)(2 )1(   . Theorem E: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some 10,1  k , 011 ......   nn k . Then all the zeros of P(z) lie in the closed disk n n j jn n n a k a kz     0 000 2)(2 )1(   . Theorem F: : Let    n j j j zazP 0 )( be a polynomial of degree n such that for some real ,,......,2,1,0, 2 arg;, nja j    and for some 10,1  k , 011 ...... aaaak nn   . Then all the zeros of P(z) lie in the closed disk n n j jnn a aaaak z      1 1 00 sin2)1sin(cos2)sincos1(  . Some questions which have been raised by some researchers in connection with the Enestrom-Kakeya Theorem are[2]: What happens , if ( i) instead of the leading coefficient n a , there is some j a with 11   jjj aaa such that for some 1k , 01111 ............    jjjnn akaaaa , j=1,2,…..,n and (ii) for some 011121 ......;1,1 2 aaakakkk nn   . In this direction, Liman and Shah [7, Cor.1] have proved the following result: Theorem G: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some ,1k , 01111 ............ aaakaaaa nn    . Then P(z) has all its zeros in n nj n j jn a aaakaaa z           )()1(00  . Unfortunately, the conclusion of the theorem is not correct and their claim that it follows from Theorem 1 in [7] is false. The correct form of the result is as follows: Theorem H: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some ,1k , 01111 ............ aaakaaaa nn    . Then P(z) has all its zeros in
3. 3. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 11 n n a akaaa z  )1(200   . In this paper , we are going to prove the following more general result: Theorem 1: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some 10,1  k , 01111 ............ aaakaaaa nn    . Then P(z) has all its zeros in n n a aaaaka z 000 2)()1(2    . Remark 1: For 1 ,Theorem 1 reduces to Theorem H. Taking in particular 1 1     a a k in Theorem 1, we get the following Corollary 1: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that 01111 ............ aaaaaaa nn    . Then P(z) has all its zeros in n n a aaaa a aa a z 000 1 2)()(2         . For 1 , Cor. 1 reduces to the following Corollary 2: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that 01111 ............ aaaaaaa jnn    . Then P(z) has all its zeros in n n a aaa a aa a z ))( 00 1         . Theorem 1 is a special case of the following more general result: Theorem 2: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some ,1k 01111 ............     knn . Then P(z) has all its zeros in n n j jn a k z     0 000 22)()1(2   . Remark 2: If njj ,......,2,1,0,0  i.e. j a is real, then Theorem 2 reduces to Theorem 1.
4. 4. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 12 Applying Theorem 2 to the polynomial –iP(z), we get the following result: Theorem 3: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some 10,1  k , 01111 ............    jjjnn k . Then P(z) has all its zeros in n n j jn a ka z     0 000 22)()1(2   . For polynomials with complex coefficients, we have the following form of Theorem 1: Theorem 4: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some 10,1  k , 01111 ............ aaaakaaa nn    . Then P(z) has all its zeros in )1sin(2)1sin(cos)sin(cos[ 1    kkaaka a z n n ]2)1sin(cos 00 aa   Remark 3: For k=1, Theorem 4 reduces to Theorem F with k=1. Next, we prove the following result: Theorem 5: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some 10,1,1 21  kk , 012121 ......    nnn kk . Then P(z) has all its zeros in n n j jnnnnnnn a kkkk z      0 011121121 22)()()(  . Remark 4: If njj ,......,2,1,0,0  i.e. j a is real, we get the following result: Corollary 3: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some 10,1,1 21  kk , 012121 ...... aaaakak nnn   . Then P(z) has all its zeros in n nnnnnnn a aaaaakakakak z 011121121 2)()()(    . Applying Theorem 2 to the polynomial –iP(z), we get the following result from Theorem 4: Theorem 6: Let    n j j j zazP 0 )( be a polynomial of degree n with jj a )Re( , nja jj ,......,1,0,)Im(   such that for some 10,1,1 21  kk ,
5. 5. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 13 012121 ......    nnn kk . Then P(z) has all its zeros in n n j jnnnnnnn a kkkk z      0 011121121 22)()()(  . Theorem 7: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some real ,,......,2,1,0, 2 arg;, nja j    and for some 10,1,1 21  kk , 012121 ...... aaaakak nnn   . Then P(z) has all its zeros in                  2 1 0 01121 sin2)1sin(cos 2)cos1()sincos1()sincos1( 1 n j j nnnn n aa aaaakak a z   . Remark 4: For 1, 21  kkk , Theorem 6 reduces to Theorem F. Taking 1 in Theorem 7, we get the following Corollary 4: Let    n j j j zazP 0 )( be a polynomial of degree n such that for some real ,,......,2,1,0, 2 arg;, nja j    and for some ,1,1 21  kk , 012121 ...... aaaakak nnn   . Then P(z) has all its zeros in nnn n aakak a z   )sincos1()sincos1([ 1 121  ]sin22)sincos1()cos1( 1 1 001      n j jn aaaa  . II. LEMMA For the proof of Theorem 6, we need the following lemma: Lemma: Let 1 a and 2 a be any two complex numbers such that 21 aa  and for some real numbers  and  , 2,1, 2 arg  ja j   ,then  sin)(cos)( 212121 aaaaaa  . The above lemma is due to Govil and Rahman [3] . 3. Proofs of Theorems Proof of Theorem 2: Consider the polynomial )......)(1()()1()( 01 1 1 azazazazzPzzF n n n n    1 1 1 211 1 )(......)()(         zaazaazaaza n nn n nn n n 0011 )(......)( azaazaa    
6. 6. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 14 ......)()( 1 211 1     n nn n nn n n zzza  })......(){()}(){( ......)}(){()}(){( 001100001 1 1 1             zziz zkkzkk n nn For ,1z we have, nj z j ,.......,2,1,1 1  so that, by using the hypothesis, 1 1 1 111 1 ......[)(          zkzzzazF n nn n nn n n 0 1 100 011 1 )1( ......)1()1(                n j j jj zz zazkzkzk 11111 1 ...... 1 {[[     nnnnnn n z k z zaz }] 1111 )1( 1 ...... 1 )1( 11 )1( 0 1 1010 10111` n n j jnjjnn nnnn zzzz z a z k z k z k            )1(......[[ 1211   kkzaz nnnnn n ])( )1(......)1( 0 1 1 00011          n j jj kk 0 ]22)()1(2[[ 0 000     n j jnn n kzaz   if n n j jn a k z     0 000 22)()1(2   . This shows that the zeros of F(z) having modulus greater than 1 lie in n n j jn a k z     0 000 22)()1(2   . But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence, it follows that all the zeros of F(z) lie in n n j jn a k z     0 000 22)()1(2   .
7. 7. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 15 Since the zeros of P(z) are also the zeros of F(z), the result follows. Proof of Theorem 4: : Consider the polynomial )......)(1()()1()( 01 1 1 azazazazzPzzF n n n n    1 1 1 211 1 )}(){(......)()(         zakakaazaazaaza n nn n nn n n ......)()}(){( 1 211        zaazakaaka 00001 )}(){( azaaaa   . For ,1z we have, nj z j ,.......,2,1,1 1  , so that, by using the hypothesis and the Lemma, 1 1 1 111 1 ......[)(         zkaazaazaazazF n nn n nn n n 00 011 1 )1( ......)1()1( aza zaazakzakazak             11211 1 ...... 1 {[    nnnnnn n z kaa z aaaazaz }] 11 )1( 1 ...... 1 )1( 11 )1( 010 10111` nn nnnn z a z a z aa z ak z aka z ak        kaaaaaazaz nnnnn n   1211 ......{[ }])1( ......)1()1( 00 011 aa aaakakaak       cos)(sin)(cos){([ 2111   nnnnnnn n aaaaaazaz }])1(sin)( cos)......(sin)(cos)( )1(sin)(cos)()1( sin)(cos)(......sin)( 0001 012121 11 1121 aaaa aaaaaa akaakaakak akaakaaa nn               )1sin(2)1sin(cos)sin(cos{[    kkaakazaz nn n }]2)1sin(cos 00 aa   0 if )1sin(2)1sin(cos)sin(cos[ 1    kkaaka a z n n ]2)1sin(cos 00 aa  
8. 8. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 16 This shows that the zeros of F(z) having modulus greater than 1 lie in )1sin(2)1sin(cos)sin(cos[ 1    kkaaka a z n n ]2)1sin(cos 00 aa   But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence, it follows that all the zeros of F(z) lie in )1sin(2)1sin(cos)sin(cos[ 1    kkaaka a z n n ]2)1sin(cos 00 aa   Since the zeros of P(z) are also the zeros of F(z), the result follows. Proof of Theorem 5: Consider the polynomial )......)(1()()1()( 01 1 1 azazazazzPzzF n n n n    001 1 211 1 )(......)()( azaazaazaaza n nn n nn n n     1 211121121 1 )()}()(){(     n nn n nnnnnn n n zzkkkkza  })( ......){()}(){(...... 001 100001      z ziz n nn For ,1z we have, nj z j ,.......,2,1,1 1  , so that, by using the hypothesis, 1 211121121 1 [)(     n nn n nn n nn n nn n n zzkzkzkkzazF  ])1(...... 0 1 10001      n j j jj zzza ...... 1 [[ 211121121 z kkkkzaz nnnnnnnnn n    }] 1111 )1( 1 0 1 1010101 n n j jnjjnnn zzzzz      ......)1()1({[ 21121121   nnnnnnn n kkkkzaz  }])()1( 0 1 10001      n j jj  ......)1()1({[ 1121121   nnnnnn n kkkkzaz  }]22)( 0 000    n j j  )()(){([ 11121121   nnnnnnnn n kkkkzaz  }]22 0 0    n j j  0 if
9. 9. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 17 )()()[( 1 11121121   nnnnnnn n kkkk a z  ]22 0 0    n j j  This shows that the zeros of F(z) having modulus greater than 1 lie in )()()[( 1 11121121   nnnnnnn n kkkk a z  ]22 0 0    n j j  . But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence, it follows that all the zeros of F(z) lie in )()()[( 1 11121121   nnnnnnn n kkkk a z  ]22 0 0    n j j  Since the zeros of P(z) are also the zeros of F(z), Theorem 4 follows. Proof of Theorem 6. Consider the polynomial )......)(1()()1()( 01 1 1 azazazazzPzzF n n n n    001 1 211 1 )(......)()( azaazaazaaza n nn n nn n n     1 211121121 1 )()}()(){(     n nn n nnnnnn n n zaazaakaakakakza 00001 )}(){(...... azaaaa   For ,1z we have, nj z j ,.......,2,1,1 1  , so that, by using the hypothesis and the Lemma, 1 211121121 1 [)(     n nn n nn n nn n nn n n zaazaakzakzakakzazF  ])1(...... 0001 azazaa   ...... 1 [[ 211121121   z aaaakaakakakzaz nnnnnnnnn n ] 11 )1( 1 010101 nnn z a z a z aa    ......{[ 211121121   nnnnnnnnn n aaaakaakakakzaz }])1( 0001 aaaa   nnnnnn n akakakakakzaz )1(sin)(cos){([ 1121121    }])1(sin)(cos)( ......sin)(cos)()1( 00101 212112 aaaaa aaaaak nnnnn      nnnn n aakakzaz   )sincos1()sincos1({[ 121  }]sin22)sincos1()cos1( 1 1 001      n j jn aaaa  >0 if
10. 10. Extensions of Enestrom-Kakeya Theorem www.ijceronline.com Open Access Journal Page 18 nnn n aakak a z   )sincos1()sincos1([ 1 121  ]sin22)sincos1()cos1( 1 1 001      n j jn aaaa  . This shows that the zeros of F(z) having modulus greater than 1 lie in nnn n aakak a z   )sincos1()sincos1([ 1 121  ]sin22)sincos1()cos1( 1 1 001      n j jn aaaa  . But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence, it follows that all the zeros of F(z) lie in nnn n aakak a z   )sincos1()sincos1([ 1 121  ]sin22)sincos1()cos1( 1 1 001      n j jn aaaa  . Since the zeros of P(z) are also the zeros of F(z), the result follows. REFERENCES [1] A.Aziz and B.A.Zargar, Some Extensions of Enestrom-Kakeya Theorem, Glasnik Math. 31(1996), 239-244. [2] N. K. Govvil, International Congress of ASIAM, Jammu University, India March 31-April 3, 2007. [3] N. K. Govil and Q.I. Rahman, On the Enestrom-Kakeya Theorem, Tohoku Math.J.20 (1968), 126-136. [4] M. H. Gulzar, Some Refinements of Enestrom-Kakeya Theorem, Int. Journal of Mathematical Archive -2(9, 2011),1512-1529. [5] M. H. Gulzar, Ph.D thesis, Department of Mathematics, University of Kashmir Srinagar, 2012. [6] A. Joyal, G. Labelle and Q. I. Rahman, On the Location of Zeros of Polynomials, Canad. Math. Bull., 10(1967), 53-66. [7] A. Liman and W. M. Shah, Extensions of Enestrom-Kakeya Theorem, Int. Journal of Modern Mathematical Sciences, 2013, 8(2), 82-89. [8] M. Marden, Geometry of Polynomials, Math. Surveys, No.3; Amer. Math. Soc. Providence R.I. 1966.