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IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-5728,p-ISSN: 2319-765X, Volume 8, Issue 3 (Sep. - Oct. 2013), PP 33-35 www.iosrjournals.org

A Redefined Riemann Zeta Function Represented via Functional Equations Using the Osborne’s Rule Adeniji A.A. Ekakitie E.A Mohammed M.A Amusa I.S Department of Mathematics, University of Lagos, Akoka, Nigeria. Department of Mathematics Yaba College of Technology, Lagos, Nigeria. Department of Mathematics Yaba College of Technology, Lagos, Nigeria. Department of Mathematics, University of Lagos, Akoka, Nigeria.

Abstract: Intrigues most researchers about the Riemann zeta hypothesis is the ability to employ cum different approaches with instinctive mindset to obtain some very interesting results. Motivated by their style of reasoning, the result obtained in this work of redefining or re-representation of Riemann zeta function in different forms by employing different techniques on two functional equations made the results better, simpler and concise new representations of Riemann zeta function. Keywords: Analytic Continuation, Osborne’s rule, Riemann Zeta Function đ??“đ??‡đ??„ đ??Žđ??’đ?? đ??Žđ??‘đ???đ??„′ đ??’ đ??‘đ??”đ??‹đ??„ đ??€đ???đ???đ??‘đ??Žđ??€đ??‚đ??‡ Osborne′s rule gave a more comprehensive method for hyperbolic functions interms of the Euler method. đ?‘’ đ?‘Ľ +đ?‘’ −đ?‘Ľ đ?‘&#x;đ?‘’đ?‘?đ?‘Žđ?‘™đ?‘™ đ?‘“đ?‘&#x;đ?‘œđ?‘š đ?‘?đ?‘œđ?‘ đ?‘•đ?‘Ľ = 2 and the fact that đ?‘’ đ?œƒđ?‘– = đ?‘?đ?‘œđ?‘ đ?œƒ + đ?‘–đ?‘ đ?‘–đ?‘›đ?œƒ đ??źđ?‘šđ?‘?đ?‘™đ?‘Śđ?‘–đ?‘›đ?‘” đ?‘Ąđ?‘•đ?‘Žđ?‘Ą â„œ đ?‘’ đ?‘Ľ = đ?‘?đ?‘œđ?‘ đ?‘•đ?‘Ľ (1.0) đ?‘ 

đ?‘

â„œ đ?‘’ 2đ?‘™đ?‘›đ?‘Ľ = coshâ Ą ( đ?‘™đ?‘›đ?‘Ľ) 2 From the Riemann zeta function đ?œ đ?‘ = đ?‘  đ?‘ âˆ’1 2

4

∞ 1

đ?‘ đ?‘  − đ?œ‹ 2Γ 2

(1.1) 5

∞ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ đ?‘›=1 đ?‘’

1

3

đ?‘

đ?‘›đ?‘› 2 đ?œ‹ 2 đ?‘Ľ 4 − 2 đ?‘›đ?‘›đ?œ‹đ?‘Ľ 4 cosh

2

��� ��

(1.2)

Substituting 1.0 , into equation 1.1 , then the equation 1.2 becomes đ?œ đ?‘ = đ?‘  đ?‘ âˆ’1 2

4

∞ 1

đ?‘ đ?‘  đ?œ‹ 2Γ 2 −

5

1

3

đ?‘

đ?‘›đ?‘› 2 đ?œ‹ 2 đ?‘Ľ 4 − 2 đ?‘›đ?‘›đ?œ‹đ?‘Ľ 4 đ?‘’ 2đ?‘™đ?‘›đ?‘Ľ đ?‘‘đ?‘Ľ

∞ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ đ?‘›=1 đ?‘’

(1.3)

considering the right arm of the r. h. s side of the equation(1.3) ∞ 1

=

5

∞ 1

5 4

3

đ?‘›đ?‘› 2 đ?œ‹ 2 đ?‘Ľ − 2 đ?‘›đ?‘›đ?œ‹đ?‘Ľ

∞ đ?‘›=1

1

3

đ?‘

đ?‘›đ?‘› 2 đ?œ‹ 2 đ?‘Ľ 4 − 2 đ?‘›đ?‘›đ?œ‹đ?‘Ľ 4 đ?‘’ 2đ?‘™đ?‘›đ?‘Ľ đ?‘‘đ?‘Ľ

∞ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ đ?‘›=1 đ?‘’

5 4

∞

1 4

đ?‘’

đ?‘ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ + đ?‘™đ?‘›đ?‘Ľ 2

đ?‘ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ + đ?‘™đ?‘›đ?‘Ľ 2

(1.4)

đ?‘‘đ?‘Ľ

(1.5) 1 4

∞3

đ?‘ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ + đ?‘™đ?‘›đ?‘Ľ 2

= ∞đ?‘›=1 1 ( đ?‘›đ?‘› 2 đ?œ‹ 2 đ?‘Ľ đ?‘’ đ?‘‘đ?‘Ľ − 1 2 đ?‘›đ?‘›đ?œ‹đ?‘Ľ đ?‘’ considering the left hand side of the equation 1.6 i. e ∞ đ?‘›=1

đ?‘‘đ?‘Ľ

(1.6)

∞ 5 −đ?‘›đ?‘›đ?œ‹đ?‘Ľ +đ?‘ đ?‘™đ?‘›đ?‘Ľ 2 đ?‘Ľ4 đ?‘’ đ?‘‘đ?‘Ľ. 1 5 đ?‘  −đ?‘›đ?‘›đ?œ‹đ?‘Ľ + đ?‘™đ?‘›đ?‘Ľ

đ?‘›đ?‘› 2 đ?œ‹ 2

đ??żđ?‘’đ?‘Ą I =

∞ đ?‘Ľ4đ?‘’ 1 14

đ?‘‘đ?‘Ľ and introducing the integration by part:

2

9 4

4

(1.7)

đ?‘

4

đ?‘Ąđ?‘•đ?‘’đ?‘› , đ??ź = (18+4đ?‘ ) 9 đ?‘Ľ đ?‘’ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ +2đ?‘™đ?‘›đ?‘Ľ − 9 (đ?‘›đ?‘›)3 đ?œ‹ 3 From equation (1.8) ∞

đ?‘†đ?‘˘đ?‘?đ?‘?đ?‘œđ?‘ đ?‘’ đ?‘€1 = 4

1 4

4

9

(1.8)

9

đ?‘Ľ 4 đ?‘’ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ +đ?‘™đ?‘›đ?‘Ľ đ?‘‘đ?‘Ľ, đ?‘?đ?‘Ś đ?‘“đ?‘˘đ?‘&#x;đ?‘Ąđ?‘•đ?‘’đ?‘&#x;đ?‘–đ?‘›đ?‘” đ?‘Ąđ?‘•đ?‘’ đ?‘–đ?‘›đ?‘Ąđ?‘’đ?‘”đ?‘&#x;đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘?đ?‘Ś đ?‘?đ?‘Žđ?‘&#x;đ?‘Ą đ?‘“đ?‘&#x;đ?‘œđ?‘š 13

đ?‘

4

đ?‘€1 = (đ?‘›đ?‘›)3 đ?œ‹ 3 đ?‘Ľ 4 đ?‘’ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ +2đ?‘™đ?‘›đ?‘Ľ + đ?‘›đ?‘›đ?œ‹ 9 13 13 đ?‘ đ?‘˘đ?‘?đ?‘ đ?‘Ąđ?‘–đ?‘Ąđ?‘˘đ?‘Ąđ?‘–đ?‘›đ?‘” 2.0 đ?‘–đ?‘›đ?‘Ąđ?‘œ 1.8 , đ?‘¤đ?‘’ đ?‘œđ?‘?đ?‘Ąđ?‘Žđ?‘–đ?‘›: 18

∞ 9 −đ?‘›đ?‘›đ?œ‹đ?‘Ľ +đ?‘ đ?‘™đ?‘›đ?‘Ľ 2 đ?‘Ľ4đ?‘’ đ?‘‘đ?‘Ľ 1

đ?‘

4

∞ 13 −đ?‘›đ?‘›đ?œ‹đ?‘Ľ + đ?‘ đ?‘™đ?‘›đ?‘Ľ 2 đ?‘Ľ4đ?‘’ đ?‘‘đ?‘Ľ 1 4đ?‘ 

4

13

−

4đ?‘ 26 đ?‘ 

(1.9),

đ?‘€1

(2.0) 4

∞

13

đ?‘

đ??ź = (18+4đ?‘ )[9 đ?‘Ľ 4 đ?‘’ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ +2đ?‘™đ?‘›đ?‘Ľ − 9 đ?‘›đ?‘› 3 đ?œ‹ 3 đ?‘€ + 26 đ?‘€ = 13 đ?‘Ľ 4 đ?‘’ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ +2đ?‘™đ?‘›đ?‘Ľ + 13 đ?‘›đ?‘›đ?œ‹ 1 đ?‘Ľ 4 đ?‘’ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ +2đ?‘™đ?‘›đ?‘Ľ đ?‘‘đ?‘Ľ (2.1) ∞ 13 đ?‘  đ?‘  26 4 13 4 đ??šđ?‘&#x;đ?‘œđ?‘š 2.0 , đ?‘¤đ?‘’ đ?‘•đ?‘Žđ?‘Łđ?‘’ đ?‘€1 = [ đ?‘Ľ 4 đ?‘’ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ +2đ?‘™đ?‘›đ?‘Ľ + đ?‘›đ?‘›đ?œ‹ đ?‘Ľ 4 đ?‘’ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ +2đ?‘™đ?‘›đ?‘Ľ đ?‘‘đ?‘Ľ , 26 + 4đ?‘  13 13 1

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33 | Page


A Redefined Riemann Zeta Function Represented Via Functional Equations Using The Osborne’s 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 2.0 𝑖𝑛𝑡𝑜 1.8 𝐼=

18(𝑛𝑛 )2 𝜋 2

4

18+4𝑠

9

18(𝑛𝑛 )2 𝜋 2

9

9 4

4

𝑠

𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 − 𝑠 2

4(𝑛𝑛 )2 𝜋 2

18

26

9

18+4𝑠

26+4𝑠

468 𝑛𝑛 2 𝜋 2

−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥

13 4

4

13

4

𝑠

[ 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 + 13

𝑠 2

−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥

= 18+4𝑠 9 𝑥 𝑒 − 1053 +396𝑠+36𝑠 2 (13 𝑥 𝑒 (2.2) 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑖𝑛𝑔 𝑡𝑕𝑒 𝑟𝑖𝑔𝑕𝑡 𝑕𝑎𝑛𝑑 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡𝑕𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1.6 𝑖. 𝑒

4 13

𝑛𝑛𝜋

144

∞ 13 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥 2 𝑥4𝑒 𝑑𝑥 1

(2.1)

4

) − 1053 +234 𝑠 (𝑛𝑛) 𝜋

4

∞ 13 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥 2 𝑥4𝑒 𝑑𝑥 1

1 𝑠 𝑠 ∞3 ∞ 1 3 −𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥 ∞ 2 𝑑𝑥 = 𝑛𝑛𝜋 1 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥 𝑛=1 1 2 𝑛𝑛𝜋𝑥 4 𝑒 2 𝑠 ∞ 1 𝑆𝑢𝑝𝑝𝑜𝑠𝑒 𝑀2 = 1 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥

=

𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑦 𝑝𝑎𝑟𝑡 𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (2.4) 5

5

4

𝑠

9

16𝑛𝑛𝜋

𝑠

(2.3) (2.4) ∞ 9 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥 2 𝑥4 𝑒 𝑑𝑥 1

16

[5 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 + 45 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 + 45 (𝑛𝑛)2 𝜋 2 16𝑛𝑛𝜋𝑠901∞𝑥54𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑑𝑥] 𝑀2 =

5+2𝑠

(2.5) 𝑟𝑒𝑐𝑎𝑙𝑙 𝑀2 =

∞ 1

10

1

2 3𝑛𝑛𝜋

𝑠

𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥 𝑡𝑕 𝑒𝑛 𝑓𝑟𝑜𝑚 (1.6) 𝑀2 =

𝑀2 = 𝑛𝑛𝜋 (15+6𝑠)

4 5

5 4

𝑥 𝑒

𝑠 2

−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥

+

16𝑛𝑛𝜋

16𝑛𝑛𝜋𝑠901∞𝑥54𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑑𝑥,

45

9 4

𝑥 𝑒

𝑠 2

−𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥

1

𝑠

𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥, 𝑤𝑒 𝑜𝑏𝑡𝑎𝑖𝑛

1

∞ 9 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥 2 𝑥4𝑒 𝑑𝑥 1

16

+ 45 (𝑛𝑛)2 𝜋 2

(2.6) 𝑓𝑟𝑜𝑚 𝑡𝑕𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1.6 , 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝐼 𝑎𝑛𝑑 𝑀2 , 𝑡𝑜 𝑜𝑏𝑡𝑎𝑖𝑛 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1.6 , 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 = 18(𝑛𝑛 )2 𝜋 2

4

18+4𝑠

9

9

𝑠

468 𝑛𝑛 2 𝜋 2

13

4

𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 − 1053 +396𝑠+36𝑠 2

13

𝑠

∞ 13 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥 2 𝑥4𝑒 𝑑𝑥 1

144

𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 − 1053 +234 𝑠 𝑛𝑛 4 𝜋 4

10𝑛𝑛𝜋15+6𝑠45𝑥54𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥+16𝑛𝑛𝜋45𝑥94𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥+1645𝑛𝑛2𝜋21∞𝑥94𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑑𝑥 −16𝑛𝑛𝜋𝑠901∞𝑥54𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑑𝑥 (2.7) ∞

𝑟𝑒𝑐𝑎𝑙𝑙 𝑀2 = =

18 𝑛𝑛 2 𝜋 2 4 18+4𝑠

9

9

1

𝑠

𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥 , 𝐼 =

1

1

𝑠

5

𝑠

𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥 , 𝑀1 =

468 𝑛𝑛 2 𝜋 2

𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 −

1053 +396𝑠+36𝑠 2

13 𝑠 4 −𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥 2 𝑥4𝑒 13

144

𝑠

1

𝑛𝑛 4 𝜋 4

1053 +234 𝑠

9

𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝑑𝑥 ∞ 13 −𝑛𝑛𝜋𝑥 +𝑠 𝑙𝑛𝑥 2 𝑥4𝑒 𝑑𝑥 1

40𝑥54𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑛𝑛𝜋75+30𝑠+160𝑛𝑛𝜋𝑥94𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑛𝑛𝜋675+3270𝑠−16𝑛𝑛2𝜋2𝑛𝑛𝜋675+270𝑠 𝑀1+16𝑛𝑛𝜋𝑠𝑛𝑛𝜋1375+540𝑠𝐼 (2.8) 𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑙𝑖𝑘𝑒 𝑡𝑒𝑟𝑚𝑠, 𝑤𝑒 𝑕𝑎𝑣𝑒: =

18(𝑛𝑛 )2 𝜋 2 4 18+4𝑠 144

1053 +234𝑠

𝑠

468 𝑛𝑛

2𝜋 2

4

13

𝑠

𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 −

1053 +396𝑠+36𝑠 2 13 13 𝑠 16 𝑛𝑛 2 𝜋 2 4 4 ∞ 4 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 𝜋 1 𝑥 𝑒 𝑑𝑥 − 𝑛𝑛𝜋 675 +270𝑠

9

𝑛𝑛

9

𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 −

𝑀1 +

5

40𝑥 4 𝑒 −𝑛𝑛𝜋𝑥

𝑠 + 𝑙𝑛𝑥 2

𝑛𝑛𝜋 75+30𝑠 16𝑛𝑛𝜋𝑠

𝑛𝑛𝜋 1375 +540𝑠

18(𝑛𝑛 )2 𝜋 2 18+4𝑠

4

9

𝑠

468 𝑛𝑛 2 𝜋 2

13

4

160 𝑛𝑛𝜋 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥

𝑠 + 𝑙𝑛𝑥 2

𝑛𝑛𝜋 675 +3270𝑠

𝐼

(2.9)

𝑟𝑒𝑐𝑎𝑙𝑙 𝑓𝑟𝑜𝑚 1.6 , 𝑏𝑦 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦𝑖𝑛𝑔 𝑤𝑒 𝑜𝑏𝑡𝑎𝑖𝑛 = ∞ 𝑛=1

9

+

5

𝑠

𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 − 1053 +396𝑠+36𝑠 2 (13 𝑥 4 𝑒 −𝑛𝑛𝜋𝑥 +2𝑙𝑛𝑥 ) − 9

40𝑥 4 𝑒

𝑠 −𝑛𝑛𝜋𝑥 + 𝑙𝑛𝑥 2

𝑛𝑛𝜋 (75+30𝑠)

+

160𝑛𝑛𝜋𝑥94𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑛𝑛𝜋(675+3270𝑠)−1441053+234𝑠(𝑛𝑛)4𝜋41∞𝑥134𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑑𝑥−16(𝑛𝑛 )2𝜋2𝑛𝑛𝜋(675+270𝑠)𝑀1+16𝑛𝑛𝜋𝑠𝑛𝑛𝜋(1375+540𝑠)𝐼 (3.0) = ∞ −𝑛𝑛𝜋𝑥 𝑛=1 𝑒

18(𝑛𝑛 )2 𝜋 2 4 18+4𝑠

9

9

𝑠

𝑥 4 𝑒 2𝑙𝑛𝑥 −

468 𝑛𝑛

2𝜋 2

4

13

𝑠

( 𝑥 4 𝑒 2𝑙𝑛𝑥 ) −

1053 +396𝑠+36𝑠 2 13

5 𝑠

40𝑥 4 𝑒 2𝑙𝑛𝑥 𝑛𝑛𝜋 (75+30𝑠)

9 𝑠

+

160 𝑛𝑛𝜋 𝑥 4 𝑒 2𝑙𝑛𝑥 𝑛𝑛𝜋 (675 +3270 𝑠)

1441053+234𝑠(𝑛𝑛)4𝜋41∞𝑥134𝑒−𝑛𝑛𝜋𝑥+𝑠2𝑙𝑛𝑥𝑑𝑥−16𝑛𝑛2𝜋2𝑛𝑛𝜋(675+270𝑠)𝑀1+16𝑛𝑛𝜋𝑠𝑛𝑛𝜋(1375+540𝑠)𝐼 (3.1)

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A Redefined Riemann Zeta Function Represented Via Functional Equations Using The Osborne’s ∞

đ?‘

đ?‘

đ?‘’ −đ?‘›đ?‘›đ?œ‹đ?‘Ľ = đ?œ“ đ?‘Ľ đ?‘Žđ?‘›đ?‘‘ đ?‘’ đ?‘™đ?‘›đ?‘Ľ 2 = đ?‘Ľ 2

đ??żđ?‘’đ?‘Ą đ?‘› =1

Then equations 3.1 becomes

= đ?œ“(đ?‘Ľ)đ?‘Ľ

đ?‘ 2

18(đ?‘›đ?‘› )2 đ?œ‹ 2 4 18+4đ?‘

144

(đ?‘›đ?‘›)4 đ?œ‹ 4 1053 +234đ?‘

9

9 4

468 đ?‘›đ?‘› 2 đ?œ‹ 2

4

5

13 4

2

4 đ?‘ đ?‘  − đ?œ‹ 2Γ 2

∞ 13 −đ?‘›đ?‘›đ?œ‹đ?‘Ľ +đ?‘ đ?‘™đ?‘›đ?‘Ľ 2 đ?‘Ľ4đ?‘’ đ?‘‘đ?‘Ľ 1

đ?‘

đ?œ“(đ?‘Ľ)đ?‘Ľ 2

160đ?‘›đ?‘›đ?œ‹ đ?‘Ľ 4

đ?‘Ľ − 1053 +396đ?‘ +36đ?‘  2 (13 đ?‘Ľ ) − đ?‘›đ?‘›đ?œ‹ (75+30đ?‘ ) + đ?‘›đ?‘›đ?œ‹ (675+3270 đ?‘ ) − 16(đ?‘›đ?‘› )2 đ?œ‹ 2

16đ?‘›đ?‘›đ?œ‹đ?‘

− đ?‘›đ?‘›đ?œ‹ (675+270đ?‘ ) đ?‘€1 + đ?‘›đ?‘›đ?œ‹ (1375 +540đ?‘ ) đ??ź

đ?‘†đ?‘˘đ?‘?đ?‘ đ?‘Ąđ?‘–đ?‘Ąđ?‘˘đ?‘Ąđ?‘–đ?‘›đ?‘” 3.0 đ?‘–đ?‘›đ?‘Ąđ?‘œ 1.2 đ?œ đ?‘  = đ?‘  đ?‘ âˆ’1

9

40đ?‘Ľ 4

18(đ?‘›đ?‘› )2 đ?œ‹ 2 4 18+4đ?‘

9

9

đ?‘Ľ4 −

468 đ?‘›đ?‘› 2 đ?œ‹ 2 1053 +396đ?‘ +36đ?‘ 

4

13

( đ?‘Ľ4 )− 2 13

5

40đ?‘Ľ 4 đ?‘›đ?‘›đ?œ‹ (75+30đ?‘ )

(3.2)

9

+

160 đ?‘›đ?‘›đ?œ‹ đ?‘Ľ 4 đ?‘›đ?‘›đ?œ‹ (675 +3270 đ?‘ )

−

1441053+234đ?‘ (đ?‘›đ?‘›)4đ?œ‹41∞ đ?‘Ľ134đ?‘’−đ?‘›đ?‘›đ?œ‹đ?‘Ľ+đ?‘ 2đ?‘™đ?‘›đ?‘Ľđ?‘‘đ?‘Ľâˆ’16(đ?‘›đ?‘›)2đ?œ‹2đ?‘›đ?‘›đ?œ‹(675+270đ?‘ )đ?‘€1+16đ?‘›đ?‘›đ?œ‹đ?‘ đ?‘›đ?‘›đ?œ‹(1375+540đ?‘ ) đ??ź . (3.3) from the above represenation the Îś s is valid for đ?‘  > 1

Conclusion This representation really shows how the complex part of the equation could be eliminated by introducing a hyperbolic function from Enoch and Adeyeye(2012) result which was the reason why employ the technique of the Osborne’s Rule approach. Also from the result obtained, by observation one see how much primes numbers are been played around the synthetic process. This actually obeys the rules guiding the Number theory (every integer is a product of prime). This method helps in removing the "�"in the result using the Euler’s Method though we could not see how the integers are been expressed in the products of primes much but this validate some of the claims of researchers like Euler (1748), Odgers (2004). Thus, the Osborne’s approach is more preferable.

References [1] [2] [3] [4] [5]

E. Bombieri, (2000), Problems of the millennium: The Riemann hypothesis, Clay mathematical Institute. O. Enoch, (2012), A New Representation of the Riemann Zeta Function Via Its Functional Equation O. Enoch, (2012), A General Representation of the Zeros of the Riemann Zeta Function via Fourier series Expansion O.O.A. Enoch and F.J. Adeyeye, (2012), A Validation of the Real Zeros of the Riemann Zeta Function via the Continuation Formula of the Zeta Function, Journal of Basic & Applied Sciences, 2012, 8, 1 P. Sarnak, (2004), Problems of the Millennium: The Riemann Hypothesis by Princeton University and courant Institute of Mathematics.

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