Σr2=0. Σ Br. Σ br. Σ r=0. br = Σ. Σa r-s b s (1.2) s=0. Σa r-s b s-t c t (1.3) t=0. cr = Σ. dr = Σ. Σa r-s b s-t c t-u d u (1.4) u =0.

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Junior Byrd
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1 0 Powe of Infinite Seie. Multiple Cauchy Poduct The multinomial theoem i uele fo the powe calculation of infinite eie. Thi i becaue the polynomial theoem depend on the numbe of tem, o it can not be applied to a eie whee the numbe of tem i infinite. Fo the powe calculation of infinite eie, multiple Cauchy poduct i ueful. Fomula.. ( Multiple Cauchy poduct of infinite eie ) The following expeion hold fo two o moe convegent infinite eie. a b = a b a - b (.2) c = a b c a, a2, Epecially, a ( ) - a = d = an, = =0 a - b -t c t (.3) 2=0 n=0 t a - b -t c t-u d u (.4) u =0 n- a, -2 a 2,2-3 a n-,n- -n a n,n (.n) 2 (-) a a 2- (.2) Poof The Cauchy poduct of two eie i expeed a follow. a b = Making a,b evee ode, we obtain (.2). Next, let Then a b - = B a b - (2 ) a b c = B c Accoding to (2 ), B c = And eplacing t, in (B), B = a t b -t Subtituting thi fo the above, B c - (B) - -

2 i.e. B c = a b c = Making a,b,c evee ode, we obtain (.3). Next, let a t b -t c - = C a t b -t c - a t b -t c - (C) Then a b c d = C d Accoding to (2 ), C d = C d - And eplacing tu, t, in (C), C = t a u b t-u c -t u =0 Subtituting thi fo the above, i.e. C d = a b c t a u b t-u c -t d - u =0 d = t a u b t-u c -t d - u =0 Making a,b,c,d evee ode, we obtain (.4). Heeafte, in a imila way, we obtain (.n). Let b = (-) a in (.2), then a ( ) = 0 ( ) - a = - a a 0- + ( ) a - ( ) - a a a = 2 ( ) ( -) a a - - a a (-) a a 3- + Thefoe, 0 = (-) 2 a a 0- + (-) 4 a a 2- + (-) a a 4- + a ( ) - a = Example Tiple Cauchy poduct 2n - (-) a a 2n-- = 0 fo n=,2,3, 2 (-) a a 2- When (.3) i expanded up to m=4 by uing fomula manipulation oftwae Mathematica, it i a follow - 2 -

3 Epecially, when a 0 =, b 0 =, c 0 =, By eplacing a, b, c, with a z, b z, c z, in Fomula.., the following fomula i obtained immediately. Fomula.2. ( Multiple Cauchy poduct of powe eie ) The following expeion hold fo two o moe convegent powe eie. a z b z = a z b z a - b z c z = a z b z c z a, z a2, z an, z = =0 2=0 d z = n=0 a - b -t c t z t u =0 a - b -t c t-u d u z n- a, -2 a 2,2-3 a n-,n- -n a n,n z (.2') (.3') (.4') (.n') Epecially, a z ( ) - a z = 2 (-) a a 2- z 2 (.2') Example ymbolic calculation When (.3') i expanded up to m =4 by uing fomula manipulation oftwae Mathematica, it i a follow

Epecially, when a 0 =, b 0 =, c 0 =, Example2 numeic calculation f() z = 5+z 5 = + - 5 z + - 5 2 z 2 + 3 - z 3 + z <5 5 z g() z = e 3 = +!3 z + 2!3 2 z 2 + 3!

4 Epecially, when a 0 =, b 0 =, c 0 =, Example2 numeic calculation f() z = 5+z 5 = z z z 3 + z <5 5 z g() z = e 3 = +!3 z + 2!3 2 z 2 + 3!3 3 z 3 + z z 2 2 z 3 3 h() z = co z = + co + co + co +! 2 2! 2 3! 2 When the poduct of thee function i calculated up to m =2 by uing fomula manipulation oftwae Mathematica, it i a follow

5 And if thi i illutated with f z g z h z the vicinity of -4., it i a follow. Both ae almot identical except fo Example3 (.2) z ( )!! Thi i conitent with e z e -z =. - z = = 2 (-) ( 2 )!! = C0 0 z 0 0! + = z 2 ( 2- )! 2 (-) C ( 2 )! 2 z 2 2 (-) C 2 z 2 2 = (-) 2C =0-5 -

6 .2 Powe of Infinite Seie (Pat) By etting a = b = c = = a in Fomula.., we obtain the following fomula immediately. Fomula.2. ( Powe of infinite eie ) The following expeion hold fo convegent infinite eie. 2 a 3 a 4 a = = = n a = =0 a - a (2.2) 2=0 a - a -t a t (2.3) n=0 t a - a -t a t-u a u (2.4) u =0 n- a -2 a 2-3 a n- -n a n (2.n) Example The 3d powe of zeta eie When (2.3) i expanded up to m=7 by uing fomula manipulation oftwae Mathematica, it i a follow Zeta eie i a follow. () z = ( +) z So, by eplacing a 0, a "/ + z " = + 2 z + 3 z + 4 z + Re() z > =,2,3,, it become a follow

7 Though the above wa ymbolic calculation, if it i numeically calculated and illutated with 3 z, it i a follow. Though it i calculated up to m =50, both ae almot ovelapped and blue 3 z cannot be een. By eplacing a with a z in Fomula.2., the following fomula i obtained immediately. Fomula.2.2 ( Powe of powe eie ) The following expeion hold fo convegent powe eie. a z 2 a z 3 a z 4 = = = a z n = =0 a - a z 2=0 a - a -t a t z n=0 t a - a -t a t-u a u z u =0 n- a -2 a 2-3 a n- -n a n z (2.2') (2.3') (2.4') (2.n') Example All 3 powe eie Let u calculate the 3 powe of the eie a follow. z 3 = ( 3 )! 3 e z +2e -z/2 co 3z 2 f() z The Maclauin expanion of the 3d powe of the ight ide i poible, but it i not o eay. So, we obediently - 7 -

8 culculate the cube of the left ide. A fit, if (2.3') i expanded up to m =6 by uing fomula manipulation oftwae Mathematica, it i a follow. Hee, by eplacing a 0, a "/3!" =,2,3,, z z 3, it become a follow. Although the above wa ymbolic calculation, when thi i numeically calculated, it i a follow. If thi i illutated with f 3 z, it i a follow. Both ae ovelapped exactly and bluef 3 z can not be een

9 - 9 -

10 .3 Powe of Infinite Seie (Pat2) Fomula.3. ( Powe of infinite eie ) The following expeion hold fo convegent infinite eie. 2 a 3 a 4 a = = = 2 a +2 a a -2-8 Poof ( a +b) 2 = a 2 +b 2 +2ab a a (3.2) =+ a =+ a a a a +4 =+ a =+t=+ Pefoming the following ubtitution to thi, we obtain (3.2). Next, Hee, a +b a, a 2 +b 2 a 2 a 2 a a a t +8, ab a a =+ =+t=+ a a =+ a a a t (3.3) =+t=+u =t+ a a a t a u (3.4) ( a +b +c) 3 = a 3 +b 3 +c 3 +3a 2 b +3ab 2 +3a 2 c+3ac 2 +3b 2 c+3bc 2 +6abc = a 3 +b 3 +c 3 +3a 2 b +ab 2 +a 2 c+ac 2 +b 2 c+bc 2 +3abc-3abc a 2 b +ab 2 +a 2 c+ac 2 +b 2 c+bc 2 +3abc = ( a +b +c )( ab+bc+ca) Subtituting thi fo the above, ( a +b +c) 3 = a 3 +b 3 +c 3 +3( a +b +c )( ab+bc+ca) -3abc Pefoming the following ubtitution to thi, we obtain (3.3). a 3 +b 3 +c 3 a 3, a+b +c In cae of the 4th degee, the following i obtained by imila calculation. a, ab+bc+ca, abc =+t=+ a a =+ a a a t ( a +b +c+d) 4 = a 4 +b 4 +c 4 +d 4-2a 2 b 2 +a 2 c 2 +a 2 d 2 +b 2 c 2 +b 2 d 2 +c 2 d 2 Pefoming the following ubtitution to thi, we obtain (3.4). a +b +c+d + 4( a +b +c+d) 2 ( ab+ac+ad+bc+bd+cd) - 8( abc+abd+acd+bcd )( a +b +c+d) + 8abcd a, a 4 +b 4 +c 4 +d 4 a 4 a 2 b 2 +a 2 c 2 +a 2 d 2 +b 2 c 2 +b 2 d 2 +c 2 d 2 a 2 a 2 =

11 ab+ac+ad+bc+bd+cd abc+abd+acd+bcd a a =+ =+t=+ a a a t, abcd =+t=+u =t+ In addition, in the cae of the 5th o highe degee, a imple expeion can not be obtained. a a a t a u Example The 3d powe of infinite eie When (3.3) i expanded up to m=5 by uing fomula manipulation oftwae Mathematica, it i a follow Since it wa long, the 4th and ubequent line wee omitted. Intead, the equivalence wa veified, and the equality wa alo veified in the cae of m=00. A the eult, it wa confimed that both ide ae equal in both cae. By eplacing a with a z in Fomula.3., the following fomula i obtained immediately. Fomula.3.2 ( Powe of powe eie ) The following expeion hold fo convegent powe eie. a z 2 a z 3 = = 2 a z 2 +2 a 3 z 3 +3 a a z + (3.2') =+ a z =+ a a z + -3 =+t=+ a a a t z ++t (3.3') - -

12 4 = a z a a z =+ -8 a z a a a t z ++t =+t=+ +8 =+t=+ a z 4 a z 2 u =t+ =+ a a z + a a a t a u z ++t+u (3.4') Example The 4th powe of infinite eie When (3.4') i expanded up to m=5 by uing fomula manipulation oftwae Mathematica, it i a follow Since it wa long, the 4th and ubequent line wee omitted. Intead, the equivalence wa veified, and the equality wa alo veified in the cae of m=50. A the eult, it wa confimed that both ide ae equal in both cae Alien' Mathematic Kano Kono - 2 -

Σk=1. g r 3/2 z. 2 3-z. g 3 ( 3/2 ) g r 2. = 1 r = 0. () z = ( a ) + Σ. c n () a = ( a) 3-z -a. 3-z. z - + Σ. z 3, 5, 7, z ! = !

Σk=1. g r 3/2 z. 2 3-z. g 3 ( 3/2 ) g r 2. = 1 r = 0. () z = ( a ) + Σ. c n () a = ( a) 3-z -a. 3-z. z - + Σ. z 3, 5, 7, z ! = ! 09 Maclauin Seies of Completed Riemann Zeta 9. Maclauin Seies of Lemma 9.. ( Maclauin seies of gamma function ) When is the gamma function, n is the polygamma function and B n,kf, f, ae Bell polynomials,