素数の規則を見つけたい。。。

素数の規則を見つけたい。。。 (701ﾚｽ)
上下前次 1-新
通常表示 512ﾊﾞｲﾄ分割ﾚｽ栞

241: １３２人目の素数さん [sage] 2023/12/23(土) 02:17:03.83 ID:O5dB6rNY

>>240

(((a^2+b^2)*e^(i*2*arctan(a/b))+2^(3/2)*a*b*e^(i*-π/4)))=-a^2+2ab+b^2
(((a^2+b^2)*e^(i*2*arctan(b/a))+2^(3/2)*a*b*e^(i*-π/4)))=a^2+2ab-b^2

(n+1)^2+2n*(n+1)-n^2=(n+1)^2+2(n+2)*(n+1)-(n+2)^2 ←nに何を入れても等しくなる

(((2^2+1^2)*e^(i*2*arctan(2/1))+2^(3/2)*2*1*e^(i*-π/4)))=1
(((2^2+1^2)*e^(i*2*arctan(1/2))+2^(3/2)*2*1*e^(i*-π/4)))=7

(((2^2+3^2)*e^(i*2*arctan(3/2))+2^(3/2)*2*3*e^(i*-π/4)))=7
(((2^2+3^2)*e^(i*2*arctan(2/3))+2^(3/2)*2*3*e^(i*-π/4)))=17

(((4^2+3^2)*e^(i*2*arctan(4/3))+2^(3/2)*3*4*e^(i*-π/4)))=17
(((4^2+3^2)*e^(i*2*arctan(3/4))+2^(3/2)*3*4*e^(i*-π/4)))=31

(((4^2+5^2)*e^(i*2*arctan(5/4))+2^(3/2)*5*4*e^(i*-π/4)))=31
(((4^2+5^2)*e^(i*2*arctan(4/5))+2^(3/2)*5*4*e^(i*-π/4)))=49

(((6^2+5^2)*e^(i*2*arctan(6/5))+2^(3/2)*5*6*e^(i*-π/4)))=49
(((6^2+5^2)*e^(i*2*arctan(5/6))+2^(3/2)*5*6*e^(i*-π/4)))=71

ζ1(s)=|ζ1(s)|*e^(i*θ)      ←素数のみのゼータ関数(s=0点の時)=1/2^s+1/3^s+1/5^s+・・・
ζ2(s)=|ζ2(s)|*e^(i*(θ+π))　←非素数のみのゼータ関数(s=0点の時)=1+1/4^s+1/6^s+1/8^s+1/9^s+・・・

ζ1(s)+ζ2(s)=ζ(s)
|ζ1(s)|=|ζ2(s)|
ζ1(s)*ζ2(s)=|ζ1(s)|*|ζ2(s)|*e^(i*(2θ+π))

ζ1(s)=√(ζ1(s)*ζ2(s))*e^(-iπ/2)
ζ2(s)=√(ζ1(s)*ζ2(s))*e^(iπ/2)

√(ζ1(s)*ζ2(s))*e^(-iπ/2)+√(ζ1(s)*ζ2(s))*e^(iπ/2)=ζ(s)

(√(ζ1(s)*ζ2(s))*e^(-iπ/2)+√(ζ1(s)*ζ2(s))*e^(iπ/2))^2=ζ(s)^2

(ζ1(s)*ζ2(s))*e^(-iπ)+(ζ1(s)*ζ2(s))*e^(iπ)+2*(ζ1(s)*ζ2(s))=ζ(s)^2 ←2*(ζ1(s)*ζ2(s))=ζ(s)^2になる

2^n*(ζ1(s)*ζ2(s))^(1/n)=2*(ζ1(s)*ζ2(s))=(ζ1(s)+ζ2(s))^2=ζ(s)^2

lim [n→∞]　2^(n-1)*(ζ1(s)*ζ2(s))^(1/n)=ζ(s) ←n→無限のとき

http://rio2016.5ch.net/test/read.cgi/math/1640355175/241

298: １３２人目の素数さん [sage] 2023/12/29(金) 21:13:50.10 ID:voXPt7J2

cos(2pi*(1/2+2/3+3/5))=cos(2pi*(7/(2*3*5)))
cos(2pi*(1/2+2/3+1/5))=cos(2pi*(11/(2*3*5)))
cos(2pi*(1/2+2/3+2/5))=cos(2pi*(13/(2*3*5)))=cos(2pi*((2*3*5-13)/(2*3*5)))=cos(2pi*(17/(2*3*5)))
cos(2pi*(1/2+2/3+2/5))=cos(2pi*(17/(2*3*5)))
cos(2pi*(1/2+2/3+1/5))=cos(2pi*(19/(2*3*5)))=cos(2pi*((2*3*5*7-19)/(2*3*5)))=cos(2pi*(41/(2*3*5)))
cos(2pi*(1/2+2/3+3/5))=cos(2pi*(23/(2*3*5)))
cos(2pi*(1/2+2/3+4/5))=cos(2pi*(29/(2*3*5)))
cos(2pi*(1/2+2/3+4/5))=cos(2pi*(31/(2*3*5)))
cos(2pi*(1/2+2/3+3/5))=cos(2pi*(37/(2*3*5)))
cos(2pi*(1/2+2/3+1/5))=cos(2pi*(41/(2*3*5)))
cos(2pi*(1/2+2/3+2/5))=cos(2pi*(43/(2*3*5)))
cos(2pi*(1/2+2/3+2/5))=cos(2pi*(47/(2*3*5)))

cos(2pi*(1/2+2/3+3/5+2/7))=cos(2pi*(11/(2*3*5*7)))=cos(2pi*((2*3*5*7-11)/(2*3*5*7)))=cos(2pi*(199/(2*3*5*7))) ←11*17以上、17^2未満なので素数
cos(2pi*(1/2+2/3+1/5+4/7))=cos(2pi*(13/(2*3*5*7)))=cos(2pi*((2*3*5*7-13)/(2*3*5*7)))=cos(2pi*(197/(2*3*5*7))) ←11*17以上、17^2未満なので素数
cos(2pi*(1/2+2/3+1/5+5/7))=cos(2pi*(17/(2*3*5*7)))=cos(2pi*((2*3*5*7-17)/(2*3*5*7)))=cos(2pi*(193/(2*3*5*7))) ←11*17以上、17^2未満なので素数
cos(2pi*(1/2+2/3+3/5+1/7))=cos(2pi*(19/(2*3*5*7)))=cos(2pi*((2*3*5*7-19)/(2*3*5*7)))=cos(2pi*(191/(2*3*5*7))) ←11*17以上、17^2未満なので素数
cos(2pi*(1/2+2/3+4/5+1/7))=cos(2pi*(23/(2*3*5*7)))=cos(2pi*((2*3*5*7-23)/(2*3*5*7)))=cos(2pi*(187/(2*3*5*7))) ←11*17
cos(2pi*(1/2+2/3+2/5+4/7))=cos(2pi*(29/(2*3*5*7)))=cos(2pi*((2*3*5*7-29)/(2*3*5*7)))=cos(2pi*(181/(2*3*5*7))) ←11^2以上、11^*17未満なので素数

http://rio2016.5ch.net/test/read.cgi/math/1640355175/298

299: １３２人目の素数さん [sage] 2023/12/29(金) 21:22:54.77 ID:voXPt7J2

cos(2pi*(1/2+2/3+3/5+2/7))=cos(2pi*(11/(2*3*5*7)))=cos(2pi*((2*3*5*7-11)/(2*3*5*7)))=cos(2pi*(199/(2*3*5*7))) ←13^2以上、17^2未満なので素数
cos(2pi*(1/2+2/3+1/5+4/7))=cos(2pi*(13/(2*3*5*7)))=cos(2pi*((2*3*5*7-13)/(2*3*5*7)))=cos(2pi*(197/(2*3*5*7))) ←13^2以上、17^2未満なので素数
cos(2pi*(1/2+2/3+1/5+5/7))=cos(2pi*(17/(2*3*5*7)))=cos(2pi*((2*3*5*7-17)/(2*3*5*7)))=cos(2pi*(193/(2*3*5*7))) ←13^2以上、17^2未満なので素数
cos(2pi*(1/2+2/3+3/5+1/7))=cos(2pi*(19/(2*3*5*7)))=cos(2pi*((2*3*5*7-19)/(2*3*5*7)))=cos(2pi*(191/(2*3*5*7))) ←13^2以上、17^2未満なので素数
cos(2pi*(1/2+2/3+4/5+1/7))=cos(2pi*(23/(2*3*5*7)))=cos(2pi*((2*3*5*7-23)/(2*3*5*7)))=cos(2pi*(187/(2*3*5*7))) ←11*17
cos(2pi*(1/2+2/3+2/5+4/7))=cos(2pi*(29/(2*3*5*7)))=cos(2pi*((2*3*5*7-29)/(2*3*5*7)))=cos(2pi*(181/(2*3*5*7))) ←13^2以上、11*17未満なので素数

cos(2pi*(1/2+2/3+2/5+2/7))=cos(2pi*(31/(2*3*5*7)))=cos(2pi*((2*3*5*7-31)/(2*3*5*7)))=cos(2pi*(179/(2*3*5*7))) ←13^2以上、11*17未満なので素数
cos(2pi*(1/2+2/3+4/5+6/7))=cos(2pi*(37/(2*3*5*7)))=cos(2pi*((2*3*5*7-37)/(2*3*5*7)))=cos(2pi*(173/(2*3*5*7))) ←13^2以上、11*17未満なので素数
cos(2pi*(1/2+2/3+3/5+3/7))=cos(2pi*(41/(2*3*5*7)))=cos(2pi*((2*3*5*7-41)/(2*3*5*7)))=cos(2pi*(169/(2*3*5*7))) ←13^2
cos(2pi*(1/2+2/3+1/5+3/7))=cos(2pi*(43/(2*3*5*7)))=cos(2pi*((2*3*5*7-43)/(2*3*5*7)))=cos(2pi*(167/(2*3*5*7))) ←11*13以上、13^2未満なので素数

http://rio2016.5ch.net/test/read.cgi/math/1640355175/299

305: １３２人目の素数さん [sage] 2023/12/30(土) 20:35:30.01 ID:jsoLHdB8

ζ(1/2+i*y)=1+1/2^(1/2+i*y)+1/3^(1/2+i*y)+1/4^(1/2+i*y)+5^(1-(1/2+i*y))/(1/2+i*y-1)+5^(-(1/2+i*y))/2
+1/6*1/2!*5^(1-(1/2+i*y)-2)*(1/2+i*y)
-1/30*1/4!*5^(1-(1/2+i*y)-4)*(1/2+i*y)*(1/2+i*y+1)*(1/2+i*y+2)
+1/42*1/6!*5^(1-(1/2+i*y)-6)*(1/2+i*y)*(1/2+i*y+1)*(1/2+i*y+2)*(1/2+i*y+3)*(1/2+i*y+4)
+1/R2k

ζ(1/2+i*0)=1+1/2^(1/2+i*0)+1/3^(1/2+i*0)+1/4^(1/2+i*0)+5^(1-1/2-i*0)/(-1/2+i*0)+5^(-1/2-i*0)/2
+1/6*1/2!*5^(1-(1/2+i*0)-2)*(1/2+i*0)
-1/30*1/4!*5^(1-(1/2+i*0)-4)*(1/2+i*0)*(1/2+i*0+1)*(1/2+i*0+2)
+1/42*1/6!*5^(1-(1/2+i*0)-6)*(1/2+i*0)*(1/2+i*0+1)*(1/2+i*0+2)*(1/2+i*0+3)*(1/2+i*0+4)
+1/42
=-1.460345326910927484773421538534732892012336163703945420633977740...≒-1.46＝ζ(1/2)

-1.464072106873427134267436827982618352404737194303297963507762570
0.0037267799624996494940152894478854603924010305993525428737848287
-9.316949906249123735038223619713650981002576498381357184462... × 10^-6
1.3975424859373685602557335429570476471503864747572035776693... × 10^-7

http://rio2016.5ch.net/test/read.cgi/math/1640355175/305

307: １３２人目の素数さん [sage] 2023/12/30(土) 21:16:29.25 ID:jsoLHdB8

ζ(x+i*y')-ζ(x+i*y)=1-1+1/2^(x+i*y')-1/2^(x+i*y)+1/3^(x+i*y')-1/3^(x+i*y)+1/4^(x+i*y')-1/4^(x+i*y)
+5^(1-(x+i*y'))/(x+i*y'-1)-5^(1-(x+i*y))/(x+i*y-1)+5^(-(x+i*y'))/2-5^(-(x+i*y))/2

ζ(x+i*y')-ζ(x+i*y)≒(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1/2^(x/2+i*y'/2)+1/2^(x/2+i*y/2))+(1/3^(x/2+i*y'/2)-1/3^(x/2+i*y/2))*(1/3^(x/2+i*y'/2)+1/3^(x/2+i*y/2))+(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1/2^(x/2+i*y'/2)+1/2^(x/2+i*y/2))*(1/4^(x/2+i*y'/2)+1/4^(x/2+i*y/2))
+5^(1-(x+i*y'))/(x+i*y'-1)-5^(1-(x+i*y))/(x+i*y-1)+5^(-(x+i*y'))/2-5^(-(x+i*y))/2

ζ(x+i*y')-ζ(x+i*y)≒(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1+(1/2^(x/2+i*y'/2)+1/2^(x/2+i*y/2))*(1/4^(x/2+i*y'/2)+1/4^(x/2+i*y/2)))+(1/3^(x/2+i*y'/2)-1/3^(x/2+i*y/2))*(1/3^(x/2+i*y'/2)+1/3^(x/2+i*y/2))+5^(1-(x+i*y'))/(x+i*y'-1)-5^(1-(x+i*y))/(x+i*y-1)+5^(-(x+i*y'))/2-5^(-(x+i*y))/2
1/4^(x/2+i*y'/2)-1/4^(x/2+i*y/2)=1/2^(x+i*y')-1/2^(x+i*y)=(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1/2^(x/2+i*y'/2)+1/2^(x/2+i*y/2))

1/2^(x/2+i*y/2+i*π/2)=-1/2^(x/2+i*y/2)

(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2+i*π/2))*(1+(1/4^(x/2+i*y'/2)+1/4^(x/2+i*y/2))))
(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2+i*π/2))=(1/2^(x/4+i*y'/4)-1/2^(x/4+i*y/4+i*π/4))*(1/2^(x/4+i*y'/4)+1/2^(x/4+i*y/4+i*π/4))
1/2^(x/4+i*y/4+i*π/4+i*π/2)=-1/2^(x/4+i*y/4+i*π/4)
(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2+i*π/2))=(1/2^(x/4+i*y'/4)-1/2^(x/4+i*y/4+i*π/4))*(1/2^(x/4+i*y'/4)-1/2^(x/4+i*y/4+i*π/4+i*π/2))
=(1/2^(x/4+i*y'/4)-1/2^(x/4+i*y/4+i*π/4))*(1/2^(x/8+i*y'/8)-1/2^(x/8+i*y/8+i*π/8+i*π/8))**(1/2^(x/8+i*y'/8)+1/2^(x/8+i*y/8+i*π/8+i*π/8))
無限に分解していく際にx=1/2でないと都合が悪い可能性がある（1/2^nで実部を表せない)

http://rio2016.5ch.net/test/read.cgi/math/1640355175/307

308: １３２人目の素数さん [sage] 2023/12/30(土) 22:03:36.06 ID:jsoLHdB8

1/2^(x+i*y+i*π/ln2)=1/2^(x+i*y)*1/e^(i*π)=-1/2^(x+i*y)

ゼータ関数をζ(x+i*y)≒1+1/2^(x+i*y)と簡略化する
ζ(x+i*y’)とζ(x+i*y)を考えて差がほぼ0になる点を探す

ζ(x+i*y')-ζ(x+i*y)≒(1/2^(x+i*y')-1/2^(x+i*y))=(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1/2^(x/2+i*y'/2)+1/2^(x/2+i*y/2))
=(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1/2^(x/2^2+i*y'/2^2)-1/2^(x/2^2+i*y/2^2-i*π/ln2^2+i*π/ln2))
=(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1/2^(x/2^3+i*y'/2^3)-1/2^(x/2^3+i*y/2^3-i*π/ln2^3+i*π/ln2^2))*(1/2^(x/2^3+i*y'/2^3)-1/2^(x/2^3+i*y/2^3*+i*π/ln2^3+i*π/ln2^2+i*π/ln2))

lim[n→∞] (1/2^(x/2^n+i*y'/2^n)-1/2^(x/2^n+i*y/2^n+i*π/ln2^n+i*π/ln2^(n-1)+i*π/ln2^(n-2)+i*π/ln2^(n-3)+・・・・+i*π/ln2))≒0
lim[n→∞]Σ[k=1→n]i*π/ln2^k=i*π/ln2^n+i*π/ln2^(n-1)+i*π/ln2^(n-2)+i*π/ln2^(n-3)+・・・・+i*π/ln2=i*π*∞ mod 2π
nの値が無限でないときlim[n→m]Σ[k=1→n]i*π/ln2^kのときΣ[k=1→n]i*π/ln2^kはmod 2πされるため0から2πの値をとる

A=2^’x/2^m)*e^(i*y') B=2^(x/2^m)*e^(i*y+lim[n→m]Σ[k=1→n]i*π/ln2^k)

AとBの角度差がlim[n→m]Σ[k=1→n]i*π/ln2^kと可変する
長さが半分になり続ける２本のベクトルの間のベクトルの積とみなせるため
初期値が1/2でないと0に収束しない可能性がある

(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1/2^(x/2^3+i*y'/2^3)-1/2^(x/2^3+i*y/2^3-i*π/ln2^3+i*π/ln2^2))*(1/2^(x/2^4+i*y'/2^4)-1/2^(x/2^4+i*y/2^4*+i*π/ln2^4+i*π/ln2^3+i*π/ln2^2))
*(1/2^(x/2^4+i*y'/2^4)-1/2^(x/2^4+i*y/2^4*+i*π/ln2^4+i*π/ln2^3+i*π/ln2^2+i*π/ln2))

http://rio2016.5ch.net/test/read.cgi/math/1640355175/308

309: １３２人目の素数さん [sage] 2023/12/30(土) 22:26:56.47 ID:jsoLHdB8

ζ(x+i*y')-ζ(x+i*y)≒(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1/2^(x/2+i*y'/2)+1/2^(x/2+i*y/2))
＝(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2+iπ/ln2)
＝(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1/2^(x/2^2+i*y'/2^2)-1/2^(x/2^2+i*y/2^2+iπ/ln2^2)*(1/2^(x/2^2+i*y'/2^2)-1/2^(x/2^2+i*y/2^2+iπ/ln2^2+iπ/ln2)
=(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1/2^(x/2^2+i*y'/2^2)-1/2^(x/2^2+i*y/2^2+iπ/ln2^2)
*(1/2^(x/2^3+i*y'/2^3)-1/2^(x/2^3+i*y/2^3+iπ/ln2^3+iπ/ln2^2)*(1/2^(x/2^3+i*y'/2^3)-1/2^(x/2^3+i*y/2^3+iπ/ln2^3+iπ/ln2^2+iπ/ln2)
=(1/2^(x/2+i*y'/2)-1/2^(x/2+i*y/2))*(1/2^(x/2^2+i*y'/2^2)-1/2^(x/2^2+i*y/2^2+iπ/ln2^2)
*(1/2^(x/2^3+i*y'/2^3)-1/2^(x/2^3+i*y/2^3+iπ/ln2^3+iπ/ln2^2)*(1/2^(x/2^4+i*y'/2^4)-1/2^(x/2^4+i*y/2^4+iπ/ln2^4+iπ/ln2^3+iπ/ln2^2)
*(1/2^(x/2^4+i*y'/2^4)-1/2^(x/2^4+i*y/2^4+iπ/ln2^4+iπ/ln2^3+iπ/ln2^2+iπ/ln2)

2ベクトルの角度差がy'-y+lim[n→m]Σ[k=a→n]i*π/ln2^k)と可変する
長さは1/2^(x/2^m)になる
初期値が1/2でないと0に収束しない可能性がある

http://rio2016.5ch.net/test/read.cgi/math/1640355175/309

311: １３２人目の素数さん [sage] 2023/12/31(日) 13:32:01.36 ID:ZQRjm/0R

ゼータ関数をζ(x+i*y)≒1+1/2^(1/2+i*y)+1/3^(1/2+i*y)+1/4^(1/2+i*y)と簡略化
ζ(x+i*y')=ζ(x+i*y)となるときゼロ点しかないと仮定する(y'≠y)

|半径1/P(n)^(x/2^m)の円内の余弦の長さ|=Π(2*1/P(n)^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/lnP(n)^k))/2)

|ζ(x+i*y')-ζ(x+i*y)|=Π(2*1/2^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln2^k))/2) +Π(2*1/3^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln3^k))/2)
+Π(2*1/4^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln4^k))/2) ←　　Π(2*1/2^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln2^k))/2)＝０の時０に収束する

Π(2*1/2^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln2^k))/2) =2^a*1/2^(x*(1/2^1+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+・・))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln2^k))/2) ←0に収束する必要がある
Π(2*1/3^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln3^k))/2) =2^a*1/3^(x*(1/2^1+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+・・))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln3^k))/2) ←0に収束する必要がある
Π(2*1/4^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln4^k))/2) =2^a*1/4^(x*(1/2^1+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+・・))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln4^k))/2) ←0に収束する必要がある

2^a*1/2^(x*(1/2)/(1/2-1/2^∞))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln2^k))/2)
2^a*1/3^(x*(1/2)/(1/2-1/2^∞))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln3^k))/2)
2^a*1/4^(x*(1/2)/(1/2-1/2^∞))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln4^k))/2)

http://rio2016.5ch.net/test/read.cgi/math/1640355175/311

312: １３２人目の素数さん [sage] 2023/12/31(日) 14:52:49.83 ID:ZQRjm/0R

ゼータ関数をζ(x+i*y)≒1+1/2^(1/2+i*y)+1/3^(1/2+i*y)+1/4^(1/2+i*y)と簡略化
ζ(x+i*y')=ζ(x+i*y)となるときゼロ点しかないと仮定する(y'≠y)

|半径1/P(n)^(x/2^m)の円内の余弦の長さ|=Π(2*1/P(n)^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/lnP(n)^k))*lnP(n)/2)

|ζ(x+i*y')-ζ(x+i*y)|=Π(2*1/2^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln2^k))*ln2/2) +Π(2*1/3^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln3^k))*ln3/2)
+Π(2*1/4^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln4^k))*ln4/2) ←　　Π(2*1/2^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln2^k))/2)＝０の時０に収束する

Π(2*1/2^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln2^k))/2) =2^a*1/2^(x*(1/2^1+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+・・))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln2^k))*ln2/2) ←0に収束する必要がある
Π(2*1/3^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln3^k))/2) =2^a*1/3^(x*(1/2^1+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+・・))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln3^k))*ln3/2) ←0に収束する必要がある
Π(2*1/4^(x/2^m))*sin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln4^k))/2) =2^a*1/4^(x*(1/2^1+1/2^2+1/2^3+1/2^4+1/2^5+1/2^6+・・))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/ln4^k))*ln4/2) ←0に収束する必要がある

|ζ(x+i*y')-ζ(x+i*y)|=lim ΣΠ2^a*1/P(l)^(x*(1/2)/(1/2-1/2^∞))*Πsin((y'-(y+lim[n→m]Σ[k=a→n]i*π/lnP(l)^k))*lnP(l)/2)=0
これが収束するときにx=1/2しかない可能性がある

http://rio2016.5ch.net/test/read.cgi/math/1640355175/312

313: １３２人目の素数さん [sage] 2023/12/31(日) 17:11:21.67 ID:ZQRjm/0R

|ζ(x+i*y')-ζ(x+i*y)|=1/2^(x+i*y')-1/2^(x+i*y)+1/3^(x+i*y')-1/3^(x+i*y)+1/4^(x+i*y')-1/4^(x+i*y)
+5^(1-(x+i*y'))/(x+i*y'-1)-5^(1-(x+i*y))/(x+i*y-1)+5^(-(x+i*y'))/2-5^(-(x+i*y))/2

1/2^(x+i*y')-1/2^(x+i*y)=2*1/2^x*sin((y'-y)*ln2/2)*e^(i*(π/2+(y'+y)*ln2/2))
1/3^(x+i*y')-1/3^(x+i*y)=2*1/3^x*sin((y'-y)*ln3/2)*e^(i*(π/2+(y'+y)*ln3/2))
1/4^(x+i*y')-1/4^(x+i*y)=2*1/4^x*sin((y'-y)*ln4/2)*e^(i*(π/2+(y'+y)*ln4/2))

5^(1-x-i*y'))/(x-1+i*y')-5^(1-x-i*y)/(x-1+i*y)=5^(1-x)/√((x-1)^2+y'^2)*e^(i*'y'*ln5-arctan(y'/(x-1)))-5^(1-x)/√((x-1)^2+y^2)*e^(i*y*ln5-arctan(y/(x-1)))
5^(-(x+i*y'))/2-5^(-(x+i*y))/2=5^(-x)/2*(e^(i*-y'ln5)-e^(i*-yln5))

ζ(x+i*y')-ζ(x+i*y)≒2*1/2^x*sin((y'-y)*ln2/2)*e^(i*(π/2+(y'+y)*ln2/2))+2*1/3^x*sin((y'-y)*ln3/2)*e^(i*(π/2+(y'+y)*ln3/2))+2*1/4^x*sin((y'-y)*ln4/2)*e^(i*(π/2+(y'+y)*ln4/2))
+5^(1-x)/√((x-1)^2+y'^2)*e^(i*'y'*ln5-arctan(y'/(x-1)))-5^(1-x)/√((x-1)^2+y^2)*e^(i*y*ln5-arctan(y/(x-1)))
+5^(-x)/2*(e^(i*-y'ln5)-e^(i*-yln5))
がx≠1/2のときy,y'をもたない(y≠y'>0)

http://rio2016.5ch.net/test/read.cgi/math/1640355175/313

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