■超越数とその仲間たち(その102)
Sj=Σ1/(8n+j)
とおくと
π=4S1−2S4−S5−S6
また,1の8乗根をζ=(1+i)/√2として,
Lj=ln(1−ζ^j/√2)
とおくと,
L0=ln(1−1/√2)
L1=1/2ln(1/2)−iarctan1=L7~
L2=1/2ln(3/2)−iarctan(1//√2)=L6~
L3=1/2ln(5/2)−iarctan(1/3)=L5~
L4=ln(1+1/√2)
−Sj/2^j/2=1/8(L0+L1/ζ^j+L2/ζ^2j+L3/ζ^3j+L4/ζ^4j+L5/ζ^5j+L6/ζ^6j+L7/ζ^7j)
π=4S1−2S4−S5−S6=2L0−(2−2i)L1+2L4+(2+2i)L7
ln2=S2+S4/2+S6/4+S8/8
ln3=2S2+S6/2
ln5=2S2+2S4+S6/2
√2ln(√2+1)=S1+S3/2+S5/4+S7/8
√2arctan(1/√2)=S1−S3/2+S5/4−S7/8
arctan(1/3)=S1−S2−S4/2−S5/4
0=8S1−8S2−4S3−8S4−2S5−2S6+S7
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A=1/8ln(1+z)/(1−z)
B=1/2^7/2ln(1+√2z+z^2)/(1−√2z+z^2)
C=1/4arctan(z)
D=1/2^5/2arctan(√2z/(1−z^2))
Σz^8n+1/(8n+1)=A+B+C+D
Σz^8n+3/(8n+3)=A−B−C+D
Σz^8n+5/(8n+5)=A−B+C−D
Σz^8n+7/(8n+7)=A+B−C−D
一般に
Σz^mn+a/(mn+a)
=−1/m{ln(1−z)+(−1)^a(m even)ln(1+z)+f(z))
f(z)=Σcos2πna/m・ln(1−2zcos2πn/m+z^2)−2sin2πna/m・arctan(zsin(2πn/m)/(1−zcos)2πn/m))
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