【√】 頼む数学詳しい奴来てくれ
■ このスレッドは過去ログ倉庫に格納されています
5chでこれが1の原始11乗根の「一つ」というのを見かけたんだが。どうなんや？
exp(i*2π/11)=cos(2π/11)+i*sin(2π/11)=
-1/10
+1/40(-1+√(5)+i√(10+2√(5)))(-11/4(89+25√(5)+(45√(5-2√(5))-5√(5+2√(5)))i))^(1/5)
+1/40(-1+√(5)+i√(10+2√(5)))(-11/4(89-25√(5)+(45√(5+2√(5))+5√(5-2√(5)))i))^(1/5)
+1/40(-1+√(5)-i√(10+2√(5)))(-11/4(89+25√(5)-(45√(5-2√(5))-5√(5+2√(5)))i))^(1/5)
+1/40(-1+√(5)-i√(10+2√(5)))(-11/4(89-25√(5)-(45√(5+2√(5))+5√(5-2√(5)))i))^(1/5)
+i/10√(55
-5(-11/4(89+25√(5)+(45√(5-2√(5))-5√(5+2√(5)))i))^(1/5)
-5/4(-1+√(5)-i√(10+2√(5)))(-11/4(89-25√(5)+(45√(5+2√(5))+5√(5-2√(5)))i))^(1/5)
-5(-11/4(89+25√(5)-(45√(5-2√(5))-5√(5+2√(5)))i))^(1/5)
-5/4(-1+√(5)+i√(10+2√(5)))(-11/4(89-25√(5)-(45√(5+2√(5))+5√(5-2√(5)))i))^(1/5)
)
=0.841253532831181168861811648919367717513292498420537898642650117...+0.540640817455597582107635954318691695431770607898113840035749889...i 別の表現
exp(i*2π/11)=cos(2π/11)+i*sin(2π/11)=
-1/10
+1/40(-1+√(5)+i√(10+2√(5)))(-11/4(89+25√(5)+(45√(5-2√(5))-5√(5+2√(5)))i))^(1/5)
+1/40(-1+√(5)+i√(10+2√(5)))(-11/4(89-25√(5)+(45√(5+2√(5))+5√(5-2√(5)))i))^(1/5)
+1/40(-1+√(5)-i√(10+2√(5)))(-11/4(89+25√(5)-(45√(5-2√(5))-5√(5+2√(5)))i))^(1/5)
+1/40(-1+√(5)-i√(10+2√(5)))(-11/4(89-25√(5)-(45√(5+2√(5))+5√(5-2√(5)))i))^(1/5)
+i√(11/20
-1/20(-11/4(89+25√(5)+(45√(5-2√(5))-5√(5+2√(5)))i))^(1/5)
-1/80(-1+√(5)-i√(10+2√(5)))(-11/4(89-25√(5)+(45√(5+2√(5))+5√(5-2√(5)))i))^(1/5)
-1/20(-11/4(89+25√(5)-(45√(5-2√(5))-5√(5+2√(5)))i))^(1/5)
-1/80(-1+√(5)+i√(10+2√(5)))(-11/4(89-25√(5)-(45√(5+2√(5))+5√(5-2√(5)))i))^(1/5)
)
=0.841253532831181168861811648919367717513292498420537898642650117...+0.540640817455597582107635954318691695431770607898113840035749889...i ■ このスレッドは過去ログ倉庫に格納されています
