Phi in anderen Zusammenhängen
Geschrieben von Gabi am 01. Januar 2003 14:06:39:
Als Antwort auf: Mathematischer Torkado geschrieben von Gabi am 27. Dezember 2002 12:07:47:
Eingabe 1.61803... :
1618033988749894 = (a007) (1+1*sqrt(5))/2
1618033988749894 = (a261) -3+2*x^2+5*x^3+6*x^4+2*x^5
1618033988749894 = (a266) Bernoulli(11,x)
1618033988749894 = (a267) Euler(4,x)
1618033988749894 = (a268) Mobius(3,x)
1618033988749894 = (a268) Mobius(4,x)
1618033988749894 = (a274) -1-x+x^2
1618033988749894 = (a312) cos(Pi*1/10)/cos(Pi*3/10)
1618033988749894 = (a315) -2+2*x+2*x^2
1618033988749894 = (b000) Golden ratio= (1+sqrt(5))/2 = phi
1618033988749894 = (f157) F(1/5,4/5;1/2;3/4)
1618033988749894 = (m001) (1/2-sr(5)*cos(Pi/5))/cos(Pi/5)
1618033988749894 = (m001) 2*cos(Pi/5)
1618033988749894 = (m192) (cos(Pi/5)+2)/(-5^(1/2)+1/2)Eingabe 0.61803... :
6180339887498948 = (a007) sqrt((3-sqrt(5))/2)
6180339887498948 = (a261) -3+2*x^2+5*x^3+6*x^4+2*x^5
6180339887498948 = (a266) Bernoulli(11,x)
6180339887498948 = (a267) Euler(4,x)
6180339887498948 = (a268) Mobius(3,x)
6180339887498948 = (a268) Mobius(4,x)
6180339887498948 = (a274) -1-x+x^2
6180339887498948 = (a312) cos(Pi*3/10)/sin(Pi*2/5)
6180339887498948 = (a313) cos(Pi*2/5)+cos(Pi*2/5)
6180339887498948 = (a315) -2+2*x+2*x^2
6180339887498948 = (d104) phi(1n+1)
6180339887498948 = (m001) 1/2/cos(Pi/5)
6180339887498948 = (m001) 2*cos(Pi/5)-sr(5)
6180339887498948 = (m192) (-5^(1/2)+1/2)/(cos(Pi/5)+2)
6180339887498948 = (w004) sol of (log(x)+log(1+x))/cos(2*x)
6180339887498948 = (w204) Rad((1 + R^(-1))^(-(1/2)), n=1..inf)MfG
Gabi