You are not logged in.
Pages: 1
i^2=-1
j^2=1
so k^2=0
e^kx=cosk x+ksink x
e^kx=1+kx+0x/2+0k^2/6...=1+kx
cosk x=1
sink x=x
so tank x=x asink x=x Acosk x=1 acosk x=anything atank x=x
e^ix is on the x^2+y^2=1
e^jx is on the x^2-y^2=1
so e^kx is on the x^2+0y^2=1 or x=+-1
e^ix x is half the area of a sector
e^jx x is half the area of a hyperbola sector thingy
then e^kx x is half the area of a triangle(does work with cosk and sink)
conclusion:the zero factor crumples stuff up a lot
Offline
Pages: 1