viernes, 2 de agosto de 2019

funcions eliptiques

k_{n!}=k_{1}·k_{2}·...·k_{n}


cos[n](t) = ∑ ( ( (-1)^{k_{n!}}/(2(k_{n!}))! )·t^{(2(k_{n!}))})


sin[n](t) = ∑ ( ( (-1)^{k_{n!}}/(2(k_{n!})+1)! )·t^{(2(k_{n!})+1})


x^{n+1}+y^{n+1} = n^{n+1} <==> ( x=cos[n](t) <==> y=sin[n](t) )
x^{n+1}+y^{n+1} = 1 <==> ( x=(cos[n](t)/n) <==> y=(sin[n](t)/n) )


cos[n](0)=n
sin[n](0)=0
sin[n](pi/2) = n
cos[n](pi/2) = 0
sin[n](pi) = n·( 1+(-1)^{n} )^{(1/(n+1))}
cos[n](pi) = (-n)
sin[n]( (-1)(pi/2) ) = (-n)
cos[n]( (-1)(pi/2) ) = n·( 1+(-1)^{n} )^{(1/(n+1))}


sin[n](pi/4) = ( n/2^{(1/(n+1))} )
cos[n](pi/4) = ( n/2^{(1/(n+1))} )
sin[n]( (-1)(pi/4) ) = ( (-n)/2^{(1/(n+1))} )
cos[n]( (-1)(pi/4) ) = ( (n/2^{(1/(n+1))})·(2+(-1)^{n})^{(1/(n+1))} )
sin[n](3pi/4) = ( (n/2^{(1/(n+1))})·(2+(-1)^{n})^{(1/(n+1))} )
cos[n](3pi/4) = ( (-n)/2^{(1/(n+1))} )
sin[n]( (-1)(3pi/4)) = ( (-1)^{n}(n/2^{(1/(n+1))}) )
cos[n]( (-1)(3pi/4) ) = ( (-1)^{n}(n/2^{(1/(n+1))}) )


d_{x}[f(x)]=( 1+(-1)( f(x) )^{n+1} )^{(1/(n+1))} <==> f(x)=(sin[n](x)/n)
d_{x}[g(x)]=(-1)·( 1+(-1)( g(x) )^{n+1} )^{(1/(n+1))} <==> g(x)=(cos[n](x)/n)


d_{x}[f(x)]=a·( 1+(-1)( f(x) )^{n+1} )^{(1/(n+1))} <==> f(x)=(sin[n](ax)/n)
d_{x}[g(x)]=(-a)·( 1+(-1)( g(x) )^{n+1} )^{(1/(n+1))} <==> g(x)=(cos[n](ax)/n)





x^{(n+1)·p}+y^{(n+1)·q} = n^{n+1} <==> ... 
... ( x = ( cos[n](x) )^{(1/p)} & y = ( sin[n](x) )^{(1/q)} )

No hay comentarios:

Publicar un comentario