sonido y serie dos-geométrica

Sonido racional:

R·d_{t}[q(t)]+i·C·q(t) = 0

q(t) = potencial[Q(x,y,z)]·(1/A)·e^{(-i)·(C/R)·t}

R·d_{t}[q(t)]+(-i)·C·q(t) = 0

q(t) = div[Q(x,y,z)]·A·e^{i·(C/R)·t}

Q(x,y,z) = aq_{0}·< x,y,z >

q(t) = aq_{0}·(1/2)·( x^{2}+y^{2}+z^{2} )·(1/A)·e^{(-i)·(C/R)·t}

q(t) = 3aq_{0}·A·e^{i·(C/R)·t}

Q(x,y,z) = (-1)·aq_{0}·< x,y,z >

q(t) = (-1)·aq_{0}·(1/2)·( x^{2}+y^{2}+z^{2} )·(1/A)·e^{(-i)·(C/R)·t}

q(t) = (-3)·aq_{0}·A·e^{i·(C/R)·t}

Q(x,y,z) = a^{2n+1}q_{0}·< x^{2n+1},y^{2n+1},z^{2n+1} >

q(t) = ...

... a^{2n+1}q_{0}·( 1/(2·(n+1)) )·( x^{2·(n+1)}+y^{2·(n+1)}+z^{2·(n+1)} )·...

... (1/A)·e^{(-i)·(C/R)·t}

q(t) = a^{2n+1}q_{0}·(2n+1)·( x^{2n}+y^{2n}+z^{2n} )·A·e^{i·(C/R)·t}

Q(x,y,z) = (-1)·a^{2n+1}q_{0}·< x^{2n+1},y^{2n+1},z^{2n+1} >

q(t) = ...

... (-1)·a^{2n+1}q_{0}·( 1/(2·(n+1)) )·( x^{2·(n+1)}+y^{2·(n+1)}+z^{2·(n+1)} )·...

... (1/A)·e^{(-i)·(C/R)·t}

q(t) = (-1)·a^{2n+1}q_{0}·(2n+1)·( x^{2n}+y^{2n}+z^{2n} )·A·e^{i·(C/R)·t}

Sonido irracional:

L·d_{tt}^{2}[q(t)]+C·q(t) = 0

q(t) = potencial[Q(x,y,z)]·(1/A)·e^{(-i)·(C/L)^{(1/2)}·t}

L·d_{tt}^{2}[q(t)]+C·q(t) = 0

q(t) = div[Q(x,y,z)]·A·e^{i·(C/L)^{(1/2)}·t}

Q(x,y,z) = aq_{0}·< x,y,z >

q(t) = aq_{0}·(1/2)·( x^{2}+y^{2}+z^{2} )·(1/A)·e^{(-i)·(C/L)^{(1/2)}·t}

q(t) = 3aq_{0}·A·e^{i·(C/L)^{(1/2)}·t}

Q(x,y,z) = (-1)·aq_{0}·< x,y,z >

q(t) = (-1)·aq_{0}·(1/2)·( x^{2}+y^{2}+z^{2} )·(1/A)·e^{(-i)·(C/L)^{(1/2)}·t}

q(t) = (-3)·aq_{0}·A·e^{i·(C/L)^{(1/2)}·t}

Sonido racional cuadrático:

R·d_{t}[q(t)]+i·C·q(t) = 0

q(t) = anti-potencial[Q(x,y,z)]·(1/A^{2})·e^{(-i)·(C/R)·t}

R·d_{t}[q(t)]+(-i)·C·q(t) = 0

q(t) = anti-div[Q(x,y,z)]·A^{2}·e^{i·(C/R)·t}

Q(x,y,z) = bq_{0}·< yz,zx,xy >

q(t) = bq_{0}·(1/4)·( (yz)^{2}+(zx)^{2}+(xy)^{2} )·(1/A^{2})·e^{(-i)·(C/L)^{(1/2)}·t}

q(t) = 3bq_{0}·A^{2}·e^{i·(C/R)·t}

Q(x,y,z) = (-1)·bq_{0}·< yz,zx,xy >

q(t) = (-1)·bq_{0}·(1/4)·( (yz)^{2}+(zx)^{2}+(xy)^{2} )·(1/A^{2})·e^{(-i)·(C/L)^{(1/2)}·t}

q(t) = (-3)·bq_{0}·A^{2}·e^{i·(C/R)·t}

Serie dos-geométrica:

( 1/(1+(-a)) )·( 1/(1+(-b)) )+(-1)·( 1/(1+(-a)) )+(-1)·( 1/(1+(-b)) )+1

sum[ i+j = k ][ a^{i}b^{j} ] = ( a/(1+(-a)) )·( b/(1+(-b)) )